1: Star like light moving in the sky, what could it be? (score 169776 in 2017)
Question
I was camping this weekend away from the city, so my friends and I decided to spend one of the nights watching the sky, since it’s impossible to see anything around where we live. After a little while, of staring up, I saw what looked like a star moving. It moved fast, but not a long distance, and various directions and back. It was about the same size of the other stars near by. It was not as bright as other stars though. I stopped looking up for a little bit, to make sure my eyes were seeing things correctly. But then I looked up again I continued to see the light moving. I thought I was going crazy, so I pointed it out to my friend. It took him a little while, but after a few minutes he also saw it. His girlfriend also saw it a few minutes later. My friend brought up something about satellite reflections and other theories, but we have no idea what we saw.
So what could it be?
Answer accepted (score 17)
There was a object, apparently flying above you, that you couldn’t identify. By definition this is an unidentified flying object. However this does not imply that it was an extra-terrestrial spacecraft.
UFO reports can be explained by a combination of:
- Not recognising a known natural object, such as Venus, or unusual clouds.
- Planes, drones, Chinese lanterns or satellites.
- Optical illusion, dreams or the effects of drugs.
- Deliberate falsification.
There is no real way to decide what it was you saw. There are a couple of points in your report that are odd. You describe it as the same size as stars, but less bright. Stars have no visible size, and have many magnitudes of brightness. Your report suggests that you are someone who doesn’t regularly observe the sky, and so is likely to be more susceptible to optical illusions and tricks that eyes can play on us.
Answer 2 (score 16)
The League of Lost Causes wrote the definitive How to identify that light in the sky? guide:
Licensed under Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License
Answer 3 (score 11)
Most likely a satellite. They look exactly like stars, but they glide across the sky smoothly.
Airplanes may have multiple light sources, some blinking lights, and you can definitely perceive how low it seems to be: An airplane somewhat seems to come from the horizon and disappear the same way, while a satellite “seems to always be at the same distance from you”, like gliding on an imaginary half-sphere or dome of a night sky covering you.
Last summer I went for a night bike ride for 6 hours and during this time, I spotted 12 satellites just by observing. They’re a very common thing. This time I got very lucky though, because one of the 12 satellites unexpectedly was something called an Iridium flare and I had never seen one of those in real life before.
There’s a lot of man-made space junk orbiting the Earth, some satellites still operated, some not. There’s a bunch of satellites known as Iridium, which happen to have a design that includes large, reflective solar panels.
I live in the North so in the summer, the sun only barely goes below the horizon and the summer nights are short. When I’m just in the dark side but an Iridium satellite happens to fly approximately above me, if everything is in a specific angle, the sun from behind the horizon hits the satellite’s solar panels, and like a mirror, reflects those rays to the dark side, into a viewer’s eyes.
It was my first time seeing that and I must say, it’s extremely impressive when you’re merely looking at a “moving star”, then unexpectedly it grows very bright, brighter than any star in the night sky, and finally reverts back to the normal looking gliding satellite.
Because people are very much aware of all of the space junk that’s up there, there are plenty of websites and mobile apps that you can use to see where the most common satellites are at the moment, and, because we know the route of each satellite, as well as the year and location specific data regarding the Earth and the Sun, Iridium flares can also be predicted, like you would predict things like Solar eclipses.
2: How much magnification is needed to see planets of solar system? (score 119191 in 2016)
Question
I have a 3inch Newtonian reflector telescope with 300 mm focal length. I can use highest magnification of 75x using a 4mm eyepiece. But in 75x I can’t see the details of Jupiter what was expected. Instead I see a little blurry image. Now I would like to know how much magnification is necessary to see a good details of Jupiter and other planets. And one more question: Is there any way to improve the vision of my 3 inch telescope?
Answer accepted (score 35)
You’re probably asking the wrong question - which I am going to answer anyway, and after that I am going to answer the question you should have asked instead.
As a general rule, there isn’t much point in pushing the magnification above 2x the diameter of the instrument, measured in mm. 3 inch, that’s 75mm, that’s 150x max. Beyond that limit, even under ideal skies the image is large but blurry.
After that, seeing (or air turbulence) pushes that limit further down. Your aperture is small enough that it almost never suffers from seeing, but larger instruments are often affected. It varies greatly with time, place and season. There are times when a 12" dobsonian, that in theory could do 600x, is clamped down by seeing to 150 … 180x. There are times when you could take a 20" dobsonian all the way up to 1000x - but that’s very, VERY rare, it’s the stuff of legends.
Assuming average seeing conditions and instruments of usual size (refractors of 3…4" aperture, reflectors 6" or larger), here are some rules of thumb:
Jupiter is seen best under mid-high magnification. It’s rare that more than 200x is beneficial. This is because it’s a very low contrast object, and additional magnification comes at the cost of less contrast, which makes things worse.
Saturn works best at high-ish magnification, bit more than Jupiter but maybe not much more. Around 200 … 250x usually works. It depends on what you do - if you’re trying to see the ring divisions, push it a bit higher.
Mars can use the highest magnification that you could generate, given the instrument and the conditions. It’s a very small object, contrast is not bad, so crank it all the way up. Most instruments are limited by seeing when observing Mars.
Moon is the same as Mars.
As you can see, magnification is never an issue for you. More magnification will not make it better. In fact, more magnification always means the image is more blurry, not more crisp - it’s always a compromise between size and blurriness that decides the optimal magnification.
Don’t worry, everyone begins thinking that more is always better. Soon enough, experience shows them what’s really going on.
That being said, I believe it’s not magnification that’s giving you trouble, but the general condition of the optical stack that you’re using. These are things that are extremely important, and yet are ignored by many, many amateurs - and the results are not optimal. Here are a few things that you should investigate:
Collimation
Is your scope collimated? In other words, are all optical elements aligned on the same axis? The likely answer is no. It makes a huge difference in the scope’s performance, especially for planets. Here’s a collimated scope, compared to the same scope out of collimation:
Further information on Thierry Legault’s site, which is extremely informative.
A series of articles and documents regarding collimation:
http://www.cloudynights.com/documents/primer.pdf
Gary Seronik: A Beginner’s Guide to Collimation
Gary Seronik: Collimation Tools: What You Need and What You Don’t
Gary Seronik: No-Tools Telescope Collimation
Note: Some telescopes (e.g. pretty much all refractors) do not require collimation; they are collimated from factory and hold collimation pretty well. But most reflectors (SCTs, all newtonians including dobsonians, etc) do require this periodic maintenance.
Thermal equilibrium
At 3" aperture, this is probably not a big issue, but there’s no reason why you should add another problem to the existing ones. Your scope should be at the same temperature as the air around it, otherwise its performance decreases. Take it outside 1 hour before you start observing, and that should be enough for you.
Larger telescopes (around 10" … 12" and larger) should use active ventilation for better cooling (a fan on the back of the mirror). More details here:
Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 1
Gary Seronik: Beat the Heat: Conquering Newtonian Reflector Thermals — Part 2
In your case, simple passive cooling for 1 hour should be enough, but it’s worth reading those articles.
Focal ratio
A 3" scope, at 300mm focal length, that’s an f/4 instrument. That’s a pretty steep f/ ratio. Most eyepieces will not do well with such a blunt cone of light, and will start to exhibit aberrations that blur the image. Only very expensive eyepieces work well at such low focal ratios - things like TeleVue Ethos, or Explore Scientific 82 degree eyepieces.
Try and keep the planet in the center - most aberrations are lower there. Even very simple eyepieces do better in the middle of the image.
Look at the stars. Are they tiny and round in the center, and large and fuzzy at the edge? Those are aberrations from various sources (eyepiece, primary mirror, etc).
Coma
Of course, at f/4 even the best eyepieces out there cannot do anything about coma - an aberration coming out of any parabolic mirror, which becomes pretty obvious around f/5, very obvious at f/4, and a major problem at f/3. Again, coma is zero in the center of the image, and increases towards the edge.
A coma corrector is used in some cases, such as the TeleVue Paracorr, but I strongly recommend that you DO NOT use one - I suspect your instrument is aberrating in ways that overwhelm coma anyway. Jupiter would not be too blurry even at full f/4 coma at the edge. This paragraph is for informational purposes only.
Coma should become a concern with large telescopes, using high quality optics, with a focal ratio of around f/5 and less. E.g., you have a 20" dob with an f/4 mirror, then you should worry about coma - provided that collimation and so on are taken care of.
Optics quality
An f/4 parabola is not super easy to make at any size. I’ve made my own optics, and the lower the f/ ratio, the more difficult the process is. Many small, cheap telescopes are made in a hurry, and the difficult focal ratio poses additional problems - as a result, many manufacturers do a poor job. There are even cases where the primary mirror is left spherical, with disastrous results.
This is something you can do nothing about. If the primary mirror is bad, then that’s just the way things are. An optician might try to correct it, but it’s a difficult process, and quite expensive. I only added this here so you are informed.
This is what I would do in your case:
I would take the scope out 1 hour before observing, every time.
I would try and learn how to collimate the scope. I would try to figure out a few simple collimation techniques, and a few simple tests. I would spend a few days / weeks practicing that. I would keep reading about collimation.
When collimation is at least partially under control, I would learn how to properly focus the scope. Seems simple, but it can be tricky. Use a bright star, and try and make it as small as possible. Use the Moon when it’s visible, and try and make it crisp and clear. Do not try this with a miscollimated scope, since it’s pointless.
After a few months, when I gain confidence that the scope is in better shape, very well collimated, very well focused, I might try to borrow a better eyepiece from a friend. I said borrow, not buy. Something like a 3 … 4mm eyepiece, good quality, that would give me a comparison for the existing eyepieces. This ONLY makes sense with a scope that is in perfect collimation, perfect temperature, perfect focus. If an improvement is seen, then get a better eyepiece - but do not spend hundreds of dollars for an expensive eyepiece that will then be used in a tiny cheap scope. Second-hand eyepieces often work exactly as well as new ones.
If you know someone in your area who makes mirrors, see if they agree to put your primary mirror on the Foucault tester, and assess its condition. But beware: the results might be very disappointing. Or not. You kind of never know with these little scopes.
EDIT: After the scope is collimated and so on, you could try to increase magnification by using a 2x barlow with your eyepieces, but do not expect miracles - the image will be bigger, but probably rather “mushy”. More magnification is not always better, there’s always a trade-off.
Good luck, and clear skies to you!
Answer 2 (score 1)
In typical seeing condition you should be able to use a magnification (see here) of about 25-30x per inch of apperture, so for your telescope that is about 100x, in exceptional condition you could push that up to maybe double that. Also the more magnification you use the less contrast you will have in the image, so really you want the lowest magnification that gives an image size compatible with being able to see the bands since you will be contrast limited.
You will find simulated images of Jupiter through a small telescope here and Saturn here. Though personal experience suggests that the simulated image of Jupiter through a 3" aperture is optimistic. IIRC a suggestion of banding is just about at the limit of what I can see on Jupiter in the small scope.
I am guessing that your telescope is this
3: Is the sun any brighter during a solar eclipse (score 76361 in 2017)
Question
I keep hearing that you can go blind during a solar eclipse.
Is the sun actually any brighter during a solar eclipse? This is how the media seems to frame it, and they suggest you need special glasses to view the eclipse.
But after some reading, it seems that the warning is about just not staring at the sun in general, and there is nothing special about the eclipse.
Is this the case?
Answer accepted (score 6)
The Sun is the same as ever. It’s just that people tend to stare at it for long times during the eclipse. A quick glance doesn’t do much damage, but prolonged staring could be bad.
Those who see the phase of totality have an additional risk. After the Sun goes completely dark, and it then finally shows up again, some people feel compelled to keep watching that tiny sliver of light without protection for a long time. Don’t. When the actual surface of the Sun is visible again after totality is over, it’s time to put the googles back on.
Do not use regular sunglasses. They typically don’t block the kinds of radiation they should block to allow you to watch the event safely. There are several options here:
Very dark welding glasses. Ask a welder or go to a welding shop. They have these special glasses to protect themselves from the radiation from the welding arc, which is similar to the Sun’s radiation. These glasses are numbered to show how dark they are. For viewing the Sun, use the darkest ones, labeled #14. A welding mask with a #14 glass is perfectly safe no matter how long you’re looking at the Sun. #13 glass might be safe too for short durations, but it’s not recommended. Anything lower than that is not safe.
Goggles made specifically for watching the Sun. There are many kinds of them. As long as they pass the ISO 12312-2 international safety standard they’re fine.
I recommend the goggles made of Baader solar film. They show the Sun in its original color (not red, green, orange, etc). There are several kinds here, too; the Baader AstroSolar Silver/Gold Film passes the ISO standard, these are what I use. You can buy them online in various places and they’re very cheap.
More info:
Answer 2 (score 3)
No, the intrinsic brightness of the sun does not change. Part of the sun light is blocked from reaching earth however making the sun appear dimmer or less intense. This has a tendency to cause people to stare at the sun to see this effect. The sun is so bright even a portion of the light reaching your eyes is intense enough to damage them. So do not look at the sun during the eclipse, or any other time, without proper eye protection.
4: What are the differences between matter, dark matter and antimatter? (score 64291 in )
Question
I thought dark and anti matter were kinda the same, but after saw a video, they mention that dark matter is not antimatter but their explanation is a little fast so I got doubts.
What are the differences between matter, dark matter and antimatter? Are they related? How they interact each other? Where can I see an example of this interaction? and also, in terms of percentage, how much of matter, dark matter and antimatter exist in the universe?
Thanks!
Answer accepted (score 12)
Matter is the stuff you are made of.
Antimatter is the same as matter in every way, looks the same, behaves the same, except its particles have electrical charges opposite to matter. E.g., our electrons are negatively charged, whereas a positron (an antimatter “electron”) is positively charged. The positron is the “anti-particle” of the electron.
When a particle meets its anti-particle, they “annihilate”: the two particles disappear, and gamma photons are released carrying off their energy. For this reason, should a lump of matter touch a lump of antimatter, they would annihilate, and a giant explosion would result because of the huge energy released (E=mc^2).
Matter and antimatter are definitely related: same thing, but with opposite signs. Twins, but opposites.
It is not clear why, but it seems like there isn’t that much antimatter out there, more like trace amounts. Definitely not as much as regular matter as far as we can tell. This is puzzling to physicists and cosmologists, because you’d expect the Big Bang to make roughly equal amounts of matter and antimatter. Scientists agree that the paradox of “excess matter” will advance physics even further once it’s solved.
Dark matter - we don’t really know what it is. It’s not even sure it’s “matter” in a conventional sense, or related to it in any way. We just know that galaxies are rotating in such a way that indicates there’s a lot more mass out there, but it is mass that we cannot see and cannot be accounted for in the usual ways. Hence the name “dark” (as in invisible) matter.
Dark matter doesn’t seem to interact much with regular matter, except gravitationally. Right now dark matter could be passing through you and you wouldn’t notice. Dark matter also does not interact with light, so you can’t see it. It doesn’t seem to interact much with itself either, so for this reason dark matter cannot form “clumps” such as planets or stars. Instead, it probably exists in a diffuse form. Bottom line, dark matter interacts pretty much only via gravity.
The shape of galaxies is a proof of the existence of dark matter, and is a result of the interaction between matter and dark matter. Without dark matter, galaxies would be much less massive, and the outer parts would rotate much more slowly compared to the center. Due to dark matter, galaxies are quite massive, and they rotate almost as solid objects - the outer parts rotate approximately as fast as the central parts.
Estimates vary, but it seems like there’s something like 5x to 6x more dark matter out there compared to regular matter.
Answer 2 (score 3)
A pretty good site for quick explanations is the Particle Adventure
Anti Matter is really quite simple and very similar to regular matter. It just happens to explode violently when it touches regular matter - like a positron (Positive Electron) and an electron (negative) will touch and evaporate into a pair of gamma rays. The Proton/anti Proton or Neutron/anti Neutron or Proton/antineutron or Neutron/antiProton (they interact because they have some quarks/anti-quarks in common). Those reactions are more complicated, but the gist is the same. They explode violently when they touch. So, because matter and anti matter tend to evaporate each other, there’s really no primordial anti-matter left in the universe, cause there was slightly more matter.
But, other than the explosive interaction, Antimatter is almost exactly the same as matter, You could, in theory build a star, a planet, trees and life out of anti-matter.
Dark Matter is a lot more different. We don’t really know what it is, but it’s a different kind of matter, it’s transparent and it’s non binding but it has mass. Regular matter can bind together gravitationally, that’s why it forms into things like stars, planets, comets, asteroids, etc. Dark matter doesn’t do that. It loosely collects around galaxies, or, perhaps more accurately, galaxies collect inside large pockets of dark matter.
5: Very bright star in the east at northern hemisphere. What is it? (score 62489 in 2014)
Question
From some time now (few weeks as far as I can remember), there is a very bright star in the eastern sky.
I first thought it was Venus, but according to this link, Venus is in the western sky and sets 2 hours after sunset. But this (very) bright star is still up even though it’s midnight (6 hours after sunset).
Does anyone know what is this star (or planet) ?
Please excuse my silly question as I know nothing about astronomy.
Thanks.
Answer accepted (score 5)
It will be very difficult to tell you without knowing where you are located. Particularly, if you are in the northern or southern hemisphere. At this time of year, most of the brightest stars are in the sky. Also where in the sky, the particular star that you are wondering about is. And any other constellations that you can identify.
For stars rising this time of year, particularly bright ones you might be thinking of Sirius. Which would be near the constellation Orion below the hourglass shape of the it (http://starchart.polaris.net/chart.php?constel=ori&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC).
Another possibility would be Aldebaran, it is located in a “V” shape of stars that make up the constellation Taurus. http://starchart.polaris.net/chart.php?constel=tau&layoutType=Key&layoutOri=Portrait&project=SANSONS&type=PNGC
To find out I would take a look at some of the online star charts that are available for your time and location. That will give you a good idea of what you are seeing. Here is a list of some online:
http://www.skyandtelescope.com/resources/internet/3305011.html?page=1&c=y
Answer 2 (score 4)
As other people have pointed out, it is hard to work out which star it is, without knowing your general location. However, after checking on Stellarium, there seem to be a couple of likely suspects:
- Sirius - the brightest star in the sky. I’ve seen it myself - and on a good, dark night, it can really stand out.
- Jupiter - the king of the planets is also rising at about the same time. It is brighter than any star in the sky, by a wide margin (though fainter than Venus), and it can really stand out.
Other than that, there aren’t really that many objects rising in the East at the time you specify that could really stand out.
There are a couple of useful ways to tell the two apart:
- Sirius is a bright white object - perhaps with a subtle bluish tinge to it, whereas Jupiter has a slight yellow tint to it.
- Jupiter is currently rising in the North-East, and can get very high in the sky at the moment from the northern hemisphere, whereas Sirius rises in the South-East, and doesn’t get that high (though that does depend on location).
- Sirius tends to twinkle, and ‘flicker’, as its light is disturbed by air currents, whereas Jupiter remains very steady - perhaps not twinkling at all.
As mentioned earlier, the best method is usually to use software like Stellarium, which will tell you exactly where everything is, and hopefully give you a definitive answer to which object it is.
Answer 3 (score 2)
A good way to get to know what is visible at your location is to use a software that maps the sky. One such free software is Stellarium available for all platforms. You can easily either find your place (or a nearby place) in its built in database of locations or alternatively provide the longitude/latitude for your location. With such software you will be able to see the sky map in real time and hence figure out what everything is.
6: How long does a sunrise or sunset take? (score 59518 in )
Question
From the time that the sun appears on the horizon, or meets it on its setting, to the time that it is fully visible, or no longer visible on its setting, how much time passes? Secondly, is there a place in the world where a sunrise/sunset occurs over a period of a few days? Meaning, that from the time it begins to appear over the horizon until it is fully visible, a period of a few days pass without night intervening (and the same for the opposite with sunset)?
Answer accepted (score 11)
As noted in http://aa.quae.nl/en/antwoorden/zonpositie.html#14 the length of sunrise/sunset varies from approximately 128/cos(latitude) seconds at the equinoxes to approximately 142/cos(1.14*latitude) at the solstices.
More specifically, here’s the length of sunrise/sunset at various latitudes:
Beyond 65 degrees north or south latitude, the sun does not rise or set daily, and the length of sunrise/sunset increases significantly.
The data plotted above is the length of sunrise, but the length of sunset is very similar.
All calculations for this program were made with this program:
https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12824.c
The raw output of sunrise/sunset times:
https://github.com/barrycarter/bcapps/blob/master/ASTRO/sun-rise-set-multiple-latitudes.txt.bz2
You can verify these results at: http://aa.usno.navy.mil/data/docs/RS_OneYear.php
The longest sunrise I found for 2015 was at 89 degrees 51 minutes south latitude, 125 degrees east longitude. There, the sun starts rising 20 Sep 2015 at 2352, bobbles up and down a bit (but never quite sets), and finally finishes rising 43 hours and 21 minutes later, at 22 Sep 2015 at 1913, but see caveat at the end of this answer.
You can “verify” this by first visiting http://aa.usno.navy.mil/data/docs/RS_OneYear.php with these parameters:
to get:
Sun or Moon Rise/Set Table for One Year
o , o , Astronomical Applications Dept.
Location: E125 00, S89 51 Rise and Set for the Sun for 2015 U. S. Naval Observatory
Washington, DC 20392-5420
Universal Time
Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
Day Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set
h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m
01 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
02 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
03 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
04 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
05 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
06 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
07 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
08 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
09 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
10 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
11 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
12 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
13 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
14 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
15 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
16 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
17 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
18 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
19 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
20 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- 2352 **** **** **** **** **** ****
21 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
22 **** **** **** **** 1842 1614 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
23 **** **** **** **** 0708 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
24 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
25 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
26 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
27 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
28 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
29 **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
30 **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
31 **** **** ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** ****
(**** object continuously above horizon) (---- object continuously below horizon)Note that the sun rises at 2352 on September 20th, and doesn’t set for the rest of the year, verifying the sunrise start time.
Verifying the end time is a little tricker. To do this, visit http://ssd.jpl.nasa.gov/horizons.cgi with the following parameters:
to get:
Revised : Jul 31, 2013 Sun 10
PHYSICAL PROPERTIES (revised Jan 16, 2014):
GM (10^11 km^3/s^2) = 1.3271244004193938 Mass (10^30 kg) ~ 1.988544
Radius (photosphere) = 6.963(10^5) km Angular diam at 1 AU = 1919.3"
Solar Radius (IAU) = 6.955(10^5) km Mean density = 1.408 g/cm^3
Surface gravity = 274.0 m/s^2 Moment of inertia = 0.059
Escape velocity = 617.7 km/s Adopted sidereal per = 25.38 d
Pole (RA,DEC in deg.) = 286.13,63.87 Obliquity to ecliptic = 7 deg 15'
Solar constant (1 AU) = 1367.6 W/m^2 Solar lumin.(erg/s) = 3.846(10^33)
Mass-energy conv rate = 4.3(10^12 gm/s) Effective temp (K) = 5778
Surf. temp (photosphr)= 6600 K (bottom) Surf. temp (photosphr)= 4400 K (top)
Photospheric depth = ~400 km Chromospheric depth = ~2500 km
Sunspot cycle = 11.4 yr Cycle 22 sunspot min. = 1991 A.D.
Motn. rel to nrby strs= apex : RA=271 deg; DEC=+30 deg
speed: 19.4 km/s = 0.0112 AU/day
Motn. rel to 2.73K BB = apex : l=264.7+-0.8; b=48.2+-0.5
speed: 369 +-11 km/s
Results
*******************************************************************************
Ephemeris / WWW_USER Fri Jan 1 21:49:19 2016 Pasadena, USA / Horizons
*******************************************************************************
Target body name: Sun (10) {source: DE431mx}
Center body name: Earth (399) {source: DE431mx}
Center-site name: (user defined site below)
*******************************************************************************
Start time : A.D. 2015-Sep-22 19:00:00.0000 UT
Stop time : A.D. 2015-Sep-22 20:00:00.0000 UT
Step-size : 1 minutes
*******************************************************************************
Target pole/equ : IAU_SUN {East-longitude +}
Target radii : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}
Center geodetic : 125.000000,-89.850000,7.057E-13 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 125.000000,16.7540774,-6356.730 {E-lon(deg),Dxy(km),Dz(km)}
Center pole/equ : High-precision EOP model {East-longitude +}
Center radii : 6378.1 x 6378.1 x 6356.8 km {Equator, meridian, pole}
Target primary : Sun
Vis. interferer : MOON (R_eq= 1737.400) km {source: DE431mx}
Rel. light bend : Sun, EARTH {source: DE431mx}
Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2
Atmos refraction: NO (AIRLESS)
RA format : HMS
Time format : CAL
RTS-only print : NO
EOP file : eop.160101.p160324
EOP coverage : DATA-BASED 1962-JAN-20 TO 2016-JAN-01. PREDICTS-> 2016-MAR-23
Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s
Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (>38.000=NO), Daylight (NO )
Table cut-offs 2: Solar Elongation ( 0.0,180.0=NO ),Local Hour Angle( 0.0=NO )
*******************************************************************************
Date__(UT)__HR:MN Azi_(a-appr)_Elev
****************************************
$$SOE
2015-Sep-22 19:00 *m 128.1772 -0.3117
2015-Sep-22 19:01 *m 127.9272 -0.3109
2015-Sep-22 19:02 *m 127.6771 -0.3101
2015-Sep-22 19:03 *m 127.4270 -0.3093
2015-Sep-22 19:04 *m 127.1770 -0.3085
2015-Sep-22 19:05 *m 126.9269 -0.3077
2015-Sep-22 19:06 *m 126.6769 -0.3069
2015-Sep-22 19:07 *m 126.4268 -0.3061
2015-Sep-22 19:08 *m 126.1767 -0.3053
2015-Sep-22 19:09 *m 125.9267 -0.3045
2015-Sep-22 19:10 *m 125.6766 -0.3037
2015-Sep-22 19:11 *m 125.4266 -0.3029
2015-Sep-22 19:12 *m 125.1765 -0.3021
2015-Sep-22 19:13 *m 124.9264 -0.3013
2015-Sep-22 19:14 *m 124.6764 -0.3005
2015-Sep-22 19:15 *m 124.4263 -0.2997
2015-Sep-22 19:16 *m 124.1762 -0.2989
2015-Sep-22 19:17 *m 123.9262 -0.2981
2015-Sep-22 19:18 *m 123.6761 -0.2973
2015-Sep-22 19:19 *m 123.4261 -0.2964
2015-Sep-22 19:20 *m 123.1760 -0.2956
2015-Sep-22 19:21 *m 122.9259 -0.2948
2015-Sep-22 19:22 *m 122.6759 -0.2940
2015-Sep-22 19:23 *m 122.4258 -0.2932
2015-Sep-22 19:24 *m 122.1757 -0.2923
2015-Sep-22 19:25 *m 121.9257 -0.2915
2015-Sep-22 19:26 *m 121.6756 -0.2907
2015-Sep-22 19:27 *m 121.4256 -0.2899
2015-Sep-22 19:28 *m 121.1755 -0.2890
2015-Sep-22 19:29 *m 120.9254 -0.2882
2015-Sep-22 19:30 *m 120.6754 -0.2874
2015-Sep-22 19:31 *m 120.4253 -0.2865
2015-Sep-22 19:32 *m 120.1753 -0.2857
2015-Sep-22 19:33 *m 119.9252 -0.2849
2015-Sep-22 19:34 *m 119.6751 -0.2840
2015-Sep-22 19:35 *m 119.4251 -0.2832
2015-Sep-22 19:36 *m 119.1750 -0.2823
2015-Sep-22 19:37 *m 118.9250 -0.2815
2015-Sep-22 19:38 *m 118.6749 -0.2807
2015-Sep-22 19:39 *m 118.4248 -0.2798
2015-Sep-22 19:40 *m 118.1748 -0.2790
2015-Sep-22 19:41 *m 117.9247 -0.2781
2015-Sep-22 19:42 *m 117.6746 -0.2773
2015-Sep-22 19:43 *m 117.4246 -0.2764
2015-Sep-22 19:44 *m 117.1745 -0.2756
2015-Sep-22 19:45 *m 116.9245 -0.2747
2015-Sep-22 19:46 *m 116.6744 -0.2739
2015-Sep-22 19:47 *m 116.4243 -0.2730
2015-Sep-22 19:48 *m 116.1743 -0.2721
2015-Sep-22 19:49 *m 115.9242 -0.2713
2015-Sep-22 19:50 *m 115.6742 -0.2704
2015-Sep-22 19:51 *m 115.4241 -0.2696
2015-Sep-22 19:52 *m 115.1740 -0.2687
2015-Sep-22 19:53 *m 114.9240 -0.2678
2015-Sep-22 19:54 *m 114.6739 -0.2670
2015-Sep-22 19:55 *m 114.4239 -0.2661
2015-Sep-22 19:56 *m 114.1738 -0.2652
2015-Sep-22 19:57 *m 113.9237 -0.2644
2015-Sep-22 19:58 *m 113.6737 -0.2635
2015-Sep-22 19:59 *m 113.4236 -0.2626
2015-Sep-22 20:00 *m 113.1735 -0.2618
$$EOE
*******************************************************************************
Column meaning:
TIME
Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in
the 1st-column denotes a B.C. date. First-column blank (" ") denotes an A.D.
date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.
Later calendar dates are in the Gregorian system.
Time tags refer to the same instant throughout the universe, regardless of
where the observer is located.
The dynamical Coordinate Time scale is used internally. It is equivalent to
the current IAU definition of "TDB". Conversion between CT and the selected
non-uniform UT output scale has not been determined for UTC times after the
next July or January 1st. The last known leap-second is used over any future
interval.
NOTE: "n.a." in output means quantity "not available" at the print-time.
SOLAR PRESENCE (OBSERVING SITE)
Time tag is followed by a blank, then a solar-presence symbol:
'*' Daylight (refracted solar upper-limb on or above apparent horizon)
'C' Civil twilight/dawn
'N' Nautical twilight/dawn
'A' Astronomical twilight/dawn
' ' Night OR geocentric ephemeris
LUNAR PRESENCE WITH TARGET RISE/TRANSIT/SET MARKER (OBSERVING SITE)
The solar-presence symbol is immediately followed by another marker symbol:
'm' Refracted upper-limb of Moon on or above apparent horizon
' ' Refracted upper-limb of Moon below apparent horizon OR geocentric
'r' Rise (target body on or above cut-off RTS elevation)
't' Transit (target body at or past local maximum RTS elevation)
's' Set (target body on or below cut-off RTS elevation)
RTS MARKERS (TVH)
Rise and set are with respect to the reference ellipsoid true visual horizon
defined by the elevation cut-off angle. Horizon dip and yellow-light refraction
(Earth only) are considered. Accuracy is < or = to twice the requested search
step-size.
Azi_(a-appr)_Elev =
Airless apparent azimuth and elevation of target center. Adjusted for
light-time, the gravitational deflection of light, stellar aberration,
precession and nutation. Azimuth measured North(0) -> East(90) -> South(180) ->
West(270) -> North (360). Elevation is with respect to plane perpendicular
to local zenith direction. TOPOCENTRIC ONLY. Units: DEGREES
Computations by ...
Solar System Dynamics Group, Horizons On-Line Ephemeris System
4800 Oak Grove Drive, Jet Propulsion Laboratory
Pasadena, CA 91109 USA
Information: http://ssd.jpl.nasa.gov/
Connect : telnet://ssd.jpl.nasa.gov:6775 (via browser)
telnet ssd.jpl.nasa.gov 6775 (via command-line)
Author : Jon.Giorgini@jpl.nasa.gov
*******************************************************************************The sun’s angular diameter is about 32 arcminutes, so the sun’s lower limb is 16 arcminutes below the sun’s center. When the center of the sun has geometric elevation -18 arcminutes (-0.3 degrees), the lower limb has geometric elevation -34 arcminutes. Since refraction near the horizon is also 34 arcminutes, the sun’s lower limb rises when the sun’s geometric elevation is -0.3 degrees.
In the table above, this occurs between 1914 and 1915, but my program uses slightly more accurate data for the sun’s angular diameter, and the sun actually finishes rising between 1913 and 1914 (and closer to 1913).
You can then fly almost halfway across the world to latitude 89 degrees 51 minutes and longitude -19 degrees to see the one-minute-shorter longest sunset, which starts at 23 Sep 2015 at 2128 and ends at 25 Sep 2015 at 1648, a length of 43 hours and 20 minutes.
In this case, you would use http://aa.usno.navy.mil/data/docs/RS_OneYear.php to verify the ending time of the sunset, and HORIZONS to verify the start time of the sunset.
Polar sunrises and sunsets are considerably shorter:
-
At the North Pole, the sun starts rising at 18 Mar 2015 at 2015, and finishes rising at 20 Mar 2015 at 0441, a length of 32 hours and 26 minutes.
-
At the South Pole, the sun starts setting at 21 Mar 2015 at 1650, and finishes setting at 23 Mar 2015 at 0117, a length of 32 hours and 27 minutes.
-
At the South Pole, the sun starts rising at 21 Sep 2015 at 0508, and finishes rising at 22 Sep 2015 at 1400, a length of 32 hours and 52 minutes.
-
At the North Pole, the sun starts setting at 24 Sep 2015 at 0243, and finishes setting at 25 Sep 2015 at 1131, a length of 32 hours and 48 minutes.
Main caveat: Like HORIZONS and the sunrise/sunset tables above, I assume 34 arcminutes of refraction at the horizon. That’s reasonable for most locations, but may be unreasonable close the pole, where the longest sunrises and sunsets occur. In particular, refraction can change rapidly at these latitudes, allowing for potentially much longer sunrises and sunsets.
I now believe that http://what-if.xkcd.com/42/ is inaccurate, and will ping the author to let him know.
Answer 2 (score 11)
As noted in http://aa.quae.nl/en/antwoorden/zonpositie.html#14 the length of sunrise/sunset varies from approximately 128/cos(latitude) seconds at the equinoxes to approximately 142/cos(1.14*latitude) at the solstices.
More specifically, here’s the length of sunrise/sunset at various latitudes:
Beyond 65 degrees north or south latitude, the sun does not rise or set daily, and the length of sunrise/sunset increases significantly.
The data plotted above is the length of sunrise, but the length of sunset is very similar.
All calculations for this program were made with this program:
https://github.com/barrycarter/bcapps/blob/master/ASTRO/bc-solve-astro-12824.c
The raw output of sunrise/sunset times:
https://github.com/barrycarter/bcapps/blob/master/ASTRO/sun-rise-set-multiple-latitudes.txt.bz2
You can verify these results at: http://aa.usno.navy.mil/data/docs/RS_OneYear.php
The longest sunrise I found for 2015 was at 89 degrees 51 minutes south latitude, 125 degrees east longitude. There, the sun starts rising 20 Sep 2015 at 2352, bobbles up and down a bit (but never quite sets), and finally finishes rising 43 hours and 21 minutes later, at 22 Sep 2015 at 1913, but see caveat at the end of this answer.
You can “verify” this by first visiting http://aa.usno.navy.mil/data/docs/RS_OneYear.php with these parameters:
to get:
Sun or Moon Rise/Set Table for One Year
o , o , Astronomical Applications Dept.
Location: E125 00, S89 51 Rise and Set for the Sun for 2015 U. S. Naval Observatory
Washington, DC 20392-5420
Universal Time
Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
Day Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set Rise Set
h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m h m
01 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
02 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
03 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
04 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
05 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
06 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
07 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
08 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
09 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
10 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
11 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
12 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
13 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
14 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
15 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
16 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
17 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
18 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
19 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** ****
20 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- 2352 **** **** **** **** **** ****
21 **** **** **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
22 **** **** **** **** 1842 1614 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
23 **** **** **** **** 0708 ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
24 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
25 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
26 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
27 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
28 **** **** **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
29 **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
30 **** **** ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** **** **** **** **** ****
31 **** **** ---- ---- ---- ---- ---- ---- ---- ---- **** **** **** ****
(**** object continuously above horizon) (---- object continuously below horizon)Note that the sun rises at 2352 on September 20th, and doesn’t set for the rest of the year, verifying the sunrise start time.
Verifying the end time is a little tricker. To do this, visit http://ssd.jpl.nasa.gov/horizons.cgi with the following parameters:
to get:
Revised : Jul 31, 2013 Sun 10
PHYSICAL PROPERTIES (revised Jan 16, 2014):
GM (10^11 km^3/s^2) = 1.3271244004193938 Mass (10^30 kg) ~ 1.988544
Radius (photosphere) = 6.963(10^5) km Angular diam at 1 AU = 1919.3"
Solar Radius (IAU) = 6.955(10^5) km Mean density = 1.408 g/cm^3
Surface gravity = 274.0 m/s^2 Moment of inertia = 0.059
Escape velocity = 617.7 km/s Adopted sidereal per = 25.38 d
Pole (RA,DEC in deg.) = 286.13,63.87 Obliquity to ecliptic = 7 deg 15'
Solar constant (1 AU) = 1367.6 W/m^2 Solar lumin.(erg/s) = 3.846(10^33)
Mass-energy conv rate = 4.3(10^12 gm/s) Effective temp (K) = 5778
Surf. temp (photosphr)= 6600 K (bottom) Surf. temp (photosphr)= 4400 K (top)
Photospheric depth = ~400 km Chromospheric depth = ~2500 km
Sunspot cycle = 11.4 yr Cycle 22 sunspot min. = 1991 A.D.
Motn. rel to nrby strs= apex : RA=271 deg; DEC=+30 deg
speed: 19.4 km/s = 0.0112 AU/day
Motn. rel to 2.73K BB = apex : l=264.7+-0.8; b=48.2+-0.5
speed: 369 +-11 km/s
Results
*******************************************************************************
Ephemeris / WWW_USER Fri Jan 1 21:49:19 2016 Pasadena, USA / Horizons
*******************************************************************************
Target body name: Sun (10) {source: DE431mx}
Center body name: Earth (399) {source: DE431mx}
Center-site name: (user defined site below)
*******************************************************************************
Start time : A.D. 2015-Sep-22 19:00:00.0000 UT
Stop time : A.D. 2015-Sep-22 20:00:00.0000 UT
Step-size : 1 minutes
*******************************************************************************
Target pole/equ : IAU_SUN {East-longitude +}
Target radii : 696000.0 x 696000.0 x 696000.0 k{Equator, meridian, pole}
Center geodetic : 125.000000,-89.850000,7.057E-13 {E-lon(deg),Lat(deg),Alt(km)}
Center cylindric: 125.000000,16.7540774,-6356.730 {E-lon(deg),Dxy(km),Dz(km)}
Center pole/equ : High-precision EOP model {East-longitude +}
Center radii : 6378.1 x 6378.1 x 6356.8 km {Equator, meridian, pole}
Target primary : Sun
Vis. interferer : MOON (R_eq= 1737.400) km {source: DE431mx}
Rel. light bend : Sun, EARTH {source: DE431mx}
Rel. lght bnd GM: 1.3271E+11, 3.9860E+05 km^3/s^2
Atmos refraction: NO (AIRLESS)
RA format : HMS
Time format : CAL
RTS-only print : NO
EOP file : eop.160101.p160324
EOP coverage : DATA-BASED 1962-JAN-20 TO 2016-JAN-01. PREDICTS-> 2016-MAR-23
Units conversion: 1 au= 149597870.700 km, c= 299792.458 km/s, 1 day= 86400.0 s
Table cut-offs 1: Elevation (-90.0deg=NO ),Airmass (>38.000=NO), Daylight (NO )
Table cut-offs 2: Solar Elongation ( 0.0,180.0=NO ),Local Hour Angle( 0.0=NO )
*******************************************************************************
Date__(UT)__HR:MN Azi_(a-appr)_Elev
****************************************
$$SOE
2015-Sep-22 19:00 *m 128.1772 -0.3117
2015-Sep-22 19:01 *m 127.9272 -0.3109
2015-Sep-22 19:02 *m 127.6771 -0.3101
2015-Sep-22 19:03 *m 127.4270 -0.3093
2015-Sep-22 19:04 *m 127.1770 -0.3085
2015-Sep-22 19:05 *m 126.9269 -0.3077
2015-Sep-22 19:06 *m 126.6769 -0.3069
2015-Sep-22 19:07 *m 126.4268 -0.3061
2015-Sep-22 19:08 *m 126.1767 -0.3053
2015-Sep-22 19:09 *m 125.9267 -0.3045
2015-Sep-22 19:10 *m 125.6766 -0.3037
2015-Sep-22 19:11 *m 125.4266 -0.3029
2015-Sep-22 19:12 *m 125.1765 -0.3021
2015-Sep-22 19:13 *m 124.9264 -0.3013
2015-Sep-22 19:14 *m 124.6764 -0.3005
2015-Sep-22 19:15 *m 124.4263 -0.2997
2015-Sep-22 19:16 *m 124.1762 -0.2989
2015-Sep-22 19:17 *m 123.9262 -0.2981
2015-Sep-22 19:18 *m 123.6761 -0.2973
2015-Sep-22 19:19 *m 123.4261 -0.2964
2015-Sep-22 19:20 *m 123.1760 -0.2956
2015-Sep-22 19:21 *m 122.9259 -0.2948
2015-Sep-22 19:22 *m 122.6759 -0.2940
2015-Sep-22 19:23 *m 122.4258 -0.2932
2015-Sep-22 19:24 *m 122.1757 -0.2923
2015-Sep-22 19:25 *m 121.9257 -0.2915
2015-Sep-22 19:26 *m 121.6756 -0.2907
2015-Sep-22 19:27 *m 121.4256 -0.2899
2015-Sep-22 19:28 *m 121.1755 -0.2890
2015-Sep-22 19:29 *m 120.9254 -0.2882
2015-Sep-22 19:30 *m 120.6754 -0.2874
2015-Sep-22 19:31 *m 120.4253 -0.2865
2015-Sep-22 19:32 *m 120.1753 -0.2857
2015-Sep-22 19:33 *m 119.9252 -0.2849
2015-Sep-22 19:34 *m 119.6751 -0.2840
2015-Sep-22 19:35 *m 119.4251 -0.2832
2015-Sep-22 19:36 *m 119.1750 -0.2823
2015-Sep-22 19:37 *m 118.9250 -0.2815
2015-Sep-22 19:38 *m 118.6749 -0.2807
2015-Sep-22 19:39 *m 118.4248 -0.2798
2015-Sep-22 19:40 *m 118.1748 -0.2790
2015-Sep-22 19:41 *m 117.9247 -0.2781
2015-Sep-22 19:42 *m 117.6746 -0.2773
2015-Sep-22 19:43 *m 117.4246 -0.2764
2015-Sep-22 19:44 *m 117.1745 -0.2756
2015-Sep-22 19:45 *m 116.9245 -0.2747
2015-Sep-22 19:46 *m 116.6744 -0.2739
2015-Sep-22 19:47 *m 116.4243 -0.2730
2015-Sep-22 19:48 *m 116.1743 -0.2721
2015-Sep-22 19:49 *m 115.9242 -0.2713
2015-Sep-22 19:50 *m 115.6742 -0.2704
2015-Sep-22 19:51 *m 115.4241 -0.2696
2015-Sep-22 19:52 *m 115.1740 -0.2687
2015-Sep-22 19:53 *m 114.9240 -0.2678
2015-Sep-22 19:54 *m 114.6739 -0.2670
2015-Sep-22 19:55 *m 114.4239 -0.2661
2015-Sep-22 19:56 *m 114.1738 -0.2652
2015-Sep-22 19:57 *m 113.9237 -0.2644
2015-Sep-22 19:58 *m 113.6737 -0.2635
2015-Sep-22 19:59 *m 113.4236 -0.2626
2015-Sep-22 20:00 *m 113.1735 -0.2618
$$EOE
*******************************************************************************
Column meaning:
TIME
Prior to 1962, times are UT1. Dates thereafter are UTC. Any 'b' symbol in
the 1st-column denotes a B.C. date. First-column blank (" ") denotes an A.D.
date. Calendar dates prior to 1582-Oct-15 are in the Julian calendar system.
Later calendar dates are in the Gregorian system.
Time tags refer to the same instant throughout the universe, regardless of
where the observer is located.
The dynamical Coordinate Time scale is used internally. It is equivalent to
the current IAU definition of "TDB". Conversion between CT and the selected
non-uniform UT output scale has not been determined for UTC times after the
next July or January 1st. The last known leap-second is used over any future
interval.
NOTE: "n.a." in output means quantity "not available" at the print-time.
SOLAR PRESENCE (OBSERVING SITE)
Time tag is followed by a blank, then a solar-presence symbol:
'*' Daylight (refracted solar upper-limb on or above apparent horizon)
'C' Civil twilight/dawn
'N' Nautical twilight/dawn
'A' Astronomical twilight/dawn
' ' Night OR geocentric ephemeris
LUNAR PRESENCE WITH TARGET RISE/TRANSIT/SET MARKER (OBSERVING SITE)
The solar-presence symbol is immediately followed by another marker symbol:
'm' Refracted upper-limb of Moon on or above apparent horizon
' ' Refracted upper-limb of Moon below apparent horizon OR geocentric
'r' Rise (target body on or above cut-off RTS elevation)
't' Transit (target body at or past local maximum RTS elevation)
's' Set (target body on or below cut-off RTS elevation)
RTS MARKERS (TVH)
Rise and set are with respect to the reference ellipsoid true visual horizon
defined by the elevation cut-off angle. Horizon dip and yellow-light refraction
(Earth only) are considered. Accuracy is < or = to twice the requested search
step-size.
Azi_(a-appr)_Elev =
Airless apparent azimuth and elevation of target center. Adjusted for
light-time, the gravitational deflection of light, stellar aberration,
precession and nutation. Azimuth measured North(0) -> East(90) -> South(180) ->
West(270) -> North (360). Elevation is with respect to plane perpendicular
to local zenith direction. TOPOCENTRIC ONLY. Units: DEGREES
Computations by ...
Solar System Dynamics Group, Horizons On-Line Ephemeris System
4800 Oak Grove Drive, Jet Propulsion Laboratory
Pasadena, CA 91109 USA
Information: http://ssd.jpl.nasa.gov/
Connect : telnet://ssd.jpl.nasa.gov:6775 (via browser)
telnet ssd.jpl.nasa.gov 6775 (via command-line)
Author : Jon.Giorgini@jpl.nasa.gov
*******************************************************************************The sun’s angular diameter is about 32 arcminutes, so the sun’s lower limb is 16 arcminutes below the sun’s center. When the center of the sun has geometric elevation -18 arcminutes (-0.3 degrees), the lower limb has geometric elevation -34 arcminutes. Since refraction near the horizon is also 34 arcminutes, the sun’s lower limb rises when the sun’s geometric elevation is -0.3 degrees.
In the table above, this occurs between 1914 and 1915, but my program uses slightly more accurate data for the sun’s angular diameter, and the sun actually finishes rising between 1913 and 1914 (and closer to 1913).
You can then fly almost halfway across the world to latitude 89 degrees 51 minutes and longitude -19 degrees to see the one-minute-shorter longest sunset, which starts at 23 Sep 2015 at 2128 and ends at 25 Sep 2015 at 1648, a length of 43 hours and 20 minutes.
In this case, you would use http://aa.usno.navy.mil/data/docs/RS_OneYear.php to verify the ending time of the sunset, and HORIZONS to verify the start time of the sunset.
Polar sunrises and sunsets are considerably shorter:
-
At the North Pole, the sun starts rising at 18 Mar 2015 at 2015, and finishes rising at 20 Mar 2015 at 0441, a length of 32 hours and 26 minutes.
-
At the South Pole, the sun starts setting at 21 Mar 2015 at 1650, and finishes setting at 23 Mar 2015 at 0117, a length of 32 hours and 27 minutes.
-
At the South Pole, the sun starts rising at 21 Sep 2015 at 0508, and finishes rising at 22 Sep 2015 at 1400, a length of 32 hours and 52 minutes.
-
At the North Pole, the sun starts setting at 24 Sep 2015 at 0243, and finishes setting at 25 Sep 2015 at 1131, a length of 32 hours and 48 minutes.
Main caveat: Like HORIZONS and the sunrise/sunset tables above, I assume 34 arcminutes of refraction at the horizon. That’s reasonable for most locations, but may be unreasonable close the pole, where the longest sunrises and sunsets occur. In particular, refraction can change rapidly at these latitudes, allowing for potentially much longer sunrises and sunsets.
I now believe that http://what-if.xkcd.com/42/ is inaccurate, and will ping the author to let him know.
Answer 3 (score 3)
OK, lets start with the simplest mathematical approach to illustrate the path to a fully analytical answer. The sun presents an angular width of 32 arcminutes to any point on earth. That is 32/60 or 0.533 degrees of arc or angular span. Lets assume the Earth does not have its 23 degrees of tilt, for this first approximation. Then as a second approximation lets assume the Earth rotates around the sun in 24 hours, you are still on the equator. Our calculation is as follows;
0.533 degrees/360 degrees) = (hours sunset/24 hours).
Solve for hours sunset and you get,
24 hrs X (0.533/360) = 0.0355 hrs, which is
0.0355 hrs X 60 min/hrs = 2.13 minutes, which is
2.13 min X 60 secs/min = 128 seconds
OK, now that is the first order approximation only and explains the minima of the nice charts previously provided.
The first and trivial correction would be to notice that the 24 hour assumption is not accurate, hence leap years! Beyond that we have actually 23:56 per year. That will get you 127.56 seconds for sunset.
The real solution for the deep divers out there is to understand that the angular width of the sun in the sky is 32 arcminutes but only for any one instant in time for any one point on the Earth. So the next calculation would be to integrate over the diameter of the earth to incorporate the angular width of you are traversing during the sunset traverse time. You the observer are moving, rotating with the surface of the Earth, and hence you are spreading out the apparent angular size of the sun to the extent you are traversing during that sunset period, and this will add time to the sunset period.
Now that is the easier side of all of this. The next calculation would add the geometric correction for latitude which the observer is located in. This introduces horizontal relative component of motion of the sun to the observer, greatly increasing the time when one is not at the summer or winter equinoxes. (The prior calculations had the sun directly perpendicular to the rotation of the Earth.) In the titled Earth Sun systems, this effect is minimized at the equinox positions of the earth sun system and asymptotes toward the prior calculation if one is on the equator and on the equinox twice per year. Again, this is seen nicely in the charts of the previous answers.
I hope that help folks understand some of basic underpinnings of the math and geometry which the actual calculations must take into effect.
No calculators allowed and you can still get there.
7: Is it safe to watch solar eclipse’s reflection in water? (score 54462 in 2016)
Question
There will be a solar eclipse soon at my area… naturally I want to watch this with my son.
Is it safe for us to watch the solar eclipse’s reflection in the water? There is a swimming pool near my house and I plan to watch it there.. is it safe to do so?
I remember vividly doing this with my dad when I was little, but some website says we are not supposed to watch it in a bucket of water..
So is it ok to watch it in the water?
Answer accepted (score 8)
If it’s not safe to look at the sun’s reflection in the water normally then it’s no safer during an eclipse because even if 1% of the sun is showing this is more than enough light to seriously damage your eyes. The best way to view the sun, if you haven’t ordered your eclipse glasses yet, is to poke a hole in a piece of paper and project an image of the sun onto another, like so:
There are instructions on how to do this here on the JPL website.
8: moonless night and lunar phase (score 44074 in )
Question
How to define moonless night? There is no moon at all during some night?
How to calculate and know whether the moon would appear during one night?
If the lunar phase is full moon, is it possible that the moon will not appear ?
Answer accepted (score 9)
A moonless night is, as you suspect, a night in which the Moon does not appear visible in the sky. This happens once per month, when the Moon is near the Sun. Due to the proximity of the Moon and the Sun in the sky, at that time the Moon is the smallest sliver possible, and therefore not a full moon.
This is because it is actually the Sun that illuminates the Moon, and when the Sun and the Moon are in the same direction in the sky we are seeing the non-illuminated side of the Moon. Note the direction of the sunlight in this image:
Obviously, the direction of the sunlight is the direction of “up” during the day. If you look at the horizon slightly after sunset or slightly before sunrise, you might actually catch a glimpse of the sliver of Moon before it set or rises slightly after or before the Sun.
9: Can this picture of the moon directly above the sun from the horizon be real? (score 43196 in 2017)
Question
I recently saw this image, but somebody told me that it is fake, possibly photoshopped. The moon and the sun appear to be on the same perpendicular from the horizon. Is there any way to tell if this shot was actually captured?
Answer accepted (score 32)
No, it’s not real. Given that the moon appears to be nearly full, it would be near to the horizon opposite the sun, which is clearly not the case. Also the framing of the sun and moon between the trees means that the sun and moon are on the same side of the trees, which is not possible when the moon is full.
Answer 2 (score 9)
No, neither is true. The moon itself does not emit light at optical wavelengths. It only reflects the sunlight, although in rare cases you can see it reflecting light from our earth (in which case the light is also from the sun). The side being illuminated should face the sun. So, if you see the sun aligned with the moon as in the photo, it must be faked. But you need the moon and the sun to be in the same line of sight in this case, e.g. a solar/moon eclipse.
Answer 3 (score 6)
The position of the moon and the sun in this picture are possible.
It’s more feasible in between the tropics and depends on the direction of the moon orbit inclination.
The issue in this picture is the homogeneous brightness of the moon.
It’s look like a full moon, which happen when the earth is between the moon and sun, which is not the case in this picture.
So what the moon should look like ? In this case the visible part of the moon will be lighted by the sun in a small crescent. And the rest of the visible part of the moon will be lighted by an almost full earth. That it, a earthlight. The brightness contrast is gigantic but you can see it by your eyes.
Here a picture of it:
And another one with tree to have a brightness reference:
We can’t see such a contrast on the original image. And the picture doesn’t look like a HDR image.
So yes. This is a photomontage.
10: What is the formula to predict lunar and solar eclipses accurately? (score 38341 in 2013)
Question
A number of ancient civilizations had devised methods to predict exact dates and times of such eclipses, marking them as important events. Hence I assume the predictions were based on calculations, which should be quite easy to do now. So what is the exact formula to predict the exact date and time(this is optional, but desirable) of lunar and solar eclipses? Also how to calculate if the solar eclipse will be visible from a particular location or not?
Answer accepted (score 12)
The NASA sites have some very useful resources for this I will list them below:
Lunar Eclipses
This Link has an index for all lunar eclipses from -1999 to +3000, predominantly a statistics page but also has this page that contains how to calculate when lunar eclipses are.
There is more than one formula depending on which time frame you are trying to look in.
This is the formula for eclipses between the year 2005 and 2050:
\[\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2\]
Where:
\[y = year + (month - 0.5)/12\]
\[t = y - 2000\]
Solar Eclipses
This Link has an index like above but for all of the solar eclipses from -1999 to +3000.
This link has the formula for calculating solar eclipses. This is the formula for between 2005 and 2050:
\[\Delta T = 62.92 + 0.32217 * t + 0.005589 * t^2\]
Where:
\[y = year + (month - 0.5)/12\]
\[t = y - 2000\]
Answer 2 (score 9)
Calculation of solar eclipses can be done using Besselian elements. The basic idea is to compute the motion of the Moon’s shadow on a plane that crosses the Earth’s center. Then, the shadow cone of the Moon can be projected on the Earth surface. The Besselian elements are the following:
- X and Y: the coordinates of the center of the shadow in the fundamental plane
- D: the direction of the shadow axis on the celestial sphere
- L1 and L2: the radii of the penumbral and umbral cone in the fundamental plane
- F1 and F2: the angles which the penumbral and umbral shadow cones make with the shadow axis
- \(\mu\): the ephemeris hour angle
what you have to do now is to compute the variation of these parameters, which are time-dependent. It happens that it can be done using polynomial expensions for a given reference time \(t_0\). The polynomial expension is of the form, for a Besselian element a:
\[a = a_0 + a_1\times t + a_2\times t^2 + a_3\times t^3\]
(a third order expension is enough in general), with \(t = t_1 - t_0\), \(t_0\) being the Terrestrial Dynamical Time (TDT) to the nearest hour of the instant of greatest eclipse.
Sources:
11: Is stacking welder’s glasses a safe way to watch at the eclipse? (score 36707 in 2015)
Question
You can find in many place on the Internet that welder’s glass #14 is good for looking at an eclipse. Tomorrow (March, 20th 2015 at 10:45 CET) there’s a solar eclipse and yesterday I could only find glasses #11 and #9. I tried briefly this morning, and combining a #11 and a #9 was giving me a good view of the sun. Using 2 #11 gave a too dark image. Now, I read different opinions on the internet.
Against (Perkins Observatory): http://perkins.owu.edu/solar_viewing_safety.htm
Be careful that you use the right kind of glass! Welder’s glass is numbered from 1 to 14 with 14 being the darkest. It is only number 14 glass that is dark enough for solar viewing! And NO STACKING! A pair of number 7’s or a 10 and a 4 together DO NOT have the same protection as a single piece of number 14 (see unsafe methods for more details).
Favorable (Royal Astronomical Society of Canada): https://www.rasc.ca/tov/safety
If SN14 filter is not available, it is possible to combine lower shade numbers to get roughly the same level of eye protection from solar radiation, e.g. combining SN 6 and SN 8 filters. However the image quality may be considerably poorer than that seen through the single SN14 filter
I could not find a table or something explaining which kind of protection gives each number; according to the Canadian website, the only concern is about how much infrared light goes through, ultraviolet does not seem to be a problem in almost any case (I was surprised to read that).
Note: It’s not my intention to open here the discussion on what could be other safe methods to watch the eclipse, this is well explained everywhere around. I read too late about the eclipse to order specific glasses.
Edit: One of the answers here report the following formula: (more insight at this link)
13 or darker is safe enough. Also, you CAN add up welding glass, using the formula S(sum) = S1 + S2 -1. S(sum) should be greater than or equal to 13
Answer accepted (score 1)
YES
You have summarized pretty much everything available on the subject. There are loads of people saying you cannot, but fortunately for us, the Royal Astronomical Society of Canada, that you quote, includes evidence to back up their claims. They are a respected institution and the author of this article appears immensely qualified. The article itself does not include evidence that welding goggles can be stacked, but the inclusion of such for all the major points adds credibility to the entire article.
It does appear like the glasses are not exactly additive, this Reddit post goes through some calculations, but they are pretty close such that aiming for 1 tint higher will more than offset it and if anything is overkill.
Answer 2 (score -3)
I know it’s a bit late for the UK eclipse, but for future reference:
Welders goggles do not cut out the same frequency of light that Mylar/Black Polymer eclipse glases do.
Arc welding glass #14 is much darker than brazing welders glass but there is no guarantee that they will block the Sun’s ultra-violet and infrared rays.
The problem is there are no pain sensors in your retina so you don’t feel your eyes cooking and the dark lenses (as do sunglasses) cause your pupils to dialate letting in more ultra-violet and infrared rays so they could make things worse.
I have used arc Welders glass on a camera and it worked fine but gave a picture of a green Sun. Project the image on paper to take it.
Do NOT look through a camera viewfinder even with the correct eclipse glasses on if you do not have the correct solar filer on the LENS end.
12: How does neutron star collapse into black hole? (score 35060 in )
Question
We know the spectacular explosions of supernovae, that when heavy enough, form black holes. The explosive emission of both electromagnetic radiation and massive amounts of matter is clearly observable and studied quite thoroughly. If the star was massive enough, the remnant will be a black hole. If it wasn’t massive enough, it will be a neutron star.
Now there’s another mode of creation of black holes: the neutron star captures enough matter, or two neutron stars collide, and their combined mass creates enough gravity force to cause another collapse - into a black hole.
What effects are associated with this? Is there an explosive release of some kind of radiation or particles? Is it observable? What physical processes occur in the neutrons as they are subjected to the critical increase of pressure? What is the mass of the new black hole, comparing to its neutron star of origin?
Answer accepted (score 24)
A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place. See Chandrasekhar limit on wikipedia for details.
A neutron star is formed during a supernova, an explosion of a star that is at least 8 solar masses.
The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star, and then into a black hole.
We know that 1 electron + 1 proton = 1 neutron;
1 neutron = 3 quarks = up quark + down quark + down quark;
1 proton = 3 quarks = up quark + up quark + down quark;
A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star.
During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back.
During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons.
In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons.
If the neutron star’s mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.
The upper/lower mass limit for a quark star is not known (or at least I couldn’t find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole.
When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged, non-rotating-charged, non-rotating-uncharged.
What we would see visually during the transformation would be a hard radiation flash. This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs).
We know that atoms break up into protons, neutrons, electrons under pressure.
Under more pressure, protons and electrons combine into neutrons.
Under even more pressure, neutrons break down into quarks.
Under still more pressure, perhaps quarks break down into still smaller particles.
Ultimately the smallest particle is a string: open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that’s how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings.
Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene.
A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity.
We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate, giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa.
So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses).
This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole.
A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star, thus gaining mass.
Here is a nice drawing showing exactly that.
Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies.
Answer 2 (score 16)
Just to focus on one part of your question. Whilst it might be possible for a neutron star to accrete material, or for two neutrons stars to collide, in order to form black holes, this kind of event must be quite rare (although see below)
The distribution of measured neutrons star and black holes masses can be fitted with an estimated true distribution. Here it is, from Ozel et al. (2012). You can see there is a distinct gap between the highest mass neutron stars (currently the record holder has a mass of about \(2M_{\odot}\) and the smallest black holes (about \(5M_{\odot}\)). This confirmed slightly earlier work by Farr et al. (2011).
The merging of neutron stars must happen though. The obvious example is the Hulse-Taylor binary neutron star system, where the two object are spiralling together, presumably by the emission of gravitational waves, and will merge in about 300 million years. The combined mass of the 2 neutron stars is \(2.83M_{\odot}\), but the mass of any black hole they create would be lower, with the difference radiated away as neutrinos and gravitational waves.
Merging neutron stars (or merging neutron star + black hole binaries) are thought to be the progenitors of short duration gamma ray bursts or so-called Kilonova eventsthat are generally seen in high redshift galaxies. These typically last a second or less, but involve an energy release of about \(\sim 10^{44}\) J. They may produce a black hole, or perhaps a more massive neutron star. There will also be a gravitational wave signature (a “chirp”) that could be detected by the next generation of gravitational wave experiments (now a reality). These black holes may be isolated and hence not represented in the mass distribution above. A further observational signature of these events may be in the form of the current levels of a number of heavy r-process elements, like Iridium and Gold, that may mostly be produced in these events.
As for accretion onto an existing neutron star - well it looks quite rare because there may be a large gap between the highest masses at which neutron stars are produced in supernovae (maybe \(1.5M_{\odot}\)) and the maximum mass of a neutron star. We know that the latter is at least \(2M_{\odot}\), but it could be higher, perhaps \(3M_{\odot}\), the maximum allowed by General Relativity. As to the outcome of this hypothetical event, well sticking to non-speculative physics, the most likely thing to happen would be the production of massive hyperons in the neutron star core at sufficiently high densities (\(>10^{18}\) kg/m\(^3\)), which would lead to an instability (due to the removal of degenerate neutrons that are providing the majority of support); the neutron star may then slip inside its event horizon (about 6km for a \(2M_{\odot}\) neutron star) and become a black hole. Some sort of explosion seems unlikely, though a gravitational wave signature might be possible.
EDIT: An update on the NS/BH mass distribution above. I saw a talk recently at a conference - the explanation of the distribution has two broad thrusts; either the black holes are not produced in this mass range because of the physics of the progenitors, or there is a strong observational bias against seeing them. An example of the former explanation can be found in Kochanek (2014), who proposes that there is a class of “failed supernovae” between 16 and 25\(M_{\odot}\) that do manage to eject their envelopes in weak transient events, but leave behind their helium cores to form the lowest mass 5-8\(M_{\odot}\) black holes. Lower mass progenitors are then responsible for the neutron stars.
The observational bias is that the companions to the lowest mass black holes in binary systems may be always overflowing their Roche lobes. The resultant accretion signature swamps the companion spectrum and prevents a dynamical mass estimate (e.g. Fryer 1999). The Chandra Galactic Bulge Survey is attempting to find examples of quiescent, relatively low X-ray luminosity, eclipsing compact binaries, with which to measure a more unbiased black hole mass distribution.
Further Edit: There continue to be challenges and claims that there are “low-mass” black holes that could be formed via accretion-induced collapse of a neutron star (pointed out by Alexandra Veledina). For instance Cygnus-X3 has a claimed mass of \(2.4^{+2.1}_{-1.1}\ M_{\odot}\) according to Zdziarski et al. 2013, but these observations lack the precision to be really sure yet.
13: Did atoms in human body indeed come from stars? (score 32511 in 2014)
Question
I think I am not alone who saw videos about that we (humans) are made of same atoms which someday were in stars. In other words, some atoms in our bodies are from stars which exploded billions of years ago.
I wonder if it is indeed true. I mean human’s life begins when sperm cell fertilize egg cell. Now does that sperm cell or egg cell indeed contains some of the exact atoms from those stars?
I know little bit strange question, but would be interesting to hear if it is indeed true, that atoms in our body are same which someday were in stars.
In case you are wondering that’s the video I am talking about: http://www.youtube.com/watch?v=9D05ej8u-gU
Answer accepted (score 7)
The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.
So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.
After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that’s all. No oxygen, iron, carbon etc.
Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).
But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.
If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with \(A<90\)) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).
Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.
All these different processes produce distinctive patterns of element abundances.
The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.
Analysis of “presolar grains” found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.
[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]
Answer 2 (score 7)
The chemical elements in our bodies are inherited from the Earth. The Earth was formed in a disc of gas and dust swirling around the protosun 4.5 billion years ago. The material that formed the Earth was a selection of the material from that protostellar nebula that was itself once part of a larger molecular cloud.
So the atoms in our body were once part of this molecular cloud, so we need to understand how they got there.
After the first ten minutes or so, the universe contained mainly hydrogen, helium and some traces of lithium, deuterium and tritium - and that’s all. No oxygen, iron, carbon etc.
Almost all of the heavier chemical elements are made inside stars. We could stop there - the atoms of carbon, oxygen, calcium etc. in our bodies must have been made in stars, and since these atoms/nuclei are stable, they must survive unchanged (you could argue about whether their electrons get swapped about in chemical reactions etc., but since electrons are indistinguishable this hardly matters).
But how do they get into a molecular cloud and what sort of stars make these elements? A couple of answers correctly identify massive stars that explode as supernovae as important. But they are by no means the only contributor, or even the most important contributor for some elements.
If we take carbon and nitrogen, these are manufactured in nuclear reactions inside stars of even a bit less than a solar mass during the horizontal branch and asymptotic giant branch stages. These stars may be less massive and produce less C and N than massive stars, but there are many more of them. The central material is mixed to the surface during thermal pulses and the outer envelope, enriched in a variety of chemical elements, is gradually lost into space via a slow wind. This is a major source of carbon, nitrogen, fluorine, lithium and a number of heavy elements - Ba, La, Zr, Sr, Pb and many others - produced in the s-process. About 50% of the elements heavier than iron are made in the s-process, which can occur in both massive stars that explode (mainly isotopes with \(A<90\)) and the less massive AGB stars with slow, massive winds (elements up to lead and bismuth).
Iron, nickel and many other elements such as sulphur and silicon are also produced during type Ia supernovae. This is the detonation of a white dwarf, the end stage of a low-mass star, after mass transfer or merger. Milder novae explosions caused by the ignition of material accreted onto a white dwarf also enrich the interstellar medium.
All these different processes produce distinctive patterns of element abundances.
The enriched material is swept up by neighbouring supernova explosions, by interactions with spiral arms and other molecular clouds. It cools, condenses and collapses to form a new generation of stars.
Analysis of “presolar grains” found inside meteorites tells us what our solar system formed from. These analyses tell us that all of the above processes were important in making the chemical elements that made up the Earth and hence those in our bodies.
[Further details on the production of elements heavier than iron (including supernovae, low-mass AGB stars, colliding neutron stars etc.) can be found in my Physics SE answer to this question. ]
Answer 3 (score 4)
Almost all hydrogen nuclei (protons), some helium atomic nuclei and traces of lithium nuclei are thought to have formed early in the universe, after the big bang. Almost all other atomic nuclei are thought to have formed in stars or have decayed from atomic nuclei, which have formed in stars. A minor fraction forms by high-energy collisions with cosmic rays.
The electrons of the hull of atoms in parts formed during the big bang, part of them come into existence, when neutrons decay to protons. These neutrons may have been free neutrons, or neutrons bound in instable atomic nuclei.
Hence our body doesn’t contain the exact same atoms as they formed in stars. But without stars most of the atoms besides hydrogen in our body wouldn’t exist.
Our body contains many of the exact same atomic nuclei, as they formed in stars, not the exact same atoms/ions.
To be a little more precise: Our body doesn’t contain many free atoms, but mainly molecules and ions.
14: What is this rapidly twinkling red, blue, and white star I saw? (score 32108 in 2017)
Question
Last night, I was on my balcony at 1AM (PST) and I looked up and saw two stars near the horizon (I’d guess ~30 degrees above the horizon), and they were “twinkling” about twice as fast as other stars higher in the sky, and I could clearly see them changing from red to white to blue repeatedly. Other stars in the sky only appeared white to me, and didn’t seem to “twinkle” as rapidly as these two stars did. The red and blue make me think of red-shift and blue-shift, but I don’t know how I would see both from the same object.
What was I seeing?
I don’t know if it helps, but I am in the Los Angeles area, and I was looking in a roughly north direction. almost exactly to the east, according to google maps.
Edit: I tried taking a picture, but light pollution from the nearby street lights wouldn’t permit me taking a decent picture. However, I noticed a group of three stars close together in nearly a perfect almost vertical line, and managed to find that in Stellarium. I think I found the two stars I am seeing: Procyon and Sirius
Is there anything about either of these stars that would make them show as red/blue?
Answer accepted (score 8)
It’s most probably Sirius. At this time of year (at 1 am local time) it’s low in the sky in the East, so there is a lot of atmosphere in the way, and as Sirius is a bright bluish star, it will show all the colours described as it twinkles.
Answer 2 (score 5)
As you have already identified the objects you were seeing, I’ll explain the effect you were seeing. In that situation there are three things to consider: atmospheric chromatic dispersion, seeing and human color perception.
Light entering the atmosphere is refracted, because of the changing speed of light in air compared to the vacuum of space. The amount of refraction depends on the wavelength, causing optical dispersion. This effect is strongest for objects appearing near the horizon. Basically the different color components of the star’s image appear at slightly different elevations in the sky (nice in-depth explanation).
Turbulence in the atmosphere causes the “twinkling” or seeing. Moving pockets of hotter and colder air act like lenses that are projecting the star’s light into varying directions, so a varying amount of light reaches your eye. Together with the dispersion this produces a colorful twinkling.
Because the color perception in the human eye doesn’t work in faint light, this colorful twinkling is observed for the brightest stars only.
The doppler shift is not the reason for the red and blue colors. At a typical velocity dispersion of 30km/s of our surrounding stars, the doppler shift changes the wavelength by an imperceptible 10-4 fraction. Also this would have to change extremely rapidly, which simply does not occur for our average night-sky stars or most other objects, for that matter :-)
Answer 3 (score 5)
As you have already identified the objects you were seeing, I’ll explain the effect you were seeing. In that situation there are three things to consider: atmospheric chromatic dispersion, seeing and human color perception.
Light entering the atmosphere is refracted, because of the changing speed of light in air compared to the vacuum of space. The amount of refraction depends on the wavelength, causing optical dispersion. This effect is strongest for objects appearing near the horizon. Basically the different color components of the star’s image appear at slightly different elevations in the sky (nice in-depth explanation).
Turbulence in the atmosphere causes the “twinkling” or seeing. Moving pockets of hotter and colder air act like lenses that are projecting the star’s light into varying directions, so a varying amount of light reaches your eye. Together with the dispersion this produces a colorful twinkling.
Because the color perception in the human eye doesn’t work in faint light, this colorful twinkling is observed for the brightest stars only.
The doppler shift is not the reason for the red and blue colors. At a typical velocity dispersion of 30km/s of our surrounding stars, the doppler shift changes the wavelength by an imperceptible 10-4 fraction. Also this would have to change extremely rapidly, which simply does not occur for our average night-sky stars or most other objects, for that matter :-)
15: What is the name of our Solar System? (score 31525 in 2015)
Question
what is our Solar System called rather than “the Solar System”?
I’ve found “Sol System” and “Monmatia”, too, but is there more to it?
Answer accepted (score 11)
It’s just called “the Solar System”. (Plenty of places and objects have names like that; it’s no different from “the Arctic” or “the Moon” or “the Sun”.)
(“Sol system” is an invention of science fiction writers; it has no general use outside some science fiction contexts. Anything else is going to be something similar, or general crackpottery of one kind or another.)
Answer 2 (score 6)
The solar system was named long before we knew that other solar systems existed. Just like the Sun is not the only Sun in the universe but we still call it the Sun.
Answer 3 (score 2)
To answer your question succinctly, the Solar System also goes by the names: The Copernican System, The Heliocentric System, and The Planetary System, in addition to the ones you have mentioned. There aren’t too many other names, actually, so just stick to Solar System since it’s the most widely accepted.
16: How many planets are there in this solar system? (score 31346 in 2015)
Question
So, in school (that’s a long time age) they have been teaching us there are 9 planets in our solar system.
- Mercury
- Venus
- Earth
- Mars
- Jupiter
- Saturn
- Uranus
- Neptune
- Pluto
But every now and then I keep reading stories about another “dwarf planet” (Eris, discovered in 2005) that - depending on what source tells the story - is another planet according to the astronomical definition, while other sources say that it isn’t a planet. Some even say Pluto isn’t a planet anymore either.
The result: I’m confused due to the contradicting stories. Even Wikipedia isn’t clear about Eris and only writes (emphasis mine):
NASA initially described it as the Solar System’s tenth planet.
Initially? So, is it a 10th planet or not? Fact is, there is another “something” out there and it surely seems to look like a planet. Yet, some people keep stating there are 9 planets in our solar system, while others say there are more than 9 planets, and then again there are people stating that the latest definition of “planet” has kicked out Pluto too so there are actually fewer than 9 planets in our solar system.
Trying to get a definite, official, and astronomically correct answer I can actually rely on, I’m therefore asking: How many planets are there in this solar system?
EDIT
The “Definition of planet” at Wikipedia doesn’t really help either, as it states:
Many astronomers, claiming that the definition of planet was of little scientific importance, preferred to recognize Pluto’s historical identity as a planet by “grandfathering” it into the planet list.*
* Dr. Bonnie Buratti (2005), “Topic — First Mission to Pluto and the Kuiper Belt;”From Darkness to Light: The Exploration of the Planet Pluto"", Jet Propulsion Laboratory. Retrieved 2007-02-22.
So, if you link somewhere to provide proof, it would be great if you could point me to a more trusted source than Wikipedia. Ideally, an astronomical trusted source and/or paper.
Answer accepted (score 68)
In addition to Undo’s fine answer, I would like to explain a bit about the motivation behind the definition.
When Eris was discovered, it turned out to be really, really similar to Pluto. This posed a bit of a quandary: should Eris be accepted as a new planet? Should it not? If not, then why keep Pluto? Most importantly, this pushed to the foreground the question
what, exactly, is a planet, anyway?
This had been ignored until then because everyone “knew” which bodies were planets and which ones were not. However, with the discovery of Eris, and the newly-realized potential of more such bodies turning up, this was no longer really an option, and some sort of hard definition had to be agreed upon.
The problem with coming up with a hard definition that decides what does make it to planethood and what doesn’t is that nature very rarely presents us with clear, definite lines. Size, for example, is not a good discriminant, because solar system bodies come in a continuum of sizes from Jupiter down to meter-long asteroids. Where does one draw the line there? Any such size would be completely arbitrary.
There is, however, one characteristic that has a sharp distinction between some “planets” and some “non-planets”, and it is the amount of other stuff in roughly the same orbit. This is still slightly arbitrary, because it’s hard to put in numbers exactly what “roughly” means in this context, but it’s more or less unambiguous.
Consider, then a quantity called the “planetary discriminant” µ, equal to the ratio of the planet’s mass to the total mass of other bodies that cross its orbital radius and have non-resonant periods (so e.g. Neptune doesn’t count as sharing Pluto’s orbit) up to a factor of 10 longer or shorter (to rule out comets, which has little effect in practice). This is still a bit arbitrary (why 10?) but it’s otherwise quite an objective quantity.
Now take this quantity and calculate it for the different bodies you might call planets, comparing it to both the objects’ mass,
and their diameter,
or with an arbitrary horizontal axis, in order of decreasing discriminant,
Suddenly, a natural hard line emerges. If you look only at the mass and the diameter of the objects (shown in the insets above the plots), then there is a pretty continuous spread of values, with bigger gaps between the gas giants and the terrestrial planets than between Mercury and Eris/Pluto. However, if you look at the planetary discriminant, on the vertical axis, you get a very clear grouping into two distinct populations, separated by over four orders of magnitude. There’s a finite set of bodies that have “cleared their orbits”, and some other bodies which are well, well behind in that respect.
This is the main reason that “clearing its orbital zone” was chosen as a criterion for planethood. It relies on a distinction that is actually there in the solar system, and very little on arbitrary human decisions. It’s important to note that this criterion need not have worked: this parameter might also have come out as a continuum, with some bodies having emptier orbits and some others having slightly fuller ones, and no natural place to draw the line, in which case the definition would have been different. As it happens, this is indeed a good discriminant.
For further reading, I recommend the Wikipedia article on ‘Clearing the neighbourhood’, as well as the original paper where this criterion was proposed,
What is a planet? S Soter, The Astronomical Journal 132 no.6 (2006), p. 2513. arXiv:astro-ph/0608359.
which is in general very readable (though there are some technical bits in the middle which are easy to spot and harmless to skip), and from which I took the discriminant data for the plots above.
Edit: I must apologize for having included, in previous versions of this post, an incorrect plot, caused by taking data from Wikipedia without verifying it. In particular, the planetary discriminant for Mars was wrong (1.8×105 instead of 5.1×103), which now puts it below Neptune’s instead of just below Saturn’s, but the overall conclusions are not affected. The Mathematica code for the graphics is available at Import["http://goo.gl/NaH6rM"]["http://i.stack.imgur.com/CQA4T.png"].
… and, as a final aside: Pluto is awesome. It was visited in July 2015 by the New Horizons probe, which found a world that was much more rich, dynamic, and active than anyone expected, including what appear to be churning lakes of solid nitrogen ringed by mountains of water ice, among other marvels.
(Note the image has been colour-enhanced to bring out the variety of surface materials; the true-colour version of this image is here.) I, personally, don’t feel it’s at all necessary to ‘grandfather’ Pluto into the list of planets to really feel the awe at the amazing place it is - it’s perfectly OK for it to be a cool place with cool science, that is also not a planet.
Answer 2 (score 68)
In addition to Undo’s fine answer, I would like to explain a bit about the motivation behind the definition.
When Eris was discovered, it turned out to be really, really similar to Pluto. This posed a bit of a quandary: should Eris be accepted as a new planet? Should it not? If not, then why keep Pluto? Most importantly, this pushed to the foreground the question
what, exactly, is a planet, anyway?
This had been ignored until then because everyone “knew” which bodies were planets and which ones were not. However, with the discovery of Eris, and the newly-realized potential of more such bodies turning up, this was no longer really an option, and some sort of hard definition had to be agreed upon.
The problem with coming up with a hard definition that decides what does make it to planethood and what doesn’t is that nature very rarely presents us with clear, definite lines. Size, for example, is not a good discriminant, because solar system bodies come in a continuum of sizes from Jupiter down to meter-long asteroids. Where does one draw the line there? Any such size would be completely arbitrary.
There is, however, one characteristic that has a sharp distinction between some “planets” and some “non-planets”, and it is the amount of other stuff in roughly the same orbit. This is still slightly arbitrary, because it’s hard to put in numbers exactly what “roughly” means in this context, but it’s more or less unambiguous.
Consider, then a quantity called the “planetary discriminant” µ, equal to the ratio of the planet’s mass to the total mass of other bodies that cross its orbital radius and have non-resonant periods (so e.g. Neptune doesn’t count as sharing Pluto’s orbit) up to a factor of 10 longer or shorter (to rule out comets, which has little effect in practice). This is still a bit arbitrary (why 10?) but it’s otherwise quite an objective quantity.
Now take this quantity and calculate it for the different bodies you might call planets, comparing it to both the objects’ mass,
and their diameter,
or with an arbitrary horizontal axis, in order of decreasing discriminant,
Suddenly, a natural hard line emerges. If you look only at the mass and the diameter of the objects (shown in the insets above the plots), then there is a pretty continuous spread of values, with bigger gaps between the gas giants and the terrestrial planets than between Mercury and Eris/Pluto. However, if you look at the planetary discriminant, on the vertical axis, you get a very clear grouping into two distinct populations, separated by over four orders of magnitude. There’s a finite set of bodies that have “cleared their orbits”, and some other bodies which are well, well behind in that respect.
This is the main reason that “clearing its orbital zone” was chosen as a criterion for planethood. It relies on a distinction that is actually there in the solar system, and very little on arbitrary human decisions. It’s important to note that this criterion need not have worked: this parameter might also have come out as a continuum, with some bodies having emptier orbits and some others having slightly fuller ones, and no natural place to draw the line, in which case the definition would have been different. As it happens, this is indeed a good discriminant.
For further reading, I recommend the Wikipedia article on ‘Clearing the neighbourhood’, as well as the original paper where this criterion was proposed,
What is a planet? S Soter, The Astronomical Journal 132 no.6 (2006), p. 2513. arXiv:astro-ph/0608359.
which is in general very readable (though there are some technical bits in the middle which are easy to spot and harmless to skip), and from which I took the discriminant data for the plots above.
Edit: I must apologize for having included, in previous versions of this post, an incorrect plot, caused by taking data from Wikipedia without verifying it. In particular, the planetary discriminant for Mars was wrong (1.8×105 instead of 5.1×103), which now puts it below Neptune’s instead of just below Saturn’s, but the overall conclusions are not affected. The Mathematica code for the graphics is available at Import["http://goo.gl/NaH6rM"]["http://i.stack.imgur.com/CQA4T.png"].
… and, as a final aside: Pluto is awesome. It was visited in July 2015 by the New Horizons probe, which found a world that was much more rich, dynamic, and active than anyone expected, including what appear to be churning lakes of solid nitrogen ringed by mountains of water ice, among other marvels.
(Note the image has been colour-enhanced to bring out the variety of surface materials; the true-colour version of this image is here.) I, personally, don’t feel it’s at all necessary to ‘grandfather’ Pluto into the list of planets to really feel the awe at the amazing place it is - it’s perfectly OK for it to be a cool place with cool science, that is also not a planet.
Answer 3 (score 11)
The correct answer is 8 (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune).
Pluto is not longer a planet since 2006 when the IAU adopted a formal definition of planet
17: Does the Sun belong to a constellation? (score 29014 in 2015)
Question
Each new star we find is generally considered to be part of the constellation it is nearest to.
Our Sun is obviously a star, just much closer. Is our Sun part of any constellation? If so, which constellation is it a part of?
Answer accepted (score 36)
Constellations are human constructs to make sense of the night sky. When you are trying to find your way around, it helps to “chunk” stars into patterns and assign those groupings names. When I want to point out a particular object in the sky (say Polaris, the North Star), I start by pointing out a familiar constellation (say Ursa Major, the Big Dipper). From there, I can tell my friend to follow this or that line to get them to look where I’m looking:
With the advent of computerized telescopes and large data sets, constellations are less important for professional astronomers. However, many stellar databases use Flamsteed or Bayer designations, which assign stars to constellations. In order to include all stars, the sky is divided into irregular regions that encompass the familiar constellations.
So, which constellations is the Sun assigned to? Well, from the perspective of someone on the Earth, the Sun moves through the constellations throughout the course of the year. Or rather, Sol moves through the region of the sky where some of the constellations would be seen if its light did not drown out distant stars. Our moon and the rest of the planets move through those same constellations. (The Greek phrase which gives us the word “planet” means “wandering star”.)
The current position of the sun against the background of distant stars changes over the course of the year. (This is important for astrology.) It’s a little easier to make sense of with a diagram:
So perhaps a better question is:
What constellation does the Sun belong to today?
Presumably an observer on an exoplanet would assign Sol to some constellation that is convenient from her perspective. But from our perspective within the Solar system our sun, moon, and planets are not part of any constellation.
Answer 2 (score 16)
No, it does not. The constellations are fixed (on time scales long enough for humans to consider as fixed, at least) patterns of stars which exist on the celestial sphere. This celestial sphere is a coordinate system which has the Earth at its center. From the Earth’s perspective, the sun rises and sets at the same rate as the constellations, but as the Earth revolves around the sun, the constellation the sun will appear to be in will change.
However, the constellation the sun appears to be in (if projected onto the celestial sphere), is exactly how the constellations of the Zodiac are determined.
Answer 3 (score 0)
We did an exercise on this particular subject at Secondary Modern School in 1958. With reference to“Patrick Moor” publications a 3-dimentional model wasmade of the nearest 50 stars. Particular reference to the “well known” ie:the plough.As the model evolved,those we are familiar with loose their familiarity because of the distance.The plough is familar if assumed all the same distance,they are not and the model highlights the lack of pattern if viewed from Orion’s Belt. Our Sun did not appear to fit into any“Constellation type pattern”from any angle and all familiar ones were lost when viewed anywhere else. An interesting “Experiment” though.Try it.Cheers Colin
18: What is the distance that the Moon travels during one orbit around the Earth? (score 28429 in 2014)
Question
Also, does it always take the same amount of time, or does it fractionally differ on each revolution?
Answer accepted (score 15)
The Moon has an orbital eccentricity of 0.0549, so its path around the Earth is not perfectly circular and the distance between the Earth and the Moon will vary from the Earth’s frame of reference (Perigee at 363,295 km and apogee at 405,503 km), see for example second animation explaining Lunar librations in this answer.
But its orbit can be said, in an oversimplified manner, to be periodic, with no significant apsidal precession (not really true, but somewhat irrelevant for my following musings here to be still close enough), so we can calculate its orbital length based on its quoted average orbital speed of 1.022 km/s and orbital period of 27.321582 days.
So, plugging our numbers in a calculator, \(l = v * t\), we get the Moon’s orbital length of 2,412,517.5 km (or 1,499,070 miles). Should be close enough. Source of all orbital elements of the Moon is Wikipedia on Moon.
Answer 2 (score 10)
Concerning your first question, a simple estimation can be done assuming the distance Earth-Moon ≅ 4·10⁵km, and the orbit circular. So you can calculate the distance as a circumference (C=2πr) like that:
2π·4·10⁵km =8π·10⁵km ≅ 2.4 millions of kilometers
Of course you can do more precise calculations, but sometimes is good to have at first an idea of the orders of magnitude.
19: Could this be a cosmic ray hit on my cameras sensor (CMOS, DSLR)? (score 28374 in )
Question
While taking pictures of the bubble nebula I noticed a very strange artifact on only one of my pictures. It can’t be a satellite since its not a straight line and this is a 90s exposure that should make a line much longer. Also I doubt this is just readout noise from the camera (image is unstretched, unedited) since hot pixels don’t form lines. So could this be a cosmic ray hit? If not does anyone have an idea what this could be?
The image was taken with my Canon 7D Mark II @ISO1600, 90s exposure. Celestron C8 SGT (XLT), Celestron Advanced VX
Answer accepted (score 8)
It’s a track about 50 pixels long, at 4.1 \(\mu m\) per pixel that’s 200 \(\mu m\) in the plane of the sensor.
It’s about 1 pixel wide, quite lumpy, and seems to change direction a bit. It really fits the description of an ionizing particle. There are multiple possible sources,
I’m pretty sure @JamesK has nailed it as an electron, but what else can we learn? There must be some way to get a very rough idea of how many \(e^- h^+\) pairs correspond to a fully exposed pixel at “@ISO1600”, then we could convert it to a dE/dx and see if it’s minimum ionizing or if it’s much higher.
Going diagonally, with a density of 2.3 g/cm\({}^3\) that’s about 1.3 mg/cm\({}^2\) areal density per pixel. A [minimum ionizing particle] at 1.5 MeV / g cm\({}^2\) would deposit only about 2000 eV which would make less than 1000 \(e^- h^+\) pairs. That might correspond to a bright pixel. It’s not out of the question that it’s a minimum ionizing particle.
20: Why don’t stars move in the night-sky as the moon does? (score 27721 in 2014)
Question
I want to know why don’t we see a change in the position of stars during the night while we do see a change in the position of the moon. I’ve checked other online sources and some answers say that it is because they are too far away, is this true?
I mean for example if i’m laying down on the earth since sunset looking at the stars, and let’s say I’m close to the equator, no matter the distance, I’ll be looking at the stars and earth will turn about 180° until sunrise, so my angle of view will be about 180°.
Answer accepted (score 12)
This is referred to as diurnal motion, due to Earth’s rotation on its axis, and it affects apparent motion of stars differently depending on their position on the skies relative to the axis of Earth’s rotation. For example, on northern hemisphere, the star that appears not to move at all is positioned so that the earth’s axis of rotation points directly towards it, and is called a Polaris due to its role as a pole star. Stars further away from true north (not to be confused with magnetic north), will appear to prescribe a circle around it as the Earth completes one rotation on its axis:
A long exposure photograph of the night skies, showing their apparent motion
So what you assert, that the stars don’t seem to move isn’t exactly true. Stars closer to celestial equator will move at equal apparent radial velocity as any object fixed on the night skies would (so at the speed of Earth’s rotation), while the stars closer to celestial poles, while maintaining this radial velocity, will prescribe a smaller circle and appear to be more stationary. And the stars exactly aligned with the Earth’s axis (like Polaris, but there isn’t any such exact counterpart on the southern hemisphere), would appear not to move at all.
Answer 2 (score 2)
As we view it, the moon moves westward across the sky due to the rotation of the earth, while at the same time it’s moving eastward in its orbit at about 1 lunar diameter in about 57 minutes, or about 1 lunar diameter per hour or 12-14 degrees per day.
That’s why the moon can be seen blocking out (or occulting) a bright star or planet as it slowly moves eastward in front of it for several minutes.
The actual motion of a star is too small to notice and it takes years to notice it without powerful telescopes. For general purpose, day-to-day astronomical observations, the actual motions of the stars in 3D space can be ignored.
21: How to calculate the expected surface temperature of a planet (score 26491 in 2016)
Question
I’m writing a program to generate solar systems but I’m having trouble calculating the expected temperature of a planet. I have found a formula to calculate this, but I haven’t been able to get a remotely correct answer out of it as it doesn’t clearly state what units your supposed to use.
This formula I found:
\[4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi R ^ 2 L_{\odot}(1 - a)}{(4 \pi d ^ 2)}\]
where \(R\) is the planet’s radius (not sure what units), \(d\) is the distance from the Sun (it mentions AU), \(a\) is the albedo, \(L_{\odot}\) is the luminosity of the Sun (which I assume can be interchanged with the luminosity of any star), \(T\) is the temperature of the planet (kelvin, this is what I’m trying to get), and \(ơ\) is the Stefan-Boltzmann constant.
The site I found it on is notes for an astronomy college course. Here is the link:
http://www.astronomynotes.com/solarsys/s3c.htm#
Any help would be very much appreciated.
Answer accepted (score 8)
The formula
\[4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi R ^ 2 L_{\odot}(1 - a)}{4 \pi d ^ 2}\]
is correct, if you want to calculate the radiative equilibrium temperature. You only need to use the right units. We can further simplify the formula to
\[T ^ 4 = \frac{ L_{\odot}(1 - a)}{16 \pi d ^ 2 ơ}\;.\]
You should input the luminosity in watts, the distance to the star in meters and the Stefan-Boltzmann constant as
\[σ = 5.670373 × 10^{−8} \;\mathrm{W}\; \mathrm{m}^{−2}\; \mathrm{K}^{−4}.\]
The albedo is dimensionless. The resulting temperature will be in Kelvins. Let me make an example for Earth:
\(d = 149,000,000,000 \;\mathrm{m}\)
\(L = 3.846×10^{26} \;\mathrm{W}\)
Albedo of Earth is 0.29. (The Bond albedo should be used.) You will get
\[ T ^ 4 = \frac{ 3.846×10^{26}(1 - 0.29)}{16 \pi \times (149,000,000,000) ^ 2 \times (5.670373 × 10^{−8})}=4,315,325,985 \;\mathrm{K}^4\;. \]
After powering this number to 1/4, we obtain temperature 256 K, which is -17° C. This looks reasonable. The real average temperature on Earth is closer to 15° C, but the greenhouse effect is responsible for the difference.
22: What will happen to life on Earth when the Andromeda and Milky Way galaxies collide? (score 25044 in 2015)
Question
It is said that the Andromeda and Milky Way galaxies are coming close to each other with a speed of approximately 400000 km/hour. They will be together in the next 4 billion years.
- What will happen to life on Earth or human beings on Earth?
- If we are about to collide in the next 4 billion years then how long before we should take action for interstellar voyage?
- Are scientists working on such projects for an interstellar voyage?
- Will we be able to get the something like Earth to go away from this galaxy/Earth/ solar system and, considering speed of human instruments/spaceships, how long will it take to go to a safe place?
Answer accepted (score 21)
What will happen to life on earth or human beings on earth?
Assuming that human beings, or life, still exists on Earth at that time, they will have survived so much due to the ongoing death of the sun, that the gravitational pertubations due to the galactic collision will be nothing.
Keep in mind that in about 1-2 billion years, the sun will be so hot and large that all the water will have boiled off the earth into space. About 3 billion years from now, the surface of the Earth will be so hot that metals will be melting.
Any life that has survived those events and still lives on Earth will surely take a galactic collision in stride.
I imagine, though, that most humans will have fled Earth - if not for distant star systems, then at least for planets in our own system that are going to be warming up enough for human habitation.
If we are about to collide in next 4 billion year then how long before we should take action for inter-stellar voyage?
As soon as possible. When interstellar voyage becomes possible, we should start sending out ships to colonize other planets and star systems. This will likely take a long time, but if we are to survive more than a billion years, it is necessary. Keep in mind that the Sun and Andromeda are events we can predict. We don’t know, and can’t predict, the next cataclysmic asteroid strike, which is likely to happen in a shorter time than a billion years. There are lots of reasons to exit the planet, we should be worried about the ones we can’t predict or see, not the ones we can predict.
Are scientists working on such projects for inter-stellar voyage?
Yes, but in small steps. Manned missions to space, to the moon, and living aboard the ISS have provided significantly valuable information that will be used in such interstellar missions. As we continue to push the boundaries of our ability to survive in space we eventually will be able to live in space, perhaps whole lifetimes will be spent in space. As engine technology progresses beyond simply lifting people out of Earth’s gravitational well, we will eventually be sending people on long voyages outside our solar system.
It’s a very, very long way off, but each advance takes us closer to that eventual goal.
Answer 2 (score 18)
4 billion years is the same timeframe of the life remaining to our Sun.
So if we have not yet invented interstellar voyages, we’re screwed, with or without Andromeda.
Besides, stars do not interact directly with each other in a galactic collision. What we will notice from the several stars we are on is that star orbits around the galactic centre will be altered by the massive gravitational perturbation made by the other galaxy. Almost all stars will change orbits from that of orbiting around the galactic centre to orbiting around the center of mass of both galaxies. Some stars will be ejected out of the new bigger galaxy. It is quite safe to think that planets will continue orbiting their stars, but not that they will not have some alterations in their orbits.
On the other hand, from the gas to gas interaction, there will be a lot of new pressure waves in the interstellar medium, which will account for the formation of billions of new stars in new nebulae.
Answer 3 (score 16)
First, note that by the time Andromeda is close enough for collisions with wandering stars to become a concern, Earth’s average temperature will have changed significantly, and the planet will be unrecognizable.
When Sol is 8.5 billion years old, it will still have hydrogen available for fusion, but as it fuses it contracts and expands differentially. The contraction causes hydrogen fusion to become more favorable, so that Sol will have 50% greater power output (\(6 \times {10}^{26}\ \mathrm{W}\)) and 3% greater effective temperature (\(6000\ \mathrm{K}\)). Fusion also causes Sol to lose mass at a prodigious rate (currently \(4 \times {10}^9 \mathrm{kg/s}\)); it will release \(6 \times {10}^{43} \mathrm{J}\) from fusion, which corresponds to \(7 \times {10}^{26}\ \mathrm{kg}\). That is about one hundred Earth masses of sunlight but only \(1 \over 3000\) the mass of Sol. Gravitation with Earth decreases proportionally, so Earth’s orbit might on average expand \(3000\ \mathrm{km}\) per billion years. Other gravitational effects might change Earth’s average distance by as much as \(6 \times {10}^5\ \mathrm{km}\), 4‰ of an astronomical unit. Expansion of Sol’s outer layers due to reduced gravitation will increase its radius by 20%, \(3 \times {10}^5\ \mathrm{km}\). Thus Earth will receive nearly 50% more power as well.
The energy balance of Earth wrt Sol gives the expected surface temperature:
\[ \begin{align} \bar{a} = & 0.7 & \small\text{(Average absorption)} \\ P_p = & 1366\ \mathrm{W/m^2} & \small\text{(Average solar flux incident on Earth at present)} \\ P_f = & P_p \cdot 1.5 \approx 2000\ \mathrm{W/m^2} & \small\text{(In future)} \\ \sigma = & 5.670373 \times {10}^{-8}\ \mathrm{W/m^2/K^4} & \small\text{(Stefan-Boltzmann constant)} \\ \\ T_p^4 = & \frac{\bar{a} P_p}{4 \sigma} \\ \approx & \frac{0.7 \cdot 1366\ \mathrm{W/m^2}}{2.268149 \times {10}^{-7}\ \mathrm{W/m^2/K^4}} \\ \approx & 4.2 \times {10}^9\ \mathrm{K^4} \\ T_p \approx & 250\ \mathrm{K} \\ \\ T_f \approx & T_p \cdot {1.5}^{1/4} \approx T_p \cdot 1.11 \\ \approx & 280\ \mathrm{K} \end{align} \]
Since the average surface temperature on Earth is not \(-20\ \mathrm{°C}\) — it is \(+15\ \mathrm{°C}\) and already around \(8\ \mathrm{K}\) warmer than in an airless future — we can see the atmosphere has a significant role in retaining heat. Assuming increasing cooling needs do not lead to the atmosphere retaining more heat, the average surface temperature can be expected to rise to \(+50\ \mathrm{°C}\).
The average temperature of Antarctica is now \(240\ \mathrm{K}\) in winter and \(270\ \mathrm{K}\) in summer. These can be expected to rise to \(270\ \mathrm{K}\) (just below freezing) and \(300\ \mathrm{K}\) (well above freezing) respectively, and this is a best-case scenario. Antarctica will melt. That will produce the largest component (60%) of sea level increase, in total around \(100\ \mathrm{m}\).
If Earth were still inhabited four billion years from now, it is extremely unlikely that Earth would fall into a star from Andromeda.
Space is big. Really big. You just won’t believe how vastly, hugely, mindbogglingly big it is.
— Douglas Adams, The Hitchhiker’s Guide to the Galaxy
The Milky Way is about 100,000 light years in diameter and contains about 400 billion stars. Andromeda is bigger and denser; it may have one trillion stars and a diameter of 140,000 light years. It is 2.5 million light years away but appears six times larger than Sol.
\[ \begin{align} d_M \approx & \frac{4 \times {10}^{11}\ \mathrm{stars}}{{10}^{10} \pi/4\ \mathrm{{ly}^2}} \\ \approx & 50\ \mathrm{stars/{ly}^2} \\ \\ d_A \approx & \frac{{10}^{12}\ \mathrm{stars}}{2 \times {10}^{10} \pi/4\ \mathrm{{ly}^2}} \\ \approx & 60\ \mathrm{stars/{ly}^2} \\ \end{align} \]
If the two galaxies were simply superposed, there would be about one hundred stars per square light year, viewed from infinitely far along the rotation axis. However, the Milky Way is a 2:1 ellipse as seen from Andromeda, while we see Andromeda as a 3:1 ellipse. Projecting both onto a plane between them, perpendicular to a line between their central black holes, would give a region of overlap with dimensions between \(50 \times 50\ \mathrm{{kly}^2}\) and \(50 \times 100\ \mathrm{{kly}^2}\), with at most half the Milky Way outside it. Sol is likely to be involved in the collision, since it is about 27,200 light years from the galactic center.
That doesn’t mean, though, that Earth will come close to another star, that Sol might collide, or that the solar system will be disrupted.
Considering probabilistically the worst-case scenario (the entire Milky Way falls through Andromeda on their first pass), there is a mean free path for stars. The actual stellar density of the colliding galaxies is:
\[ \rho \approx 1.4 \times {10}^{12}\ \mathrm{stars}\ /\ V_{A \cup M} \]
where the union of the two galaxies’ volumes would be a very complicated expression. Very roughly, their volumes can be described as joined cones, ignoring their spheroidal dark matter halos (which are mostly harmless).
\[ \begin{align} \rho \approx & \frac{1.4 \times {10}^{12}\ \mathrm{stars}} {\left( \frac{1}{2} \cdot \left( {10}^3\ \mathrm{ly} \cdot {10}^{10} \pi/4\ \mathrm{{ly}^2} + 1.4 \times {10}^3\ \mathrm{ly} \cdot 2 \times {10}^{10} \pi/4\ \mathrm{{ly}^2} \right) \cdot \frac{1}{3} \right)} \\ \approx & 0.28\ \mathrm{stars/{ly}^3} \\ \\ V_\star \approx & 3.6\ \mathrm{{ly}^3} \\ \\ r_\star \approx & {\left( V_\star \cdot \frac{3}{4 \pi} \right)}^{1/3} \\ \approx & 0.95\ \mathrm{ly} \end{align} \]
At a distance of 1.9 light years, Betelgeuse would look a lot like Mars. If we assume disaster results from a star closer than the diameter of the heliosphere (about 200 AU), then:
\[ \begin{align} m = & \frac{1\ \mathrm{star}}{\rho \cdot \pi \cdot 4 \times {10}^4\ \mathrm{{AU}^2}} \\ \approx & 1.1 \times {10}^{21}\ \mathrm{m} \\ \approx & 7.2 \times {10}^9\ \mathrm{AU} \\ \approx & 1.1 \times {10}^5\ \mathrm{ly} \end{align} \]
On average, a star can travel 110 thousand light years before it grazes past another, slightly less than the diameter of Andromeda. The proportion of stars from the Milky Way that do not approach within 200 AU of stars in Andromeda is at least \(1/e^{1.4 / 1.1} \approx 100/400\ \mathrm{billion\ stars}\). For Earth to approach within 4 AU of another star (one Betelgeuse radius), it can be expected to travel at least 2500 times farther, which at a relative velocity of 300 km/s would take \(9 \times {10}^{18}\ \mathrm{s} \approx 300\ \mathrm{billion\ years}\).
23: Why did Venus not lose its atmosphere without magnetic field? (score 24393 in 2019)
Question
It is often stated that the magnetosphere not only shields the planet from cosmic radiation, but also prevents atmospheric loss. Why then did Venus not lose most of its atmosphere if it doesn’t have a strong magnetic field? Is there another mechanism at play, or is the statement about the importance of magnetosphere to atmospheric loss prevention wrong?
Answer accepted (score 14)
There is an interesting article on the magnetosphere of Venus on the ESA Science and Technology site. You can find the article here and it will probably answer your question.
The article states, like you did, that there are planets, like Earth, Mercury, Jupiter and saturn, have magnetic fields interland induced by there iron core. These magnetic fields shield the atmosphere from particles coming from solar winds. It also confirms your statement that Venus lacks this intrinsic magnetosphere to shield its atmosphere from the solar winds.
The interesting thing, however, is that spacecraft observations, like the ones made by ESA’s Venus Express, have shown that the ionosphere of Venus direct interaction with the solar winds causes an externally induced magnetic field, which deflects the particles from the solar winds and protects the atmosphere from being blown away from the planet.
However, the article also explains that the Venus magnetosphere is not as protective as earth’s magnetosphere. Measurements of the Venus magnetic field show several similarities, such as deflection of the solar winds and the reconnections in the tail of the magnetosphere, causing plasma circulations in the magnetosphere. The differences might explain the fact that some gasses and water are lost from the Venus atmosphere. The magnetic field of Venus is about 10 times smaller as the earth’s magnetic field. The shape of the magnetic field is also different. Earth has a more sharp magnetotail facing away from the sun and Venus has a more comet shaped magnetotail. During the reconnections most of the plasma is lost in the atmosphere.
The article explains therefore that although Venus does not have an intrinsic magnetic field, but the interaction of the thick atmosphere with the solar winds causes an externally induced magnetic field, that deflect the particles of the solar winds. The article suggests, however, that the different magnetic field may cause that lighter gasses are not that much protected and therefore are lost into space.
I hope this sufficiently answers the question.
Kind regards, MacUserT
Answer 2 (score 8)
There are other ways to lose atmosphere. For example Jean’s Escape. If average velocity of a gas molecule exceeds escape velocity, the planet will lose atmosphere.
Venus’ atmopshere is mostly \(CO_2\) which has a higher molecular weight than the \(0_2\) and \(N_2\) of our atmosphere. So, for a given temperature and pressure, the carbon dioxide molecules have a slower speed. Venus’ gravity is about the same as earth’s and about twice Mars’ gravity.
In summary, Venus’ steep gravity well and massive gas molecules might be helpful in letting Venus hold on to an atmosphere.
24: Why do objects burn when they enter earth’s atmosphere? (score 24084 in 2014)
Question
Why do all objects burn when they enter our atmosphere? is this because of our atmosphere composition? and does this happen on other planets as well?
Answer accepted (score 8)
You’ll often hear that it’s because of friction, but that’s often not the main factor. For larger objects it’s more likely the pressure they create.
In both cases the reason is the enormous speed, often tens of kilometers per second. When a larger object enters the atmosphere at these speed the air in front of it gets compressed so much that it becomes extremely hot. (Think of pumping up a tire; you’re also compressing air and you can feel the valve becoming hot.) The compressed air will often disintegrate the object in the air, and then the debris may burn because of the heat. This is exactly what happened to the asteroid above Russia last year: it exploded with an enormous flash in the air, and left little traces on the ground.
This happens on other planets as well, if they have a sufficiently dense atmosphere. In 1994 the comet Shoemaker-Levy crashed into Jupiter. It disintegrated before entering Jupiter’s atmosphere due to the strong gravitation, but when the fragments entered the atmosphere they could easily be seen lighting up as they burned up.
edit
Remember the Space Shuttle? It had heat resistant tiles on the bottom of the craft to protect it from burning when it entered the atmosphere, even though its speed is only a fraction of a meteorite’s speed when that enters the atmosphere.
During the last launch of the Space Shuttle Columbia some material from the external fuel tank damaged this heat shield, and upon re-entry the heat and the highly pressurized air under the craft could enter it, causing the craft to disintegrate and kill all crew.
Answer 2 (score 0)
An object hitting the Earth at high speed has a lot of kinetic energy, \[\begin{align} \frac{1}{2}mv^2 \end{align}\] where m is the object’s mass and v is it’s velocity. If the velocity goes from lots to nothing relative to the Earth, that kinetic energy has to go somewhere. That somewhere ends up as heat.
25: How powerful a telescope would allow me to view the astronauts aboard ISS do a space-walk? (score 23470 in 2017)
Question
This arose from a comment posted against a question elsewhere on the stackexchange
How powerful a telescope/binoculars would allow me to view the astronauts aboard the ISS do a space-walk? Aperture? Magnification?
Answer accepted (score 41)
This has been done before, so I don’t have to go through all the heavy calculations using Rayleigh criterion accounting for atmospheric diffraction and visible light wavelength. Ralf Vandebergh, a Dutch astronomer, professional photographer and veteran satellite spotter has been busy trying to do exactly this since the 2007 and has indeed succeeded on several occasions by now using a 10 inch (25.4 cm) Newtonian reflecting telescope that has a resolving power (angular resolution on the CCD sensor) of roughly one pixel per meter at the distance to the International Space Station (ISS) that is currently in a 230 miles (370 km) orbit above the Earth:
Ralf Vandebergh’s detail of an image he took on Mar. 21, 2009 showing astronauts working outside the ISS. Credit: R. Vandebergh
Vandebergh’s personal page also hosts all kinds of other successful observations of the ISS through his telescope and recorded in both photographs as well as some short videos. Why short? Because targeting the ISS as it moves at the speed of 4.8 mi/s (7.7 km/s) is rather tricky, and the atmospheric conditions and times at which the ISS passes over some area on the surface of the Earth don’t make it any easier either. But perseverance and hard work have paid well for this individual astronomer.
Raw video of the ISS as seen through the air turbulence. Note the good visibility of the Lira antenna at the Russian
Zvezda Module in the lower part of the image. Credit: R. Vandebergh
So again, skipping the math to calculate the required angular resolution of a telescope and apply that to some arbitrarily selected image sensor size and resolution, we can see that using a well collimated 10 inch Newtonian or a Dobsonian telescope on a clear night can, with some near perfect targeting, produce a direct proof of a spacewalker doing his or her job 230 miles (370 km) high during an EVA. More powerful telescopes would of course produce better resolution images, but the atmospheric effects limit their use and are of course much harder to target an object moving fast over the skies with.
Answer 2 (score 22)
Ralf Vandebergh is one of the best amateur astronomy photographers out there who does spacecraft photography. He is using a 10" (25.4cm) Newtonian telescope, as far as I know, so this is pretty much an off the shelf telescope.
He supposedly has imaged spacewalkers on previous ISS and STS missions. Though they are only a few pixels in size, and you cannot make out any details. Nevertheless, his images are definitely, absolutely stunning:
Answer 3 (score 2)
Note that its more or less symmetric, so hardware of a caliber sufficient to observe human activity on the ground from space, would, if mounted on the ground, be more or less of the same size and technology required to observe human sized activity in space. So you can cross check both theoretical and experimental results vs what parameters are known about military surveillance satellites. So you “know” it doesn’t have to be 200 feet in diameter, and has to be way larger than binoculars, in fact about the size of a spy satellite seems to be adequate for military levels of reliability and illumination.
26: Why does the Earth have a tilt of ~23°? (score 21954 in 2013)
Question
- Is there a reason that the Earth has the tilt that it does (~23°)?
- How do we know which way is supposed to be 0°?
- Does this tilt have major consequences on the planet?
- Has it changed and will it change in the future? (If so, would there be any consequences?)
Answer accepted (score 23)
First up, the tilt is exactly 23.45 degrees.
The reason for Earth’s tilt is still not yet really proven, but scientists at Princeton stated on August 25, 2006 that planet Earth may have ‘tilted’ to keep its balance. Quote:
By analyzing the magnetic composition of ancient sediments found in the remote Norwegian archipelago of Svalbard, Princeton University’s Adam Maloof has lent credence to a 140-year-old theory regarding the way the Earth might restore its own balance if an unequal distribution of weight ever developed in its interior or on its surface.
The theory, known as true polar wander, postulates that if an object of sufficient weight – such as a supersized volcano – ever formed far from the equator, the force of the planet’s rotation would gradually pull the heavy object away from the axis the Earth spins around. If the volcanoes, land and other masses that exist within the spinning Earth ever became sufficiently imbalanced, the planet would tilt and rotate itself until this extra weight was relocated to a point along the equator.
Same goes for the lack of proof about the exact consequences on the planet or if it’ll happen again. All that is still being researched and debated, but those Princeton scientists have thrown in some interesting perspectives which you’ll discover when you visit the link I provided and completely read what they wrote.
Besides “consequences” for Earth as a planet, it should be noted that the tilt of Earth is the reason why we have seasons. So even when we yet have to find out what the consequences for Earth (as a planet) have been and/or will be, we do know that the tilt surely has consequences for all beings that live on planet Earth… the seasons that influence us all: summer, fall, winter, and spring.
And - last but not least - as for your question how they know where 0 degrees would be, you just have to look at the Earth’s rotation around the sun. Take a 90° angle of the planet’s orbit and you know where 0° would be. But I guess a picture explains more than a thousand words, so I quickly created the following graphic:
Answer 2 (score 5)
The tilt of the rotation axis of the Earth is probably twofold. First, it is probably due to Earth formation history. Even though, by angular momentum conservation, planets should have a rotation axis aligned with that of the Sun, their formation history, characterized by accretion and collision of planetisimals, should have perturbated the initial axis. Second, it is due to gravitational perturbations. And actually, regarding this effects, Earth axis should be much more tilted; hopefully, the Moon is there to help.
27: How was the Earth-Sun distance originally calculated? (score 21903 in 2014)
Question
AFAIK it was possible long before the first interplanetary probes.
Who did it?
Answer accepted (score 20)
The book The Transits of Venus, by Sheehan and Westfall, describes how Aristarchus used Hipparchus’ calculation of the Earth-Moon distance, who in turn used Eratosthenes’ calculation of the Earth’s circumference, to calculate the Earth-Sun distance.
Aristarchus of Samos was the first to seriously calculate the distance to the Sun, using geometry. When the Moon is exactly half illuminated when seen from the Earth (first or last quarter phase), then there is a right triangle between the Earth, Moon, and Sun, with the Moon at the right angle. Then he could measure the angular distance in the sky between the Sun and the Moon, plus the Earth-Moon distance and geometry, to get the Earth-Sun distance.
The most famous ancient estimate of the earth’s circumference as made by Eratosthenes of Cyrene (c. 276-196 BCE), the librarian at the great library at Alexandria. By using a simple gnomon, he found that at Syene, … the sun at the summer solstice cast no shadow at all: it was exactly overhead. … At the same moment, at Alexandria, the shadow cast by the sun shows that it stood 7.2 degrees from vertical. This difference is equal to 1/50 of a circle.
Using the distance between the cities, Earth’s circumference could be calculated.
Once the earth’s radius is known, the earth itself can be used as a baseline for determining still greater distances – the distance to the moon.
[I]t becomes possible to work out the earth-moon distance indirectly from the geometry of [lunar] eclipses. Using this method, Hipparchus of Rhodes (fl. 140 BCE) worked out that the distance of the moon was 59 earth radii. It’s a good approximation - with 1 1/2 or 2 earth radii of the modern value.
Using the Earth-Moon distance and the separation of the Moon from the Sun in the sky when the Moon was at exactly half-phase, Aristarchus calculated the Earth-Sun distance.
Aristarchus put forward a geometrical argument, based on determining the sun-earth-moon angle at the time the moon’s phase is exactly half. For this angle, which is actually 89.86 degrees, Aristarchus used 87 degrees; the disagreement is more significant that it might appear because the critical quantity is the difference between the angle and 90 degrees.
Because of this Aristarchus only got a value of the equivalent of “5 million miles”, much too small.
Phil Plait has, on his old Bad Astronomy site, an article answering a question about how astronomers originally calculated the distance from the Earth to the Sun (the AU, or astronomical unit).
Huygens was the first to calculate this distance with any kind of accuracy at all.
So how did Huygens do it? He knew that Venus showed phases when viewed through a telescope, just like our own Moon does. He also knew that the actual phase of Venus depended on the angle it made with the Sun as seen from the Earth. When Venus is between the Earth and Sun, the far side is lit, and so we see Venus as being dark. When Venus is on the far side of the Sun from the Earth, we can see the entire half facing us as lit, and Venus looks like a full Moon. When Venus, the Sun and Earth form a right angle, Venus looks half lit, like a half Moon.
Now, if you can measure any two internal angles in a triangle, and know the length of one of its sides, you can determine the length of another side. Since Huygens knew the Sun-Venus-Earth angle (from the phases), and he could directly measure the Sun-Earth-Venus angle (simply by measuring Venus’ apparent distance from the Sun on the sky) all he needed was to know the distance from Earth to Venus. Then he could use some simple trigonometry to get the Earth-Sun distance.
This is where Huygens tripped up. He knew that if you measured the apparent size of an object, and knew its true size, you could find the distance to that object. Huygens thought he knew the actual size of Venus using such unscientific techniques as numerology and mysticism. Using these methods he thought that Venus was the same size as the Earth. As it turn out, that is correct! Venus is indeed very close to being the same size as the Earth, but in this case he got it right by pure chance. But since he had the right number, he wound up getting the about the correct number for the AU.
Basically, Huygens used good methods, except for using “numerology and mysticism” to determine the size of Venus. He was lucky that Venus was almost the size of Earth; that made his estimate for the AU pretty close.
Not long after, Cassini used the parallax of Mars to determine the AU. (Same article as linked above.)
In 1672, Cassini used a method involving parallax on Mars to get the AU, and his method was correct.
Parallax is the apparent difference in angle observed due to the differing observing positions. The smaller the parallax, the larger the distance.
However, the precision of the resultant calculation depends on the precision of the observations, and measurements of the parallax aren’t that precise.
In 1716, Edmond Halley published a way to use a transit of Venus to accurately measure the solar parallax, i.e. the difference in the Sun’s position in the sky due to observers at different latitudes.
Because of the latitude difference of the observers, Venus would appear to move along chords of different length over the disk of the sun. The motion of Venus being nearly uniform, the length of each chord would be proportional with the duration of the transit. Thus, observers would not actually have to measure anything; they would only have to time the transit. Fortunately, existing pendulum clocks were more than sufficiently accurate for this purpose.
They could time the transit, which would last hours, with great precision. But they had to wait until the next transit of Venus in 1761. Then, observers observed the black drop effect, which made it very hard to time the event from start to finish precisely.
The black drop effect cannot be eliminated altogether, but it is much more ever in observations made with telescopes of imperfect optical quality (as many of those used at the 1761 transit were) and in boiling or unsteady air. Confusion about the times of the internal contacts … yielded contact times that differed among observers, because of the black drop, by as much as 52 seconds.
In the end, there as a wide range of published values, from 8.28 arc-seconds to 10.60 arc-seconds.
But then there was the transit of 1769. Observations in Norway and in the Hudson Bay were made for the northern observations, and Captain James Cook was sent to what is now Tahiti to make a southerly observation. Jérôme Lalande compiled the figures and calculated a solar parallax of 8.6 arc-seconds, close to the modern figure of about 8.794 arc-seconds. That calculation yielded the first fairly accurate calculation of the Earth-Sun distance, of 24,000 Earth radii, which given the Earth’s radius of 6,371 km, about 153,000,000 km, the accepted value being about 149,600,000 km.
28: What is the exact time Earth takes to revolve around its axis? (score 21098 in 2019)
Question
I read that our day is not exactly 24 hours long. If it is not then why weather seasons come at the same time (Summer, Winter, Autumn Spring) ? how it is equalized ?
Update: Sorry I could not make it clear. My meaning was if 3.8 seconds are shorter of everyday the how is the calculation made that summer starts in May and winter start december everytime why it does not move further since. How time is neutralized ? I hope you guys understood this time.
Answer accepted (score 7)
Our day is 23 hours and 56 minutes long, and slowing by an infinitesimal (but measurable) amount each year due to tidal losses.
Our day has a connection with the weather, in that the sun drives all our weather systems, so heating over each part of the globe happens every day, but aside from that, your question doesn’t make much sense.
Weather changes may come at the same time where you live (on a 24 hour cycle) but here in Scotland, we still have low reliability on even a 3 day weather forecast, because the weather systems that impact the UK are so complex as heating from the sun drives various air flows.
After your update, I still cannot understand what you mean. There is slippage in accuracy, but this is counteracted by leap seconds and leap years. It has nothing to do with weather. The shortest and longest days happen when they happen and are measurable. They help us define the year.
Answer 2 (score 6)
The length of a solar day is very close to 24 hours. A solar day is the time it takes the Sun to return to the same position in the sky. The exact length varies slightly over the course of the year, because the Earth’s orbit around the Sun is not perfectly circular; the tilt of the Earth’s axis with respect to its orbit also has an effect. See the linked article for details.
A sidereal day is slightly shorter. It’s the time it takes for a distant star to return to the same position in the sky, and it differs from the solar day because the Earth revolves around the Sun as it rotates on its own axis; there is one more sidereal day than solar day each year. A sidereal day is approximately 23 hours, 56 minutes, 4.0916 seconds.
Weather can actually have a minuscule effect on the Earth’s rotation, as the mass of the atmosphere is distributed differently. This effect is barely measurable. I don’t think that’s the kind of “weather changes” you were asking about; can you clarify? Obviously it gets colder at night, and that’s controlled by the length of the solar day.
29: Average amount of annual daylight at any place on earth (score 20809 in )
Question
If this is the wrong group please direct me to the correct one.
It seems intuitively obvious that the amount of daylight per annum should be the same for any latitude on earth. For example, 12 hours per day at the equator. The poles have daylight for half the year and darkness for the other half (crudely).
Is there any way to get an answer to this apparently simple question - is the annual amount of daylight the same at any point on earth?
Answer accepted (score 10)
Wikipedia strikes again:
The naive expectation is that, for every place on Earth, the Sun will appear to be above the horizon for exactly half the time. Thus, for a standard year consisting of 8760 hours, apparent maximal daytime duration would be 4380 hours. However, there are physical and astronomical effects which change that picture. Namely, atmospheric refraction allows the Sun to be still visible even when it physically sets below the horizon line. For that reason, average daytime (disregarding cloud effects) is longest in polar areas, where the apparent Sun spends the most time around the horizon. Places on the Arctic Circle have the longest total annual daytime of 4647 hours, while the North Pole receives 4575. Because of elliptic nature of the Earth’s orbit, the Southern Hemisphere is not symmetrical: Antarctic Circle at 4530 hours receives 5 days less of sunshine than its antipodes. The Equator has the total daytime of 4422 hours per year.
Further details here on astronomical causes of average daytime variation, and here on Insolation, the solar radiation received at the top of the atmosphere and its effects on the energy received at ground level.
30: What would the effects be on Earth if Jupiter was turned into a star? (score 20076 in 2014)
Question
In Clarke’s book 2010, the monolith and its brethren turned Jupiter into the small star nicknamed Lucifer. Ignoring the reality that we won’t have any magical monoliths appearing in our future, what would the effects be on Earth if Jupiter was turned into a star?
At it’s closest and furthest:
How bright would the “back-side” of the earth be with light from Lucifer?
How much heat would the small star generate on earth?
How many days or months would we actually have night when we circled away behind the sun?
How much brighter would the sun-side of earth be when Lucifer and the sun both shine on the same side of the planet?
Answer accepted (score 13)
Before I start, I’ll admit that I’ve criticized the question based on its improbability; however, I’ve been persuaded otherwise. I’m going to try to do the calculations based on completely different formulas than I think have been used; I hope you’ll stay with me as I work it out.
Let’s imagine that Lucifer becomes a main-sequence star - in fact, let’s call it a low-mass red dwarf. Main-sequence stars follow the mass-luminosity relation:
\[\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^a\]
Where \(L\) and \(M\) are the star’s luminosity and mass, and \(L_\odot\) and \(M_\odot\) and the luminosity and mass of the Sun. For stars with \(M < 0.43M_\odot\), \(a\) takes the value of 2.3. Now we can plug in Jupiter’s mass (\(1.8986 \times 10 ^{27}\) kg) into the formula, as well as the Sun’s mass (\(1.98855 \times 10 ^ {30}\) kg) and luminosity (\(3.846 \times 10 ^ {26}\) watts), and we get
\[\frac{L}{3.846 \times 10 ^ {26}} = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3}\]
This becomes \[L = \left(\frac{1.8986 \times 10 ^ {27}}{1.98855 \times 10 ^ {30}}\right)^{2.3} \times 3.846 \times 10 ^ {26}\]
which then becomes
\[L = 4.35 \times 10 ^ {19}\] watts.
Now we can work out the apparent brightness of Lucifer, as seen from Earth. For that, we need the formula
\[m = m_\odot - 2.5 \log \left(\frac {L}{L_\odot}\left(\frac {d_\odot}{d}\right) ^ 2\right)\]
where \(m\) is the apparent magnitude of the star, \(m_\odot\) is the apparent magnitude of the Sun, \(d_\odot\) is the distance to the Sun, and \(d\) is the distance to the star. Now, \(m = -26.73\) and \(d(s)\) is 1 (in astronomical units). \(d\) varies. Jupiter is about 5.2 AU from the Sun, so at its closest distance to Earth, it would be ~4.2 AU away. We plug these numbers into the formula, and find
\[m = -6.25\]
which is a lot less brighter than the Sun. Now, when Jupiter is farthest away from the Sun, it is ~6.2 AU away. We plug that into the formula, and find
\[m = -5.40\]
which is dimmer still - although, of course, Jupiter would be completely blocked by the Sun. Still, for finding the apparent magnitude of Jupiter at some distance from Earth, we can change the above formula to
\[m = -26.73 - 2.5 \log \left(\frac {4.35 \times 10 ^ {19}}{3.846 \times 10 6 {26}}\left(\frac {1}{d}\right) ^ 2\right)\]
By comparison, the Moon can have an average apparent magnitude of -12.74 at full moon - much brighter than Lucifer. The apparent magnitude of both bodies can, of course, change - Jupiter by transits of its moon, for example - but these are the optimal values.
While the above calculations really don’t answer most parts of your question, I hope it helps a bit. And please, correct me if I made a mistake somewhere. LaTeX is by no means my native language, and I could have gotten something wrong.
I hope this helps.
Edit
The combined brightness of Lucifer and the Sun would depend on the angle of the Sun’s rays and Lucifer’s rays. Remember how we have different seasons because of the tilt of the Earth’s axis? Well, the added heat would have to do with the tilt of Earth’s and Lucifer’s axes relative to one another. I can’t give you a numerical result, but I can add that I hope it wouldn’t be too much hotter than it is now, as I’m writing this!
Second Edit
Like I said in a comment somewhere on this page, the mass-luminosity relation really only works for main-sequence stars. If Lucifer was not on the main sequence. . . Well, then none of my calculations would be right.
Answer 2 (score 7)
I think it’s a fun question, if impossible. The only way to turn Jupiter into a star that’s even remotely practical is to add to it’s mass. Ignoring brown dwarfs that are very limited in energy output, to get a red dwarf going, you’d need to add at least 75-80 or so Jupiter masses. (a bit more than 24,000 earth masses). You’d want to add a fair percentage of hydrogen, but some rocky debris wouldn’t hurt the mix.
Anyway, assuming the impossible is done, there’s several things to consider. The greater gravity (75-80 times) would significantly alter all the planets orbits. Predicting exactly how is hard, but that much more mass and the planets orbits, certainly all the inner ones, would wobble a lot more and some might get pulled completely out of their orbit, likely thrown out of the solar system.
You might think that the planets nearer to Jupiter would be the most affected, but it really has more to do with tidal synch than anything else. Any of the 4 inner planets could get tugged into a new orbit. You’d also likely see the earth’s orbit elongate in resonance with Jupiter perhaps increasing the ice age/ice melt cycle. Precise answers are hard, and none of these things would happen over 1 orbit, but over time, certainly. Orbital changes to all the inner planets and perhaps Saturn as well would be inevitable if Jupiter becomes a red-dwarf. Imagine if Saturn was pulled closer to the earth, into an orbit between Mars and Jupiter, or Mercury was pulled out past the earth. Odds are it wouldn’t hit us, but we might want to keep an eye on it.
http://en.wikipedia.org/wiki/Stability_of_the_Solar_System#Mercury.E2.80.93Jupiter_1:1_resonance
2nd thing to consider is magnetism and solar flares. Young stars tend to spin very fast due to conservation of angular momentum when the stars form and this creates enormous magnetic fields and huge solar flares, much bigger than we get from the sun. It’s strange to think that a tiny red dwarf, 4 times as far from our sun as the sun would create solar flares to worry about but it’s possible. Whether it would need high angular momentum for this to happen, I’m not sure, but we could see larger solar flares from star-Jupiter than from the sun.
http://en.wikipedia.org/wiki/Flare_star
Brightness, heat and visibility was covered above, but I’ll touch on that. Brightness of -6.25 would be 5-6 times brighter than Venus and you’d see it at night, Venus isn’t seen in peak darkness, so it would be significantly brighter than any other star/planet in the sky, but significantly less bright than the moon, like, you couldn’t make your way with just that star’s light the way you can see things around you in moonlight. But when I run the numbers, I think it would be quite a bit brighter than that.
Mass to Luminosity is to the power of 3.5 - quick estimate, so, lets say the red dwarf has a mass of 80 Jupiters. That’s 0.076 Suns. 0.076^3.5 = about 1/8,000, so 4.2 times as far away at closest point (square of that), 1/8000th as bright, we’re looking at 1/140,000 times the light we get from the sun - not very much and likely less than that in it’s early stages and because the smaller stars tend to fall off, so lets estimate 1/200,000 - 1/300,000 the apparent brightness of the sun as a ballpark estimate. That’s not enough to heat the earth at all, but that’s still brighter, (a little bit) than the full moon which is about 1/400,000 the brightness of the sun. it would be enough light to see your way around, but I wouldn’t want to try to read by it. It would also be distinctly reddish light. Not the white light we’re used to getting from the day or night sky.
Finally size - a red-dwarf star of 80 Jupiter masses would actually be slightly smaller than Jupiter due to the gravitation so it would appear like a planet - not quite a point in the sky, but almost a point, but a bit brighter than the full moon and red. That’s likely bright enough to see during the day too. I don’t think it would be hard to look at or hurt your eyes, but it would shine like a tiny bright red flashlight in the distance.
http://www.space.com/21420-smallest-star-size-red-dwarf.html
I don’t think I like star-Jupiter. Lets not plan on doing this. :-)
Answer 3 (score 6)
Ignoring the impossibility of Jupiter going solar:
Assume that Jupiter turns into duplicate of the Sun in terms of energy output. Energy transmitted to the earth follows an inverse-square law. Since Jupiter is, at best, 4 times farther from the Earth than the Sun, Jupiter will supply the Earth with, at most, 1/16 the energy that the Sun supplies, for an increase of a bit more than 6%, at the most.
By comparison, between aphelion and perihelion, the Sun-Earth distance increases from around 147 million kilometers to around 152 million kilometers. This implies a seasonal energy input change of about 7%, that we experience now every year…
31: What’s the fastest moving object in the universe? (score 20013 in 2013)
Question
We know that nothing can have proper velocities larger than the speed of light in vacuum. But are there any objects in space that get close to it? Any comets, or other objects thrown by gravity or supernova explosions that were hurled to incredible speeds?
Answer accepted (score 34)
The answer to this is surprising:
We are.
And many (if not all) other galaxies.
And they move faster than light.
See, the universe is expanding, at an accelerating rate. The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster than light. While local space is flat and the local speed of light must be upheld, this need not hold at a global scale, so it is possible to have frames which move away from each other faster than \(c\). Indeed, there are some galaxies that are moving away from us faster than light (the only reason we see them is that they used to be closer and moving at a slower speed). Any pair of galaxies that are 4200 Mpc away from each other (that is, with a redshift of 1.4), are moving away from each other faster than light in each other’s frames (numbers stolen from the linked page).
Since the only consistent way to talk about motion is relative, one can say that we are moving away from other galaxies faster than light, since the reverse is also true. This can put galaxies in the bucket of the fastest moving objects in the universe. As for which is the fastest, I don’t know, we would have to find a pair of galaxies which are the farthest apart (distance measured in the frame of the galaxy, of course), but since the universe is probably more than what we observe1, we can’t pinpoint the pair of galaxies for which this is true.
For those who think that it is cheating2 to short-circuit the question with space expansion, there are other objects that go faster than light (they are not the fastest objects in the universe though), and these can be found on good ’ol Earth.
Electrons:
In nuclear reactor cooling pools3, we have a phenomenon known as Cerenkov radiation. Basically, emitted beta particles move faster than the speed of light in water. This creates an effect of similar origin as the sonic boom, where strong light emanates from the medium.
Saywhat? You think I’m cheating again2 by putting everything relative to the speed of light in a medium?
Alright, fine. Here are some fast objects that don’t require space expansion to be fast, nor do they involve any trickery of semantics where the medium in which they are being measured is not mentioned. Many have already been mentioned by astromax.
-
Tachyons:These are particles which go faster than \(c\) — this does not violate relativity as long as they never decelerate to subluminal speeds. However, there isn’t much (any?) experimental evidence for these. A lot of BSM models do predict their existence, though. So there’s still some cheating here, on to bradyonic matter:
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Gluons: These are massless, and though they don’t occur freely (except possibly in glueballs, though these most probably have mass) they do travel at \(c\). But these can’t move at any other speed, so again, this is slightly cheating. On to fermionic matter:
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Neutrinos: Now these are viable candidates. The electron neutrino is known to have very, very little mass (we have an upper bound for it, which gives ), and as a result it can easily attain very high speeds. Put it in a gravitational field, and it goes even faster. If you want macroscopic objects, however:
- Stuff spiraling around spinning black holes: Black holes have a strong gravitational field, and when rotating, they can impart angular momentum (lots of it) to nearby objects like accretion disks. Objects close to a black hole are accelerated to pretty high speeds. In fact, if an object is within the ergosphere, it moves faster than light from the point of view of certain frames of reference.
- Stuff falling into black holes: From the faraway frame, an object speeds up and approaches the speed of light as it approaches the horizon of a black hole. Arbitrarily large speeds bounded by \(c\) can be attained here.
- Black hole plasma jets: Jets being flung out of black holes can get pretty fast relative to each other.
1. Due to cosmic expansion, there can be galaxies that are no longer visible to us. Some galaxies may never have been visible to us, if we start watching from when galaxies started forming.
2. I, for one, agree with you.
3. And other places where you have massive particles being emitted really fast into a medium
Answer 2 (score 15)
There is also another mediator particle which moves at the speed of light other than the photon. This is the gluon, which is the exchange particle for the strong force. The odd thing about the gluon is that it’s never seen by itself (that is, outside of collections of other gluons).
Also, though neutrinos do in fact have mass, they are neutral particles. Why I’m bringing this up is because in supernovae explosions neutrinos can arrive before the photons in some circumstances - they do not interact with charged particles. Also, because they are weakly interacting particles, they pass through considerable amounts of mass (namely dust and gas) before an interaction might occur. What this means is if you could detect the neutrinos coming from a supernova it could potentially give you early warning that the photons would soon follow. This would give you time to measure its light curve (see: SNEWS: The SuperNova Early Warning System).
Answer 3 (score 13)
There are plenty of rapidly moving objects in astrophysics.
A good place where one can get moving relativistically is near an event horizon of a black hole. A simple Newtonian estimate illustrates the point. Black hole has all its mass \(M\) hidden under an event horizon of the radius of order \(r_{g}=\dfrac{2GM}{c^2}\). An object moving circularly in the gravitational field of a black hole at the radius \(\alpha r_{g}\), where \(\alpha>1\), would have Newtonian orbital velocity \(v\) equal to \(v=\sqrt{\dfrac{GM}{\alpha r_g}}=\dfrac{c}{\sqrt{2\alpha}}\).
This is a quilitative estimate of the velocity scale. From general relativity there are no stable circular orbits at \(\alpha<3\), but any body will have additional acceleration when inspiralling into the black hole. To add a bit of complexity, when one starts thinking in terms of general relativity, one has to wonder, what do we really mean by objects’ velocity and about such kinds of questions.
Nevertheless, the above conclusion is correct: In the field of black holes intact objects can obtain relativistic velocities, which are comparable to the speed of light.
There are many physical examples of such systems: binary megring black holes, black holes merging with neutron stars, supermassive black holes and white dwarfs, etc. While all these systems are driven to eventual merger at relativistic velocities, it is hard for any of their components to get ejected and become free floating. To my knowledge, there are no known free floating relativistic astrophysical bodies, but some of them are indeed probably produced from the pieces of material ejected away at mildly relativistic speeds during mergers involving black holes.
One other rare possibility is to have a compact binary system in the field of a supermassive black hole, which is being disrupted due to interaction with it. However, the probability for such a disruption happening when the compact binary is just about to merge is vanishingly low.
Another ubiquitous class of objects are relativistic jets, which are ultrarelativistic streams of plasma, produced mainly when some accretion onto a black hole is taking place. Particles in such jets move at highly relativictic velocities, though the exact nature of jet formation is yet not completely understood. Finally, there is plenty of relativistic particles present in the background, such as cosmic ray particles and neutrinos.
One last thing to mention would be plasmas which are at relativistic temperatures (of order \(10^9 K\)), and which hence contain particles (mainly, electrons) moving relativistically. It is rare that plasmas get temperatures that high, but it is definitely possible during core-collapse supernova.
Finally, at sufficiently early stages of the Big Bang, absolutely everything in the Universe was moving relativistically!
Edit: A few more things which came to my mind afterwards: 1) Man-made particle beams in particle accelerators are relativistic, macroscopic, but not astrophysical objects. 2) If there exists intelligent life in the Universe, it might have also produced relativistic objects of macroscopic, but again probably not astrophysical, scale (like spacecrafts).
32: Will Earth lose the Moon before the Sun goes into supernova? (score 19204 in )
Question
Ive read on some sites and saw on youtube videos that the moon is getting away from earth by 1-3 cm a year.
Is this enough to make the Earth lose the Moon before the Sun goes into Supernova?
Im asking because I would like to do the calculations for Earths magnetic pull on this subject. It appears to me that it should be a non-linear function (there are also other factors, like if the Moon could be captured by another planet or having asteroids messing with its orbit on Earth).
Thank you.
Answer accepted (score 27)
As HDE 226868 noted in his answer, the Sun is not going to go supernova. That’s something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It’s not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.
The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won’t. It’s all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I’ll have more to say on this in the third part of my answer.
The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that \[\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}\] Here, \(a\) is the semi major axis length of the Moon’s orbit, \(k\) is the Earth-Moon tidal Love number, and \(Q\) is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.
That inverse \(a^{5.5}\) factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases \(k/Q\) by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase \(k/Q\) even further.
If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That’s less than 1/3 of the Earth’s Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.
What about after a billion years? I chose a billion years because that’s about when the Moon’s recession should more or less come to a standstill. If the Earth hasn’t already died before this billion year mark, this is when the Earth dies.
Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth’s oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.
Almost all of the Moon’s recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.
Answer 2 (score 13)
The Sun does not have nearly enough mass to become a supernova. Instead, it will swell to become a red giant, enveloping Mercury, Venus, and possibly Earth. After that, it will shed its outer layers as a planetary nebula, and settle down to become a white dwarf. Wikipedia, apparently, says the exact same things I had though of:
The Sun does not have enough mass to explode as a supernova. Instead it will exit the main sequence in approximately 5.4 billion years and start to turn into a red giant. It is calculated that the Sun will become sufficiently large to engulf the current orbits of the Solar System’s inner planets, possibly including Earth.
Still, that does put a deadline for the Earth and the Moon to evolve to the point you’re looking for. Wikipedia addresses the possibility that the Earth and Moon survive:
Today, the Moon is tidally locked to the Earth; one of its revolutions around the Earth (currently about 29 days) is equal to one of its rotations about its axis, so it always shows one face to the Earth. The Moon will continue to recede from Earth, and Earth’s spin will continue to slow gradually. In about 50 billion years, if they survive the Sun’s expansion, the Earth and Moon will become tidally locked to each other; each will be caught up in what is called a “spin–orbit resonance” in which the Moon will circle the Earth in about 47 days and both Moon and Earth will rotate around their axes in the same time, each only visible from one hemisphere of the other.
So the Moon will certainly still be around by the time the Sun becomes a red giant, and for many billions of years after that.
Here’s the math: \[\frac{1 \text{ cm}}{1 \text{ year}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times 5,400,000,000 \text{ years}=54,000 \text{ km}\]
That’s about one-seventh the current distance to the Moon - at its closest pass.
33: How much gold is there in our sun? (score 19129 in 2018)
Question
XKCD 1944 claims that there is “more gold in the sun than water in the oceans”. Is this really true?
Answer accepted (score 111)
The mass of the sun is 1.989 × 1030 kg.
Abundance in the Sun of the elements gives a percentage 1 × 10-7 % for gold *, so that leaves you with a mass of 1.989 × 1021 kg of gold.
HowStuffWorks states that there is 1.26 × 1021 kg water on Earth, of which 98% is in the oceans, i.e. 1.235 × 1021 kg.
This would mean the XKCD statement is true: there is 1.6 times as much gold in the sun as there is water in the oceans.
* They cite WolframAlpha as their source. Executing SolarAbundance “Gold” there confirms this (mass) percentage.
Answer 2 (score 24)
“Element Abundances in the Sun - The Elements Handbook”, KnowledgeDoor claims that the base 10 log of the number of atoms of gold in the Sun for every \(10^{12}\) atoms of hydrogen is \(1.01 \pm 0.15\). If I’m reading their references correctly, that’s from Abundances of the Elements: Meteoritic and Solar, Anders, Edward, and Nicolas Grevesse, Geochimica et Cosmochimica Acta, volume 53, number 1, 1989, pp. 197–214, doi:10.1016/0016-7037(89)90286-X
The atomic mass of gold is \(197\) times the atomic mass of hydrogen (more precise figures are available, but irrelevant given the accuracy of the atomic proportions). So \(2020\) kg of gold for every \(10^{12}\) kg of hydrogen, meaning that ignoring all other elements and running with \(1.99 \times 10^{30}\) kg for the mass of the Sun, it contains \(4 \times 10^{21}\) kg of gold. Taking other elements into account - helium is actually significant - reduces that value to \(3 \times 10^{21}\) kg.
This is about twice as much as the mass of the ocean, which corresponds to 2 standard deviations (\(\log_{10} 2 \approx 0.3\) vs the standard deviation of \(0.15\) in the log 10 value of the abundance).
34: Why do sunspots appear dark? (score 19118 in 2013)
Question
Sunspots, such as this one, appear dark:
Why?
Answer accepted (score 24)
Typical sunspots have a dark region (umbra) surrounded by a lighter region, the penumbra. While sunspots have a temperature of about 6300 °F (3482.2 °C), the surface of the sun which surrounds it has a temperature of 10,000 °F (5537.8 °C).
From this NASA resource:
Sunspots are actually regions of the solar surface where the magnetic field of the Sun becomes concentrated over 1000-fold. Scientists do not yet know how this happens. Magnetic fields produce pressure, and this pressure can cause gas inside the sunspot to be in balance with the gas outside the sunspot…but at a lower temperature. Sunspots are actually several thousand degrees cooler than the 5,770 K (5496.8 °C) surface of the Sun, and contain gases at temperature of 3000 to 4000 K (2726.9 - 3726.8 °C). They are dark only by contrast with the much hotter solar surface. If you were to put a sunspot in the night sky, it would glow brighter than the Full Moon with a crimson-orange color!
Sunspots are areas of intense magentic activity, as is apparent in this image:
You can see the material kind of getting stretched into strands.
As for the reason it is cooler than the rest of the surface:
Although the details of sunspot generation are still a matter of research, it appears that sunspots are the visible counterparts of magnetic flux tubes in the Sun’s convective zone that get “wound up” by differential rotation. If the stress on the tubes reaches a certain limit, they curl up like a rubber band and puncture the Sun’s surface. Convection is inhibited at the puncture points; the energy flux from the Sun’s interior decreases; and with it surface temperature.
All in all, the sunspots appear dark because the are darker than the surrounding surface. They’re darker because they are cooler, and they’re cooler because of the intense magnetic fields in them.
Answer 2 (score 4)
Sunspots are cooler because their magnetic fields inhibit the replenishing of heat from convective flows (due to the interaction between plasma and magnetic fields). This allows them to cool radiatively. The rest of the solar surface is constantly being replenished by convective cells that re-heat it.
Solar plasma at the photosphere radiates roughly as a black-body, meaning that the energy (and wavelength) spectrum of radiation follows the Plank function. The Plank function scales up and down with temperature at all energies (and wavelengths). The brightness of a blackbody at a given energy (or wavelength) is determined by the value of the plank function at that energy.
The image you showed is taken in the ‘photospheric continuum’, which is a black-body dominated part of the radiation spectrum. So, because Sunspots are cool (compared to their surroundings), this means that their Plank function is lower than their surroundings, and hence their brightness is lower, causing them to appear dark in the image.
35: What will happen when landing on Jupiter? (score 18781 in 2018)
Question
Jupiter is a gas giant, so landing on it will not be like landing on Earth, our Moon or Mars etc., as it does not have a solid surface like these.
If we have a hypothetical spaceship or probe landing on Jupiter, and it can withstand the enormous pressure, what or how will be the sequence of events?
I already read this:
A major problem in sending space probes to Jupiter is that the planet has no solid surface on which to land, as there is a smooth transition between the planet’s atmosphere and its fluid interior. Any probes descending into the atmosphere are eventually crushed by the immense pressures within Jupiter.
What I do not understand:
- Is this enormous or immense pressure of atmosphere or Jupiter’s surface?
- While descending, will we know that we have transitioned from the atmosphere to the surface?
- Why does animated footage (real or conceptual) of Jupiter appear like a high viscous liquid (e.g. lava) rotating in an opposite direction, rather that gas(es)?
- Does Jupiter have an inner solid core?
I’ve already gone through these:
Answer accepted (score 14)
Jupiter does not have a “surface” and nor is there anything but an arbitrary division between interplanetary space and where its atmosphere begins.
The crushing pressure is its atmospheric pressure. The deeper into the atmosphere you go, the greater the column of gas that lies above you. It is the weight of this column of gas that is responsible for the rapid increase in pressure with depth.
The answer to your last question is most definitely addressed in the duplicate question about whether Jupiter is entirely made of gas. There is quite likely to be a liquid phase nearer the centre and there may be a solid core of order ten times the mass of the Earth. It is not a settled question.
The gas motions you talk about are essentially belts of weather systems in the upper layers of Jupiter’s atmosphere. It is all most definitely gas that you can see.
36: How to make a telescope for viewing planets, moon and DSOs using a convex lens of aperture 100 mm and focal length 200 mm and other lenses at home? (score 18472 in 2016)
Question
I wanted to make a telescope with DIY things lying around in home. I read up that the aperture was a very important aspect and thus bought a convex lens with 100mm aperture and 200mm focal length, to use it as an objective lens. Apart from that, I have the following lenses lying around:-
- 1 convex lens - focal length 20 cm aperture - 10 cm
- 2 convex lenses - focal length 17 cm aperture - 5 cm
- 1 convex lens - focal length 2.2 cm aperture - 2.5 cm
- 1 concave lens - focal length 8.33 cm aperture - 5.5 cm
- 1 convex lens - focal length 200 cm aperture - 5.5 cm
I can also buy lenses of range of Power(In Dioptres) from (-14D) - (+14D) of aperture range of 5.5 cm to 7 cm, if suggested.
I made a telescope using lens 1(as objective) and lens 3(as eyepiece) which on calculation, gave a magnification of 9x, but according to stellafane.org and http://skyandtelescope.com the minimum magnification is 14x and that I was ‘wasting’ the light gathered by the objective lens. This telescope showed me a bright Mars, Jupiter and Saturn, but as dots. Upon seeing the Moon, I could just see it a little bigger than what would be seen with the naked eye.
I would like to know the way to make the telescope, which I would like to use to observe planets and DSOs if possible. I am ready to make separate telescopes also, if suggested.
I would like to know which lenses should I use, or buy and the way to use it. I live in a suburb and it is not very lighted up.
Is there any way I could use the lens 1 in a telescope which I could use to view the planets?
[Edit]
I have accepted @JamesScreech ’s answer and am on my way to making it. Apart from that, I would also like to use the big aperture of lens 1, for viewing DSOs, But, it has a very short focal length(just 200mm), and therefore, I am finding it hard to get any eyepiece lens which can “completely accommodate the exit pupil of my scope”(as said on skyandtelescope )
Any way to get such a lens, or make an eyepiece barrell sort of, using a combination of lenses mentioned above or buying. And how to use it?
Thanks.
Answer accepted (score 1)
I generally agree with the answer above, but have a couple more insights which might help you if you decide to proceed with trying to make your own scope… The lens pairs that James mentioned (crown and flint) are known as a doublet. Glass has two key properties in play here - its index of refraction (how much it bends light) and its dispersion (how much that bending changes over color). The lens pair balances a strongly convex crown (low index and low dispersion) with a weakly concave flint (high index and high dispersion). The dispersions are designed to cancel out, while you want the curvature of the convex crown to overpower the concave flint in terms of index, so it still has some ability to focus. The design also inherently lends itself toward long focal lengths which are desirable in telescope objectives.
Eyepieces, due to their short-desirable focal lengths necessitate more lenses which allow you to balance the chromatic aberration, and also address other optical aberrations which come into play with such a short focal length (distortion, astigmatism, coma, and spherical aberration being the main concerns). There are well-established design forms which are often used for making well-corrected eyepieces, some of which can be found right on wikipedia: https://en.wikipedia.org/wiki/Eyepiece#Eyepiece_designs
One thing to consider, which I don’t think has been mentioned is selection of focal lengths and apertures. A 5cm aperture is plenty sufficient to view the Galilean moons, and probably some bright DSOs if you’re well-corrected, and if your focal lengths are well-chosen. The system magnification is the ratio of focal lengths between the objective lens and eyelens/eyepiece. (200cm/2.2cm = 90.9x). This means that something like the Galilean moons, which have a max extent of about 1/8 degree, would be magnified to have an apparent extent of 11 degrees (much easier to resolve). Your aperture selection (particularly of your objective) will determine the light-gathering ability. But the magnification applies here too, so if you have a 5cm objective at 91x, your “exit pupil” will only be 0.55mm diameter, which is tiny compared to your eye’s aperture. You’d still be able to see the object, but your eyes will easily accommodate up to a 30cm objective aperture (3mm exit pupil). Keep in mind, there is a tradeoff between aperture and aberrations, so unless you’re designing a very well-aligned 2- or 3-element objective, you may want to stick with a max objective aperture of 50-75mm.
In terms of alignment, don’t just set the lenses a certain distance apart and expect to see an image. You will need to allow for some adjustment, which is probably easier looking at a distant object during the day. After you form an image, you may need to adjust the centration and tilt of the eyelens to form the sharpest image to optimize your alignment.
All that said, a small aperture, high-end refractive (glass) telescope can perform better than a reflective telescope of the same aperture. But as aperture increases, the cost of the materials and impact of aberrations makes refractive telescopes vastly inferior to reflective (mirror) telescopes. Since the design for these only necessitates 1 powered mirror and an off-the-shelf eyepiece for $50 or less, the best bang for your buck will definitely be a reflective telescope. Sorry if that’s not what you’re hoping to hear, but it’s why most telescopes on the market today are reflective.
Answer 2 (score 1)
I generally agree with the answer above, but have a couple more insights which might help you if you decide to proceed with trying to make your own scope… The lens pairs that James mentioned (crown and flint) are known as a doublet. Glass has two key properties in play here - its index of refraction (how much it bends light) and its dispersion (how much that bending changes over color). The lens pair balances a strongly convex crown (low index and low dispersion) with a weakly concave flint (high index and high dispersion). The dispersions are designed to cancel out, while you want the curvature of the convex crown to overpower the concave flint in terms of index, so it still has some ability to focus. The design also inherently lends itself toward long focal lengths which are desirable in telescope objectives.
Eyepieces, due to their short-desirable focal lengths necessitate more lenses which allow you to balance the chromatic aberration, and also address other optical aberrations which come into play with such a short focal length (distortion, astigmatism, coma, and spherical aberration being the main concerns). There are well-established design forms which are often used for making well-corrected eyepieces, some of which can be found right on wikipedia: https://en.wikipedia.org/wiki/Eyepiece#Eyepiece_designs
One thing to consider, which I don’t think has been mentioned is selection of focal lengths and apertures. A 5cm aperture is plenty sufficient to view the Galilean moons, and probably some bright DSOs if you’re well-corrected, and if your focal lengths are well-chosen. The system magnification is the ratio of focal lengths between the objective lens and eyelens/eyepiece. (200cm/2.2cm = 90.9x). This means that something like the Galilean moons, which have a max extent of about 1/8 degree, would be magnified to have an apparent extent of 11 degrees (much easier to resolve). Your aperture selection (particularly of your objective) will determine the light-gathering ability. But the magnification applies here too, so if you have a 5cm objective at 91x, your “exit pupil” will only be 0.55mm diameter, which is tiny compared to your eye’s aperture. You’d still be able to see the object, but your eyes will easily accommodate up to a 30cm objective aperture (3mm exit pupil). Keep in mind, there is a tradeoff between aperture and aberrations, so unless you’re designing a very well-aligned 2- or 3-element objective, you may want to stick with a max objective aperture of 50-75mm.
In terms of alignment, don’t just set the lenses a certain distance apart and expect to see an image. You will need to allow for some adjustment, which is probably easier looking at a distant object during the day. After you form an image, you may need to adjust the centration and tilt of the eyelens to form the sharpest image to optimize your alignment.
All that said, a small aperture, high-end refractive (glass) telescope can perform better than a reflective telescope of the same aperture. But as aperture increases, the cost of the materials and impact of aberrations makes refractive telescopes vastly inferior to reflective (mirror) telescopes. Since the design for these only necessitates 1 powered mirror and an off-the-shelf eyepiece for $50 or less, the best bang for your buck will definitely be a reflective telescope. Sorry if that’s not what you’re hoping to hear, but it’s why most telescopes on the market today are reflective.
37: Why don’t astronomers use meters to measure astronomical distances? (score 18352 in 2017)
Question
In astronomy distances are generally expressed in non-metric units like: light-years, astronomical units (AU), parsecs, etc. Why don’t they use meters (or multiples thereof) to measure distances, as these are the SI unit for distance? Since the meter is already used in particle physics to measure the size of atoms, why couldn’t it be used in astrophysics to measure the large distances in the Universe?
For example:
- The ISS orbits about 400 km above Earth.
- The diameter of the Sun is 1.39 Gm (gigameters).
- The distance to the Andromeda Galaxy is 23 Zm (zettameters).
- At its furthest point, Pluto is 5.83 Tm (terameters) from the Sun.
Edit: some have answered that meters are too small and therefore not intuitive for measuring large distances, however there are plenty of situations where this is not a problem, for example:
- Bytes are used for measuring gigantic amounts of data, for example terabytes (1e+12) or petabytes (1e+15)
- The energy released by large explosions is usually expressed in megatons, which is based on grams (1e+12)
- The SI unit Hertz is often expressed in gigahertz (1e+9) or terahertz (1e+12) for measuring network frequencies or processor clock speeds.
If the main reason for not using meters is historical, is it reasonable to expect that SI-unites will become the standard in astronomy, like most of the world switched from native to SI-units for everyday measurements?
Answer accepted (score 81)
In addition to the answer provided by @HDE226868, there are historical reasons. Before the advent of using radar ranging to find distances in the solar system, we had to use other clever methods for finding the distance from the Earth to the sun; for example, measuring the transit of Venus across the surface of the sun. These methods are not as super accurate as what is available today, so it makes sense to specify distances, that are all based on measuring parallaxes, in terms of the uncertain, but fixed, Earth-Sun distance. That way, if future measurements change the conversion value from AU to meters, you don’t have to change as many papers and textbooks.
Not to mention that such calibration uncertainties introduce correlated errors into an analysis that aren’t defeatable using large sample sizes.
I can’t speak authoritatively on the actual history, but solar system measurements were all initially done in terms of the Earth/sun distance. For example, a little geometry shows that it’s pretty straightforward to back out the size of Venus’s and Mercury’s orbit in AU from their maximum solar elongation. I don’t know how they worked out the orbital radii of Mars, etc, but they were almost certainly done in AU long before the AU was known, and all of that before the MKS system existed, let alone became standardized.
For stars, the base of what is known as the “cosmological distance ladder” (that is “all distance measures” in astronomy) rests on measuring the parallax angle: \[\tan \pi_{\mathrm{angle}} = \frac{1 AU}{D}.\] To measure \(D\) in ‘parsecs’ is to setup the equation so that the angle being measured in arcseconds fits the small angle approximation. That is: \[\frac{D}{1\, \mathrm{parsec}} = \frac{\frac{\pi}{180\times60\times60}}{\tan\left(\pi_{\mathrm{angle}} \frac{\pi\, \mathrm{radians}}{180\times60\times60 \, \mathrm{arcsec}}\right)}.\] In other words, \(1\operatorname{parsec} = \frac{180\times 3600}{\pi} \operatorname{AU}\).
Astronomers also have a marked preference for the close cousin of mks/SI units, known as cgs. As far as I can tell, this is due to the influence of spectroscopists who liked the “Gaussian units” part of it for electromagnetism because it set Coulomb’s constant to 1, simplifying calculations.
Answer 2 (score 24)
I would suggest it also makes the material more reachable for the human mind.
I just can’t work with insanely large or small numbers. They convey no meaning.
But 1 AU is easy, even if I don;t know exactly what that is in meters, I know what it means and it is a convenient scale for the mind.
Likewise when we talk about stellar distances, what use is the distance in meters (or AU) ? It makes more sense to work with light years. Again most people know what that means even if they don’t know exactly what it is in meters.
And when we go cosmic you’re also talking about colossal times in the past, so light years do convey a double meaning here. If I told you the distance in meters, that doesn’t instantly tell you how far back in time it is as well.
So I think it’s a matter on convenience and comprehension.
Answer 3 (score 9)
Along with the other answers, there is one other reason, specifically when measuring the distances to other galaxies.
When stating the distance to other galaxies, Astronomers rarely ever state the distance in any unit of length, they tend to use redshifts (z). This unit is not actually a unit of length (it is a dimensionless ratio of wavelengths), nor does it linearly convert to a distance (z=2 is not twice as far as z=1), nor is there an excepted conversion between redshift and distance (it depends on what model of the universe you assume).
Redshift is used because it can be very accurately measured. There are features in a star or a galaxies spectra that we know the exact wavelength that they are emitted at and so the redshift can be calculated exactly by:
\[ z=\frac{\lambda_{obs}}{\lambda_{em}}-1 \]
This is an observed, exact (within experimental error) property. Converting this to a distance is confusing: are you talking about the distance the object is away from us instantaneously now, or instantaneously when the photon that you see was emitted, or the distance the photon you see travelled? Do you wish to take into consideration local movement as well as Hubble (universe) expansion? Add on to this the shape of the universe, the rate of expansion of the universe, the rate of change of the expansion of the universe (dark energy/Hubble constants/other effects), and you see that any conversion to an actual distance is problematic and would require that you define exactly what type of conversion and with what assumptions. It is easier to stay with the well-defined easy-to-measure redshift.
A good (degree-level) work that summarises all the different types of cosmological distances and their calculations is Hogg 2000.
38: Why does Jupiter have so many moons? (score 18203 in 2013)
Question
Jupiter has a great many moons - in the hundreds, and they’re still being discovered.
What is the current theory for where all these moons came from? Are they rocks flying through space captured by Jupiter’s gravity?
Answer accepted (score 23)
Mass.
The more massive a body, the larger the gap between its lowest and highest orbit; the range of speeds at which a random body entering its gravity is likely to remain as its satellite. Sun has millions of satellites if you count all the asteroids; smaller planets tend to have one or two moons at most (Pluto with five being a notable and not fully explained exception)
To a lesser degree there’s a matter of shape too. A regularly round body will have more regular and stable orbit than a potato-shaped one. Jupiter, being a gas giant is perfectly round. This doesn’t play that much of a role though, especially with higher orbits.
And last but not least, no destabilizing influence of other bodies. It’s very hard to maintain a lunar orbit - artificial satellites around our Moon last only a couple years each, because relatively close neighborhood of Earth tends to destabilize orbit of anything orbiting the Moon. Jupiter being a single massive planet with relatively tiny (relatively to its mass) moons doesn’t have them influence each other all that much.
Answer 2 (score 18)
Bigger is better.
Most moons, especially those of gas giants, are not “formed”, they are just “captured” (unlike our Moon, which could have been captured, but probably was formed in a much more exciting way).
Jupiter is the most massive planet in the solar system. It stands to reason that it has a larger region of gravitational influence (where its influence outweighs the force due to the other planets and the sun). So, it’s easy for it to capture rocky masses.
If you have a look at the contours on the following image (Ignore the Lagrange points marked on it, I only want the contours)
the circular area around the Earth is more or less the area (there’s a velocity dependence here which I’m not getting into) in which a moon-like body can form a reasonably stable orbit. The size of the small “well” will increase as the planet moves farther from the sun, and also when the planet is more massive.
Jupiter is both pretty far away from the Sun, and is very massive. This leads to a huge sphere of influence.
The asteroid belt may have something to do with this too, but I doubt it (it’s pretty far away). However, if we assume the “half-baked planet formation” theory for the formation of the belt, Jupiter may have leeched off much of the mass that would have otherwise become part of that planet during the formative period.
39: Is the light we see from stars extremely old? (score 17964 in 2013)
Question
Our nearest star Proxima Centauri is 4.243 light years away from Earth.
Does that mean we are seeing light that is 4.243 years old everyday?
Answer accepted (score 21)
Yes, the speed of light in vacuum (or c) is 299,792,458 m/s and one light-year is the distance the light travels in one Julian year (365.25 days), which comes out as 9.4605284 × 1015 meters. Since c is the maximum speed at which all energy, matter, and information in the Universe can travel, it is the universal physical constant on which the light-year (ly) as one of the astronomical units of length is based.
That means that visible light as an electromagnetic radiation cannot travel faster than c and in one Julian year it can traverse a maximum distance of
d = t * c
d is distance in meters
t time in seconds
c the speed of light in vacuum in meters per second
If we calculate this distance for a 4.243 ly distant object, that comes out as 4.243 * 365.25 * 86,400 s * 299,792,458 m * sˉ¹ or exactly 40,141,879,395,160,334.4 meters (roughly 40 trillion kilometers or 25 trillion miles).
That is the distance the light traveled since it was last reflected of (or in our case emitted from, since Proxima Centauri is a red dwarf star) the surface of a celestial object to be 4.243 Julian years later visible at our observation point, in this case our planet Earth from where the distance to Proxima Centauri you quoted was measured.
The more powerful the telescope, the further into the past we can see because the light is much older! This goes the same regardless of the distance of the object you’re observing, but astronomy is particularly neat in this regard and we can observe objects that are so distant that we see them from the time when they were still forming.
For further reading on other units used to measure far away objects you might be interested in reading this question on the parsec.
Answer 2 (score 11)
A deeper answer is “yes and no”. In the frame of reference of the light itself the journey from Proxima to here is instantaneous. In our frame of reference it takes four years - this is all bound up in relativity and the nature of spacetime.
But in the everyday sense we are indeed looking back in time at light from the stars.
Answer 3 (score 7)
Actually, the light hitting us from Proxima Centauri is not necessarily 4.243 years old. Perhaps some of the photons arriving here were created in the photosphere of Proxima. But some of them will have been created in the center of the star, and these photons may take many years to arrive at the photosphere, where they are then “emitted”.
For our sun, it is written (in Wikipedia’s article about our Sun):
“The gamma rays (high-energy photons) released in fusion reactions are absorbed in only a few millimeters of solar plasma and then re-emitted again in a random direction and at slightly lower energy. Therefore it takes a long time for radiation to reach the Sun’s surface. Estimates of the photon travel time range between 10,000 and 170,000 years.”
Similarly, many of the photons arriving from Proxima may be many tens of thousands of years old. Their travel time from Proxima’s photosphere is only a small part of their journey to Earth.
40: What is the current theory for the formation of the Earth’s Moon? (score 17933 in 2017)
Question
Given that the Earth and Moon are often said in the literature to have very similar geochemistry, what is the current theory as to how the Earth’s Moon formed?
Answer accepted (score 10)
The current accepted theory is known as the Giant Impact Hypothesis, where according to this NASA webpage “Origin of the Earth and Moon” (Taylor) a Mars sized object collided into the early Earth.
This theory allows explanations of (from the link above):
The chemical makeup:
The giant impact hypothesis is consistent with our ideas for how planets were assembled and explains some important features of the Earth-Moon system, such as why the Moon has only a tiny metallic core.
and in terms of orbits:
To account for the amount of angular momentum in the Earth-Moon system, Cameron estimated that the object would need to be about 10% the mass of Earth, about the size of Mars. (Angular momentum is the measure of motion of objects in curved paths, including both rotation and orbital motion. For the Earth and Moon this means the spin of each planet plus the orbital motion of the Moon around the Earth.)
However, recently, there has been some revisions to this theory.
According to NASA’s page “NASA Lunar Scientists Develop New Theory on Earth and Moon Formation”, based on concerns that the Mars-sized would likely to have had a different composition to the Earth (inconsistent with the current similarities in geochemistry).
The new hypothesis is (from the link above):
After colliding, the two similar-sized bodies then re-collided, forming an early Earth surrounded by a disk of material that combined to form the moon. The re-collision and subsequent merger left the two bodies with the similar chemical compositions seen today.
This is also discussed in “Huge Moon-Forming Collision Theory Gets New Spin” (Wall, 2012), basing the revised hypothesis on the rotation rate, which is theorised to have been very fast, from the article:
Earth’s day had been just two to three hours long at the time of the impact, Cuk and Stewart calculate, the planet could well have thrown off enough material to form the moon (which is 1.2 percent as massive as Earth).
Further discussion, based on the geochemical makeup is discussed in “Geochemical Constraints on the Origin of the Earth and Moon” (Jones and Palme) conclude that:
Although none of these observations actually disproves the giant impact hypothesis, we find it disquieting that the obvious consequences expected of a giant impact are not observed to be fulfilled.
So, there is still some question as to precisely how the Moon was formed.
Answer 2 (score 4)
A recent study reported in the Phys.org article “Study crashes main Moon-formation theory” , based on the Nature Geoscience paper “A multiple-impact origin for the Moon” performed numerical modelling that suggests that
the Moon could instead be the product of a succession of a variety of smaller collisions.
The specific strength of such a theory is that “Numerous”impactors" would have excavated more Earth material than a single one" according to the authors, allowing for a stronger composition-based correlation with the Earth.
The authors also contend that
sub-lunar moonlets are a common result of impacts expected onto the proto-Earth in the early Solar System and find that the planetary rotation is limited by impact angular momentum drain.
The process is simplified as follows (from the article):
An interesting consequence of this model, according to the authors is that:
“Building the Moon in this way takes many millions of years, implying that the Moon’s formation overlapped with a considerable portion of Earth’s growth,”
41: Does the Moon have any oxygen in its atmosphere? (score 17282 in 2013)
Question
Since the moon has gravity, it’s almost impossible that there aren’t some gasses trapped on the surface by the moon’s gravity. Has any free-floating oxygen been found on the Moon? If so, in what concentration?
Answer accepted (score 20)
The Moon’s atmosphere is very thin compared to the Earth, so thin that it is usually said to have no atmosphere. The Moon’s gravity is not strong enough to retain lighter elements, so they escape into space.
Apollo 17 carried an instrument called the Lunar Atmosphere Composition Experiment (LACE). Oxygen is not listed as one of the elements it found on this NASA web page. The principle gases found were neon, helium and hydrogen. Others included methane, carbon dioxide, ammonia, and water. Additionally, ground instruments have detected sodium and potassium in the lunar atmosphere.
The recently launched Lunar Atmosphere and Dust Environment Experiment (LADEE - pronounced laddie, not lady) should be able to tell us more soon.
Update: According to this article at Spaceflight Now, LADEE did detect oxygen in the Moon’s atmosphere.
Answer 2 (score 10)
Just to add to GreenMatt’s answer, according to the article “The Lunar Atmosphere: History, Status, Current Problems and Context” (Stern, 1999), the lunar atmosphere is in fact a tenuous exosphere, which the authors describe as being composed of
“independent atmospheres” occupying the same space.
This is further elaborated in “The Lunar Dusty Exosphere: The Extreme Case of an Inner Planetary Atmosphere” (NASA), that
in direct response to these intense and variable environmental drivers, the Moon releases a low density neutral gas forming a collisionless atmosphere. This ~100 tons of gas about the Moon is commonly called the lunar surface-bounded exosphere
There is also an ionosphere, due to (from the ANSA article):
Ions are also created directly either by surface sputtering or subsequent neutral photoionization, forming a tenuous exo-ionosphere about the Moon.
The authors also suggest that ionic oxygen may be present due to surface sputtering.
Due to solar radiation and the solar wind, dust particles become charged as well, and can be subsequently listed from the lunar surface.
42: A good telescope for the viewing of Nebulae, Stars and Planets (score 17162 in 2017)
Question
So, I am a 15 year old interested in astronomy. My father used to have a reflector telescope, but was never so serious about it that he could recommend me a good telescope. I am looking to buy a telescope up to £400 (possibly £500) and would like one that is mainly great for looking at nebulae. Stars and Planets would be great also, but are not necessary. Light Pollution isn’t too bad. I have been recommended a Newtonian Reflector, but am not too sure if this would be suitable for my conditions. Also, does anyone recommend a good, British website where I could buy such a scope from? Finally, is there much point in me buying a computerized scope, will this allow me to align the scope better, or is it just going to be a waste of money.
Answer accepted (score 20)
This is a very common question, yet very hard to answer if you prefer a clear, concise, uncontroversial answer that applies to all situations. So I’m not going to do that.
Instead, I’m going to describe your main options, and let you choose. Be aware that you’ll make the choice while still not knowing much about optics. So, in a sense, it will be just the beginning of a learning journey, and you’ll have to keep learning and trying things out. Hope you like that sort of thing, because it’s kind of mandatory in this hobby.
Also, there are two kinds of people in this field: those who enjoy using the instruments, and those who enjoy making the instruments. I’m 25% of the former, 75% of the latter kind (I make telescopes and mirrors, and I do some optics design). So that’s the perspective you’ll get.
First off, a few words about the objects you’ll observe. There are two kinds of objects:
HIGH RESOLUTION, HIGH BRIGHTNESS
These are the planets, the Moon, many double stars, and the like. These are typically pretty bright, so the amount of light you capture is not essential. But these objects show high-resolution detail, so the resolving power of your scope is crucial.
You get more resolving power if the aperture (the diameter) of the scope is bigger. But there’s a problem - seeing (the inverse of turbulence).
http://en.wikipedia.org/wiki/Astronomical_seeing
When air is turbulent, it will reduce the resolving power of your instrument. If your scope is 200 mm (8“) or bigger, this will happen A LOT. If your instrument is 100 mm (4”) or smaller, it’s not so noticeable because you don’t have that much resolving power to begin with.
There are ways to forecast seeing, because it depends on weather:
So, for high resolution targets, the point of having a large instrument (like a huge dobsonian) is because once in a while seeing is so good (turbulence is so low) that you can actually use all its huge power - but most of the time you’ll be using it at much less than its theoretical power, as if it’s a 4" instrument or whatever the seeing is at the moment.
These objects are not affected by light pollution. You could observe the Moon from a brightly illumined parking lot in the middle of the city. You don’t need to be in deep darkness while observing them - in fact, it’s beneficial if you’re not adapted to very deep darkness.
LOW BRIGHTNESS
These are most of the nebulae, star clusters, and galaxies. These are distant, faint objects. Their light is very faint, so it’s essential you capture as much of it as possible. Again, bigger aperture is better.
These are the kind of objects that huge dobsonians (half-meter aperture, or more) are made for. In a very large scope, there’s galaxies everywhere.
But again, there’s a catch. These objects are affected by light pollution.
http://www.jshine.net/astronomy/dark_sky/
You need to take your scope far and away from any city, town and industrial complex (at least a 1 hour trip at highway speed). This is pretty hard to do in Europe; in the US, population density is low and this is somewhat easier. Anyway, just stay away from city lights and the view will improve.
A few of these objects will be visible from the city: the Andromeda galaxy, the M13 cluster, the Orion nebula, etc. But many of them look much, much better away from light polution. In any case, you should be able to observe the whole Messier catalogue (over 100 objects) from the city - but out in the boondocks these things look better.
http://en.wikipedia.org/wiki/List_of_Messier_objects
It’s important that your eyes are adapted to very deep darkness when observing these things, so stay away from any source of light. Even the screen of your phone is far too bright. Street lights 500 meters away are too bright. And turn off that flashlight and keep it off. One second of exposure to a bright light destroys a dark adaptation that took 20 minutes to build.
Since these objects are so faint, the eye doesn’t see a whole lot of high-resolution detail, so they are not affected by seeing (as far as the human eye is concerned).
Okay, now here are the main types of instruments that you’re most likely to encounter:
NEWTONIAN REFLECTORS IN GENERAL (DOBSONIANS IN PARTICULAR)
Many people will say “just get an 8 inch dob as your first scope”, and that’s usually correct. This design gives the biggest aperture per price. Biggest bang for the buck. It scales up easily to very, very large sizes. It’s the ideal instrument for observing the “faint fuzzies” (nebulae, galaxies). It’s great for learning, because you can easily observe its guts. When seeing is good, a large dob will open up the sky for you - you can crank up the magnification a lot (depending on aperture).
On the flip side, due to the low price, you will see many dobs out there that are shoddily made.
You need to remember two things with a dob: collimation, and thermal management.
Collimation means keeping all the optics aligned. The mirrors will move a little, due to the way a dob is built, and they’ll lose collimation. So don’t forget to re-adjust it once in a while. In time, you’ll get so good at it, you’ll give it a two-minute quick adjustment every time you observe, and it’ll be good to go; that’s what I do.
This is why collimation is important:
http://legault.perso.sfr.fr/collim.html
This is a good introduction to newtonian collimation (which is a vast topic, you could write a whole book on it):
http://www.cloudynights.com/documents/primer.pdf
Thermal management means keeping the scope at the same temperature as the ambient air. If you don’t do this, internal convection will destroy the image. This is why:
http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-1/
http://garyseronik.com/beat-the-heat-conquering-newtonian-reflector-thermals-part-2/
Small dobs (up to 6…8“), just take it outside at least 1 hour before you observe - or better yet, 2 hours. Let it”breathe“. Dobs 10” or larger, you MUST use a fan on the primary mirror to force cooling.
REFRACTORS
These instruments are very convenient and very low maintenance. Take it out of the box, and it’s good to go. It’s collimated from factory, and it doesn’t care about thermal issues. A refractor makes a great “travelscope”, and a great grab-and-go scope.
On the flip side, they are the most expensive, in terms of how much aperture you get for the money you spend. And they are typically made in small apertures; most commercial refractors are not bigger than 6" or so; those that are bigger are extremely expensive. The small aperture limits the maximum performance of these instruments.
There is this myth out there that refractors are somehow magically better for planets (and other high-resolution targets), whereas dobs, even very large ones, are not so great. This is false; load up the designs in Zemax, do a basic performance analysis, and you’ll see that aperture is the dominant factor, with everything else being secondary. But there are several reasons why this myth is around, and it’s instructive to see what these reasons are:
- refractors perform at 100% of their potential no matter what, right out of the box. Dobs require maintenance and fiddling to perform at 100%.
- many dob owners simply don’t do the required maintenance - they don’t collimate their scope, and don’t give it time to reach thermal equilibrium. Then of course it will perform miserably on planets. There are 12" dobs out there that perform worse than 4" refractors because of how out of collimation they are and how hot the mirror is.
- many refractors out there are quite well made (they ought to be, given the higher price you’re paying), whereas many dobs are cheaply made in a sweatshop. So you’re not comparing apples to apples.
Bottom line: don’t buy a refractor because you “want to watch planets”. But do buy a refractor if simplicity of use and the low maintenance are very important to you (and you don’t care much about price).
A decent refractor on an excellent mount can do pretty good astrophotography. More on that below.
CATADIOPTRICS (CASSEGRAIN, MAKSUTOV, DALL-KIRKHAM, RITCHIE-CHRETIEN)
Broadly speaking, these are for astrophotography (not always, but usually). But can you do visual astronomy with them? Sure. It’s just that it’s easier to optimize them for photo, with this design.
Usually, they come installed on a motorized mount. All but the very cheapest ones have a computer on-board; you just punch the name of the object into the remote control, and the scope will turn around automatically to face the object. Sounds cool, right?
Well, I am going to say something a little controversial here: this is not a good place to start.
First off, astrophotography has an extremely steep learning curve. It’s very easy to take a little blurry photo of Jupiter that will impress nobody. It’s very hard to take a great, high resolution photo like these:
http://www.acquerra.com.au/astro/
The skills and knowledge required to take that kind of photos are more easily acquired by doing purely visual astronomy, on a dob or whatever, for a couple years or more. Also, the equipment required for that sort of stunt costs many thousands of dollars - your cheap little Cassegrain cannot do it.
So why buy a small Cass? Well, if you have an overwhelming interest in astrophoto, and are prepared to deal with the difficult learning, then go ahead. But realize that the first images are not going to be very impressive, and the motorized mount will not teach you some valuable skills that you would acquire more easily pushing a dob around and keeping your eye plugged into the ocular for a while. Also, money spent on the motorized mount is money NOT spent on aperture, and the aperture is so important for overall performance.
For astrophotography, the mount is of huge importance; the scope per se is relatively less important (I’m generalizing). A so-so scope on a great mount is much better than the other way around. But a great GEM mount can be very expensive.
Well, that’s it. Do some more research on your own, then make a decision. Don’t forget that learning only begins with the moment you acquire the scope.
Above all, have fun, and keep looking up.
Clear skies to you!
Answer 2 (score -1)
I have had all three “styles” of scopes. The “Best” scope is the one you get out and go outside with:
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The 4.5" (114mm) reflector on a very cheap equatorial mount was my first scope. Not very good at all. It worked and I did see brightest planets, stars, Andromeda, Pleiades, and of course the Moon. No it wasn’t a great scope, but I used and abused it.
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The next was a cheap 90mm refractor on a cheap equatorial mount. Same viewing but I think just a tad better as far as clarity, but it had the exact same mount, and just as unsteady.
-
Next was the 8" Dob, which I love, and still have, and the last scope I’ll probably ever buy is my Stellarvue Access SV102. The mount I purchased separate, which is the Vixen Optics Advanced Polaris-M Motorized EQ Mount, with Tripod. I sold the 4.5" (114) Reflector.
My experience has been that I stayed around long enough, went to enough star parties and evens, checked out all the astronomers in the area, went to McDonalds Observatory on Star Party Nights, and eventually got around to putting together a scope package that I use every night….well almost. There are pro’s and con’s to all scopes, but if I was a beginner again, I’d take the Dobsonian hands down for a starter scope. I still use it frequently. Just my opinion…
43: Why do Black Holes in the middle of galaxies not suck up the whole galaxy? (score 17090 in 2017)
Question
As stated in several sources, it’s supposed that in every galaxy there is a black hole in the middle.
My question is, why do these black holes in the middle of galaxies not suck up all the surrounding matter in the galaxy?
Answer accepted (score 84)
You shouldn’t think of black holes as “sucking things in”. Black holes interact with matter through gravity, just the same as any other object. Think of our Solar System. All the planets orbit around the sun because it has a lot of mass. Since the planets have some lateral motion (they’re not moving directly towards or away from the sun), they circle around it. This is known as conservation of angular momentum.
When talking about gravity, all that matters is the mass of the objects involved. It doesn’t really matter what kind of object it is*. If you were to replace the sun with a black hole that had the same mass as our sun, the planets would continue on the exact same orbits as before.
Now, the black holes at the centers of most spiral galaxies do accumulate mass. Some of these black holes have accretion disks around them. These are swirling disks of gas and dust that is slowly falling into the black hole. These gas and dust particles lose their angular momentum through interactions with gas and dust nearby and by radiating energy as heat. Some of these black holes have very large accretion disks, and can generate huge amounts of electromagnetic radiation. These are known as active galactic nuclei.
So, long story short, black holes don’t “suck”. They just interact with things gravitationally. Stars, gas, and other matter in the galaxy has angular momentum, so it stays in orbit around the center of the galaxy. It doesn’t just fall straight in. This is the same reason the Earth orbits around the Sun.
*Disclaimer: When you talk about things like tidal forces, you do need to take into account the size of the objects. But for orbital mechanics, we don’t need to worry about it because the distances between the objects are generally much larger than the objects themselves.
Answer 2 (score 11)
I once heard of a Japanese cartoon/movie/show where space pirates threatened to compress the planet Jupiter into a black hole and thus destroy half the Milky Way galaxy.
It sounds like an interesting idea, but… even if you could compress Jupiter into a black hole, its mass would remain the same, meaning that Jupiter (now a black hole) would still continue to move around our sun in its same orbit, and Jupiter’s moons would still continue to orbit Jupiter as they did before.
Many people think that once a star collapses into a black hole, its “sucking power” (its gravitational force) increases. This is simply not the case. Believe it or not, many stars are less massive after they turn into a black hole than before, when they were shining stars. This is because, at the end of their lives, some stars shed a significant portion of their outer layer into space right before they collapse into a black hole.
I’ve read that if you compressed the Earth to the size of a cherry, its density would be so great that it would turn into a black hole. Assuming that were true and it was actually done, the black hole of Earth would still continue orbiting the sun once every year, and Earth’s moon would continue to orbit the Earth about once every 29.5 days. (Now, the spin of the new black-hole-Earth about its axis would probably be different, but the time it would take to orbit the sun would not change.)
Surprisingly, once the Earth got compressed into a cherry-sized black hole, less space debris would fall into it than before (when the Earth was the size of… well, Earth). This is because the newly-formed black-hole-Earth would take up much less space (volume) and asteroids and comets would be more likely to miss the cherry-sized (or slightly-larger-than-cherry-sized) volume that, if not missed, would cause the debris to be sucked into the black hole.
If the debris missed the black-hole-Earth by even a kilometer (which might seem like a large distance to us, but is very miniscule in astronomical terms), it would be flung off in a different direction, possibly never to return.
So, basically, a common misconception people have regarding black holes is that nothing has more gravity than a black hole, and that stars that form into black holes all of a sudden have increased gravity and therefore get more “sucking power.” This is simply not true. Black holes still have the same mass as before (sometimes less, depending on how they’re formed), and how much “sucking power” they have is still dependent on how much mass they’re made up of.
While it may be true that the most massive stars in the universe are indeed black holes (if you’d even call them stars at that point), there exist many stars that are more massive (and therefore have more “sucking power”) than many black holes.
So the fact that our galaxy’s center probably contains a super-massive black hole doesn’t mean that the black hole would suck up any more matter than if it were the same amount of mass which happened to not be in black-hole-form.
Answer 3 (score 4)
Gravity follows the law of inverse square. To put it simply if you double the distance from a gravity source you quarter is effect. So if you double the distance you are from the earth you feel 1/4g. It’s important to note that as the distance increases it will never be 0, it will always be some non-zero value no matter the distance.
So at galactic distances the force of gravity the central black hole has very little effect.
This only explains part of it. The other part is conservation of angular momentum.
The force of gravity and the angular momentum is what is responsible for orbits. In orbital mechanics you raise your orbit by adding speed, not altitude. Your adding angular momentum which raises your orbit. To lower your orbit you reduce you speed which reduces your angular momentum and your altitude.
So for things to “fall” into a black hole they must be travailing at a speed where their orbit intersects the event horizon. This is rarely the case or those “things” wouldn’t truly be in orbit to begin with. So the very fact that all the “stuff” that makes up the galaxy orbits the central black hole means that it cant just fall into it.
These 3 things are always in balance in a stable orbit, force of gravity, speed and altitude ( or distance from the gravity source ). If you change one of them the other 2 must also changes. If you decrease speed your altitude goes down, and the gravity increases. If you increase the gravity the speed must also increase or the altitude will decrease.
So you see things can’t just fall into the black hole. That said it is my view that eventually everything in the galaxy will fall into the central black hole, however this will take many billions of years.
Of course this is overly simplifying things, and I am by no means an expert on this stuff. But it’s something I can picture in my mind, the balance between momentum and gravity.
v
44: Why did the moon abruptly change positions in the sky? (score 17087 in )
Question
Three or four weeks ago I was outside showing my neighbors how cool the moon was through a ‘backyard’ telescope in broad daylight (~8PM PDT). It was 3/4 of the way across the sky (setting behind the houses behind us.) Two weeks later the moon was not even up yet at 8PM. The planets I’ve been watching (Jupiter, Saturn, Mars) are all rising and setting a little earlier each day as I would expect.
Why did the moon abruptly change its position in the sky a couple of weeks ago?
Answer accepted (score 11)
Compared to the planets the moon changes it’s rise and set times very quickly. Using the calculator on this page it can be seen that if one were looking at the sky on May 1st, 2014 from Irvine, California the moon set at 10:08PM (which at 8PM would have had the moon most of the way across the sky as noted in the question.)
The moon rises and sets approximately one hour later each day. So, by May 14th the moon did not even rise until nearly 8PM - which also correlates with the observations noted above.
Contrasting that with the other context point made in the OP, using this page we can see that on May 1st Saturn rose at 8:07PM and on May 14th rose at 7:11PM.
So while the moon changes its rise a set times by almost an hour each day, the planets change much more slowly by comparison, almost an hour over two weeks.
If one weren’t continually watching the moon’s progression it would appear to abruptly change positions when comparing it to the planets’ movements.
Answer 2 (score 5)
The moon didn’t abruptly change position, it is gradually and continually changing position. You’ve obviously never really noticed what it is doing until now. You would really benefit from getting a planetarium application for your PC or device and adjusting the time base to see how everything moves. You’ll get a much better understanding of the mechanics of it all, why the moon goes through phases, and why it rises and sets at different times.
Answer 3 (score 4)
The moon orbits the Earth in roughly 29 days. So in 14 days, it would be on the other side of the Earth.
45: Is the Sun in our solar system moving or stationary? (score 17073 in 2017)
Question
When I was small, I read that Sun is fixed at the center of the solar system and that all the other planets rotate around it.
But later I heard that even the Sun is not fixed; it moves. Is this true?
Why had people previously thought that the Sun is fixed?
Is this correct planets in helix path or spring like path?
Answer accepted (score 15)
The Sun moves, even in the context of the solar system. Gravity of the planets (mostly Jupiter) pulls the Sun out of position with respect to the centre of gravity of the solar system. This wikipedia entry explains it in a lot more detail, and also explains that their common centre of gravity lies outside of the sun.
This wobbling of a star due to planets orbiting them is also one method of how we detect planets around other stars.
Assuming that the Sun is fixed within the context of the solar system is a fairly good approximation. You need accurate and long term observations to detect the wobble.
Answer 2 (score 8)
The Sun orbits the Galactic center once approximately every 226 million years as the galaxy spins, which is also travelling through space towards an unknown source of gravity (that is currently blocked from view by the luminosity of the galaxy itself) known as “The Great Attractor”.
Answer 3 (score 5)
The Sun moves in many ways. For one the sun does wobble, as stated above, due to Jupiter’s pull, this is actually how astronomers are able to find new planets! They can use physics and Mathematics to figure out a planets size and number of planets orbiting a star just by studying the stars wobble.
Secondly the Sun along with the Solar System orbits the center of the Milky Way Galaxy. The length of this process is called a Galactic Year. The Solar System’s Galactic year ranges somewhere from 225 to 250 million years.
Lastly our Galaxy and the Sun move as a whole through space, which is what will eventually cause the Milky Way Galaxy to collide with the Andromeda Galaxy.
46: What is the angular diameter of Earth as seen from the Moon? (score 16623 in )
Question
…today I must have left my brain home. I know it’s a simple calculation but I keep failing simplest calculations today.
What is the angular diameter of Earth as seen from the surface of the Moon?
Answer accepted (score 8)
You can calculate the angular diameter of the Earth using the equation: \[a = \arctan \frac{D}{d}\] where \(a\) is the angular diameter, \(D\) is the physical diameter of the Earth, and \(d\) is the distance from the Moon to the Earth.
The equatorial radius of the Earth is \(r_E = 6378.1 \textrm{km}\), the diameter is therefore \(D= 2 \times r_E = 12756.2\).
- The mean distance to the Moon is \(d=384399\textrm{km}\). This gives a (mean) angular diameter of \(a=1.90065\)°.
- At Perigee the distance is \(d=362600\textrm{km}\). This gives an angular diameter of \(a=2.01482\)°
- At Apogee the distance is \(d=405400\textrm{km}\). This gives an angular diameter of \(a=1.80226\)°
These Moon-Earth distances are as seen from the centre of the Moon. To calculate the diameter from the surface of the Moon, you’ll have to subtract the position of the observer along the Moon-Earth axis.
If the observer is on the Moon’s equator and the Earth is at zero hour angle (i.e. on the local meridian), the distance to the Earth needs to be subtracted by \(r_M=1738.14\textrm{km}\). This gives the following values:
- The mean distance to the Moon is \(d=384399-1738 = 382661\textrm{km}\). This gives a (mean) angular diameter of \(a=1.90928\)°.
- At Perigee the distance is \(d=362600-1738 = 360862\textrm{km}\). This gives an angular diameter of \(a=2.02452\)°
- At Apogee the distance is \(d=405400-1738 = 403662\textrm{km}\). This gives an angular diameter of \(a=1.81001\)°
The angular diameter of the Earth from the surface of the Moon is, therefore, between \(a=1.80226\)° (at apogee and the Earth is near the horizon) and \(a=2.02452\)° (at perigee and for an observer at the equator and when the Earth is at maximum altitude on the meridian).
Or about 2 degrees.
Answer 2 (score 5)
The Earth viewed from moon will appear larger, in proportion to how much larger the Earth’s diameter is versus the moons diameter.
- Earth diameter 7900mi
- Moon diameter 2100mi
So the Earth-view from the moon would appear 3.75 times as large as the Moon appears in the sky on earth.
I looked up the moon’s typical angular diameter, it is 0.5 degree. So the Earth’s typical angular diam would be 1.9 degree.
Answer 3 (score 4)
The average angular diameter of the Moon, as seen from the Earth, is about 31 arcminutes.
The angular diameter depends on the distance between the two objects and the diameter of the object being viewed. Specifically, for small angles, it is the diameter divided by the distance. When the distance is the same, the angular size is proportional to the diameter.
The distance remains the same when viewing the Earth from the Moon, but the Earth is larger. According to NASA, the diameter of the Moon is 3,476 km, and the diameter of the Earth is 12,756 km.
So, because it’s proportional, the angular diameter can be calculated as follows:
\(a_{Earth} = a_{Moon} \times {d_{Earth}\over d_{Moon}}\)
$ = 31 arcminutes $
$ 114 arcminutes$, or just under 2 degrees.
This is approximate, because not only is this valid only for small degrees, where the tangent of an angle can be approximated by the angle itself (in radians), the Earth-Moon distance varies because the Moon’s orbit around the Earth is an ellipse.
47: What is the difference between our time and space time? (score 16438 in 2014)
Question
I am trying to understand the phenomenon of space-time. But, everything on internet seems to be too complicated for me since I do not have a background in physics. Can anyone give me simple explanations for the following:
- What is the difference between time and space-time?
- How does gravity affect the passage of time?
- What is the speed of light and how does it relate to time?
- How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?
- How is time, or for example the age of the universe, actually measured experimentally?
Answer accepted (score 13)
What is the difference between time and space-time?
Space-time is time plus space.
How does gravity affect the passage of time?
The higher the gravity of a planet or star and the closer to that body the slower the time.
What is the speed of light and how does it relate to time?
The speed of light is 299,792.4580 km/s in vacuum, the speed at which light propagates, roughly 1.3 seconds from Earth to Moon. Velocity is distance divided by time; this applies also to the speed of light.
How do scientists deal with timescales on the order of billions of years if time is not constant for all observers in the universe?
They treat time dependent of the observer. For different observers on Earth variations are tiny, in many cases neglectable in comparison to measurement errors, although not for precision measurements.
How is time, or for example the age of the universe, actually measured experimentally?
There are many ways to measure or estimate ages. The age of the earth can be estimated by ratios of certain radionuclides in the oldest rocks. Estimates of the age of the universe are obtained by simulations based on observations (redshift and distance) of distant galaxies, and on observations of the cosmic microwave background. The redshift is used for a velocity estimate; together with a distance estimate (obtained by stars of known brightness) one can calculate back, when all objects of the universe should have been close together.
Answer 2 (score 5)
Space, as we experience it, is simply three-dimensional Euklidean (flat) space. A flat space is one in which parallel lines never intersect. Consider the two-dimensional space of your kitchen table: (if your table is new) this space is flat. But the surface of a sphere, also a two-dimensional space, is not flat. In the same way a three-dimensional space can be curved (though this is hard/impossible to imagine).
Time as we experience it is simply linearly progressing. You can combine space and time to a four-dimensional space-time, one dimension of which is time-like. The theory of relativity now says that this four-dimensional space-time is not always exactly flat and that any curvature of space-time is equivalent to a force. Any form of energy (and rest-mass energy is the most compact such form) curves space-time and hence exerts a force – known as gravity.
This curving of space time affects both space and time, hence gravity also affects clocks etc. For weak forces (any we experience as humans), the curvature effects are extremely minute.
Answer 3 (score 0)
Particle like, molecule or atoms or any smallest particle behaves differently in different environment. For example, these particle behaves different on earth or on moon or on earth’s orbit or near to black hole horizon.
This was actually proved by the theory of relativity given by Albert Einstein. We (Human) have to adjust artificial satellite’s clock everyday 0.0000000001 second. Otherwise we will get wrong data from satellite on earth computers and all.
Now IMO, Any object in Gravitational environment or the object travelling at a speed of light must be getting some effect at atomic level and all electromagnetic level also. Because of that clock become slow ( because electromagnetic/atomic movement which requires to produce output in clock becomes slow ).
In the same way, IMO (not validated ), when we travel at the speed of light or roaming near to the black-holes horizon then our biological activity or any electromagnetism which is required to THINK becomes slow in such a way that difference produced by this effect is dramatically big. It is said that, a person on earth become old but the person travelling at a speed of light will think ‘I traveled only for a week!’(see, this person’s biological/atomic activity become slow, even about thinking) but actually from earth’s person’s point of view he/she (traveller) traveled for at least 75 year at a speed of light which is real fact. Normally after 75 years anybody will become old on earth. This is all IMO.
What is the speed of light and how does it relate to time?speed of light is 300000 Km/sec.
48: If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface? (score 15930 in 2019)
Question
I understand this is a silly hypothetical but I’m asking for a 7 year old so please bear with me.
Imagine an interstellar stray gas giant comes flying through our solar system.
If we were not concerned that it would also steal our atmosphere and create tidal forces that destroyed everything… How close would it need to come to us to exert enough gravity to lift people off the ground and pull them into its own orbit?
Answer accepted (score 74)
TL:DR Jupiter isn’t dense enough for its gravity gradient over Earth’s radius to produce a 1g tidal acceleration, even right at Jupiter’s surface.
Jupiter’s gravity will pull on the Earth itself, as well as everything on it.
It’s not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will scale with the mass of each object; if the Earth is a zillion times more massive than we are, then Jupiter’s gravitational force will also be about a zillion times larger.
What that means is that Earth will accelerate towards Jupiter, and we will accelerate along with it, and so we won’t “feel the tug” anywhere near as strongly as one might suspect.
Instead, let’s think about the size of the Earth, and the fact that people on the near side will be closer to Jupiter than the center of mass of the Earth, and people on the far side will be farther away.
Since people nearer to Jupiter will feel a slightly stronger acceleration than the center of mass of the Earth, they will feel a quite gentle tug. We’ll calculate that in a minute.
But believe it or not, people on the far side of the Earth, feeling less of a tug than the Earth’s center of mass, will believe they are being pulled in the opposite direction! They won’t really be pulled away from Jupiter, but they will not accelerate towards Jupiter as fast as the Earth, and so it will feel like they are being repelled.
This kind of force is called a tidal force and this is the picture that’s often used with the concept:
Source Replace “Satellite” with “Jupiter”
The acceleration we feel due to gravity is expressed as
\[a_G = \frac{GM}{r^2}\]
where \(G\) is the gravitational constant and equal to about \(6.674 \times 10^{-11}\) m^3/kg s^2 and M is each mass that’s pulling on you.
If you put in 6378137 meters and the mass of the Earth (\(5.972 \times 10^{+24}\) kg) you get the familiar 9.8 m/s^2.
If Jupiter were 114,000,000 meters or 114,000 kilometers away, the Earth would accelerate at 1 g towards it, but people on the close and far side would accelerate very differently. On the close side, being 6,378 kilometers closer, would feel an acceleration 1.2 m/s^2 greater, so they would feel that they weighed 12% less. And people on the far side would also feel about the same amount lighter because they felt less acceleration than the Earth.
If Jupiter were so close that it were practically touching the Earth, it still wouldn’t pull is off of Earth, assuming that Earth remained intact. But that wouldn’t last very long!!! Earth would be accelerating towards Jupiter at about 20.9 m/s^2, and people on the near side would feel acceleration of 24.8 towards Jupiter, but relative to Earth that’s only 3.9 m/s^2, so not enough to overcome Earth’s gravity of -9.8 m/s^2.
On the far side of Earth it’s similar; the acceleration towards Jupiter would be 17.8 m/s^2 but minus Earth’s acceleration of - 20.9 it’s -3.0 m/s^2 away, but that’s also not enough to overcome the attraction to Earth of in this case +9.8 m/s^2.
When Earth touches Jupiter, we will feel about 40% lighter on the near side and 31% lighter on the far side of Earth, but we would not leave the surface.
However, in just minutes we’d be pulled so deep into Jupiter that we would be crushed by Jupiter’s internal atmospheric pressure.
It would certainly be fun, but it wouldn’t last long!
Answer 2 (score 17)
About 70,000 km. If the Earth orbited Jupiter (or flew by) at a closer distance, not just we would leave the surface, but the whole Earth would disintegrate since all its mass will be leaving too.
70,000 km is Jupiter’s Roche limit (although its actual value varies a lot depending on the other involved body), the radius where tidal forces (already explained in uhoh’s answer) overwhelm gravitational forces and any orbiting body can’t hold itself by its own weight. In that context, people on the surface don’t behave differently than rocks.
Btw., this scenario is also explored in a youtube video. I wouldn’t say it’s very good but it may be helpful to explain the Roche limit to a 7 year old.
Answer 3 (score 9)
As other answers point out, Jupiter is not quite massive enough to pull a planet of Earth’s density apart. But we can use a slightly heavier object instead – say, a small cold brown dwarf massing 13 Jupiters, or about 4000 Earths. According to the rigid-body Roche formula, its Roche limit is then \(\sqrt[3]{2\cdot 4000}=20\) earth radii, or 130,000 km. The radius of the brown dwarf is not much larger than Jupiter’s (since they’re both made of compressible gas), so smaller than 100,000 km, and there’s some room for Earth to be destroyed without actually colliding with it.
Our brown dwarf, wandering sedentarily around in the galaxy, spots our sun, and decides to take a closer look. It comes screaming through the inner solar system on a hyperbolic orbit, which means that it will be moving at somewhat more than solar escape velocity when it narrowly misses Earth – call it 100 km/s. Switching perspective, we can say that Earth comes at the brown dwarf at 100 km/s and just about misses. This speed allows us to spend a almost half an hour inside the Roche limit, if we almost touch the brown dwarf at the time of closest approach.
But that sounds like it will be unnecessarily dramatic, so it’s instead have the brown dwarf give is a slightly wider berth such that at the time of closest approach, the acceleration of gravity at ground zero will be a more pedestrian \(-0.1\;\rm m/s^2\). That will be the case when our distance to the brown dwarf is \(\sqrt[3]{\frac{9.82}{0.1+9.82}} = 0.997\) of the Roche limit, or 128,000 km. The length of our path through the Roche zone is then about 20,000 km, which means the encounter takes 200 seconds. Call it three minutes.
(The sharp-eyed reader will notice that these numbers mean that the far side of the earth is actually never inside the 130,000-km limit, but what really counts is the first derivative of the brown dwarf’s gravitational field, so if you’re standing on the antipodal point you’ll still have the Earth’s center-of-mass pulled away under you even if you yourself is outside the limit. The numbers are all approximate anyway).
(On the other hand, a few minutes clearly not enough time for the molten inside of the Earth to flow into a hydrostatic equilibrium in the new situation, so using the rigid-body formula is appropriate).
What happens then?
First, of course, it is an awesome sight. The brown dwarf dominates the sky with an angular diameter of somewhere between 60° and 100°.
Then, it may be getting uncomfortably hot. Not necessarily “the mountains are melting” hot or even “the seas boil away” hot. But this says that the coldest brown dwarves have about the temperature of a baking oven, and having a significant part of the sky at 150 °C can make anyone sweat. No worries, though – it will all be over in a few hours, so just go inside and crank up the AC; that’ll deal with it fine.
Right at ground zero gravity decreases smoothly while we approach Roche. When it passes zero G you’re in free fall and start floating gently upwards. Except that everything around you – cars, houses, trees, the soil itself – is also in free fall since the only thing that kept them down was gravity. So to a first approximation your local experience is not about bring ripped off Earth, but just of weightlessness. (Or is it? See below.)
Ditto at the antipodal point.
One problem that shows up here is that the atmosphere is escaping into space. Since there is no gravity to keep it down, it escapes rather faster than the gentle floating of cars, trees, and people, propelled by its own pressure. Even before we reach Roche, the air may have become too thin to breathe. On the other hand fresh air will rush in from the surrounding areas to fill the void, creating the great-great-grandmother of all hurricanes. (And a great-great-grandfather around the antipode, of course).
On a great circle 90° from ground zero, gravity increases to about 1.7 G. You feel heavy. Ho hum.
Between these areas dramatic things happen. At about 45° from ground zero (or the antipode) the tidal force is at right angles to vertical, so the strength of gravity is about what we’re used to – but its direction is different. It’s as if the world is tilted by tens of degrees, rather like how bad sci-fi movies pretend “entering a gravitational field” works. Tall buildings tip over; many not-so-tall ones just collapse. Lakes and seas do things that make the word “tsunami” pack up and go home, hopelessly outclassed. What the water doesn’t get, unstoppable rockslides will. And don’t forget the hypercane-force gales as the atmosphere slides “downwards” almost unimpeded.
This assumes that the ground below is rigid, of course. It isn’t quite, though it probably has enough structural integrity that the preceding paragraph is still true. In any case, the entire crust of Earth starts sliding “down” towards ground zero (or, as always, the antipode). Different parts of the crust slide at different velocities, though. Near the “ho hum” zone the crust is stretched; at ground zero or antipode, crust piles up. Nothing actually has time to move more than (very roughly) some tens of kilometers from its starting position at best, but that is quite sufficient to get cataclysmic hyper-earthquakes at every tectonically active zone on earth. Where there is no active zone to take up the stress, new ones open up.
I’m not entirely sure what the mantle is doing, but it probably isn’t something nice.
One thing the mantle is doing happens around ground zero. Without any net gravity to keep the crust down, hydrostatic pressure in the lower lithosphere drops towards zero. Dissolved volatiles in magmas everywhere attempt to outgas, forming bubbles and expanding the magma until the sheer inertia of the overlying rocks resists it. The effect is to push the crust upwards faster than it is being pulled by the tides. So standing at ground zero you may not get to experience weightlessness after all. Instead you get to stand right on top of the greatest volcano eruption in the history of the planet. Truly the experience of a lifetime.
Then the three minutes are up and the brown dwarf recedes again.
At ground zero you’re now at least a kilometer higher than you started out, together with everything around you, and still moving upwards at tens of kilometers an hour. That’s far less than escape velocity, so what goes up must come down again. Except “down” is now most likely a boiling volcanic inferno. Aren’t you glad you didn’t get roasted by the brown dwarf to start with?
There’s still time for the sliding tectonic plates to slide to a halt, and for the new rifts in the “ho hum” zone to start rivaling the ground zero volcano. Unless the interior of the earth deformed elastically so everything now tries to slide back.
The planet still exists, though. No mass was actually lost. On the other hand, the encounter did change our collective velocity by several tens of kilometers per second, which is broadly comparable to our usual orbital motion. That’s going to wreak total havoc on the seasons.
Oh well. It’s not as if any of us would be around to complain about that.
(The sharp-eyed reader from before will note that most of these calamities would happen even without getting all the way to the Roche limit. So if the world has to end, Jupiter’s gravity field might be capable enough, after all.)
49: Does the Milky Way move through space? (score 15880 in 2018)
Question
Does our galaxy moves through space? Or does it stay in a single location? If it does move, what causes it to move?
Answer accepted (score 73)
Does the Milky Way move through space?
Yes it does.
I’m very fascinated with space, although I don’t have a degree or any formal education, I’m still very in love with everything about it and want to learn constantly.
Good man Mike.
One thing I ask myself is if our galaxy moves through space?
It does. When we look at the Cosmic Microwave Background Radiation we see a “dipole anisotropy” due to the motion of the Earth relative to it:
Image courtesy of William H. Kinney’s Cosmology, inflation, and the physics of nothing
See Wikipedia for more:
“From the CMB data it is seen that the Local Group (the galaxy group that includes the Milky Way galaxy) appears to be moving at 627±22 km/s relative to the reference frame of the CMB (also called the CMB rest frame, or the frame of reference in which there is no motion through the CMB) in the direction of galactic longitude l = 276°±3°, b = 30°±3°.[82][83] This motion results in an anisotropy of the data (CMB appearing slightly warmer in the direction of movement than in the opposite direction).[84]”
627 km/s is quite fast. See this article, which says it’s 1.3 million miles an hour. The speed of light is just under 300,000 km/s or 670 million miles per hour, so the Milky Way is moving through the Universe at circa 0.2% of the speed of light. Also see the CMBR physics answer by ghoppe which talks about the CMBR reference frame, which is in effect the reference frame of the universe.
Or does it stay in a single location? If it does move, what causes it to move?
I’m afraid I don’t know why it’s moving. Perhaps it’s because the Universe is full of things moving in fairly random directions. Like a gas.
Hopefully the question makes sense, if not I can elaborate.
It certainly makes sense to me!
Edit 13/09/2017 : as Dave points out in the comments, there are other motions, including the motion of the solar system around the galaxy, which is circa 514,000 mph. (See the Wikipedia Galactic Year article). And the motion of galaxies isn’t neat and tidy either.
Answer 2 (score 34)
Galaxies move through space with velocities of the order of a several 100 km per second; small velocities for small groups (~100 km/s; e.g Carlberg et al. 2000) and large velocities for rich clusters (~1000 km/s; e.g Girardi et al. 1993).
In addition to this so-called “peculiar velocity”, galaxies also also carried away from each other due to the expansion of the Universe, at a velocity proportional to the distance from each other (the “Hubble flow”). But this is not a motion through space; rather it is space itself that is expanding (and hence the velocity may exceed the speed of light for sufficiently large distances).
One may define a “global reference frame” with respect to which velocities are measured. Any reference is valid, but it makes sense to use the frame in which all galaxies are, on average, in rest (when the Hubble flow is subtracted)\(^\dagger\). In this frame, the Local Group that the Milky Way is a part of moves with some \(620\, \mathrm{km \, s}^{-1}\) (as noted in John Duffield’s answer above), whereas the center of the Milky Way has a velocity\(^\ddagger\) of \(\mathbf{565 \pm 5 \, km \, s^{-1}}\) (Planck Collaboration et al. 2018).
What causes this movement of galaxies? Galaxies that are not too far from each (i.e. closer), “feel” each others mutual gravitational forces. A galaxy in a group or cluster moves around in the common gravitational field, but for galaxies that are farther away, the Hubble flow carries them away from each other too fast for them to attract each other.
This movement can be traced back to the tiny quantum mechanical fluctuations in the primordial soup of particles during cosmic inflation, i.e. less than \(\sim10^{-32}\,\mathrm{s}\) after the Big Bang. As time went by, ever-so-slight overdensities grew in amplitude, until they collapsed to form the structure we see in the Universe today. During this collapse, matter grew turbulent, whirling clumps around that eventually became the galaxies that orbit each other.
\(^\dagger\)Formally, one uses the frame in which the cosmic microwave background is isotropic.
\(^\ddagger\)Taking into account our motion around the Galactic center, our Sun (currently) moves through space at \(369.82\pm0.11\,\mathrm{km\, s^{-1}}\).
Answer 3 (score 9)
NO, if don’t want to consider it to be doing so.
Everything in the universe only “moves” when considered from a different frame of reference. From the observational frame of reference of the Milky Way galaxy, every other object in the universe is moving, and the Milky Way is stationary.
When I walk to work, what I am actually doing is dragging the surface of the Earth, and everything on it, towards me with my feet, until my office arrives.
50: Life planets orbiting black-holes. Can/Do they really exist? (score 14627 in 2019)
Question
So, I watched Interstellar and if you watched it too you know that there’s a planet orbiting a black-hole, they call it Miller’s Planet. According to the movie, every hour on Miller’s Planet is equivalent to 7 years on Earth due to gravitational pull from the black-hole.
Question: Assuming there are other life forms in the universe, is it really possible for it to be near a black-hole? Is it possible that they came to existence thousands (or even millions) of years before us, but are not as advanced as us because time on our planet is way quicker than that of theirs? is a lot more for us? If they have an assignment due tomorrow, we have, say, 100 years more time to do it (assume what else we can do in that 100 years). Or they actually are more advanced than us, but from Earth, they somehow live in the past?
Answer accepted (score 9)
Well, first things first. It’s not likely to have a planet orbiting near a black hole and in significant time dilation because the tidal effects would likely tear anything that close apart. Certainly a planet orbiting a stellar mass black hole would need to be quite far away so as to not be torn apart, so any time dilation would be pretty small.
Around a super-massive black hole, the tidal effects are smaller and a nearish orbit with some measure of time dilation is possible. (see link below for more specifics),
But a stable planetary orbit, you probably max out at about 20% time dilation and only around a super-massive black hole, where there’s only 1 per galaxy. The idea of 1 day to 100 years isn’t practical, it’s 80 days to 100 days if you’re talking about a stable planetary orbit.
and, I’m not sure you’d want to be that close to the black hole in the center of the galaxy, not because the orbit isn’t stable, but because stars are in that orbit too. It might not be a safe place to be.
So, in reality, you’d want a stellar mass black hole and a distant orbit, where the time dilation would be quite small, and in that scenario, yes, life is possible because of tidal energy, so a planet could have liquid surface water and an atmosphere, even if the black hole gave off very little light and heat.
Such a planet in a tidal-energy orbit would probably be tidally locked which would protect the far side from any gamma rays the black hole spits out when it eats, so it would in theory be a good place for life. No significant light source, unless it was a binary system, so, plants would have it harder, but there would be heat.
There’s another problem. The creation of black holes tends to blow everything appart in a huge explosion. It’s unclear that a planet would survive a black hole’s birth, so you might need a captured planet.
Finally, intelligent life . . . we really don’t know enough about how common intelligent life is on other planets. Life might be fairly enough, but intelligent life is far less clear and there’s other factors than just time.
Today, we simply don’t know enough to predict if there’s intelligent life out there or not. There probably is life elsewhere in the universe, though even that isn’t 100% certain, but regarding intelligent life, there’s far too many unknowns in that equation. I think a black hole might not be optimal for the formation intelligent life because of the lack of light so, much less photosynthesis, so, slower oxygen formation (if it follows the same pattern the earth does) and the unlikeliness that a planet would survive the black hole’s creation.
Answer 2 (score 5)
Interstellar’s “Miller’s Planet” is utter rubbish…
First of all, black holes don’t start out as black holes. Black holes form at the end of the life cycle of very big star (at least 25 solar masses but more frequently over 35 or 40 SMs) when it goes supernova or hypernova. Any planets in orbit around that star are going to be obliterated before the black hole becomes a black hole. We need to remember that black holes are not stars which happen to be so dense light cannot escape their gravity. Black holes are the REMNANTS of stars. And, the process of becoming a black hole obliterates planets.
Also, that planet in the movie orbiting a black hole had liquid water. There is no such thing as a goldilocks zone around a black hole like you have around a star where water is not frozen and not boiled off. That means it must be just the right distance to a heat source. The black hole isn’t going to give it that heat or light.
Finally, for that “planet” to have gravitational time dilation equivalent to 7 years per earth hour, it’ll have to be so close to the black hole that mile high waves will be the last thing you worry about. If you are not fried by radiation in seconds if you are that close. In fact, the planet will be ripped apart, pulverized, irradiated and basically become part of the (glowing) accretion disc depicted in the movie. That disc, BTW, can be light years across in massive black holes.
But, hey, it’s a movie and it’s supposed to be entertainment not science. So I can forgive all that. What I could not forgive was the blatant stupidity in the plot, such as the crew KNOWING about the time dilation but not knowing that Miller would have just landed even though they had gotten his signal for many years. They landed, had their harrowing encounter with the wave and THEN realize that as far as Miller is concerned he had just landed and only just died!?!! Bugs Bunny Science can nonetheless be entertaining. But, silly and illogical plots make a movie hard to follow.
Answer 3 (score 2)
According to Opatrný et al. (2016) “Life under a black sun”, it might be possible to have warm temperatures around an isolated supermassive black hole thanks to the blue-shifted cosmic background radiation. Their calculations lead to an estimated equilibrium temperature of 890°C for Miller’s planet (this is without the additional radiation coming from the accretion disc), which does not bode well for the watery environment shown in the movie. A planet orbiting further out might be able to support liquid water. In the past the planet would need to be located further away from the black hole because of the higher temperature of the background radiation in the early universe.
Whether such systems actually exist is another matter. The inhabitants of such a planet would have to hope that nothing came too close to the black hole, as accretion would make the environment rather hostile.
51: How do we have photos of galaxies so far away? (score 14496 in )
Question
A possible answer for this is that, light emitted from the galaxies travelled a billion miles all the way to earth, where the hubble space telescope picked up this light through its sensors, and was able to construct an image of the galaxy
but if this is true, and galaxies are billions of miles away, shouldn’t the light particles emitted from the galaxies be scattered all over the place? after all they have been travelling from millions of years, and have probably collided with asteroids and other foreign objects. What were the chances that about 95% of the photons actually reached earth, giving us a very detailed image.
Consider the andromeda galaxy which has a distance of 1.492 × 10^19 mi from earth. If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy, evident from the photo below?
Shouldn’t like half of the galaxy be missing since photons could have hit other objects, and “never have reached earth”?
Answer accepted (score 51)
There are two reasons that often — but not always — light from galaxies millions and even billions of lightyears away make it through the Universe and down to us:
Particle number and particle size
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First, the intergalactic medium (IGM) is extremely dilute. The number density of particles out there is of the order \(n\sim10^{-7}\,\mathrm{cm}^{-3}\), or roughly 26 orders of magnitude lower that the air at sea level! That means that if you consider a tube from Andromeda to the Milky Way with cross-sectional area of \(1\,\mathrm{cm}^{2}\), it will contain roughly one microgram of matter (thanks to Rob Jeffries for catching a factor \(10^6\) error).
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Second, even if a photon comes close to an atom, it will only be absorbed if its energy matches closely some transition in the atom. Since most of the atoms are ionized (and thus should be called plasma instead, but in astronomy the distinction if often not made), there are no electrons to absorb the photon. The photons are more likely to interact with the free electrons via Thomson scattering, but the Thomson cross section is immensely small \((\sim10^{-24}\,\mathrm{cm}^{2})\), so even if you consider the CMB photons — which have traveled through the Universe almost since the Big Bang — only around 5% of them have interacted with electrons on their way.
In other words: The amount of transmitted light depends on two factors: 1) The amount of matter along the line of sight, and 2) that matter’s ability to absorb the light. In the IGM, both are tremendously small. When the light enters the interstellar medium (ISM) inside our galaxy, it may encounter denser clouds with atoms that are able to absorb the light. But usually (although not always) “dense” is still very dilute compared to Earth’s atmosphere.
Mathematical expression
In general, if a beam of light traverses a region of particles, each with a cross section \(\sigma\) (measured e.g. in cm\(^2\)), passing \(N\) particles per area of the beam (measured e.g. in cm\(^{-2}\)), then the opacity of the medium is given by the optical depth \(\tau\), defined by \[ \tau \equiv N \, \sigma. \] The transmitted fraction \(f\) of photons is then \[ f = e^{-\tau}. \] In general \(\sigma\) depends on the wavelength, and thus part of the spectrum may pass unhindered, while another part may be completely absorbed.
The figure below (from here) shows the spectrum of a quasar lying at a distance of 22 billion lightyears, i.e. \(10\,000\) times farther away than Andromeda. You see that there are several thin absorption lines (caused by intervening hydrogen clouds whose densities are a factor of 10-100 higher than the IGM), but still most of the light makes it down to us.
Because the light we see from this quasar was emitted so long ago, the Universe was considerably smaller at that time, and thus the density was larger. Nonetheless, only a small fraction is absorbed. The farther away the light is emitted, the longer ago it was, which means smaller Universe, and higher density, and thus the more light is absorbed. If you consider this quasar (from here) which lies 27 billion lightyears away, you see that much more light is absorbed in part of the spectrum. Still, however, much light make it through to us.
The reason that it is only the short wavelengths that are absorbed is quite interesting — but that’s another story.
Answer 2 (score 24)
As Rob Jeffries says, the universe is mostly empty space. A photon can easily travel thousands of light years without interacting with anything. Most of the interaction would occur when photons entered the earth’s atmosphere. The Hubble avoids this. These photos were most likely from combining several viewing sessions giving basically an extended time period for observing the galaxy.
Answer 3 (score 23)
There’s a misconception in your question I don’t think the other answers have addressed.
If light emitted from the galaxy travels in all directions, then how is it that we can still map out the entire galaxy
Light is emitted from the galaxy in all directions. Only a tiny, tiny fraction of it is directed to Earth, and of that, an even tinier fraction is collected by any given telescope. But we can still see it, because galaxies are very, very bright. Andromeda contains about a trillion stars.
52: app to locate star by coordinates (score 14423 in )
Question
I’ve named a star and would love to be able to find it in the sky - is there an app I can use to locate stars given a specific coordinate?
Answer accepted (score 3)
There are apps which will help guide you to more significant stars (Google skymap comes to mind if you’re looking for a simple mobile system - Stellarium may also be a good option), but it’s highly unlikely that any app will include the location and names of so-called ‘personal’ stars, as the bodies who sell such things don’t have any the authority to do so and are not generally recognized in the actual world of Astronomy.
Note that the IAU (International Astronomy Union) is the body generally recognized to have the responsibility of naming and classification of stars and other bodies. They also come out strongly against such practices and services:
As an international scientific organization, the IAU dissociates itself entirely from the commercial practice of “selling” fictitious star names or “real estate” on other planets or moons in the Solar System. Accordingly, the IAU maintains no list of the (several competing) enterprises in this business in individual countries of the world. Readers wanting to contact such enterprises despite the explanations given below should search commercial directories in their country of origin.
Answer 2 (score 2)
Alternatively, if you were to get the Right Ascension and Declination coordinates from the company that sold you the star you could use a database like SIMBAD to find the actual designation of the star you purchased. You could then probably use the SIMBAD designation to find the star using Stellarium.
Here’s a link to the SIMBAD database: http://simbad.u-strasbg.fr/simbad/sim-fcoo
I hope this helps!
53: How loud would the Sun be? (score 14319 in 2015)
Question
Sound can’t travel through outer space. But if it could, how loud would the Sun be? Would the sound be dangerous to life on Earth, or would we barely hear it from this distance?
Answer accepted (score 99)
The Sun is immensely loud. The surface generates thousands to tens of thousands of watts of sound power for every square meter. That’s something like 10x to 100x the power flux through the speakers at a rock concert, or out the front of a police siren. Except the “speaker surface” in this case is the entire surface of the Sun, some 10,000 times larger than the surface area of Earth.
Despite what “user10094” said, we do in fact know what the Sun “sounds” like – instruments like SDO’s HMI or SOHO’s MDI or the ground-based GONG observatory measure the Doppler shift everywhere on the visible surface of the Sun, and we can actually see sound waves (well, infrasound waves) resonating in the Sun as a whole! Pretty cool, eh? Since the Sun is large, the sound waves resonate at very deep frequencies – typical resonant modes have 5 minute periods, and there are about a million of them going all at once.
The resonant modes in the Sun are excited by something. That something is the tremendous broadband rushing of convective turbulence. Heat gets brought to the surface of the Sun by convection – hot material rises through the outer layers, reaches the surface, cools off (by radiating sunlight), and sinks. The “typical” convection cell is about the size of Texas, and is called a “granule” because they look like little grains when viewed through a telescope. Each one (the size of Texas, remember) rises, disperses its light, and sinks in five minutes. That produces a heck of a racket. There are something like 10 million of those all over the surface of the Sun at any one time. Most of that sound energy just gets reflected right back down into the Sun, but some of it gets out into the solar chromosphere and corona. No one can be sure, yet, just how much of that sound energy gets out, but it’s most likely between about 30 and about 300 watts per square meter of surface, on average. The uncertainty comes because the surface dynamics of the Sun are tricky. In the deep interior, we can pretend the solar magnetic field doesn’t affect the physics much and use hydrodynamics, and in the exterior (corona) we can pretend the gas itself doesn’t affect the physics much. At the boundary layers above the visible surface, neither approximation applies and the physics gets too tricky to be tractable (yet).
In terms of dBA, if all that leaked sound could somehow propagate to Earth, well let’s see… Sunlight at Earth is attenuated about 10,000 times by distance (i.e. it’s 10,000 times brighter at the surface of the Sun), so if 200 W/m2 of sound at the Sun could somehow propagate out to Earth it would yield a sound intensity of about 20 mW/m2. 0dB is about 1pW/m2 , so that’s about 100dB. At Earth, some 150,000,000 kilometers from the sound source. Good thing sound doesn’t travel through space, eh?
The good folks at the SOHO/MDI project created some sound files of resonant solar oscillations by speeding up the data from their instrument by 43,000 times. You can hear those here, at the Solar Center website. Someone else did the same thing with the SDO/HMI instrument, and superposed the sounds on first-light videos from SDO. Both of those sounds, which sound sort of like rubber bands twanging, are heavily filtered from the data – a particular resonant spatial mode (shape of a resonant sound) is being extracted from the data, and so you hear mainly that particular resonant mode. The actual unfiltered sound is far more cacophonous, and to the ear would sound less like a resonant sound and more like noise.
Answer 2 (score 26)
While Sir Cumference’s post is a very intriguing answer, but I’m afraid it’s wrong. The sun’s surface is clearly in motion, but that does not necessarily result in the radiation of audible sound, even if the sun and earth where in a fluid medium (such as a air) that would allow sound transfer.
To explain why, we can actually apply the same line of analysis to the earth’s ocean. The surface moves a lot, so sound should be radiated. However, we hear nothing unless you are really close by and have breaking waves.
Let’s run the math with rough numbers: The ocean has a surface area of about 510 million square kilometers. \(150 \cdot 10^{12} m^2\). Let’s say the average wave height is 1m and the average wave frequency is 0.1 Hz (1 wave every 10 s). If the ocean were a spherical source this would create a sound power of \(5 \cdot 10^{24} W\) and the sound pressure at 1000 km away would be 240 dB SPL. That’s obviously not the case, otherwise we’d all be dead.
So why not? In order for sound to actually radiate, the surface must move uniformly. For every ocean wave that moves air up there is a wave nearby that moves air down and so the contributions simply cancel. Technically speaking, we need to calculate the power by integrating the normal intensity over the entire surface, the intensity has equal amounts of positive and negative components and the sum over those is zero.
That’s the same reason why you put a loudspeaker in a box: in open air, the air motion from the front of the cone and from the rear of the cone will simply cancel out, so you put it in a box to get rid of the sound from the rear.
So I think the real answer here is: you would hear absolutely nothing since the sound contributions from different parts of the sun’s surface would cancel each other out. Sound radiation over that distance would only occur if the sun’s surface moves uniformly, i.e. the whole sun expands or contracts. That does happen to some degree but only at very, very low frequencies which are inaudible and where sound radiation is a lot less efficient.
Answer 3 (score 4)
Along with the other answers, which differ, about the loudness of the Sun there is information available about what it actually sounds like. I would describe it as as varying humming with static.
Listen to the raw audio in this NASA video: “NASA | Sun Sonification (raw audio)”, a narrated version by NASA Goddard: “Sounds of the Sun”, or visit Goddard Media Studios’ webpage: “Sounds of the Sun”. The article says nothing about the “loudness”.
Another webpage at NASA, with an identical name to the one at GMS: “Sounds of the Sun”, provides some additional information:
"The Sun is not silent. The low, pulsing hum of our star’s heartbeat allows scientists to peer inside, revealing huge rivers of solar material flowing around before their eyes — er, ears. NASA heliophysicist Alex Young explains how this simple sound connects us with the Sun and all the other stars in the universe. This piece features low frequency sounds of the Sun. For the best listening experience, listen to this story with headphones.
…
These are solar sounds generated from 40 days of the Solar and Heliospheric Observatory’s (SOHO) Michelson Doppler Imager (MDI) data and processed by A. Kosovichev. The procedure he used for generating these sounds was the following. He started with doppler velocity data, averaged over the solar disk, so that only modes of low angular degree (l = 0, 1, 2) remained. Subsequent processing removed the spacecraft motion effects, instrument tuning, and some spurious points. Then Kosovichev filtered the data at about 3 MHz to select clean sound waves (and not supergranulation and instrumental noise). Finally, he interpolated over the missing data and scaled the data (speeded it up a factor 42,000 to bring it into the audible human-hearing range (kHz)). For more audio files, visit the Stanford Experimental Physics Lab Solar Sounds page. Credits: A. Kosovichev, Stanford Experimental Physics Lab.".
As is explained on the Stanford webpage: “Solar Sound Speed Variations” they have been able to analyze these sounds to produce a density plot of the Sun. Further information is available on Stanford’s webpage: “Helioseismology” where they explain:
"Waves
The primary physics in both seismology and helioseismology are wave motions that are excited in the body’s (Earth or Sun) interior and that propagate through a medium. However, there are many differences in number and type of waves for both terrestrial and solar environments.For the Earth, we usually have one (or a few) source(s) of agitation: earthquake(s).
For the Sun, no one source generates solar “seismic” waves. The sources of agitation causing the solar waves that we observe are processes in the larger convective region. Because there is no one source, we can treat the sources as a continuum, so the ringing Sun is like a bell struck continually with many tiny sand grains.
On the Sun’s surface, the waves appear as up and down oscillations of the gases, observed as Doppler shifts of spectrum lines. If one assumes that a typical visible solar spectrum line has a wavelength of about 600 nanometers and a width of about 10 picometers, then a velocity of 1 meter per second shifts the line about 0.002 picometers [Harvey, 1995, pp. 34]. In helioseismology, individual oscillation modes have amplitudes of no more than about 0.1 meters per second. Therefore the observational goal is to measure shifts of a spectrum line to an accuracy of parts per million of its width.
Oscillation Modes
The image below was generated by the computer to represent an acoustic wave ( p mode wave) resonating in the interior of the Sun.
The three different kinds of waves that helioseismologists measure or look for are: acoustic, gravity, and surface gravity waves. These three waves generate p modes, g modes, and f modes, respectively, as resonant modes of oscillation because the Sun acts as a resonant cavity. There are about 10^7 p and f modes alone. [Harvey, 1995, pp. 33]. Each oscillation mode is sampling different parts of the solar interior. The spectrum of the detected oscillations arises from modes with periods ranging from about 1.5 minutes to about 20 minutes and with horizontal wavelengths of between less then a few thousand kilometers to the length of the solar globe [Gough and Toomre, p. 627, 1991].
The figure above shows one set of standing waves of the Sun’s vibrations. Here, the radial order is n = 14, angular degree is l = 20, and the angular order is m = 16. Red and blue show element displacements of opposite sign. The frequency of this mode determined from the MDI data is 2935.88 +/- 0.2 microHz.
The wikipedia webpage on Helioseismology offers this power chart: 
An analysis of the Sun’s p-modes was offered in: “Activity-related variations of high-degree p-mode amplitude, width, and energy in solar active regions” (Jan 21 2014), by R. A. Maurya, A. Ambastha and J. Chae. In section 3 they provide a formula to convert the 3 dimensional resonance to amplitude:
…
"1. Introduction
Photospheric five-minute oscillations, probably first observed by Leighton et al. (1962), are caused by trapped acoustic waves (p-modes) inside the solar interior (Ulrich 1970; Leibacher & Stein 1971) and are well known and have been studied extensively. It is believed that the energy of p-modes is contributed by convective or radiative fluxes. A precise determination of the p-modes properties provides a powerful tool to probe the solar interior. High-degree (\(ℓ\) > 200) acoustic oscillations are vertically trapped in a spherical shell with the photosphere as the upper boundary and the lower boundary depending on the horizontal wavenumber, \(\small{k^2_h = \frac{l(l+1)}{r^2}}\), and the frequency (\(ω\)),
\[\frac{l(l+1)}{r^2_t} = \frac{w^2}{c^2_s(r_t)}, \tag{1}\]
where \(r_t\) is depth of the lower turning point. Lifetimes of high-degree modes are much shorter than the sound travel time around the Sun, therefore local effects are more important for these modes than for the low-degree modes, which have longer horizontal wavelengths and longer lifetimes. It is likely that high-degree acoustic waves are not global modes, that is, they do not remain coherent while travelling over the circumference to interfere with themselves. Therefore, they can locally be considered as horizontally travelling, vertically trapped waves. These are observed as photospheric motions inferred from the Doppler shifts of photospheric spectral lines.
…
3. Analysis techniques
3.1. Ring diagrams and p-mode parametersTo estimate the p-mode parameters corresponding to a selected area over the Sun, the region of interest is tracked over time. This spatio-temporal area is defined by an array (or data cube) of dimension \(N_x × N_y × N_t\). Here, first two dimension (\(N_x,N_y\)) correspond to the spatial size of the active region (AR) along \(x\)- and \(y\)-axes, representing zonal and meridional directions, and the third (\(N_t\)) to the time \(t\) in minutes. The data cubes employed for the ring diagram analysis have typically duration of 1664 min and cover area of 16° × 16° centred around the location of interest. This choice of area is a compromise between the spatial resolution on the Sun, the range of depth and the resolution in spatial wavenumber of the power spectra. A larger size allows accessing the deeper sub-photospheric layers, but only with a coarser spatial resolution. On the other hand, a smaller size not only limits access to the deeper layers, but also renders the fitting of rings more difficult.
The spatial coordinates of pixels in tracked images are not always integer. To apply the three-dimensional Fourier transform on tracked data cube, we interpolated the coordinates of tracked images to integer values, for which we use the sinc interpolation method. Three-dimensional Fourier transformation of data cube truncates the rings near the edges due to the aliasing of higher frequencies toward lower side. To avoid the truncation effects, we apodized the data cube in both the spatial and temporal dimensions. The spatial apodization was obtained by a 2D-cosine bell method, which reduces the 16° × 16° area to a circular patch with a radius of 15° (Corbard et al. 2003).
The observed photospheric velocity signal \(v(x,y,t)\) in the data cube is a function of position (\(x,y\)) and time (\(t\)). Let the velocity signal in frequency domain be \(f(k_x,k_y,ω)\), where, \(k_x\) and \(k_y\) are spatial frequencies in \(x\)- and \(y\)- directions, respectively, and ω is the angular frequency of oscillations. Then the data cube \(v(x,y,t)\) can be written as
\[v(x,y,t) = \int \int \int f(k_x,k_y,\omega)e^{i(k_x x + k_y y + \omega t)} dk_x dk_y d\omega.\tag{2} \]
The amplitude \(f(k_x,k_y,ω)\) of p-mode oscillations is calculated using three-dimensional Fourier transformation of Eq. (2). The power spectrum is given by \[P(k_x,k_y,\omega) = \left | f(k_x,k_y,\omega) \right |^2. \tag{3} \]
5. Summary and conclusions
We studied the high-degree p-mode properties of a sample of several flaring and dormant ARs and associated QRs, observed during solar cycles 23 and 24 using the ring-diagram technique, assuming plane waves, and their association with magnetic and flare activities. The changes in p-mode parameters are the combined effects of duty cycles, foreshortening, magnetic and flare activities, and measurement uncertainties.
The p-mode amplitude (\(A\)) and background power (\(b_0\)) of ARs were found to be decreasing with their angular distances from the disc centre, while the width increases slowly. The effects of foreshortening on the mode amplitude and width are consistent with reports by Howe et al. (2004). The decrease in mode amplitude \(A\) with distance arises because with increasing distance from the disc centre we measure only the cosine component of the vertical displacement. Moreover, foreshortening causes a decrease in spatial resolution of the Dopplergrams as we observe increasingly closer toward the limb. This reduces the spatial resolution determined on the Sun in the centre-to-limb direction, and hence leads to systematic observational errors.
The second-largest effects on p-mode parameters are caused by duty cycle. We found that the mode amplitude increases with increasing duty cycle, while the mode width and background power show the opposite trend. Similar results were reported previously for the global p-mode amplitude and width, for example by Komm et al. (2000a). These authors reported the strongest increase in mode width and reduction in amplitude with duty cycle when its values are lower. These changes in mode parameters may be caused by the increase in signal samples in data cubes. However, we found that for a few modes in the five-minute and in higher-frequency bands, the mode amplitudes do not increase significantly with duty cycle. The effect of the duty cycle decreases with increasing harmonic degree \(ℓ\). To study the relation of mode parameters with magnetic and flare activities, we corrected the mode parameters of all the ARs and QRs for foreshortening. …".
The exact loudness, as calculated above, is a function of where and when you measure.
The Wikipedia webpages: Chladni figures (flat), mechanical resonance and Helmholtz resonance (air-filled sphere) provide some related information about the difficulty and complexity of the calculations. The paper: “A review on Asteroseismology” (Nov 7 2017), by Maria Pia Di Mauro discusses standing waves travelling inside the star which interfere constructively with themselves giving rise to resonant modes.
54: Do all planets have a molten core? (score 14240 in )
Question
As we know, according to Wikipedia on Earth’s inner core:
The Earth’s inner core is the Earth’s innermost part and according to seismological studies, it is primarily a solid ball with a radius of about 1220 kilometers, or 760 miles (about 70% of the Moon’s radius). It is believed to consist primarily of an iron–nickel alloy and to be approximately the same temperature as the surface of the Sun: approximately 5700 K (5400 °C).
Now the question is, do all planets have molten inner cores?
Answer accepted (score 4)
The short answer is no. Take Mercury for example in this comparison of Earth Mercury core. Mercury is thought to have a liquid outer core and solid inner core. The gas giants like Jupiter are thought to have a relatively tiny rocky core but the convective motion in the metallic hydrogen is what gives them the strong magnetic fields.
See also: Is Mercury's core liquid? for more on Mercury’s core.
Answer 2 (score 1)
A lot of the data we have on the inner structure of the Earth comes from seismology - the study of earthquake waves travelling through the ground. This requires sensors placed at different points around the planet to pick up the waves and infer the structures they passed through. I haven’t looked into it but I would think that this would be very difficult to do on other planets in terms of getting lots of sensors there and positioning them correctly, so maybe we just don’t know this sort of thing about other planets yet!
Answer 3 (score 0)
Depends on what you call “planet.” For example, in comparison to Earth, the Earth and Moon is a “binary planet system.” And the moon doesn’t have any molten core. Also, you said inner. Inner core of Earth (and many other planets) are solid because of (what I think is pressure). Finally, if it is outer core you meant, and you didn’t count the moon as a “planet”, would you count extrasolar planets as “planets”. Please clarify, but to your current question, no.
55: Why is it okay to watch a sunset but not an eclipse? (score 14077 in 2017)
Question
People watch sunsets all the time. You don’t see people using special “sunset glasses”.
Yet with an eclipse, warnings are posted everywhere not to view it without special glasses.
Why is this?
Answer accepted (score 19)
During a sunset, the Sun is lower in the sky than during most of the day - much lower. Therefore, light from the Sun travels through about 120 miles of dense atmosphere, compared to the roughly 2 miles it travels through from straight up. Here’s a rough sketch (not to scale) to demonstrate this. It is clear that \(B>A\):
Light scatters in the atmosphere; in fact, one type of scattering is why the sky is blue. The longer travel distance means that there is much more scattering of ultraviolent light, which in turn means that the light you see is less intense.
Okay, you say. But doesn’t the eclipse still block a lot of light? Well, unless there’s 100% coverage - totality - there’s still plenty of light coming from the uncovered part of the Sun, and that matters. The uncovered part is as bright as it normally is, and looking at that part is just as dangerous with or without the eclipse.
There’s one more thing to consider, which is that people watching a sunset don’t look at the Sun; they look at the clouds and sky around the Sun. If you look directly at the Sun, your eyes will be damaged, no matter what’s happening, an eclipse or a sunset.
Answer 2 (score 5)
There is also a second reason for this. The density of the atmosphere decreases as you go up in the atmosphere. At sunset, the sun’s rays hit the atmosphere at an angle and refract through the atmosphere. The refraction is proportional to the incident angle, so it happens more at the horizon than during the day when the sun is high in the sky. The refraction also causes dispersion of light. This is part of the reason why the sun becomes red at sunset. The redness is also caused by the increasing scattering of blue light as the light passes through more atmosphere. The other colors, including ultraviolet radiation are systematically removed from the sun as it drops below the horizon. Because the light is refracted, the sun’s image appears above the horizon even though the sun has actually already gone below the horizon. At this point, the damaging ultraviolet light can no longer reach your eyes.
56: How would Earth’s climate differ if it’s axis were tilted around 90 degrees like Uranus? (score 13942 in 2013)
Question
As the title states, what would the effect on Earth’s climate be if it’s axis were tilted approximately 90 degrees like Uranus?
I’m specifically wondering about the effects on seasons, temperature, and life on Earth. However, I’m happy to hear about other effects as well.
Answer accepted (score 8)
This is a good question, and has been somewhat a topic of study by a few. So below is some consequences of if Earth had a tilt like that of Uranus (for this explanation, I am disregarding the wobbles in the Earth’s current axis, just focussing on the 90 degree axis):
According to the article Not All Habitable Zones Are Created Equal (Moomaw, 1999), each day and night would be 6 months each (including an epically long dusk and dawn), with daytime temperatures reaching up to 80C, but the night time side may not reach freezing - owing to the time to take to cool down from the 6 month day. Parts of the equatorial regions would be, however, permanently encased in ice.
Pretty much like this diagram of Uranus’ rotation (from Source: University of Hawaii:
A major consequence of this according to the author is:
In such an environment, life could almost certainly still appear, but it would have much more difficulty evolving into forms that could survive such grotesque temperature extremes – which would greatly slow down its evolution into more complex forms, maybe by billions of years.
A great consequence of this, particularly the 6 month night is that photosynthesis would have been stilted, if it could have started at all. This has a great consequence on oxygen levels of the atmosphere.
Interestingly, the authors claim that if the Earth had an axial tilt of 90 degrees, but at 210 million km from the sun, then:
its climate would be positively balmy – the equator would be 11 deg C (52 deg F), and the poles would never rise above 46 deg C (115 deg F) or fall below 3 deg C (37 deg F). Earth would have no ice anywhere on its surface, except on some of its highest mountains.
According to the article High Planetary Tilt Lowers Odds for Life? (Hadhazy, 2012), has a great way of putting it:
“Your northern pole will be boiled during part of the year while the equator gets little sunlight,” said Heller. Meanwhile, “the southern pole freezes in total darkness.” Essentially, the conventional notion of a scorching hell dominates one side of the planet, while an ultra-cold hell like that of Dante’s Ninth Circle prevails on the other.
Then, to make matters worse, the hells reverse half a year later. “The hemispheres are cyclically sterilized, either by too strong irradiation or by freezing,” Heller said.
They also describe that if life were to evolve, extremophiles (specifically, thermophiles) would be dominant - seasonally.
57: Why doesn’t the sun pull the moon away from earth? (score 13937 in 2017)
Question
If the suns gravitational pull is strong enough to hold much larger masses in place (all the planets) and at much greater distances (all planets further away from the sun then earth) why does it not pull the moon away from earth?
Answer accepted (score 18)
Why doesn’t the sun pull the moon away from earth?
Short answer: Because the Moon is much closer to the Earth than it is to the Sun. This means the gravitational acceleration of the Earth toward the Sun is almost the same as is the gravitational acceleration of the Moon toward the Sun.
The Moon’s acceleration toward the Sun, \(-GM_\odot\frac{\boldsymbol R+\boldsymbol r}{||\boldsymbol R+\boldsymbol r||^3}\) is indeed about twice that of the Moon toward the Earth, \(-GM_\oplus\frac{\boldsymbol r}{||\boldsymbol r||^3}\). This is irrelevant. What is relevant is the Moon’s earthward acceleration due to gravitation compared to the difference between the Moon’s and Earth’s sunward gravitational acceleration, \[\boldsymbol a_{\odot,\text{rel}} = -GM_{\odot}\left(\frac{\boldsymbol R + \boldsymbol r}{||\boldsymbol R + \boldsymbol r||^3} - \frac{\boldsymbol R}{||\boldsymbol R||^3}\right)\] This relative acceleration toward the Sun is a small perturbation (less than 1/87th in magnitude) on the Moon’s gravitational acceleration toward the Earth. Given the current circumstances, the Sun can’t pull the Moon away from the Earth.
Longer answer:
The gravitational force exerted by the Sun on the Moon is more twice that exerted by the Earth on the Moon. So why do we say the Moon orbits the Earth? This has two answers. One is that “orbit” is not a mutually exclusive term. Just because Moon orbits the Earth (and it does) does not mean that it doesn’t also orbit the Sun (or the Milky Way, for that matter). It does.
The other answer is that gravitational force as-is is not a good metric. The gravitational force from the Sun and Earth are equal at a distance of about 260000 km from the Earth. The short-term and long-term behaviors of an object orbiting the Earth at 270000 km are essentially the same as those of an object orbiting the Earth at 250000 km. That 260000 km where the gravitational forces from the Sun and Earth are equal in magnitude is effectively meaningless.
A better metric is the distance at which an orbit remain stable for a long, long, long time. In the two body problem, orbits at any distance are stable so long as the total mechanical energy is negative. This is no longer the case in the multi-body problem. The Hill sphere is a somewhat reasonable metric in the three body problem.
The Hill sphere is an approximation of a much more complex shape, and this complex shape doesn’t capture long-term dynamics. An object that is orbiting circularly at (for example) 2/3 of the Hill sphere radius won’t remain in a circular orbit for long. Its orbit will instead become rather convoluted, sometimes dipping as close to 1/3 of the Hill sphere radius from the planet, other times moving slightly outside the Hill sphere. The object escapes the gravitational clutches of the planet if one of those excursions beyond the Hill sphere occurs near the L1 or L2 Lagrange point.
In the N-body problem (for example, the Sun plus the Earth plus Venus, Jupiter, and all of the other planets), the Hill sphere remains a reasonably good metric, but it needs to be scaled down a bit. For an object in a prograde orbit such as the Moon, the object’s orbit remains stable for a very long period of time so long as the orbital radius is less than 1/2 (and maybe 1/3) of the Hill sphere radius.
The Moon’s orbit about the Earth is currently about 1/4 of the Earth’s Hill sphere radius. That’s well within even the most conservative bound. The Moon has been orbiting the Earth for 4.5 billion years, and will continue to do so for a few more billions of years into the future.
Answer 2 (score 4)
The Moon is in orbit about the Sun, much as the Earth is. Although this is not the usual perspective from the Earth, a plot of the Moon’s trajectory shows the Moon in an elliptic orbit about the Sun. Essentially the Earth, Moon, Sun system is (meta) stable, like that of other planets orbiting the Sun.
Answer 3 (score 2)
If we “hold” the Earth and “move” the Sun away, the Moon wouldn’t stay with the Earth, but would follow the Sun. It is the only satellite in the Solar System that is attracted to the Sun stronger than to its own host planet:
our Moon is unique among all the satellites of the planets, is so far as it is the only planetary satellite whose orbital radius exceeds the threshold value, which means it is the one satellite on which the Sun’s gravitational acceleration exceeds the host planet’s gravitational acceleration. Consequently, it is the only moon in the solar system that is always falling toward the Sun.
The Moon Always Veers Toward the Sun
58: If light has no mass, why is it affected by gravity? (score 13677 in 2019)
Question
Light doesn’t accelerate in a gravitational field, which things with mass would do, because light has a universally constant velocity. Why is that exception?
Answer accepted (score 49)
Another way to answer this question is to apply the Equivalence Principle, which Einstein called his “happiest thought” (so you know it has to be good). The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead. So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now).
Given that rule, it is easy to see how light would be affected by gravity– simply imagine shining a laser horizontally. In the “left behind” reference frame, we see what would happen– the beam would start from a sequentially higher and higher point, and that raising effect is accelerating. So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser. Therefore, this must also be what is perceived from inside the box– the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward. Ergo, light “falls.”
Indeed, this is the crucial simplification of the Equivalence Principle– you never need to know what the substance is, all substances “fall the same” because it’s nothing happening to the substance, it is just the consequences of being “left behind” by whatever actually does have forces on it and is actually accelerating.
Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would “fall the same” as those with mass, but to see it requires taking a limit. Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass. All objects fall the same under Newtonian gravity. So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit. Nevertheless, Newtonian gravity doesn’t get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn’t treat the speed of light correctly.
Answer 2 (score 31)
There are a couple of ways one could approach your question:
Black holes are regions of space that have been deformed by a sufficiently concentrated mass. Light waves/particles always travel in a straight line at a constant velocity (\(c\)). Although a photon approaching a black hole will continue traveling in a straight line through space, space itself has curved so the photon’s path will curve.
While photons don’t speed up in the presence of a gravity well, they are affected by it in other ways. In specific, photons entering a gravity well are blue-shifted while photons leaving one are red-shifted. This red/blue-shifting happens because time passes slower within a gravity well than without. In all frames of reference, though, the speed of light remains constant. There’s more info on this on the wiki.
Note: The question originally referred specifically to black holes. The above hold for any concentration of matter (of which black holes are an extreme example).
Answer 3 (score 3)
TL;DR Light is affected by gravity because it travels along the space-time grid and its curvature which IS gravity. This gets very visible in black holes. also: Einstein > Newton
Black holes are black because no light that crosses the “Event Horizon” can escape ever again. Mass bends the “grid” of space-time. Light - 2-dimensionally speaking - travels along the floor of the space-time grid and follows its curvature i.e. it goes down a cone created by a presence of mass, and moves along the shortest path outwards again. This makes the journey of the light take longer. Now for a black hole things are more extreme: A black hole forms, when a lot of matter is crammed into a space that is at or smaller than the Schwarzschild Radius. The Schwarzschild Radius of any stellar object is determined solely by its mass. Any mass with a high enough denisty turns into a black hole:
rs = 2 * G / 2 c
Schwarzschild Radius =2* the gravitational constant / 2 * the speed of light.
Multiply that withM, the mass of an object in kg and you got the rs for that mass.To understand however how black holes curve space so much that they let no light escape, we must look at only a small part of Schwarzschilds equation.
To paint an image for understanding black holes, we only need this middle section:
1)2)
3)
4)
We’ve already established rs as being the Schwarzschild Radius of a particular object, r is the radius of the stellar object. When r becomes as small as rs you get a singularity1 and weird stuff starts happening, most importantly to OPs question, the space-time curvature at the black hole becomes infinite(!), this means that any light that intersects the event horizon at any point will take an infinite amount of time to travel across the black holes funnel. Even at a very flat angle relative to the event horizon, where its just ever so slightly poking it, it is lost because set theory teaches us: any subset of infinity is also infinite.
Here are some extra visalisiations:
Gravity space-time cone of earth:
Gravity space-time funnel of a black hole:
- Singularity: A singularity is basically, in calculus/algebra terms, just when you divide by zero (which you shall never do!). A 2D singularity might just look like this:
f(x) = 1/x(the singularity is there in the middle at x=0).
A 3D singularity can look like this /, singularity at x=1 (this is Riemanns zeta function).
59: How fast is a comet moving when it crosses Earth’s orbit? (score 13613 in 2014)
Question
Is it about the same as Earth’s orbital speed?
Answer accepted (score 5)
-
Comets don’t cross Earth’s orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.
-
What’s the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy \[ E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*) \] which is conserved along the orbit (\(v\) and \(r\) the Heliocentric speed and distance). For a returning comet, \(E<0\) and the speed cannot exceed the escape speed (which occurs for \(E=0\)) \[ v_{\rm escape}^2 = 2\frac{GM_\odot}{r}. \] The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, \(T=\frac{1}{2}v^2\) and \(W=-GM_\odot/r\), satisfy $ 2T+ W=0. $ For a (near-)circular orbit (such as Earth’s), \(r\) is constant and we have $ v^2_{} = GM_/r. $ Thus, at \(r\)=1AU \[ v_{\rm escape} = \sqrt{2} v_{\rm Earth} \] as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.
-
Can a comet near Earth have a speed similar to Earth’s orbital speed?. Let’s assume a comet with the same speed as Earth at \(r\)=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since \[ E = -\frac{GM_\odot}{2a}\qquad\qquad(**) \] with \(a\) the orbital semimajor axis, must also have \(a=1AU\) and the same orbital period as Earth, i.e. one year. Moreover, the comet’s apohelion satisfies \[ r_{\rm apo}\le 2a = 2{\rm AU}. \] Such comets don’t exist AFAIK. Most returning comets have much longer periods than 1 year.
-
What’s the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let’s parameterise the comet’s orbit by its period \(P=2\pi\sqrt{a^3/GM_\odot}\). From this relation we immediately get \[ \frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}. \] From equations (\(*\)) and (\(**\)), we can then find \[ v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}. \] In the limit of \(P\to\infty\), this recovers our previous result \(v_{\rm comet}\to v_{\rm escape}\). For typical period of $$70yr, the speed of the comet is close to this number.
-
Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at \(r\)=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed \(|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|\) of a comet with respect to Earth can be anyting between about 10 and 70 km/s.
Answer 2 (score 1)
There are relatively big varieties, but most of them is between 10 and 70 km/s.
If a comet is a periodic comet, that means it needs to have an elliptic orbit around the Sun. That gives an upper limit to its speed of the escape speed from the solar system on the orbit of the Earth. That is around 40 km/s.
But this 40 km/s is in the reference frame of the Sun. The Earth is moving in this reference frame with around 30 km/s, on a nearly circular orbit.
Between the escape speed and the mean speed of a circular orbit there is always a \(\sqrt{2}\) relation. It is a physical law.
Theoretically it were possible to find extrasolar comets (if the speed of it were bigger as around 70 km/s, it were a clear signature of its remote origin), but they aren’t coming.
60: Why can we sometimes see the moon during the day? (score 13499 in )
Question
At some points in the year, the moon at my location can be clearly visible during the day. Why is this?
If it is any help, I live in Alberta, Canada.
Answer accepted (score 10)
The Moon makes a full orbit around the Earth once per month. From full moon to full moon, it is 29.53058886 days (synodic month).
Full moon happens when the Moon is at one side of the Earth (right side in the image), and the Sun is at the other side. This way, Moon is fully iluminated by Sun (except in the cases where it is exactly in the opposite side of the same line, where you have a Moon eclipse). In this phase, Moon is at its highest altitude at midnight.
Then, as the Moon orbits, it comes closer to the Sun in the sky. This means both that it shows us less iluminated percentage and that its rise and set get delayed. At Last Quarter, Moon is at its highest altitude at sunrise. This means that you will be able to see the half moon figure on the sky during the morning (and last half of the night).
Then it comes New Moon, where the Moon is nearest to the Sun in the sky, and almost invisible, since it shows us the dark side during the day. Its highest altitude is at noon, but you will not be able to see it. You can, if you are lucky, see a Sun Eclipse.
Finally it comes First Quarter. To put it short, highest altitude is at sunset and you are able to see Moon during all the afternoon (and first half of the night).
Answer 2 (score 9)
Well, the Moon reflects quite a bit of the Sun’s radiation. In fact, this person states that:
The visual geometric albedo of the full moon is 12.5%, but much less at other phases.
In combination with this, the Moon is also the closest astronomical object to us, and since the intensity of light falls off as \(\propto \frac{1}{r^{2}}\) (inverse-square law), this plays arguably the biggest role in being able to see the Moon during the day. In fact, so much light from the Moon gets reflected to Earth that its apparent magnitude must be larger than the sky during the day (otherwise it would get washed out by sky noise).
According to this article, the apparent magnitude of the daytime sky is somewhere between 1.5 and 3. When you compare this to the full moon (–12.74; though you really can’t see this during the day due to the geometry of the Earth/Sun/Moon system), or even a new moon at its brightest (-2.50), the Moon can almost always be seen in the sky.
61: How old is the oldest light visible from Earth? (score 13457 in 2018)
Question
Because light can only travel so fast, all of the light we see in the sky was emitted at a previous moment in time. So if for example we see a supernova or some other great stellar event, by the time we see it, it maybe long over. That made me kind of curious, what is the most ancient light we can see from earth?
The universe is supposedly ~13+ billion years old, but we are probably not at the very edge of the known universe so all the light we see is probably less than 13 billion years old. So what is the oldest light we can see? and as an optional follow-up question how do we know the age of that light?
I guess the light itself may not actually be literally ‘old’, but its probably obvious what I’m asking here, put another way: what’s the longest distance that now earth visible light emitted has traveled to reach the earth? Though that reformation of the question gets kind of tangled with lensing effects.
Answer accepted (score 64)
The oldest light in the universe is the cosmic microwave background. Roughly 380,000 years after the Big Bang, protons and electrons “recombined”1 into hydrogen atoms. Before this, any photons scattered off the free electrons in the plasma filling space, and the universe was essentially opaque to light. Once recombination occurred, however, photons were able to “decouple” from the electrons and move through space unimpeded. This relic radiation is still observable today; it has been redshifted and cooled.
We can detect light from very distant objects, and we have. It makes more sense to talk about distance in terms of redshift; the larger the redshift, the farther away an object is. There are a number of extremely high-redshift objects, some of which have had their measurements confirmed, and others of which have not. Candidates include
- MACS0647-JD (redshift \(z=10.7^{+0.6}_{-0.4}\)), a very distant galaxy.
- UDFj-39546284 (redshift from \(z=10\) to \(z=12\)), another galaxy or protogalaxy (although its distance is uncertain, and its exact nature is disputed).
- GN-z11 (redshift \(z=11.09^{+0.08}_{-0.12}\)), a very luminous distant galaxy.
All of these objects would have formed some hundreds of millions of years after the Big Bang, however, so the light we see from them is much “younger” than that of the cosmic microwave background.
1 I’ve never liked the usage in this context, as this was the first time they combined; the “re” is kind of misleading.
Answer 2 (score 11)
What’s the oldest light that we can see?
The Cosmic Microwave Background is considered to be the oldest E-M radiation detectable to us. It’s in the microwave spectrum, so it can’t be seen with the naked eye but is picked-up by “radio telescopes”. We call it “light” in the broad sense.
One remarkable aspect about this background radiation is its near-uniformity in all directions. Astronomers reason that the uniformity is too strong for the source to be a really big thing like a huge balloon… but that would be the case if it was all actually as far apart as it seems to be.
If it were really as big as it looks, it would take twice the age of the universe for one side to be affected by the other side! Instead, astronomers believe that what we see was a very small body, which has become bigger; that’s why it looks the same in each direction. Some of the growth is called metric expansion of space and has a different meaning than ordinary growth.
How do we know the age of that light?
The age of the cosmic background light can only be determined indirectly, first by knowing how long ago the Big Bang happened, then by figuring when the light was emitted in the course of the Big Bang.
By comparing the rate at which everything seems to be getting bigger with how big everything seems to be, in the same way that you might estimate how long it would take to drive to a place given the speed of the road and the distance, we calculate the Hubble Constant. This helps us calculate how long ago the Big Bang happened.
Also, there are certain “sound waves” (baryonic acoustic oscillations) where old things we see, including the cosmic microwave background, get brighter and dimmer with a rhythm, like a clock’s pendulum. They can be measured either left-right (for moving things) or by monitoring a video (for stationary things). Measuring these rhythms and comparing them to the Hubble Constant also helps to calculate how long ago the Big Bang happened.
Finally, the microwave background has physical qualities (like temperature and density) that make it possible for us to determine when it was emitted during the expansion and cooling of the Big Bang. Together using all these calculations is how we figure the age of the cosmic microwave background light.
Astronomers believe that this combined calculation (called “LCDM”, “Lambda-CDM”, or “Big Bang Cosmology”) is very good because the different numbers do line up, for the most part*. They were pleased to report more good findings as recently as 2018 when a study called the Dark Energy Survey finished. Nevertheless, since LCDM includes certain assumptions that may never be validated, and since there are still some unexplained discrepancies, we don’t know whether another kind of calculation would be better, provided that it still fits the measurements.
How do we know that this is the oldest light?
It is only by thinking about the physical qualities of the cosmic microwave background, and thinking about when during the Big Bang it must have emitted its light, that astronomers identified it as the oldest possible light in the universe, older than any stars or galaxies. It doesn’t tell us how old it is by itself; in fact, astronomers are always making sure that it’s not in fact just a layer of dust on the telescope!
How far away is the cosmic microwave background?
This is a really hard question to answer. According to Big Bang Cosmology, the cosmic microwave background wasn’t “somewhere” but instead it was everywhere. And the distance it has traveled since the Big Bang is different than the time multiplied by light speed, because of metric expansion of space. This is a result of the relativistic length-contraction due to the speed at which everything is moving.
Is the observable universe younger than the greater universe, assuming that that exists?
Calculating the amount of time from the Big Bang to now gives the same result whether you consider our observable universe or the greater universe that may exist. That’s why the age of “our” universe is the same as the age of “the” universe.
*Some different studies to determine the Hubble constant have given cosmologists pause (link 1, link 2); depending on which part of the universe you look, it may be close to 67 or it may be closer to 73 in the standard units.
Answer 3 (score 3)
Scientists have discovered a galaxy, named GN-z11 (already mentioned by HDE 226868), which existed a mere 400 million years after the Big Bang, or about 13.3 billion years ago:
Farthest Galaxy Yet Smashes Cosmic Distance Record
The discovery of a 10 billion year old star was announced just last week:
Hubble spots farthest star ever seen
Here is a list of distant astronomical objects on Wikipedia.
62: Calculation of right ascension and declination (score 13412 in 2013)
Question
I am confused about this problem:
If I see an object from Mount Teide (longitude is 16“30’E and latitude is 28”18’N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43"40’, the stellar time at Greenwich at 0h UTC is 22h20min.
How can I calculate the right ascension and declination?
Answer accepted (score 5)
The answers to this question are very clear and much better than I could have written.
I’m not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I’m assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30’W not E.
We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-
diff in RA = Long/15
= 16° 30'/15
= 1H 06mThen:-
RA = RA at Greenwich - diff in RA
= 22H 20m - 1H 06m
= 21H 14mNow for Declination:-
The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked ‘H’. Now draw upwards 9 lines and mark the top ,‘Z’ and downwards 9 lines and mark the bottom ‘N’. You should end up with something that looks like this:-
Which is, basically a picture of your meridian as you sit and gaze south at it. ‘Z’ is your zenith, the point above you in the sky and ‘N’ is the nadir, it’s opposite out on the celestial sphere through the earth beneath your feet.
Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we’ll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-
Now we put in the body we are looking at. We know it’s elevation, just over 43°, we can just call it x as you didn’t specify a body. We can also label some of the distances we know, to end up with:-
We already knew that Lat was H -> Q, that’s how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-
dec = Alt - Lat
= 43° 40' - 28° 18'
= 15° 18'NAs ‘x’ is to the north of ‘Q’, then Declination is north.
So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18’.
This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you’ll be doing it in your head. I hope this has helped a bit.
I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.
63: Should I focus more on Aperture or Focal Length for a telescope? (score 13266 in 2013)
Question
Let’s assume there are three types of telescopes with the following specifications:
Telescope Aperture Focal Length
--------------------------------------------------------------
Example 1 70 mm 400 mm
Example 2 60 mm 700 mm
Example 3 60 mm 900 mmAs a beginner, should I look for a higher value in the aperture or the focal length?
Answer accepted (score 13)
There is one rule that is generally true for all deep sky objects (nebulae, stars, galaxies,…): Aperture matters!
For solar system objects, aperture is not that important.
The second most important thing is: What size are the objects you want to look at: Small objects need long focal lengths and high magnifications, large objects need short aperture for low magnifications.
With 400mm you could watch objects like:
- Andromeda galaxy core
- Orion nebula, other large emission or reflective nebulae (e.g. Pleiades)
- large star clusters
- low magnification lunar observations
With 900mm you could watch objects like
- Planets (Saturn, Jupiter, Mars, …)
- high magnification lunar observations
- planetary nebulae (e.g. ring nebula)
Note that 60 and 70mm aperture are still very small for telescopes! The aperture influences two things:
- Light sensitvity: The larger the aperture, the more light you can collect. Very important if you live in a city!
- Maximum resolution: Rule of thumb is that you can do aperture in mm times two as maximum magnification. I.e. for 60mm a 120x magnification is the absolute maximum which still is feasible.
The magnification is created by the eyepiece. E.g. when you have a 400mm focal length telescope and use a 10mm eyepiece, you get 400mm/10mm = 40x magnification.
Note: the shorter the eyepiece focal length, the more difficult it is to build. Good 5mm eyepieces can cost 100 USD and up. I personally started with a 750mm Newtonian with 150mm aperture and 25mm and 10mm eyepieces. That’s a good allrounder, even though planets will appear rather small with the 10mm eyepiece. But you can later invest more money in good eyepieces, which you can re-use on better telescopes which you may buy later on.
Edit: One more thing – the telescope mount is equally important as are the eyepieces and the telescope itself. A mount that fits the telescope easily is as expensive as the optical tube assembly itself. Hence many beginners start out with a Dobson telescope, which uses a very, very simple yet sturdy mount.
Answer 2 (score 9)
It depends on what you plan on using it for. For dimmer, deep sky objects you should be concerned with the balance of aperture and focal length (with a focus on aperture for light gathering power). For brighter objects, like the Moon or the solar system planets you probably don’t need much light gathering power (aperture) so a long focal length for imaging finer details should be ok. Just a reminder though: never look at the Sun without proper filters, you’ll go blind instantly, and looking at the Moon without filters will wreck your night vision and may hurt at first.
Answer 3 (score 1)
Right or wrong, myself as a beginner I started my decision based on aperture to get the appropriate light gathering ability for my area. At that point focal length became pretty much a non-issue. My choices were down to 2, and 1 of them would have been long enough that getting it outside easily would have been difficult.
64: Saw a bright orange object in the sky. What was it? (score 13200 in 2018)
Question
Some friends and I were backpacking in the mountains, and at night we saw a very bright orange object in the sky. It looked brighter than Venus. After a couple minutes we realized it was moving horizontally very slowly. Its movement wasn’t noticeable unless looking at it next to a reference point like a tree. This was way brighter and slower than any satellite I’ve seen. Anyone know what this was?
Answer accepted (score 6)
If the bright orange object was low in the southeast around midnight, it may have been Mars. In late July and early August 2018, it will be only 0.39 AU away from Earth and shine with apparent magnitude -2.7, slightly brighter than Jupiter. This is not as bright as Venus but may stand out more strongly in a midnight sky than Venus does in twilight. For most of Mars’s 2.1 year synodic period, it is farther from Earth and appears fainter.
As the Earth rotates, celestial objects appear to move about 1/4 degree per minute from east to west. For observers in the northern hemisphere, objects near the southern horizon appear to move horizontally left to right.
Answer 2 (score 0)
I came here to ask the same question on July 8 but I found you had asked already 3 hours ago. So I waited for answers. When Mike G assured it’s mars, I felt better, because when I saw the object I felt it’s never a satellite or UFO like something. But we are in different Geo location, right? So how can that answer be authentic?(although, Mike has provided some technical info for Geo locations, but I’m a bit lazy to measure and find that way) So I had to search for any reliable source.
Yesterday, searching on google, I found https://www.fourmilab.ch/cgi-bin/Yourhorizon
I corrected Latitude and Longitude with my location( the city I lived), which I found in wikipedia or you can use google maps( but that case you have to convert the decimal coordinates to degree). And then Updating the data with real time, it’s just almost accurate (with eye level measurement) for the red planet and else. Yes, it’s mars. But wait, there are more, I found Saturn, Jupiter- I was blown away, identifying them as what they are for the first time in my life. (planets are shown with symbols- the legends you can find on https://www.fourmilab.ch/yoursky/help/icons.html)
I think as it’s relative with Geo location, you can get the actual answer, trying this site, the way I found my answer.
Thanks
65: What is the hottest thing in the universe? (score 13151 in 2019)
Question
Straight from my 7 year old to you, exactly what it says on the cover:
What is the hottest thing in the universe?
To make it Stack Exchange-friendly, I’ll add the following caveats:
- it should be bounded, as in an actual compact object, or class of objects, or part of an object
- it should be observable
- it should be an astronomical object, ie a Quark Gluon Plasma created by collisions at the Large Hadron Collider doesn’t count.
Thanks, Bruce
Answer accepted (score 43)
Energetic neutrinos have been observed from the core of a supernova (SN 1987A). The inferred temperature at the “neutrinosphere” is about 4 MeV (equivalent to 50 billion K - (\(5\times 10^{10}\) K, Valentim et al. 2017). Hence it is observable and has been observed.
The very centre of the proto-neutron star that is responsible for the neutrino emission is likely to be a factor of two or so hotter, but cannot be observed, even with neutrinos, because the “neutrinosphere” is opaque to neutrinos. By the time this “clears”, the proto-neutron star is much cooler - its surface would be orders of magnitude cooler.
Arguably we could study the very core of a supernova through gravitational waves if one were to explode in our own Galaxy. Whether this counts as “observing” a hot object, I’m not sure.
In a similar vein, we have observed “kilonova” that appear to be due to the merger of two neutron stars. The temperatures generated in these events are also likely to be of order 100 billion K (\(10^{11}\) K), but again these temperatures are not observed directly - the gravitational waves and gamma rays produced in these events are caused by “non-thermal” mechanisms.
Answer 2 (score 22)
Note that while we haven’t observed anything even close, there is a theorized Absolute Hot along the lines of absolute zero. Its theorized value is ~ \(1.416 \cdot 10^{32}\) Kelvin. Above this temperature, it would be impossible to pump more energy into a system, even gravitationally.
That gives an upper bound on the maximum temperature we could measure.
Answer 3 (score 3)
If you’re ruling out the big bang, then the most extreme releases of energy in our universe should be cases of runaway gravitational collapse. There is a rigorous theorem in general relativity (Penrose singularity theorem) showing that these will generically lead to the creation of singularities. For realistic gravitational collapse, it’s expected that in the end state of this process you will have a black hole, which has an event horizon surrounding a certain specific type of singularity described as spacelike and not a strong curvature singularity (not s.c.s.).
However, during the initial process of formation of the black hole, it’s not really fully established what kind of singularity you would have. It could be timelike rather than spacelike, could be an s.c.s., and might even not be surrounded by an event horizon (which would violate the cosmic censorship hypothesis – but we don’t know whether the CCH is true or even the best way to state it). If it’s an s.c.s., then general relativity predicts that the infalling matter will be infinitely compressed, and therefore probably heated to infinite temperature. GR is a classical theory, so this should probably be interpreted as a statement that an s.c.s. would heat the matter to the Planck temperature.
So if an observer were to jump into a black hole during its initial formation, and if the observer was able to withstand the temperatures, then they might get a millisecond during which they could observe the matter around them being heated up to very high temperatures. Whether these temperatures would rise to the Planck temperature is not really known (probably not), and whether any of this might ever be observable from far away, without suicide, is not really known (but probably not).
Straight from my 7 year old to you, exactly what it says on the cover: What is the hottest thing in the universe?
So at this level, scientists don’t really know for sure, but they think if you jump into a black hole while it’s in the process of being born, you might be able to see matter heated to extremely high temperatures, probably hotter than anything else in the universe since the big bang.
66: In my mid-forties, I believe I have seen Pluto for the first time with the naked eye (score 13150 in )
Question
Early this morning going out on the balcony, I looked up on a star chart app to verify it was Jupiter I was seeing. Then I noticed the alignment of Mars, Saturn, and Pluto on the app.
Never being able to identify it before, I stared at where Pluto should be and I’m pretty sure I saw it.
My only question is - since it’s said that planets shine and stars twinkle, it did seem that Pluto was flicking a bit. Is this normal? Something to do with the relatively low luminosity and greater length of space?
Answer accepted (score 99)
Pluto is something like magnitude 14. The limit for the human vision is somewhere between magnitude 6 (widely accepted) and 8-ish (highly trained observers with perfect vision in ideal conditions using special techniques - and it’s a bit controversial anyway).
There’s zero chance that was Pluto. It was definitely a fixed star.
Answer 2 (score 16)
As Florin correctly stated, it can’t have been Pluto. You have probably looked at it and you have even gotten its light in your eyes. That little itty bitty shine just has no chance to make your retina do anything (edit: Interesting link in the comments. Might be that people actually can sense single photons. Doesn’t help at all to see Pluto though).
Stellarium is a nice tool in order to check which star you might have confused with Pluto. This is what it has to offer:
The red crosshair is where Pluto should be and Stellarium doesn’t bother to color a single pixel because Pluto is about 14.28 mag. 5 mag difference mean 100 times dimmer, so Pluto is at least 10000 times less bright than many of the stars that you see in this image, let alone the two planets (Mars is about 26 times closer than Pluto atm.).
My guess is that you have seen Pi Sagittarii (HIP 94141) which would be 2.85 mag and lines up nicely with the planets. Unless I’m wrong, it’s roughly 37000 times brighter than Pluto
67: when are you able to see Halley’s comet, and for how long? (score 13138 in 2018)
Question
I’ve been hearing about Halley’s comet for a while now, and I’m just curious as to when and where to see it, and for how long. :)
Answer accepted (score 13)
Halley’s comet orbits the sun and its orbit lasts 75 years. The orbit is a long elliptical orbit. For most of those 75 years it is a cold black dot, and frozen solid in the outer solar system. But a short period during that orbit it gets close to the sun. The ice and gas begin to boil of its surface and gets blown back by the solar wind of the sun. It then starts to look like a comet.
The last time Halley’s comet was close to the sun was in 1986, and so the next time will be in 2061. You won’t be able to see Halley’s comet until then.
(image from University of Wisconsin)
I remember 1986, and the comet was visible but hard to see. It was a very dim fuzzy dot, that was just about visible in my binoculars. This is because, unfortunately, the Earth was a long way from the comet. In 2061 Earth will be even further from the comet, however Earth will be better positioned when Halley is closest to the sun, so it should make a good show, and be clearly visible to the naked eye. Comet predictions are hard as much depends on exactly how much water vapour boils off, and this is hard to predict. Comets can have sudden changes in activity.
Moreover, Halley’s comet isn’t the only comet. There are others that can be as visible as Halley will be in 2061. There are also long period comets that are undiscovered until they come close to the sun. These can be very bright. Comet Hale-Bopp was a “great” comet in 1997 and was visible for many months. Another great comet could come along any time but we can’t predict when.
68: When we see the Sun, do we actually see its past? (score 12971 in 2015)
Question
The Sun’s rays hit our eyes around 8 minutes after they are emitted from the Sun. Does this mean that the Sun that we see is always the Sun as it was some 8 minutes before? I strongly think this must be happening; is it really a fact? Do we always see the Sun’s past?
Answer accepted (score 17)
Yes, you are right. We don’t only see the Sun 8 minutes in the past, we actually see the past of everything in space. We even see our closest companion, the Moon, 1 second in the past.
The further an object is from us the longer its light takes to reach us since the speed of light is finite and distance in space are really big.
Answer 2 (score 2)
The speed of light in a vacuum is 299,792,458 meters per second, not infinite. Let’s say, for example, particle of a beam of light, the photon, is emitted. It takes ~8 minutes to get to us; when it hits our eyes, we see it. This means that we see a photon that was emitted from the sun 8 minutes ago. We aren’t, per se, looking “back in time”, but we’re looking at a photon that is ~8 minutes old.
Answer 3 (score -1)
In fact past, present and future is still beyond of our consciousness. In reality they exist all together as spacetime. For example, when you say, we are seeing 10 years past of a star which is 10 light years away from our planet, similarly an alien right now is seeing our 10 years past earth. A nice video here https://youtu.be/vrqmMoI0wks can be helpful to understand how mind boggling reality we are living in.
69: Can you still see Polaris even if you are in the south pole? (score 12839 in )
Question
I haven’t been to south pole but can the Polaris still be viewed if the viewer is in the south pole? Or this question makes no sense at all?
Answer accepted (score 9)
Currently Polaris is at a declination of a bit over 89 degrees, which means that no one south of 1 degree south latitude can see Polaris. That’s almost all of the Southern hemisphere, let alone the South Pole.
Polaris won’t be the North Star forever, thanks to axial precession. In about 13000 years or so, Polaris will have a declination of about 46 degrees or so (twice the 23 degree axial tilt). Polaris will thus be visible in 13000 years or so as a wintertime star to all of Africa, all of Australia, and most of South America, but none of Antarctica.
After millions of years, proper motion may make Polaris visible over Antarctica. But then again, being a yellow supergiant, its unlikely that Polaris will be visible anywhere (without a telescope). It will instead be dead.
Answer 2 (score 7)
While the majority of the celestial sky is visible on both hemispheres, you are not able to see Polaris on the south pole, since Polaris is pointing directly towards the north pole. I know that during winter time, you can definitely just see the plough/big dipper (part of the Ursa Major constellation) as far south as Uluru/Ayers Rock in Australia, but that is not enough to see the northern star. The northern star will generally speaking disappear below the horizon when you are at around the equator. There is no “southern star” to orient yourself on the southern hemisphere, but there are a number of rules of thumb to find the south celestial pole on the southern hemisphere as well, several of them involving the “southern cross”, or the Crux constellation.
Answer 3 (score 3)
At the South pole no stars north of the celestial equator will be visible.
The general equations of the declination limit at a given latitude:
declination limit = \(latitude - 90\) (for the northern hemisphere, \(latitude > 0\))
declination limit = \(90 + latitude\) (for the southern hemisphere, \(latitude < 0\))
70: Can an astronomical telescope view objects on Earth? (score 12699 in 2018)
Question
I am new to telescopes and I’m planning to buy an astronomical telescope like the “Celestron AstroMaster 114 EQ Reflector”. Just curious to know whether it can be used to view objects on Earth (e.g. traffic and people miles away from tall buildings). Telescopes are used to observe stars; they must have higher power than binoculars. Right?
Answer accepted (score 8)
Yes, but many (not all) telescopes made for astronomy will invert the image. Presenting the image upright is important for terrestrial instruments, but it doesn’t matter for astronomy. Also, it usually requires fewer optical components to make an instrument that inverts the image.
Also, terrestrial instruments are made to operate in an environment where there’s plenty of ambient light (daylight), whereas many telescope designs (such as truss dobsonians) are built to operate in a dark environment and are usually hampered by excess ambient light.
So, in a nutshell, yes, you could turn a telescope towards objects here on Earth, and it would work, but for best results and easiest operation just get some good binoculars.
Finally, “power” is a much misunderstood characteristic of optical instruments. It is technically true that a large dobsonian could in theory outperform any pair of binoculars you could care to try, but in practice a large optical stack is difficult to swing around and is fickle with regard to collimation, thermal equilibrium, etc. If you’re willing and able to put up with the operating difficulties, with the necessity to flip the image around, etc, then I guess that indeed a large dobsonian could surpass anyone’s little binoculars in a terrestrial context.
But very large magnification always reduces the field of view. Go ahead and crank magnification all the way up, and it’s like looking through a straw. It’s hard to locate anything. Astronomers have specific methods to deal with this difficulty, but for a terrestrial observation the narrow field of view is a liability, not an advantage.
There’s a reason why various optical instruments have evolved into different shapes and designs to operate in different environments. “Horses for courses”, as they say.
Answer 2 (score 3)
Yes. Most consumer telescopes can be used for terrestrial viewing (or at least the ones I have used) and are often utilized for wildlife viewing or other uses for extreme telephoto requirements.
The manual doesn’t list the minimum focus distance (usually the information provided for photographic lens to describe how near they can focus), but the manual does include a short passage (pg 25) on terrestrial viewing.
Terrestrial Photography
Your telescope makes an excellent telephoto lens for terrestrial (land) photography. You can take images of various scenic views, wildlife, nature, and just about anything. You will have to experiment with focusing, speeds, etc. to get the best image desired. You can adapt your camera per the instructions at the top of this page.
Answer 3 (score 2)
An astronomical telescope can view objects on Earth if they’re far enough away to be focused, and if the telescope can be aimed at them. Most often, the targets will appear to be inverted, but you can get prisms to correct that.
Space satellites cannot place on Earth they are around the Earth but they can see things on the ground less than a foot across.
71: Why do the planets orbit in the same direction? (score 12661 in )
Question
Theoretically, planets would have an approximately equal chance of going one way in their orbit or another but in reality, this is not the case (at least in our solar system). Why is this?
Answer accepted (score 36)
The same reason (almost) all of them rotate in the same direction: because of the conservation of angular momentum.
Before a star and its planets exist, there’s just a cloud of disorganized gas and small molecules. The Solar System formed from such a cloud around 4.6 billion years ago.
On that scale, there is some small amount of rotation within the cloud. It could be caused by the gravity of nearby stellar objects, local differences in mass as the cloud churns, or even the impact of a distant supernova. The point is, all molecular clouds have at least a little rotation.
In a large system like a molecular cloud, each particle has some angular momentum, and it all adds together across a very wide area. That’s a lot of momentum, and it is conserved as the cloud continues to collapse under its own gravity. That angular momentum also flattens the cloud, which is the reason why the Solar System is near-planar.
When the cloud finally collapses, it forms a star and shortly after planets. However, angular momentum is always conserved. That’s why planets all follow the same orbit, and why almost all of them rotate in the same direction. There’s nothing to turn them the other direction, so they will continue spinning in the same direction as the original gas cloud.
There are a few exceptions, though. Whenever objects formed in such a way that sent them orbiting the opposite direction, they usually collided with objects going in the same direction as the original cloud. This destroyed any outlying objects or sent them in the same direction as the original cloud.
Still, two huge exceptions are planets Venus and Uranus. Uranus spins on an axis of almost 90-degrees (on its side). Venus meanwhile spins the opposite direction as Earth and the other planets.
In both cases there is strong evidence that these planets were struck by large objects at some point in the distant past. The impacts were large enough to overcome the angular momentum of the bodies, and give them a different spin. There are also a range of other theories; for example, some astronomers think that Venus may have been flipped upside-down. Point is, there were irregular events that happened to both of these planets.
Answer 2 (score 4)
Sir Cumference’s answer is great. Molecular clouds are generally thousands of times more massive than the Solar System, and since they’re less dense they’re much much larger in volume.
We don’t know where our Solar System originated from, and we don’t know how many other stars were born in the same cloud, probably hundreds or even thousands (just recently 1 or 2 stars were suggested to be sisters of Sol, but the jury is, as far as I know, still out on that).
Anyway, either due to interstellar winds, magnetic fields, supernovae explosions, or some other difference in average density, a volume of our mother molecular cloud began to collapse due to gravity being just a bit more in some areas.
The more the cloud became concentrated, the more the gravitational attraction increased, so the faster it collapsed. While dust and gas collide, the whole system conserves energy and momentum (as it is an isolated system) , and thus is naïve to assume that planet orbits should be random — which means any which way, you seem to have assumed that space is two dimensional, and the most random arrangement would be a flat disk.
Nope. It would be a sphere… like a swarm of flies around something stinky. When we program a computer to model a swarm of random dust and gas collapsing, it turns out that due to chance it will select a preferred direction. A random dust cloud will collapse into a disk with most of the particles orbiting in the same direction (this ignores possible effects from the Milky Way influencing the process, so even without the molecular cloud orbiting the center of the Milky Way, disk formation will occur).
Keep in mind that these answers are tentative: most of the gravity in the Milky Way is of dark matter, and we’re still working on understanding how that influences star formation and until we know a lot more about dark matter, we can’t be sure our computer models are correct. Generally, we prefer models that give results similar to the actual way our Solar System is.
But guess what? The thousands of exoplanets we’ve discovered have far more “hot Jupiters” (gas giants very close to their stars) than we expected. So we are adjusting our models. One popular idea is that planets had a lot more collisions than we used to think. This means more planets in very close to the Star, and more planets actually ejected from the star system. Who knows, perhaps that’s where Theia came from.
72: Exercise: 2D orbital mechanics simulation (python) (score 12621 in 2014)
Question
Just a little disclaimer beforehand: I have never studied astronomy or any exact sciences for that matter (not even IT), so I am trying to fill this gap by self-education. Astronomy is one of the areas that has captured my attention and my idea of self-education is head on applied approach. So, straight to the point - this is orbital simulation model that I am casually working on when I have time/mood. My major goal is to create complete solar system in motion and ability to plan spacecraft launches to other planets.
You are all free to pick this project up at any point and have fun experimenting!
update!!! (Nov10)
- velocity is now proper deltaV and giving additional motion now calculates sum vector of velocity
- you can place as many static objects as you like, on every time unit object in motion checks for gravity vectors from all sources (and checks for collision)
- greatly improved the performance of calculations
- a fix to account for interactive mod in matplotlib. Looks like that this is default option only for ipython. Regular python3 requires that statement explicitly.
Basically it is now possible to “launch” a spacecraft from the surface of the Earth and plot a mission to the Moon by making deltaV vector corrections via giveMotion(). Next in line is trying to implement global time variable to enable simultaneous motion e.g. Moon orbits Earth while spacecraft tries out a gravity assist maneuver.
Comments and suggestions for improvements are always welcome!
Done in Python3 with matplotlib library
import matplotlib.pyplot as plt
import math
plt.ion()
G = 6.673e-11 # gravity constant
gridArea = [0, 200, 0, 200] # margins of the coordinate grid
gridScale = 1000000 # 1 unit of grid equals 1000000m or 1000km
plt.clf() # clear plot area
plt.axis(gridArea) # create new coordinate grid
plt.grid(b="on") # place grid
class Object:
_instances = []
def __init__(self, name, position, radius, mass):
self.name = name
self.position = position
self.radius = radius # in grid values
self.mass = mass
self.placeObject()
self.velocity = 0
Object._instances.append(self)
def placeObject(self):
drawObject = plt.Circle(self.position, radius=self.radius, fill=False, color="black")
plt.gca().add_patch(drawObject)
plt.show()
def giveMotion(self, deltaV, motionDirection, time):
if self.velocity != 0:
x_comp = math.sin(math.radians(self.motionDirection))*self.velocity
y_comp = math.cos(math.radians(self.motionDirection))*self.velocity
x_comp += math.sin(math.radians(motionDirection))*deltaV
y_comp += math.cos(math.radians(motionDirection))*deltaV
self.velocity = math.sqrt((x_comp**2)+(y_comp**2))
if x_comp > 0 and y_comp > 0: # calculate degrees depending on the coordinate quadrant
self.motionDirection = math.degrees(math.asin(abs(x_comp)/self.velocity)) # update motion direction
elif x_comp > 0 and y_comp < 0:
self.motionDirection = math.degrees(math.asin(abs(y_comp)/self.velocity)) + 90
elif x_comp < 0 and y_comp < 0:
self.motionDirection = math.degrees(math.asin(abs(x_comp)/self.velocity)) + 180
else:
self.motionDirection = math.degrees(math.asin(abs(y_comp)/self.velocity)) + 270
else:
self.velocity = self.velocity + deltaV # in m/s
self.motionDirection = motionDirection # degrees
self.time = time # in seconds
self.vectorUpdate()
def vectorUpdate(self):
self.placeObject()
data = []
for t in range(self.time):
motionForce = self.mass * self.velocity # F = m * v
x_net = 0
y_net = 0
for x in [y for y in Object._instances if y is not self]:
distance = math.sqrt(((self.position[0]-x.position[0])**2) +
(self.position[1]-x.position[1])**2)
gravityForce = G*(self.mass * x.mass)/((distance*gridScale)**2)
x_pos = self.position[0] - x.position[0]
y_pos = self.position[1] - x.position[1]
if x_pos <= 0 and y_pos > 0: # calculate degrees depending on the coordinate quadrant
gravityDirection = math.degrees(math.asin(abs(y_pos)/distance))+90
elif x_pos > 0 and y_pos >= 0:
gravityDirection = math.degrees(math.asin(abs(x_pos)/distance))+180
elif x_pos >= 0 and y_pos < 0:
gravityDirection = math.degrees(math.asin(abs(y_pos)/distance))+270
else:
gravityDirection = math.degrees(math.asin(abs(x_pos)/distance))
x_gF = gravityForce * math.sin(math.radians(gravityDirection)) # x component of vector
y_gF = gravityForce * math.cos(math.radians(gravityDirection)) # y component of vector
x_net += x_gF
y_net += y_gF
x_mF = motionForce * math.sin(math.radians(self.motionDirection))
y_mF = motionForce * math.cos(math.radians(self.motionDirection))
x_net += x_mF
y_net += y_mF
netForce = math.sqrt((x_net**2)+(y_net**2))
if x_net > 0 and y_net > 0: # calculate degrees depending on the coordinate quadrant
self.motionDirection = math.degrees(math.asin(abs(x_net)/netForce)) # update motion direction
elif x_net > 0 and y_net < 0:
self.motionDirection = math.degrees(math.asin(abs(y_net)/netForce)) + 90
elif x_net < 0 and y_net < 0:
self.motionDirection = math.degrees(math.asin(abs(x_net)/netForce)) + 180
else:
self.motionDirection = math.degrees(math.asin(abs(y_net)/netForce)) + 270
self.velocity = netForce/self.mass # update velocity
traveled = self.velocity/gridScale # grid distance traveled per 1 sec
self.position = (self.position[0] + math.sin(math.radians(self.motionDirection))*traveled,
self.position[1] + math.cos(math.radians(self.motionDirection))*traveled) # update pos
data.append([self.position[0], self.position[1]])
collision = 0
for x in [y for y in Object._instances if y is not self]:
if (self.position[0] - x.position[0])**2 + (self.position[1] - x.position[1])**2 <= x.radius**2:
collision = 1
break
if collision != 0:
print("Collision!")
break
plt.plot([x[0] for x in data], [x[1] for x in data])
Earth = Object(name="Earth", position=(50.0, 50.0), radius=6.371, mass=5.972e24)
Moon = Object(name="Moon", position=(100.0, 100.0), radius=1.737, mass = 7.347e22) # position not to real scale
Craft = Object(name="SpaceCraft", position=(49.0, 40.0), radius=1, mass=1.0e4)
Craft.giveMotion(deltaV=8500.0, motionDirection=100, time=130000)
Craft.giveMotion(deltaV=2000.0, motionDirection=90, time=60000)
plt.show(block=True)How it works
It all boils down to two things:
-
Creating object like
Earth = Object(name="Earth", position=(50.0, 50.0), radius=6.371, mass=5.972e24)with parameters of position on grid (1 unit of grid is 1000km by default but this can be changed too), radius in grid units and mass in kg. -
Giving object some deltaV such as
Craft.giveMotion(deltaV=8500.0, motionDirection=100, time=130000)obviously it requiresCraft = Object(...)to be created in the first place as mentioned in previous point. Parameters here aredeltaVin m/s (note that for now acceleration is instantaneous),motionDirectionis direction of deltaV in degrees (from current position imagine 360 degree circle around object, so direction is a point on that circle) and finally parametertimeis how many seconds after the deltaV push trajectory of the object will be monitored. SubsequentgiveMotion()start off from last position of previousgiveMotion().
Questions:
- Is this a valid algorithm to calculate orbits?
- What are the obvious improvements to be made?
- I have been considering “timeScale” variable that will optimize calculations, as it might not be necessary to recalculate vectors and positions for every second. Any thoughts on how it should be implemented or is it generally a good idea? (loss of accuracy vs improved performance)
Basically my aim is to start a discussion on the topic and see where it leads. And, if possible, learn (or even better - teach) something new and interesting.
Feel free to experiment!
Try using:
Earth = Object(name="Earth", position=(50.0, 100.0), radius=6.371, mass=5.972e24)
Moon = Object(name="Moon", position=(434.0, 100.0), radius=1.737, mass = 7.347e22)
Craft = Object(name="SpaceCraft", position=(43.0, 100.0), radius=1, mass=1.0e4)
Craft.giveMotion(deltaV=10575.0, motionDirection=180, time=322000)
Craft.giveMotion(deltaV=400.0, motionDirection=180, time=50000)With two burns - one prograde at Earth orbit and one retrograde at Moon orbit I achieved stable Moon orbit. Are these close to theoreticaly expected values?
Suggested exercise: Try it in 3 burns - stable Earth orbit from Earth surface, prograde burn to reach Moon, retrograde burn to stabilize orbit around Moon. Then try to minimize deltaV.
Note: I plan to update the code with extensive comments for those not familiar with python3 syntax.
Answer accepted (score 11)
Let the two bodies involved have masses \(m_1, m_2\). Start with Newton’s second law \[F = ma\] where \(a\) is acceleration. The gravitational force on body 2 from body 1 is given by
\[F_{21} = \frac{G m_1 m_2}{|r_{21}|^3}r_{21}\]
where \(r_{21}\) is the relative position vector for the two bodies in question. The force on body 1 from body two is of course \(F_{12}=-F_{21}\) since \(r_{12}=-r_{21}\). If we denote the positions by \((x_1,y_1)\) and \((x_2,y_2)\), then
\[r_{21} = \begin{pmatrix} x_1-x_2 \\ y_1-y_2 \end{pmatrix}.\]
and
\[|r| = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}.\] Then Newton’s law \(a=F/m\) becomes
\[\begin{align} x_1''(t) & = \frac{G m_2 (x_2-x_1)}{|r|^3} \\ y_1''(t) & = \frac{G m_2 (y_2-y_1)}{|r|^3} \\ x_2''(t) & = \frac{G m_1 (x_1-x_2)}{|r|^3} \\ y_2''(t) & = \frac{G m_1 (y_1-y_2)}{|r|^3}. \end{align}\]
Together with the initial positions and velocities, this system of ordinary differential equations (ODEs) comprises an initial value problem. The usual approach is to write this as a first-order system of 8 equations and apply a Runge-Kutta or multistep method to solve them.
If you apply something simple like forward Euler or backward Euler, you will see the Earth spiral out to infinity or in toward the sun, respectively, but that is an effect of the numerical errors. If you use a more accurate method, like the classical 4th-order Runge-Kutta method, you’ll find that it stays close to a true orbit for a while but still eventually goes off to infinity. The right approach is to use a symplectic method, which will keep the Earth in the correct orbit – although its phase will still be off due to numerical errors.
For the 2-body problem it’s possible to derive a simpler system by basing your coordinate system around the center of mass. But I think the formulation above is clearer if this is new to you.
73: What would happen if an ice cube is left in space? (score 12317 in 2016)
Question
Recently I boarded a flight and noticed outside air temperature as -53°C at an altitude of 36860ft (11.23km). I don’t know what causes such a freezing temperature in that altitude but was wondering higher altitudes (space) may have even freezing temperatures. Here I got a doubt i.e what happens if an ice cube is left in space? Would it be melting or stay as it is?
Answer accepted (score 87)
It depends on where in outer space you are.
If you simply stick it in orbit around the Earth, it’ll sublimate: the mean surface temperature of something at Earth’s distance from the Sun is about 220K, which is solidly in the vapor phase for water in a vacuum, and the solid-vapor transition at that temperature doesn’t pass through the liquid phase. On the other hand, if you stick your ice cube out in the Oort Cloud, it’ll grow: the mean surface temperature is 40K or below, well into the solid phase, so it’ll pick up (or be picked up by) gas and other objects in space.
A comet is a rough approximation to an ice cube. If you think of what happens to a comet at various places, that’s about what would happen to your ice cube.
Answer 2 (score 12)
It would sublimate. The frozen mass of water would decrease in size as the water converts from a solid to a gas (without becoming a liquid) and drifts away.
Answer 3 (score 5)
In the vacuum of space the most important consideration is to consider how much radiation an ice cube would absorb from, for example, nearby stars and how fast the ice cube itself would radiate away energy (using Wien’s law), finding what ice cube temperature would produce an equilibrium (the temperature at which the ice cube radiate energy at the same rate it absorbed energy) and then determining if that temperature is above or below the melting point of the ice cube. If it is above the melting point (of water in a vacuum), then as the other answers have said the ice cube would sublimate; if it is below the melting point then the ice cube would stay frozen.
Specifically for an ice cube that is a cube in orbit around the sun with one side facing the sun you would need to calculate how much energy the side facing the sun absorbs from the sun as well as how much energy radiates away from all six sides of the cube and then find the equilibrium temperature.
74: If Earth didn’t rotate, would we feel heavier? (score 12314 in )
Question
Suppose the Earth’s rotation slowed for some reason. Would the lack of centrifugal force cause us to feel heavier than normal?
Likewise, if Earth’s rotation increased, would we feel lighter as centrifugal force lifts us from the ground?
Answer accepted (score 5)
If the rotation would stop, we would feel an additional gravity of \(0.034 \mbox{ m}/\mbox{s}^2\), or about 0.35%, at the equator, (incorrectly) assuming the shape of Earth isn’t changed by the changing rotation.
The centrifugal acceleration is \(v^2/r\), with \(v=465.1 \mbox{ m}/\mbox{s},\) and \(r=6378100 \mbox{ m}.\)
With shorter rotation period, the other way, as you say.
Polar regions wouldn’t be affected, with the exception, that the ellipsoid of Earth would change due to changing rotation period. Respecting this, gravity would increase at the poles with faster rotation, since the poles would get closer to Earth’s center. In equatorial regions the change of surface gravity would be affected more than in the simplified calculation above. In contrast to the poles, faster rotation would reduce surface gravity more than just by the centrifugal force.
Answer 2 (score 5)
If the rotation would stop, we would feel an additional gravity of \(0.034 \mbox{ m}/\mbox{s}^2\), or about 0.35%, at the equator, (incorrectly) assuming the shape of Earth isn’t changed by the changing rotation.
The centrifugal acceleration is \(v^2/r\), with \(v=465.1 \mbox{ m}/\mbox{s},\) and \(r=6378100 \mbox{ m}.\)
With shorter rotation period, the other way, as you say.
Polar regions wouldn’t be affected, with the exception, that the ellipsoid of Earth would change due to changing rotation period. Respecting this, gravity would increase at the poles with faster rotation, since the poles would get closer to Earth’s center. In equatorial regions the change of surface gravity would be affected more than in the simplified calculation above. In contrast to the poles, faster rotation would reduce surface gravity more than just by the centrifugal force.
75: Why not take a picture of a closer black hole? (score 12308 in )
Question
There are closer galaxies than Messier 87 for sure, even ours! It sparked my curiosity that they went with one 53 million light years away. Is there a reason for this?
Answer accepted (score 78)
I was surprised too when I first heard they were trying to image M87’s black hole.
The short answer is because it’s really, really big. It is 1500 times bigger (diameter) than our Sagittarius A, and 2100 times farther away. This makes its apparent size about 70% of that of Sgr A, which they are also attempting to image.
A cursory search of wikipedia’s List of Largest black holes shows that there’s no other black holes with a combination of size and closeness greater than these two.
A couple of other candidates are not too far off. Andromeda’s black hole is 50x the size of ours, and at 100x the distance, it would appear half the size of Sgr A. The Sombrero galaxy is 380 times farther way than Sgr A, and has a black hole estimated to be 1 billion solar masses, which is 232 times Sr A, resulting in an angular diameter about 60% of Sgr A.
There appear to be many other considerations to which black holes were chosen, as explained in this similar question. At a guess these would include how obscured each black hole is with foreground dust/stars etc, how active (and therefore bright) the nuclei are, and their inclination w.r.t earth affecting which observatories could observe them at which times.
Edit: I’ve found another plausible candidate. NGC_1600 is 200 M light years away with a central black hole estimated to be 17 billion solar masses heavy. This would put it at about 40% the apparent diameter of Sgr A*.
Comparison of the apparent size of the largest nearby black holes
And of course obligatory XKCD to remind us how small these objects really appear.
Answer 2 (score 33)
There are a few criteria necessary to see a black hole with the Event Horizon Telescope. They are, in importance:
- Active Feeding: you need a thick accretion disk with lots of matter accreting onto the black hole. M87 fits this criteria, and is a glut, consuming about 90 Earth masses a day.
- Apparent size. Even though it is 53 million light-years away, M87 is 6.5 billion solar masses. Since the radius of the event horizon scales linearly with mass, its distance is made up for by sheer scale.
Answer 3 (score 18)
As Ingolifs says, Sgr A* and M87* are the obvious candidates. At the press conference, Heino Falcke explained why they got a picture of M87* first:
But it would take some more time because Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear.
76: If Venus and Mars changed places, would we then have 3 habitable planets? (score 12173 in 2014)
Question
This is the supposition: Venus is too hot, Mars is too cold. If they switched places then Solar energy would change to make both more Earth like. We might as well have had three instead of just one “Earth” in the Solar system, if the roulette of early planet formation had played out just a little bit different.
That’s of course too simple. But wouldn’t it help alot if it were the case from the beginning that Venus had had Mars’ mass, and Mars had had Venus’ mass? Or do other factors dominate mass and distance from the Sun?
Would Venus still have had a thick, but not too thick, atmosphere if it were at 1.5 AU, because it has mass enough to keep one, and wouldn’t that have kept it warmer than Mars is today? Including flowing water on its surface under its atmospheric pressure?
Would a Mars at 0.7 AU have been warmer and maybe have had a passing atmosphere created from melting volatiles during a longer era in its history?
Answer accepted (score 12)
This is a very interesting question. Of course, as you noted, you have simplified things quite a bit; there are other factors besides temperature that affect habitability.
Regarding Venus, you probably know that Venus is extremely hot at its surface not just because it is closer to the Sun, but because it has a thick CO2 atmosphere and is warmed by the greenhouse effect. There are, in fact, two things about Venus that would prevent it from being habitable wherever you put it. One is the lack of a magnetosphere, which is necessary to prevent ionizing radiation (particularly from the solar wind) from reaching the planet’s surface. The lack of a magnetosphere appears to be due to the lack of a geodynamo on Venus, which has to do with the structure of its core. Second, Venus appears to lack tectonic plates, which you may know are responsible for earthquakes here on Earth. Interestingly, tectonic plates play a major role in controlling the buildup of CO2 in the atmosphere (see here for details). Lacking tectonic plates, Venus is doomed to have a large CO2 atmosphere wherever you put it, which would not make it a nice place to live.
Mars, on the other hand, is a very different matter. It has both a magnetosphere (albeit it is very weak) and it likely has tectonic plates (although last I heard it is thought to only have two). One of the reasons that NASA has sent so many probes to Mars is that it was thought to be habitable at one point. It is thought that Mars’s growth was stunted because of gravitation effects from Jupiter and Saturn. So, in another universe, Mars could have ended up very much like another Earth.
Answer 2 (score 5)
If, Venus had started out at Mars’ position, it is possible it would’ve ended up evolving into an ice giant. At the greater distance of Mars the Sun’s solar wind is weaker and would’ve stripped off much less of Venus’ atmosphere, even in comparison to the amount Earth lost, as Earth is considerably closer to the Sun than Mars. So, Venus would perhaps have been left with a thick helium envelope and become a mini-Neptune/super-Earth rather than a second Earth. And also, if Mars had evolved at the distance of Venus, closer in, it would’ve been stripped of any appreciable atmosphere and possibly would’ve evolved into another Mercury, a grey and airless world.
Answer 3 (score 2)
Simply “swapping” position would not in and of itself be cause for a habitable planet. The key is “changing” position. If Venus was to undergo orbital migration it could conceivably move into Earths current position and cool down, continue to develop a more stable atmosphere, and produce oceans. Then further orbital migration could cause the atmosphere to collapse and marine succession. The question then is would it become like Ceres and Jupiter. Simulating a vector reversal, where Mars undergoes orbital migration and moves into the position currently taken by Earth, would warming it up cause it to further develop an atmosphere, and a marine environment. Then, when its vector moves it to the spot currently taken by Venus would marine succession and atmospheric reduction appear as Venus is to day. From physical appearance alone I would say there is a natural progression where the planets evolve as the orbital migration causes their vector to move away from the Sun.
77: Is there any planet/star bigger than VY Canis Majoris? (score 12143 in 2015)
Question
Compared to VY Canis Majoris, our Sun is more like a speck of space dust. Is there any planet or star which is known to us that is bigger than the VY Canis Majoris?
Answer accepted (score 22)
Part 1: Assuming “larger” means greater in diameter.
Stars
Estimates on the size of stars are just that, estimates, and estimates based on rather fuzzy observations. VY Canis Majoris has been bumped down to size. The current thinking is that there are seven known stars larger than VY Canis Majoris, the largest of which is UY Scuti.
Current models indicate that the first generation of stars were much, much larger than anything we see now. It will be quite some time before we can resolve a first generation star. So far, they are just theoretical objects.
Planets
Jupiter-mass planets are about as large as a planet can get. There are some exoplanets that are larger than Jupiter, but that’s because they orbit much closer to their parent star than does Jupiter. This makes them puff up a bit. The reason Jupiter-mass planets are deemed to be the largest possible is that planets of this mass are presently assumed to have a core of degenerate hydrogen. A funny thing happens to degenerate masses when mass is added to them: They shrink in diameter. (The shrinkage becomes catastrophic as the mass approaches the Chandrasekhar limit.)
This means that assuming all other things are equal (temperature, composition), a planet more massive than Jupiter will be small in diameter than Jupiter is. Even if all other things aren’t equal, a Jupiter-diameter planet is (give or take) about as large as planet can get.
Part 2: Assuming “larger” means greater in mass.
Stars
In terms of mass, VY Canis Majoris doesn’t even make the top ten, not even close! The most massive known star is R136a1. Again, these are estimates, but mass is a bit easier to pin down than is radius (or diameter).
As is the case with physical extent, the first generation stars are presently modeled as being much, much more massive than anything we see now.
Planets
There’s not much difference between the largest planet and the smallest brown dwarf. I would argue there’s very little difference. It’s a spectrum with no distinguishing characteristic that lets one say “this is a planet” and “that is a brown dwarf”. Ignoring the distinction between super-Jupiters and brown dwarfs the largest is about 80 Jupiter masses. V1581 Cygni C is 79 Jupiter masses. (More massive than that and they start burning hydrogen, thus making them a small red dwarf.)
The current factor that is used to distinguish between brown dwarfs and super Jupiters is mass. Anything larger than 13 Jupiter masses is a brown dwarf, anything smaller, a planet. That boundary is very arbitrary.
Answer 2 (score 4)
Looking at the primary literature (and the source of the wikipedia information), it appears that Arroyo-Torres et al. (2013) have estimated a radius of \(1708\pm 192R_{\odot}\), using an interferometric angular diameter and a highly uncertain distance (where I’m not sure the distance uncertainty has been adequately included). Using similar methods Wittkowski et al. (2012) estimated a radius of \(1420\pm 120 R_{\odot}\) for VY CMa.
Taking these numbers at face value then UY Scuti is probably larger than VY CMa. However, I wouldn’t take these numbers at face value. I don’t believe the distances to these stars are known to better than 10% as claimed in these papers (they soon will be with Gaia trigonometric parallaxes). You also have to bear in mind that these are pulsational variables (hence the variable star designations), so their radii are variable.
78: Is the moon moving further away from Earth and closer to the Sun? Why? (score 12131 in )
Question
According to The NASA Moon Facts page:
The moon is actually moving away from earth at a rate of 1.5 inches per year.
Why is the moon moving further away from the Earth? Is this a result of the moons formation that set it in motion to spiral away from us? Or is this a resultant force of the gravity from the Sun and other large bodies?
Answer accepted (score 20)
Yes, the moon is moving away from Earth at around 1.48" per year. According to the BBC:
The Moon is kept in orbit by the gravitational force that the Earth exerts on it, but the Moon also exerts a gravitational force on our planet and this causes the movement of the Earth’s oceans to form a tidal bulge.
Due to the rotation of the Earth, this tidal bulge actually sits slightly ahead of the Moon. Some of the energy of the spinning Earth gets transferred to the tidal bulge via friction.
This drives the bulge forward, keeping it ahead of the Moon. The tidal bulge feeds a small amount of energy into the Moon, pushing it into a higher orbit like the faster, outside lanes of a test track.
So, tidal forces are ultimately what causes this to happen.
Also, there is a Wikipedia article on tidal forces:
Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. the Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary’s rotation. The process eventually leads to tidal locking of first the smaller, and later the larger body. The Earth–Moon system is the best studied case.
79: Could the dinosaurs have seen the asteroid that killed them? (score 12117 in 2018)
Question
Wikipedia says the Chicxulub impactor is thought to have been a 10-15 km diameter object. Would it have been visible to a (human*) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night?
* I know, there were no humans at the time.
Answer accepted (score 88)
Impacting solar system objects would have relative closing speeds from around 11 to 72 km/s.
We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which I think precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid Vesta. This has a diameter of around \(a=520\) km, gets as close as \(d=1.14\) au from the Earth and has a maximum brightness of about \(m=5.2\) apparent magnitude (and hence just visible to the naked eye) and an observed flux \(f = f_0 10^{-0.4m}\), where \(f_0\) is a zeropoint for the magnitude scale.
Thus the flux \(f_n\) received by a near-Earth asteroid of diameter \(a_n\), at a distance \(d_n\) from Earth and with the same reflectivity would be \[ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2 \left(\frac{d}{d_n}\right)^2\]
The magnitude of the dinosaur killer would then be \[m_n = m -2.5\log (f/f_n)\]
To be an at all conspicuous naked eye object, \(f_n \geq f\). If we assume \(a_n=10\)km, then \[ d_n^2(1+d_n)^2 \leq 0.0022\]
An approximate solution is obtained by assuming \(d_n \ll 1\) and thus we find $ d_n 0.047$ au or 7 million km.
Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus hitting the Earth about 3 days after becoming a naked eye object. Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun.
It thus seems to me that there is a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a few nights, but probably not much longer than that.
Answer 2 (score 9)
A carbonaceous condrite has the same reflectivity as the moon at around 7-13%.
If there was ice, if the tail was 10 times smaller than hail bopp, it would have auspiciously covered half of the sky. it could have made an incredible display in the 1-2 days preceding the collision, because it was as close to the sun as hale bopp, the brightest astronomical apparition in history, although it was 100 times closer to the earth in the final days than Hale Bopp. Hale Bopps trail was 1.5 million km long, so the KT asteroid would have needed a tail 100 times smaller, to appear as bright as Hale Bopp in the final 2 days! https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html
moon distance / moon diameter = the moon is 107 moon diameters away, so you can cover the moon with a thumb at arms length ( ~1cm/107cm) and the impactor would be nearly thumb sized at: comet diameter * 107 = 1500km away.
If you were 1000km south of Chixculub, it would stay fairly constantly thumb sized for 10-20 seconds as it arrived from the South-East at 60 degrees, and in that time it would travel from 1500km above to 1000km on the horizon. The graph suggests it had a rotation period of about 2 to 10 seconds, so it would have revolved noticeably.
If you were 100km away from the impact, it would grow from thumb size to hand sized for 10-20 seconds. Then it would hit the atmosphere at 50km and the sky would ignite about 5 seconds prior to landfall.
For most of the dinosaurs far away it would be like holding a pea or a matchstick at arms length.
Perhaps it would be a brighter than venus from 75000 km away that’s about 125 minutes prior to the impact.
It would have been like a fire fly landing slowly on a tennis court if the dino was on the other side of the world.
Mammals were nocturnal previous to the impact which is why mammals have developed whiskers, awesome hearing and only bichromatic photoreceptors of G.B, and mammal skin burns from UV. Only primates have R.G.B. Dinosaurs had R.G.B because birds have the same gene to encode red photoreceptors as tortoises who can also see red, so scientists think that dino’s had vision as good as birds, and have better vision than mammals, especially used for detecting movement and for flying.
EDIT: On wiki, the estimated diameter of the comet has been changed to 10-80km. Hale-Bopp comet was 20-40km, and it was visible for months as one of the brightest comets in human history, with a similar aphelion of 0.914AU. That implies that the dinosaur comet may have been visible for many days prior to impact, because it would have been 5 times closer than Hale-Bopp at the end. It’s possible that it was very bright in the sky for a week prior to impact.
80: Would we have more than 8 minutes of light, if the sun “went out”? (score 11994 in 2019)
Question
The common theory is, that if the Sun “shut down”, we would see the light for eight more minutes (the time that it takes the photons to reach the Earth).
However recently I have read that photons need around 100 000 years to reach the Earth, since the reactions are happening at Sun’s core, and gamma rays can’t leave the Sun without interacting with other particles, unlike neutrinos for example.
Is that theory correct? If the Sun’s core “shut down”, would we still receive photons (light) for another 100 000 years, with only neutrinos disappearing immediately?
Answer accepted (score 105)
If nuclear fusion were to suddenly stop in the centre of the Sun, then the only clear signature we would have of this is the lack of detectable neutrinos received at Earth, starting about 8 minutes after the reactions ceased. The Sun however would continue to shine for tens of millions of years at roughly its current luminosity.
The power source is not “stored” photons. The Sun itself would simply resume the slow gravitational contraction that was halted about 4.5 billion years ago when nuclear reaction rates at the centre were able to increase sufficiently to supply the radiative losses from the surface of the Sun.
The characteristic timescale for the contraction is about \[\tau_{\rm KH} = \frac{GM^2}{RL},\] which is 30 million years. i.e. The Sun has enough gravitational potential energy to supply its current luminosity for tens of millions of years.
While this is happening, the Sun would approximately maintain its current luminosity, but decrease in radius, meaning that its surface temperature would increase.
Once the Sun had contracted to a few times the size of Jupiter (so about 30% of its current radius), the contraction would begin to slow, because the electrons in the core become degenerate and the pressure increases with density by more than expected for a perfect gas. The slowing contraction decreases the rate of potential energy release and hence the solar luminosity. The contraction continues at a slow rate until the Sun becomes a hot “hydrogen white dwarf” a few times the size of the Earth, which then cools to a glowing cinder, with no further contraction, over billions of years (see What would the Sun be like if nuclear reactions could not proceed via quantum tunneling? for some more details).
Even if you were to not allow the Sun to contract, it would take some time to radiate it’s thermal energy. This timescale is approximately \[\tau_{\rm therm} \simeq \frac{3k_B T M}{m_H L},\] which assumes the Sun is a perfect gas of protons plus electrons, with an average temperature \(T\). If we take \(T =10^7\) K and the current solar luminosity, then \(\tau_{\rm therm}=\) 40 million years.
On the other hand, if your scenario is just that light from the Sun stops being emitted, then of course it goes dark on Earth about 8 minutes later.
Answer 2 (score 5)
The “common theory” you’re reading is not about the processes that produce light in stars, it’s just intended as a demonstration of the speed of light through space. When it talks about the Sun “shutting down”, it’s not talking about the nuclear fusion processes stopping, it means that the Sun as a whole stops shining. I’m not a physicist, but I don’t think there’s anything that would actually cause this to occur at a sudden time, such that we could measure the 8 minute time difference for the last photons to reach Earth.
It’s just a thought experiment, using a simplified description of an impossible event, to make a point about some other process. In this model, the Sun either gives off light or it doesn’t, and we’re measuring the time since it changed that state; we’re not concerned with what the photons do before they’re emitted from the Sun (any more than measuring time that light travels from a light bulb cares about how the electricity that powered the bulb was produced – we could measure that all the way back to the dinosaurs dying and eventually turning into fossil fuels that were used in the power plant, or even further back to the solar energy that powered life on earth).
81: How many stars and galaxies can be seen by the naked eye? (score 11975 in 2015)
Question
How many of the luminous dots that we see naked are galaxies and not stars from our galaxy?
I imagine that the majority of the luminous points that we see naked eye during the night, are actually stars from our galaxy. But how many of them are other objects (other galaxies, nebula, etc.), excluding planets from our Solar System?
Answer accepted (score 8)
In the best sky conditions, the naked eye (with effort) can see objects with an apparent magnitude of 8.0. This reveals about 43,197 objects in the sky.
There are 9 galaxies visible to the naked eye that you might see when observing the sky, and there are about 13 nebulae that you might see.
Sources:
- The Bortle Dark-Sky Scale - John E. Bortle
- How many stars are in the sky? - NASA
- Naked-eye galaxies - Wikipedia
- List of planetary nebulae - Wikipedia
- List of diffuse nebulae - Wikipedia
82: How to calculate the temperature of a star (score 11879 in 2016)
Question
I need a way to calculate the effective temperature (surface temperature) of a star for a stellar model. I need something in the form Te=….
I have:
- Radius in m
- mass in kg
- the composition of particles (eg H 90%, He 8% etc)
- the combined stored thermal energy of the body in J
Constants (any really but I’m using these for now):
- G=gravity constant=6.67408E-011
- k=kbolzmann=1.3806485279E-023
- s=sbolzmann=5,67036713E-008
- PI=pi ~3.14…
Example of the sun:
- mp=average mass of a particle=1,7E-027
- M=total mass of the body=2E30
- r=radius of the body=700000000
I’m using this equation to estimate the core temperature :
(G*mp*M)/(r*(3/2)*k)
which nets 15653011 for the sun which is close enough given that that is the only star core temperature known (afaik).
I’m using this to estimate the luminosity L:
4*PI*(r^2)*s*(Te^4)
which results in an error of ~1-5% with 90% of my sample stars which is close enough. For the sun this results in 3,95120075975041E+026 W which is only 2,7% off.
The problem is I need Te for the 2nd formula which I don’t have in my scenario.
Due to the formula for L being dependent on the surface temperature to the power of 4 this value has to be relatively precise.
Assumptions of my model:
- uniform distribution of particles: so every slice of the body has the same composition as the entire body.
- perfect sphere: every body is a perfect sphere, no handling for elliptic bodies needed.
My sample values (first line is the sun with a core temp of 15000000):
emitted energy Surface temp radius mass
(in Lsun) (in K) (in m) (in Msun)
1 5800 700000000 1
8700000 53000 25200000000 265
6300000 50100 23100000000 110
2900000 42000 23660000000 132
2000000 44000 16800000000 80
1260000 13500 140000000000 45
57500 3600 618100000000 12.4
78 5700 6440000000 2.56
78.5 4940 8540000000 2.69
15100 7350 51100000000 9.7
1.519 5790 858900000 1.1
0.5 5260 605500000 0.907
370000 3690 994000000000 19.2
123000 33000 7560000000 56
2200000 52500 12600000000 130
200000 10000 151900000000 22
446000 19000 43330000000 42.3
25.4 9940 1197700000 2.02Errors in luminosity to actual value (the maximum error is about 100% which I can live with since it might just be inaccurate measurements for the sample stars)
2.74%
6.71%
-1.13%
11.29%
-2.00%
-4.27%
106.76%
3.99%
2.51%
-6.50%
1.12%
4.00%
-8.27%
2.10%
1.57%
113.75%
1.64%
2.15%Answer accepted (score 1)
Empirically (I fit a regression on log(mass) vs log(surface temp)), using the table of values in the article on Main Sequence stars, I get a fairly well-fitting formula: \(\mathrm{estTemp} = 5740*\mathrm{mass}^{0.54}\), where estTemp is in C and mass is in multiples of the sun’s mass. Seems to work very well for all but the largest and smallest main sequence stars (and not TOO bad for those).
83: Astrophotography - Unable to achieve focus with current setup (score 11851 in 2018)
Question
I am new to astrophotography and am looking for guidance. I have the following equipment:
- Orion StarBlast 9814 4.5" Altazimuth Reflector Tabletop telescope
- 1.25in Telescope Adapter Extension Tube T Ring for Canon DSLR SLR Camera DC618
- 17mm and 6mm lenses
When I use the lenses in the Telescope Adapter Extension, focus is not achievable. My theory is that the focal plane of most DSLR are 55mm back focus plane. Reflectors are positioned farther back than where the human eye would normally be placed, and hence it is usually difficult if not impossible to bring the camera into focus.
Is this true? Also, would a Barlow lens help achieve focus or be a waste of money? Any other suggestions?
Answer accepted (score 8)
Yes, the vast majority of small telescopes made for visual observations (like most dobsonians and other newtonian-based architectures) have very limited back focus. In other words, with the focuser retracted all the way in, the focal plane of the primary mirror is sticking out only a short distance. A large back focus is not typically seen, nor is it desirable, in a visual scope. (However, most astrographs have a large back focus.)
This is compounded by the fact that most DSLR have a flange distance (focus-flange distance) which is pretty large - over 45…50 mm in many cases.
Furthermore, an additional distance is wasted by the classic T-ring adapters, which tend to push the camera way back from the focuser’s shoulder.
All these factors conspire in your case to prevent you from achieving focus.
Some ideas and solutions:
Adapters
I tell everyone who is trying to do prime focus photography with visual scopes (dobsonians, regular newtonians) to ditch the old T-ring system and just buy dedicated adapters. All you need is a piece of metal that goes in the bayonet mount on the camera on one side, and into the focuser on the other side. No need for T- components.
Some such adapters are made by telescopeadapters.com:
http://www.telescopeadapters.com/
I have two adapters like this, they save a very large amount of back focus, compared to the classic T-ring / T-adapter system. In other words, they allow the camera to get much closer to the focuser, and therefore catch the focal plane on the sensor. You just need to choose the kind of adapter that works with your camera.
Most of these adapters are 2“, so in a 1.25” focuser a size adapter is needed - but that would waste some back focus distance.
Camera
Classic DSLR cameras have a very large flange distance, so a large back focus is needed to overcome it.
Other removable-lens, large-sensor cameras have a much smaller flange distance, if they are mirrorless. Examples: Micro Four Thirds cameras (Panasonic G series), Sony NEX, Samsung NX, etc. All these systems have large sensors (comparable to DSLR) which work well for astrophotography, but don’t have moving mirrors, so their flange distances are tiny.
https://en.wikipedia.org/wiki/Mirrorless_interchangeable-lens_camera
I have a Panasonic G1, used with the adapter mentioned above. I only need 25 mm of back focus for this to work, which is an extremely small value.
Use a barlow
A barlow would increase the amount of back focus available. Try and see what happens. Keep in mind it cannot work miracles - if you need a very large amount of back focus, it may not work.
Use eyepiece projection
With some eyepieces, you can thread adapters on the user side, and then put the camera body (without lens) on the adapter. Move the eyepiece WAY out of focus, and it will generate a real image which can be projected on the sensor. This is not prime focus photography anymore, but it works and it’s better than nothing.
However, such adapters are hard to find, and many eyepieces won’t support them.
Use a dedicated USB “eyepiece camera”
You can find relatively cheap USB cameras made for imaging, some are less than $100. With one of those, you could achieve focus relatively easy. Example:
Would not offer the same performance like a DSLR, but it’s better than nothing.
Use a telescope made for imaging
Visual scopes, like a small newtonian, are not really that great for photography, for a variety of reasons. A dedicated astrograph would always work better. There are many Cassegrain systems out there made specifically for imaging, some are relatively inexpensive (not much more expensive than a DSLR).
In many cases, you need very long exposure, and you need to track the sky while the shutter is open. Many such scopes come with a tracking mount, or are made to be installed on a tracking mount. That would make it very easy to do tracking well.
It would be very hard to do tracking with other kinds of scopes (tracking platforms for dobsonians do exist, but are not very common).
Move the primary mirror up
Think about it - if you move the primary mirror up in its cell, the focal plane will stick out. This is a way to gain back focus only if you know what you’re doing. Additional spacers in the primary mirror cell could do it. Modifying the cell is also doable.
Beware, you could ruin the scope if you make a mistake. Also, if the secondary mirror is too small to begin with, you’ll increase the amount of clipping.
Afocal photography
Put the eyepiece in the scope, just like for a normal visual observation. Now put the camera at the eyepiece and move it around until you capture the image. Click the shutter. This is called afocal photography.
You’ll need a camera mount, that will attach to the focuser, holding the camera at the eyepiece. Something like this:
Keep in mind that your camera is very likely too heavy for this trick to work. Also, afocal is usually the lowest quality of all techniques, and rarely works well enough.
(There are special adapters that go on the eyepiece on one end, and thread into the camera lens like a filter on the other end. These would work better, but are hard to find, only a few eyepieces support them, and your camera is too heavy anyway.)
Keep in mind that your scope has a 1.25" focuser. Even if you do get the sensor in focus, the small size of the focuser will clip the image; you will only be able to use the central part of the sensor. It is recommended to use 2" focusers for prime focus imaging with large sensors.
Also, your f/4 scope would exhibit strong aberrations at the edge anyway, which would blur the image. For prime focus imaging, longer focal ratios are usually desirable.
All in all, your scope and your camera are mismatched. You have a pretty good camera, but a very small scope that is not made for imaging. Either use a dedicated astrograph with that camera, or keep the scope and use a cheap USB “eyepiece camera”.
84: How did our ancestors discover the Solar System? (score 11717 in 2015)
Question
I wonder, how did our ancestors discover the Solar System? They did not have any telescopes to see distant objects, right? Even a planet looks like a star from a distance.
They discovered the rotations of different planets without having much technology.
Answer accepted (score 17)
- Ancient cultures observed the sky
Night skies are naturally dark and there was no light-pollution in ancient times. So if weather permits, you can easily see a lot of stars. No need to tell about the Sun and the Moon.
Ancient people had good reasons to study the night skies. In many cultures and civilizations, stars (and also the Sun and the Moon) where perceived to have religious, legendary, premonitory or magical significance (astrology), so a lot of people were interested in them. It did not took long to someone (in reality a lot of different people independently in many parts of the world) to see some useful patterns in the stars that would be useful to navigation, localization, counting hours, counting days and relating days to seasons, etc. And of course, those patterns in the stars were also related to the Sun and the Moon.
So, surely all ancient cultures had people who dedicated many nights of their lifes to study the stars in detail right from the stone age. They would also perceive meteorites (falling stars) and eclipses. And sometimes a very rare and spetacular comet.
Then there are the planets Mercury, Venus, Mars, Jupiter and Saturn. They are quite easily to notice to be distinct from the stars because all the stars seems to be fixed in the celestial sphere, but the planets don’t. They are very easily to notice to be wandering around in the sky with the passage of the days, specially for Venus, which is the brightest “star” in the sky and is also a formidable wanderer. Given all of that, the ancient people surely become very aware of those five planets.
About Mercury, initially the Greeks thought that Mercury were two bodies, one that showed up only in the morning a few hours before the sunrise and another only a few hours after the sunset. However, soon they figured out that in fact it was only one body, because either one or the other (or neither) could be seen in a given day and the computed position of the unseen body always matched the position of the seen body.
- The Earth seems to be round
Now, out of the stone age, already into ancient times, navigators and merchants who travelled great distances perceived that the Sun rising and setting points could variate not only due to the seasonal variation, but also accordingly to the location. Also, the distance from the polar star to the horizon line also variates accordingly with the location. This fact denounces the existence of the concept nowadays known as latitude, and this was perceived by ancient astronomers in places like Greece, Egypt, Mesopotamia and China.
Astronomers and people who dependend on astronomy (like navigators) would wonder why the distance from the polar star to the horizon varied, and one possibility was that it is because the Earth would be round. Also, registering different Sun angles in different locations of the world on a given same day and at a given same hour, also gives a hint that the Earth is round. The shadow on the Moon during a lunar eclipse also gives a hint that the Earth is round. However, this by itself is not a proof that the Earth is round, so most people would bet on some other simpler thing, or simply don’t care about this phenomenon.
Most cultures in ancient times presumed that the world was flat. However the idea of the world being round exists since the ancient Greece. Contrary to the popular modern misconception, in the Middle Ages, almost no educated person on the western world thought that the world was flat.
About the Earth’s size, by observing different Sun positions and shadows angles in different parts of the world, Erasthotenes in ancient Greece calculated the size of Earth and the distance between the Earth and the Sun correctly for the first time as back as the third century B.C. However, due to the confusion about all the different and inconsistent unit measures existent back then and the difficulty to precisely estimate long land and sea distances, confusion and imprecision persisted until the modern times.
Ancient cultures also figured out that the shiny part of the Moon was illuminated by the Sun. Since the Full Moon is easily seen even at midnight, this implies that the Earth is not infinite. The fact that the Moon enters in a rounded shadow when exactly in the opposite side of the sky as the Sun also implies that it is the Earth’s shadow on the Moon. This also implies that Earth is significantly larger than Moon.
- Geocentrism
So, people observed the Sun, Moon, Mercury, Venus, Mars, Jupiter, Saturn and the fixed sphere of stars all revolving around the sky. They naturally thought that the Earth would be the center of the universe and that all of those bodies revolved around the Earth. This culminated with the work of the phylosopher Claudius Ptolemaeus about geocentrism.
Altough we now know that the ptolomaic geocentric model is fundamentally wrong, it could be used to compute the position of the planets, the Sun, the Moon and the celestial sphere of stars, with a somewhat acceptable precision at the time. It accounted to include the observation of planets velocity variations, retrograde motions and also for coupling Mercury and Venus to the Sun, so they would never go very far from it. Further, based on the velocity of the motion of those bodies in the sky, then the universe should be something like:
- Earth at the center.
- Moon orbiting the Earth.
- Mercury orbiting the Earth farther than the Moon.
- Venus orbiting the Earth farther than Mercury.
- Sun orbiting the Earth farther than Venus.
- Mars orbiting the Earth farther than Sun.
- Jupiter orbiting the Earth farther than Mars.
- Saturn orbiting the Earth farther than Jupiter.
- The celestial sphere of stars rotating around the Earth, being the outermost sphere.
In fact, the ptolomaic model is a very complicated model, way more complicated than the copernic, keplerian and newtonian models. Particularly, this could be compared to softwares that are based on severely flawed concepts but that are still working due to a lot of complex, tangled and unexplainable hacks and kludges that are there just for the sake of making the thing work.
- The discovery of the Americas
Marco Polo, in the last years of the 1200’s, was the first European to travel to China and back and leave a detailed chronicle of his experience. So, he could bring a lot of knowledge about what existed in the central Asia, the East of Asia, the Indies, China, Mongolia and even Japan to the Europeans. Before Marco Polo, very few was known to the Europeans about what existed there. This greatly inspired European cartographers, philosophers, politicians and navigators in the years to come.
Portugal and Spain fough a centuries-long war against the invading Moors on the Iberian Peninsula. The Moors were finally expelled in 1492. The two states, were looking for something profitable after so many years of war. Since Portugal ended its part of the war first, it had a head start and went to explore the seas first. Both Portugal and Spain were trying to find a navigation route to reach the Indias and the China in order to trade highly profitable spices and silk. Those couldn’t be traded by land efficiently anymore due to the fact that the lands on West Asia and North Africa were dominated by Muslim cultures unfriendly to Christian Europeans, a situation that were just made worse after the fall of Constantinople in 1453.
Portugal, were colonizing the Atlantic borders of Africa and eventually they managed to reach the Cape of Good Hope in 1488 (with Bartolomeu Dias).
A Genovese navigator called Cristoforo Colombo believed that if he sailed west from the Europe, he could eventually reach the Indies from the east side. Inspired by Marco Polo and subestimating the size of Earth, he estimated that the distance between the Canary Islands and the Japan to be 3700 km (in fact it is 12500 km). Most navigators would not venture in such voyage because they (rightly) tought that Earth was larger than that.
Colombo tried to convice the king of Portugal to finance his journey in 1485, but after submitting the proposal to experts, the king rejected it because the estimated journey distance was too low. Spain, however, after finally expelling the Moors in 1492, were convinced by him. Colombo’s idea was far-fetched, but, after centuries of wars with the Muslims, if that worked, then Spain could profit quickly. So, the Spanish king approved the idea. And just a few months after expelling the Moors, Spain sent Colombo to sail west towards the Atlantic and then he reach the Hispaniola island in Central America. After coming back, the news about the discovery of lands in the other side of the Atlantic spread quickly.
Portugal and Spain then divided the world by the Treaty of Tordesillas in 1494. In 1497, Amerigo Vespucci reached the mainland America.
Portugal would not be left behind, they managed to navigate around Africa to reach the Indies in 1498 (with Vasco da Gama). And they sent Pedro Álvares Cabral, who reached the Brazil in 1500 before crossing the Atlantic back in order to go for the Indies.
After that, Portugal and Spain quickly started to explore the Americas and eventually colonize them. France, England and Netherlands also came to the Americas some time later.
- The Earth IS round
After, the Spanish discovered and settled into the Americas (and Colombo’s plan in fact didn’t worked). The question that if it was possible to sail around the globe to reach the Indies from the east side remained open and the Spanish were still interested on it. They eventually discovered the Pacific Ocean after crossing the Panama Ishtums by land in 1513.
Eager to find a maritime route around the globe, the Spanish crown funded an expedition leadered by the Portuguese Fernão de Magalhães (or Magellan as his name was translated to English) to try to circle the globe. Magellan was an experienced navigator, and had reached what is present day Malaysia traveling through the Indian Ocean before. They departed from Spain in September 20th, 1519. It was a long and though journey that costed the lives of most of the crew. Magellan himself did not survived, having died in a battle in the Phillipines on 1521. At least, he lived enough to be aware that they in fact reached East Asia by traveling around the globe to the west, which also proves that the Earth is round.
The journey was eventually completed by the leadership of Juan Sebatián Elcano, one of the crewmen of Magellan. They reached Spain back through the Indian and Atlantic Oceans on September 6th, 1522 after traveling for almost three years a distance of 81449 km.
- Heliocentrism
There were some heliocentric or hybrid geo-heliocentric theories in ancient times. Notably by the Greek philosopher Philolaus in the 5th century BC. By Martianus Capella around the years 410 to 420. And by Aristarchus of Samos around 370 BC. Those models tried to explain the motion of the stars as rotation of the Earth and the position of the planets, specially Mercury and Venus as translation around the Sun. However those early models were too imprecise and flawed to work appropriately, and the ptolomaic model still was the model with the better prediction of the positions of the heavenly bodies.
The idea that the Earth rotates was much less revolutionary than heliocentrism, but was already more-or-less accepted with reluctancy in the middle ages. This happens because if the stars rotated around Earth, they would need do so at an astonishing velocity, dragging the Sun, the Moon and the planets with it, so it would be easier if Earth itself rotated. People were uncomfortable with this idea, but they still accepted it, and this became easier to be accepted after the Earth sphericity was an established concept.
In the first years of the 1500’s, while the Portuguese and Spanish were sailing around the globe, a polish and very skilled matemathical and astronomer called Nikolaus Kopernikus took some years thinking about the mechanics of the heavenly bodies. After some years making calculations and observations, he created a model of circular orbits of the planets around the Sun and perceived that his model were much more simpler than the ptolomaic geocentric model and was at least as precise. His model also features a rotating Earth and fixed stars. Further, his model implied that the Sun was much larger than the Earth, something that was already strongly suspected at the time due to calculations and measurements and also implied that Jupiter and Saturn were several times larger than Earth, so Earth would definitively be a planet just like the other five then-known planets were. This could be seen as the borning of the model today know as Solar System.
Fearing persecution and harsh criticism, he avoided to publish many of his works, sending manuscripts to only his closest acquaintances, however his works eventually leaked out and he was convinced to allow its full publication anyway. Legend says that he was presented to his finally fully published work in the very day that he died in 1543, so he could die in peace.
There was a heated debate between supporters and oppositors of Copernic’s heliocentric theory in the middle 1500’s. One argument for the opposition was that star parallaxes could not be observed, which implied that either the heliocentric model was wrong or that the stars were very very far and many of them would be even larger than the Sun, which seemed to be a crazy idea at the time.
Tycho Brache, which did not accepted heliocentrism, in the latest years of the 1500’s tried to save geocentrism with an hybrid geo-heliocentric model that featured the five heavenly planets orbiting the Sun while the Sun and the Moon orbited Earth. However, he also published a theory which better predicted the position of the Moon. Also, by this time, the observation of some supernovas showed that the celestial sphere of the stars were not exactly immutable.
In 1600, the astronomer William Gilbert provided strong argument for the rotation of Earth, by studing magnets and compasses, he could demonstrate that the Earth was magnetic, which could be explained by the presence of enourmous quantities of iron in its core.
- With telescopes
All of what I wrote above happened without telescopes, only by using naked eye observations and measurements around the globe. Now, add even some small telescopes and things change quickly.
The earliest telescopes were invented in 1608. In 1609, the astronomer Galieu Galilei heard about that, and constructed his own telescope. In January of 1610, Galieu Galilei, using a small telescope, observed four small bodies orbiting Jupiter at different distances, figuring out that they were Jupiter’s “moons”, he also could predict and calculate its positions along their orbits. Some months later, he also observed that Venus had phases as seen from the Earth. He also observed Saturn’s rings, but his telescope was not powerful enough to resolve them as rings, and he tought that they were two moons. These observations were incompatible with the geocentric model.
A contemporary of Galilei, Johannes Kepler, working on Copernicus’ heliocentric model and making a lot of calculations, in order to explain the differing orbital velocities, created an heliocentric model where the planets orbits the Sun in elliptic orbits with one of the ellipse’s focus in the Sun. His works were published in 1609 and 1619. He also suggested that tides were caused by the motion of the Moon, though Galilei was skeptical to that. His laws predicted a transit of Mercury in 1631 and of Venus in 1639, and such transit were in fact observed. However, a predicted transit of Venus in 1631 could not be seen due to imprecision in calculations and the fact that it was not visible in much of the Europe.
In 1650 the first double star were observed. Further in the 1600’s, the Saturn rings were resolved by the use of better telescopes by Robert Hooke, who also observed a double star in 1664 and developed microscopes to observe cellular structures. From them on, many stars were discovered to be double. In 1655, Titan were discovered orbiting Saturn, putting more confidence on the heliocentric model. More four Saturnian moons were discovered between 1671 and 1684.
- Gravitation
Heliocentrism was reasonably well-accepted in the middle 1600’s, but people was not confortable with it. Why the planets orbits the Sun afterall? Why the Moon orbits Earth? Why Jupiter and Saturn had moons? Although Keplerian mechanics could predict their moviment, it was still unclear what was the reason that makes them move that way.
In 1687, Isaac Newton who was one of the most brilliant physic and mathematic that ever lived (although he was also an implacable persecutor of his opponents), provided the gravitational theory (based on prior work by Robert Hooke). Ideas for the gravitation theory and the inverse square law already were developed in the 1670’s, but he could publish a very simple and clear theory for gravitation, very well-fundamented in physics and mathematics and it explained the motions of the celestial bodies with a great precision, including comets. It also explained why the planets, the Moon and the Sun are spherical, explained tides and it also served to explain why things falls to the ground. This made heliocentrism to be definitely widely accepted.
Also, Newton gravitational law predicted that Earth rotation would make it not exactly spherical, but a bit ellipsoidal by a factor of 1:230. Something that agreed with measures done using pendulums in 1673.
- What are the stars and the Solar System afterall?
In the early 1700’s, Edmund Halley, already knowing about newtonian laws (he was a contemporary of Newton) perceived that comets who passed near Earth would eventually return, and he found that there was a particular instance of sightings every 76 years, so he could note that those comets in reality were all the same comet, which is called after him.
The only remaining problem with the heliocentric model was the lack of observation of parallax to the stars. And nobody knew for sure what the stars were. However, if they in fact are very distant bodies, most of them would be much larger than the Sun. At the first half of the 1700’s, trying to observe parallax, James Bradley perceived phenomena like the aberration of light and the Earth’s nutation, and those phenomena also provides a way to calculate the speed of light. But the observation of parallax remained a challenge during the 1700’s.
In 1781, Uranus were discovered orbiting the Sun beyond Saturn. Although barely visible to the naked eye in the darkest skies, it was so dim that it escaped observation from astronomers until then, and so were discovered with a telescope. The first asteroids were also discovered in the early 1800’s. Investigation on pertubations on Uranus’ orbit due to the predicted newtonian and keplerian movement eventually leaded to the discovery of Neptune in 1846.
In 1838, the astronomer Friedrich Wilhelm Bessel who measured the position of more than 50000 stars with the greatest precision as possible, could finally measure the parallax of the star 61 Cygni successfully, which proved that stars were in fact very distant bodies and that many of them were in fact larger than the Sun. This also demonstrates that the Sun is a star. Vega and Alpha Centauri also had their parallaxes measured successfully in 1838. Further, those measurements permitted to estimate the distance between those stars and the Solar System to be on the order of many trillions of kilometers, or several light-years.
Answer 2 (score 8)
In the absence of the lighting levels in cities, there are a number of objects in the night sky that are easily seen to move against the background of stars: the Sun, the Moon, Venus, Jupiter, Saturn, Mars and Mercury. These are obvious to anyone who looks at the sky regularly from any dark sky site on clear nights, like most places in the pre-industrial world. These objects are the Planets or Wanderers.
The term Solar System is anachronistic when used of ancient astronomy. You have the Fixed Stars which maintain their relative positions but appear to rotate as if fixed to a sphere centred on the observer about a celestial axis about once per day. then you have the Planets which move relative to the Fixed Stars. Together with the Earth itself these constitute the World, Universe or what-have-you.
Answer 3 (score 0)
By very carefully tracking the movements of the planets across the sky. And then coming up with the counterintuitive idea that they move just like apples move as they fall from trees.
85: I’m having trouble achieving sharp telescope focus (score 11715 in 2015)
Question
I’ve been having trouble getting really sharp focus on my telescope. Using Saturn as a reference, with a 20mm eye piece it is very sharp. But when I go to 10mm or 6mm, it gets bigger and blurrier. I re-adjust focus each time I change eye pieces, but the best I can get at the stronger magnifications is still very blurry compared to 20mm. The 20mm will have nice sharp edges on the planet and rings, whereas 10mm and 6mm have fuzzy edges on the rings, and the planet itself.
I’m using the Celestron NexStar Evolution 8 Schmidt-Cassegrain Telescope. I’m in a suburban area so I’m sure there is light pollution, could that be a factor?
Last night I was observing for about an hour, which I believe would be enough time for the temperature of the telescope to match the temperature outside.
I also tried combining the 20mm with a Celestron 2x Barlow lens, just with an idea that maybe it would make a difference, but it was pretty much the same.
Is there something I’m doing wrong?
Answer accepted (score 6)
Your telescope has a 2000 mm focal length, and 200 mm aperture. With the 20 mm eyepiece, you get 100x magnification. With the 10 mm lens, you get 200x. Several things could cause what you describe:
Seeing (turbulence) might be bad. It is common in many places that less than great seeing means everything at 200x and over might start looking blurry. But seeing changes with the seasons, or day to day, hour to hour, or indeed from one second to another. It’s random. If you get the exact same results all the time, and the image never improves even after many weeks of trying, then perhaps it’s not seeing that’s the culprit.
Your telescope might not be collimated. Look in the user’s manual. Is there a collimation procedure you need to follow? (EDIT: yes, there is - page 20) A miscollimated scope indeed behaves as you described - seems “sharp” at low-ish magnification, but quickly becomes blurry if you push magnification up.
Point the scope at a bright star and defocus. Are the rings perfectly circular? If not, it might be miscollimated.
http://www.astrophoto.fr/collim.html
Another factor is optics quality. Optics with lots of aberrations cause “mushy” images and a difficulty to find the true focus. High quality optics produce a scope that is very “snappy” when achieving perfect focus, and make very sharp images when seeing is good. There’s nothing you can do about this, but the good news is - this problem is not very common nowadays; optics quality, while not always great, tends to be at least passable in many cases. Quality could be evaluated either on the test bench (which requires equipment), or via direct observation of stars (which requires a very experienced observer and takes time).
Finally, have you compared views with another instrument of similar aperture? The image at higher magnification will always seem a bit more soft, compared to lower magnification - even in perfect seeing with a collimated instrument. Also, the contrast always decreases as you increase magnification. This is why there is always an optimal magnification for each view - sometimes higher, sometimes lower, depending on many factors. The best scope in the world will look mushy and blurry if you push magnification up too much.
Can you easily see the Cassini division in Saturn’s rings? If yes, then you’re probably in okay shape (collimation and seeing), you’re just not used to the softness at higher magnification. I have a very high quality reflector (self-made), similar aperture class to yours (a bit smaller), and Cassini is immediately visible at 180x and looks sharp; it’s also visible at 255x but the whole thing starts to look mushy and faded. It looks super-sharp and high-contrast at 136x but it’s kind of small. This is in perfect collimation, perfect thermal equilibrium, high quality optics throughout the stack (primary and secondary mirrors, eyepieces), and good seeing typical of N. California.
When seeing is bad, Cassini starts to fade no matter what I do.
Does Saturn look equally soft in the center of the image, as well as near the edge? Good optics (mirrors, eyepieces) make a good image everywhere, cheap optics make an okay image in the center and a blurry image at the edge.
My bet is that it’s either bad collimation, or you haven’t compared the view at 200x with another scope, or both. But it’s hard to diagnose things over the Internet.
You should probably get a 15mm eyepiece as well. You’ll use it a lot, probably more than the 10mm.
Light pollution is never a factor for: planets, the Moon, the Sun, most double stars. These things are just too bright to care about it, so observing them from the city is fine. What does matter is seeing (turbulence). Also, it’s important that the scope is in perfect collimation.
Light pollution only matters for the “faint fuzzies”: nebulae and galaxies. These are low-brightness, low-contrast objects, that are difficult to see from the city. Being low contrast, your eye will not perceive a lot of detail anyway, so for these objects seeing doesn’t matter.
Congratulations for being aware of, and following, the rule that says you need to acclimate the scope to ambient temperature. A lot of people are not aware of it. At least you’re removing one unknown from the equation. It’s always a good idea to let the scope “breathe” outside for an hour before you even begin to observe.
One way to be 100% sure that your scope is focused perfectly is to use a Bahtinov mask. Just make (or buy) one to match the scope’s diameter, put it over the scope’s mouth, point scope at a bright star, and tweak the focuser until all spikes intersect exactly in the center of the star. Then remove the mask - the scope is in perfect focus. This will probably not help in your case, but it’s just one way to remove another uncertainty. It also helps with regular observations - I use it almost every time I observe, it makes focusing so much easier.
The mask could be made from a piece of cardboard and a sharp knife in like 20 minutes. It doesn’t need to be a marvel of engineering precision to work well.
http://www.deepskywatch.com/Articles/make-bahtinov-mask.html
One thing that is certainly NOT a factor is the central obstruction. You’ll often hear on the Internet how Cassegrain instruments, or other scopes with a large central obstruction (secondary mirror) are supposedly less “sharp” than scopes without a central obstruction, or ones with a very small obstruction.
Yes, the obstruction does matter a little, but not to the extent that Internet mythology would have you believe. A Cassegrain instrument, with good optics, in perfect collimation, can be a superb telescope even at a 40% obstruction. It simply performs like a somewhat smaller instrument, that’s all (the physics of the central obstruction are very complex, and would require an entirely different discussion on this forum to fully elucidate).
(Added in reply to a comment below)
Miscollimation is quite likely the #1 performance killer of all amateur telescopes. Some well-made refractors are more or less immune to it, but most reflectors need periodic collimation.
It’s a good idea to google around for various collimation techniques, there’s so many of them. Some require the observation of stars, some require special devices known as collimators. The technique is also different for different types of telescopes (newtonian, cass, etc).
If you own a reflector, it’s a good idea to collimate it once in a while, just in case, like changing the oil on a car. Dobsonian telescopes probably require most frequent collimation.
Check Howie Glatter’s site - he makes some of the most precise collimators currently available:
Regardless, just learn any method you can and apply it. If the telescope is massively miscollimated, you’ll get an improvement immediately.
Not being able to see Cassini is not a good sign. Maybe seeing is persistently bad (it can happen), or maybe the scope is off-collimation. Can anyone else in your area see Cassini these days in a similar aperture, in a scope known to be collimated well?
Surface detail on Saturn is not easy to see, but should be doable with apertures over 100mm. It’s like zones of latitude similar to Jupiter’s equatorial belts, but very, very pale and faded. You have to keep looking for a while to see it. Don’t worry if you don’t see that stuff for now.
Jupiter’s Great Red Spot is harder to see nowadays. It’s been shrinking and fading for decades now. Go on Wolfram Alpha and type in Great Red Spot. It will tell you if it’s visible at the moment. Usually it looks more like a dent in the equatorial belts, rather than an actual visible spot.
http://www.wolframalpha.com/input/?i=great+red+spot
Jupiter works best at not so high magnification, otherwise the contrast gets too low. Around 140x … 180x should give you decent contrast in your aperture.
86: By putting a mirror in space, would we be able to see into the past? (score 11642 in 2014)
Question
I only ask this because of how fast light travels. The question remains in the title. Why, or why not, would this work?
Answer accepted (score 14)
I think the question is referring to situating a very large mirror in space facing earth. If we were to put it several light minutes away, then events occurring opposite the mirror could be reviewed de novo with more preparation upon the warning we received upon the first light of the event arriving at earth.
For example, a supernova going off in M31 might not be under observation at the moment its light first arrives, and so the initial observations might be lost. However, with a mirror facing M31, we would be able to observe that mirror as the event unfolded, having been warned that there was something worth watching, in advance.
Nice idea! But it would likely be much less expensive to simply have multiple telescopes always watching “prime” starscape for unexpected events.
Answer 2 (score 9)
Yes, we always look into the past, when looking somewhere. There is for instance a mirror on the moon. When sending a laser beam to that mirror, we can detect the reflected light about 2.5 seconds later. This could be interpreted as looking 2.5 seconds into the past, when the laser has been fired. Details here.
Answer 3 (score 3)
Here are some thoughts adapted to an answer I placed on Phyiscs SE to a similar question some time ago. In order to observe the past we need to detect light from the Earth, reflected back to us from somewhere distant in space.
The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant (\(F \sim 1.3 \times 10^3\) Wm\(^{-2}\)) integrated over a hemisphere. Thus the total reflected light from the Earth is about \(L=5\times 10^{16}\) W.
If this light from the Earth has the same spectrum as sunlight and it gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. then, roughly speaking, the incident flux on a reflecting body will be \(L/2\pi d^2\) (because it is scattered roughly into a hemisphere of the sky).
Now we have to explore some divergent scenarios.
- There just happens to be a large object at a distance that is highly reflective. I’ll use 1000 light years away as an example, which would allow us to see 2000 years into the Earth’s past.
Let’s be generous and say it is a perfect reflector, but we can’t assume specular reflection. Instead let’s assume the reflected light is also scattered isotropically into a \(2\pi\) solid angle. Thus the radiation we get back will be \[ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},\] where \(r\) is the radius of the thing doing the reflecting.
To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of \(-26.74\). The apparent magnitude of the reflected light will be given by \[ m = 2.5\log_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 \]
So let’s put in some numbers. Assume \(r=R_{\odot}\) (i.e. a reflector as big as the Sun) and let \(d\) be 1000 light years. From this I calculate \(m=85\).
To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around \(m=30\) (http://arxiv.org/abs/1305.1931 ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So \(m=85\) is about 22 orders of magnitude fainter than detectable by HST. What’s worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.
- A big, flat mirror 1000 light years away.
How did it get there? Let’s leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case: \[ f = \frac{L}{2\pi [2d]^2} \] with \(d=1000\) light years, which will result in an apparent magnitude at the earth of \(m=37\).
OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST might be required to pick it out even if we had the sensitivity.
- Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.
Of course this doesn’t help you see into the past because we would have to send the telescope there. But it could help those in the future see into their past.
Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to \(m \sim 35\) so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.
Note also that these calculations are merely to detect the light from the whole Earth. To extract anything meaningful would mean collecting a spectrum at the very least! And all this is for only 2000 years into the past.
87: Is it possible to achieve a stable “lunarstationary” orbit around the moon? (score 11559 in 2017)
Question
Is there a stable geostationary orbit around the moon?
My feeling is, that the orbit would collide with earth, because of the moon’s slow rotation.
Answer accepted (score 66)
First off, such an orbit wouldn’t be a geostationary orbit since geo- refers to the Earth. A more appropriate name would be lunarstationary or selenostationary. I’m not sure if there is an officially accepted term since you rarely hear people talk about such an orbit.
You can calculate the orbital distance of a selenostationary orbit using Kepler’s law:
\[a = \left(\frac{P^2GM_{\text{Moon}}}{4\pi^2}\right)^{1/3}\]
In this case, \(a\) is your orbital distance of interest, \(P\) is the orbital period (which we know to be 27.321 days or 2360534 seconds), \(G\) is just the gravitational constant, and hopefully it is obvious that \(M_{\text{Moon}}\) is the mass of the Moon. All we have to do is plug in numbers. I find that
\[a = 88,417\:\mathrm{km}=0.23\:\mathrm{Earth\mathit{-}Moon\:Distance}\]
So I at least match your calculation pretty well. I think you were just relying on Wolfram Alpha a bit too much to get the units right. The units do work out right though.
If you want to determine if this orbit can exist however, you need to do a bit more work. As a first step, calculate the Moon’s Hill Sphere. This is the radius at which the Moon still maintains control over it’s satellite, without the Earth causing problems. The equation for this radius is given by
\[r \approx a_{\text{Moon}}(1-e_{\text{Moon}})\sqrt[3]{\frac{M_{\text{Moon}}}{3M_{\text{Earth}}}}\]
In this equation, \(a_{\text{Moon}} = 348,399\:\mathrm{km}\) is the Moon’s semi-major axis around the Earth and \(e_{\text{Moon}} = 0.0549\) is the Moon’s orbital eccentricity. I’m sure you can figure out that the \(M\)’s are the masses of the respective bodies. Just plug and chug and you get
\[r \approx 52,700\:\mathrm{km}\]
A more careful calculation, including the effects of the Sun is slightly more optimistic and provides a Hill radius of \(r = 58,050\:\mathrm{km}\). In either case though, hopefully you can see that the radius for a selenostationary orbit is much farther than the Hill radius, meaning that no stable orbit can be achieved as it would be too much perturbed by the Earth and/or the Sun.
One final, semi-related point. It turns out almost no orbits around the Moon are stable, even if they’re inside the Hill radius. This is primarily to do with mass concentrations (or mascons) in the Moon’s crust and mantle which make the gravitational field non-uniform and act to degrade orbits. There are only a handful of “stable” orbits and these are only achieved by orbiting in such a way as to miss passing over these mascons.
Answer 2 (score 54)
As the answer by zephyr describes very well, there are very few stable orbits around the moon, and none of them are stationary.
But the moon is tidally locked to Earth. That means that all of the Lagrangian points of the Earth-Moon system are stationary relative to the Moon surface.
88: How do we find the exact temperature of a star? (score 11540 in 2015)
Question
This is a very basic question, but I am a little confused. As far as I know, the temperature of a star is analyzed based on the color of the light it emits. So, if a star is moving away from us, then the light emitted by it will be redshifted(or if it is stationary with respect to us and the light undergoes gravitational redshift), then how do we know the exact temperature of that star or any other object because it is possible that we observe red light but actually the star might be emitting yellow light.
Answer accepted (score 4)
This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do of course have line of sight velocities which give their spectrum a redshift (or blueshift). It is a reasonably simple procedure to correct for the line of sight velocity of a star, because the redshift (or blueshift) applies to all wavelengths equally and we can simply shift the wavelength axis to account for this. i.e. We put the star back into the rest-frame before analysing its spectrum.
Gerald has talked about the blackbody spectrum - indeed the wavelength of the peak of a blackbody spectrum is inversely dependent of temperature through Wien’s law. This method could be used to estimate the temperatures of objects that have spectra which closely approximate blackbodies and for which flux-calibrated spectra are available that properly sample the peak. Both of these conditions are hard to satisfy in practice: stars are in general not blackbodies, though their effective temperatures - which is usually what is quoted, are defined as the temperature of a blackbody with the same radius and luminosity of the star.
The effective temperature of a star is most accurately measured by (i) estimating the total flux of light from the star; (ii) getting an accurate distance from a parallax; (iii) combining these to give the luminosity; (iv) measuring the radius of the star using interferometry; (v) this gives the effective temperature from Stefan’s law: \[ L = 4\pi R^2 \sigma T_{eff}^4,\] where \(\sigma\) is the Stefan-Boltzmann constant. Unfortunately the limiting factor here is that it is difficult to measure the radii of all but the largest or nearest stars. So measurements exist for a few giants and a few dozen nearby main sequence stars; but these are the fundamental calibrators against which other techniques are compared and calibrated.
A second major secondary technique is a detailed analysis of the spectrum of a star. To understand how this works we need to realise that (i) atoms/ions have different energy levels; (ii) the way that these levels are populated depends on temperature (higher levels are occupied at higher temperatures); (iii) transitions between levels can result in the emission or absorption of light at a particular wavelength that depends on the energy difference between the levels.
To use these properties we construct a model of the atmosphere of a star. In general a star is hotter on the inside and cooler on the outside. The radiation coming out from the centre of the star is absorbed by the cooler, overlying layers, but this happens preferentially at the wavelengths corresponding to energy level differences in the atoms that are absorbing the radiation. This produces absorption lines in the spectrum. A spectrum analysis consists of measuring the strengths of these absorption lines for many different chemical elements and different wavelengths. The strength of an absorption line depends primarily on (i) the temperature of the star and (ii) the amount of a particular chemical element, but also on several other parameters (gravity, turbulence, atmospheric structure). By measuring lots of lines you isolate these dependencies and emerge with a solution for the temperature of the star - often with a precision as good as +/-50 Kelvin.
Where you don’t have a good spectrum, the next best solution is to use the colour of the star to estimate its temperature. This works because hot stars are blue and cool stars are red. The colour-temperature relationship is calibrated using the measured colours of the fundamental calibrator stars. Typical accuracies of this method are +/- 100-200 K (poorer for cooler stars).
Answer 2 (score 2)
Spectral lines occur at defined wavelengths. By their redshift you can calculate the radial velocity (or gravitational redshift) of the star, or the absorbing medium, and hence the amount you’ve to shift the black body radiation to obtain the surface temperature (and the radial velocity of a possibly absorbing medium between the star and Earth).
Schematic example: Assume, you measure the following two stellar spectra, and you’re able to identify the typical H-alpha spectral emission line. This line should be at 565.3 nm:
In the second spectrum, H-alpha is at the correct position: no redshift. In the first spectrum, it’s redshifted (towards longer wavelength).
Although the measured intensity may be the same elsewhere in the spectrum, you’ll know, that the first spectrum is of a hotter star, since the maximum intensity (besides the H-alpha line) is left (towards blue) of the H-alpha wavelength, whereas the maximum intensity in the second spetrum is right (towards red) of the H-alpha line.
Both stars would look reddish, but the first one is the hotter one, and it’s redshifted, either due to Doppler shift, due to gravity, or due to cosmic expansion.
89: Why is the Solar Helical (Vortex) model wrong? (score 11529 in )
Question
Could someone give a lay description of why the Solar System’s depiction of motion as a vortex is wrong?
ref: Youtube video The helical model - our solar system is a vortex
Please don’t take for granted that because of my previous question about why planets or asteroids don’t have cometary tails, that I believe the ‘tails’ shown in the video are real. I do not. I know that the tails as shown are simply a graphic representation of the path ‘flown’ through space by the planets. It seems that the solar system we inhabit has to be moving through interstellar space in some fashion. If the paths of the planets could be shown in a video, it seems to me (a lay person) that there would have to be motion ‘forward’ in addition to the eliptical path around Sol.
Answer accepted (score 3)
It isn’t correct, because a vortex is not a helix, and so while the planets do trace a helical path as they move through the galaxy, this is not evidence of a vortex.
Yes, the sun actually is moving through space, as it traces a path around the centre of the galaxy. The whole mass of the solar system moves with it, so the planets are not left behind as the sun moves.
Rhys Taylor and Phil Plait have comprehensive smackdowns debunking this vortex idea and other misunderstandings/delusions by the author.
Answer 2 (score -4)
Solar Vortex is real if you observe it in every 250 years the movement of the sun and its planet towards red dwarf you can come to the clear pictures alas we human age is within 90 years of life span so we can’t get clear picture of move ment of the the sun in vortex. If you want to get clear pictures please use 3D effect of solar movements in years via super computer otherwise use other method called meditation concentrate on red dwarf you will get clear pictures GURU
90: What’s the likelihood of the existence of unknown elements in the Solar System? (score 11434 in 2014)
Question
What’s the chance that there might be undiscovered chemical elements in the Solar System - either on planets or around the Sun or on asteroids of the Oort-cloud?
Answer accepted (score 17)
As far as elements (e.g. on the periodic table) go, I would say the odds are very slim. We already discovered or produced all the elements of the Periodic Table up to atomic number 112 at least. As the number increases, the half lives of the elements generally decreases, and is very short for elements above 102. If this trend holds true as the number increases, practically all the “undiscovered” elements should have turned into the lower known atomic number elements.
However, there is hope. There is a theorized “island of stability” where a narrow range of yet to be discovered high atomic number elements may be stable: http://en.wikipedia.org/wiki/Island_of_stability I would say there is a slight chance this element could be discovered in the solar system.
Answer 2 (score 9)
Further to the answer of @Jonathan, the thing that distinguishes one chemical element from another is the number of protons in the nucleus, which in turn determines the number of orbital electrons in the uncharged atom.
But we already know the element that corresponds to any given number of protons between 1 and 112; that’s the atomic number. And you can’t have a fraction of a proton. The only room for possible new elements is on the end.
Answer 3 (score 5)
A chemical element is defined by the number of protons it contains, this largely defines its chemical properties. Elements can, within certain bounds, have varying number of neutrons (elements with the same number of protons but a different number of neutrons are called isotopes). Number of neutrons can have a subtle effect on chemical properties and a more significant effect on stability ie rate of radioactive decay.
But the big chemical differences which define an element are determined by the proton number and a given element will only have a handful of isotopes within a fairy narrow range.
So, elements are classified by the periodic table which lists the elements in groups according to atomic number (number of protons). When the periodic table was first proposed there were a number of gaps between known elements (at this point the existence of protons was not known). These gaps have subsequently been filled so there is no space for new elements untill you get to high atomic numbers.
The periodic table is full in terms of what could be considered reasonably stable elements. There is no fundamental reason why you can’t propose elements with ever increasing atomic numbers. However the trend so far is that with increasing atomic number elements become more and more unstable. They can be created in particle accelerators but only exist for a tiny amount of time and don’t exist in nature in any way which you might consider being a ‘real’ material like iron or copper.
There have been various predictions of theoretical islands of stability but even then we are talking about very short lived elements.
So in terms of the way that we tend to understand the term there are no new elements to discover, as all reasonably stable possibilities are accounted for.
Having said that there could well be entirely new materials composed of known elements or indeed previously unknown states of matter.
91: Is a white dwarf hotter than a Red Giant? (score 11393 in )
Question
From what I’ve read, white stars are hotter than red ones. But a white dwarf would have just heavy elements to fuse, so shouldn’t it be less bright?
Answer accepted (score 6)
“White” stars are typically much brighter than Red stars, as both the “color & brightness” of a star are directly proportional to the temperature. The only reason there are “bright” red stars is that their radius is incredibly large. Note that the “color” of a star is directly linked to the temperature.
The equation that best demonstrates this is the luminosity equation of a black body. Stars aren’t perfect black bodies, but they are close enough that they are treated as such.
L = 4πR²σT⁴
this equation tells us that the Luminosity (L) is proportionate to the Radius Squared (R²) and the Temperature to the Fourth power (T⁴). The bigger the brighter, or, the hotter the brighter. Meaning that for a given radius the hotter the star, the more luminous, and the same goes for stars of the same temperature, the larger the radius the more luminous.
White dwarfs on the other hand are not stars in the sense that they do not fuse anything, they simply glow due to the lingering heat that was generated during their time as stars.
As shown in the HR-Diagram, White Dwarfs are some of the hottest objects in the universe, and as stated by agtoever there has not been enough time for even the oldest white dwarf to have cooled passed something like 4800K.
Answer 2 (score 3)
White dwarfs start hotter when they are created (up to billions degrees Kelvin), but in the end, they end as black dwarfs, which only happens after a few billion years. As per the age of the universe, it is currently the assumption that there are no black dwarfs yet.
Red giants are (on the surface) typically below 5000 K. Their core is up to a billion degrees Kelvin.
This might be an interesting link too.
As a final note: white dwarfs are the end stage of a star’s life. Depending on the mass of the star, red giants can end up as white dwarfs. About 97% of the milky way stars will end as white dwarfs (including our sun).
92: How many sun-like stars are there in the universe? (score 11385 in 2013)
Question
After yesterday’s announcement of the Kepler telescope finding a huge amount of newly observed exoplanets, i saw a headline claiming that as much as 22% of sun like stars in the universe have planets in their habitable zone. There are loads of stars in the universe, so the number of planets in the habitable zone has to be enourmous. But how many of the stars in the universe are about the same size as our sun?
Answer accepted (score 14)
This is a question that concerns the initial mass function (IMF) - an empirical (that is, defined by observations rather than theory) function that describes the statistical distribution of stellar masses.
Edwin Salpeter (1955) was the first to describe the IMF, though if you read Chabrier (2003) there are some reasonably comprehensive explanations of the theory and history. However, these lecture notes are a fair bit more accessible.
From the approximations in the UCSC lecture notes I linked above, I get that around 4% of stars are between 0.7 and 1.3 solar masses (92% are between 0.1 and 0.7 solar masses!).
There are perhaps 100 billion stars in a galaxy and 100 billion galaxies in the observable universe, giving something on the order of \(4\times 10^{20}\) (400 billion billion) stars that are about (\(\pm 30\%\)) one solar mass.
93: If two black hole event horizons overlap (touch) can they ever separate again? (score 11376 in 2019)
Question
Hypothetical question based on my understanding that two event horizons that overlap (touch) can’t ever separate again:
Imagine a 1 billion solar mass black hole (so the event horizon is massive and very gravitationally weak) is travelling at a velocity of 0.9c through empty flat intergalactic space; now imagine an identical 1 billion solar mass black hole travelling at 0.9c but in exactly the opposite direction so the two are heading roughly towards each other. The black holes’ paths, once all the space time warping is taken into account, aren’t on a direct collision but the outermost edges of the event horizons will just ‘clip’ each other, ordinarily only overlap for a fraction of a nanosecond as these two bodies are travelling at such incredibly fast velocities and in opposite directions to each other.
So firstly, am I right in thinking that if two event horizons overlap they can never ‘unlap’?
Secondly, what would happen to this incredible amount of momentum of each other the black holes? Would it just get instantly turned into gravitational energy? Bearing in mind when black holes normally merge, it happens very slowly as black holes slowly move closer and closer together over millions of years giving off gravitational energy as that happens, so not in a fraction of a nanosecond as in this case.
And thirdly, what would this look like? Would the event horizons remain fairly spherical and the radiated energy just insane or would they stretch and warp into a kind of long thin elastic event horizon as they shoot past each other and then over time slow down and snap back to each other?
Answer accepted (score 54)
You have already got some good answers, but I’ll just try to provide one more intuitive solution on why the event horizons will never separate again if overlapping each other:
First, imagine a speck of dust that comes inside the EH of a black hole. I believe we’ll agree this speck can never escape the black hole, because nothing can come back from behind the event horizon.
Now, imagine the same speck of dust, but inside the overlapping parts of the EH of two black holes passing each other. This speck of dust will never escape any of those two black holes, because it is inside the EH of them both. If these black holes would be able to separate again, the speck caught between them would obviously escape at least one of the black holes, after being behind it’s event horizon.
Since this can not happen, the two black holes will be united from the point their event horizons are overlapping, no matter their speed.
Answer 2 (score 52)
If the event horizons ever touch and become one continuous surface, their fate is sealed - the two black holes will merge all the way in. They can never separate again, no matter what.
There are several possible ways to explain it, with varying degrees of rigorousness.
An intuitive explanation is that escape velocity at the event horizon equals the speed of light. But nothing can move as fast as light, not even a black hole. In order for the two black holes to separate, parts of one would have to “escape” the other, or move faster than light, which is impossible.
EDIT: Another intuitive “explanation” (a.k.a. lots of handwaving) - inside the event horizon, all trajectories lead to the center. There is no possible path from any place within the horizon to the outside. Whichever way you turn, you’re looking at the center. Whichever way you move, you move towards the center. If the event horizons have merged, for the black holes to split up again, parts of them would have to move “away from center” (or away from one of the centers), which is not possible.
All of the above is about as “rigorous” as “explaining” general relativity with steel balls on a rubber sheet. It’s just metaphor.
More rigorously, see this paper by Stephen Hawking:
Black holes in general relativity
As time increases, black holes may merge together and new black holes may be created by further bodies collapsing but a black hole can never bifurcate. (page 156)
EDIT: Event horizons don’t really “just clip each other”. Perfectly spherical event horizons are a theoretical abstraction (a non-rotating black hole in an otherwise empty universe). In reality, anything near a BH will deform the event horizon, which will “reach out” towards that mass. If it’s a small mass, the effect is negligible.
But if two black holes get close to each other, the EHs become egg-shaped, as if trying to touch each other. If they’re close enough, then eventually a very narrow bridge will form in between, and the EHs will merge. At that moment, the full merger is decreed and will procede with absolute certainty until it’s complete. Nothing can stop it.
See this answer:
Are black holes spherical during merger?
what would happen to this incredible amount of momentum of each other the black holes?
The resulting black hole after the merger is going to have a heck of a lot of spin, if the collision is not perfectly frontal. Whatever energy cannot be stuffed into spin, is probably going to be radiated away as gravitational waves (as others have indicated already in comments to your question).
94: What is the fastest spinning rotation of a Neutron star? (score 11226 in 2014)
Question
What is the fastest spinning rotation of a Neutron star? I have heard that Neutron stars have a specially fast spinning rotation. What is the fastest?
Answer accepted (score 11)
As it turns out, the fastest spinning neutron star found yet is a pulsar 18000 light years away in the constelation of Sagittarius which scientist catalogued as PSR J1748-2446ad. Pulsars are neutron stars that rotate, are highly magnetic and emit a strong perpendicular beam of electromagnetic radiation.
This pulsar’s speed is such that:
At its equator it is spinning at approximately 24% of the speed of light, or over 70,000 km per second.
PSR J1748-2446ad rotates a little over 700 times a second, and scientists have this to say on the theoretical limits of the rotation speed of a pulsar (from here):
Current theories of neutron star structure and evolution predict that pulsars would break apart if they spun at a rate of ~1500 rotations per second or more, and that at a rate of above about 1000 rotations per second they would lose energy by gravitational radiation faster than the accretion process would speed them up.
Answer 2 (score 1)
The fastest celestial body we know so far is neutron star XTE J1739-285. It rotates 1122 times per second. Source: https://arxiv.org/abs/0712.4310?fbclid=IwAR2g9RYhybbBlxxzKOfquBa9C33H8mMwuZ_mHZbm1RbcsetrxO3vbvXq7gA
95: Simple experimental evidence that Earth revolves around Sun (score 11167 in 2018)
Question
What are the simplest experiments or calculations that give evidence that the earth revolves around the sun? Can you please explain them and reference the history? Many simple explanations such as this cite observations such as that relative position of two stars are observed from earth vary every night - which would not be true if the stars orbited the earth. But isn’t the observation also consistent with a model where the stars orbit the earth but do so at different speeds, while the earth still orbits the sun? Simple explanations would be helpful.
Answer accepted (score 41)
The answer is ironic: Without good instruments, there is no evidence. The people who thought that the Sun went around the Earth were perfectly correct as far as the actual evidence went until the early 1700s and mid-1800s when two lines of evidence opened up that showed that the Earth moved.
Aberration of Starlight
Wikipedia has a correct but over-complicated explanation. The easiest way to think about it is to imagine yourself at a stop sign in a car in the rain, and the rain is falling straight down. When you start moving, the rain’s apparent direction of fall changes so that it appears to be falling from ahead of you and slanting down towards you. That’s aberration.
In the early 1700s, the stars were discovered to be shifting position, and in 1727, James Bradley correctly identified it as abberation of starlight due to the motion of the Earth around the Sun. (For any star in the ecliptic, the Earth is moving towards it at some time of the year and away from it six months later.)
Parallax
Wikipedia’s article on parallax is better, and I refer you to it for details. Basically, if you hold your finger up before you and look at it with your left eye closed, and then with your right eye closed, it appears to jump with respect to the background – the wall beyond or the trees outside or whatever. Switch back and forth between your eyes quickly to see it clearly.
As the Earth circles around the Sun, nearby stars also appear to shift their position relative to the more distant stars. A key point here is that there were good scientific reasons to suppose that the stars were much smaller than the Sun. Seen through a telescope, stars showed disks and if they were like the Sun, their distance could be deduced from those disks. And they were close enough that if the Earth really went around the Sun, parallax should have been observed. But it wasn’t and the lack of any noticeable parallax was a strong empirical argument against Heliocentric theories.
In reality of course parallax exists, but the parallax of all stars is small, because they are much further away than was estimated from their disks. (The visible disks were actually diffraction disks and not true disks at all – but it was not until nearly a century later that diffraction began to be understood.) Friedrich Bessel first measure the real parallax of a star in 1838.
Answer 2 (score 19)
You cannot prove that the Earth orbits the Sun rather than vice versa because this goes very much against the grain of all frames of reference being equally valid (but some make a lot more sense than others). For example, it makes much more sense to use an Earth-centered, Earth-fixed point of view rather than a non-rotating geocentric, heliocentric, barycentric, or galactocentric point of view when modeling the weather or the tides. One could, for example, use a heliocentric or even galactocentric point of view to model the Earth’s weather, but doing so would be beyond stupid.
On the other hand, when modeling the behavior solar system it makes much more sense to use a heliocentric, or even better, a solar system barycentric point of view. One could however use an Earth-centered, Earth-fixed point of view because all frames of reference are equally valid (in theory). Doing so would of course make the equations of motion quite ugly, and uglier yet on trying to make those equations of motion relativistically correct. A geocentric point of view nonetheless remains theoretically valid – even for modeling the behavior of the Milky Way.
The problem with a geocentric point of view isn’t that it’s invalid (which it isn’t). The problem is that advocates of geocentricism argued (and sadly, continue to argue) that this is the one and only valid point of view. This argument is invalid, because once again, all frames of reference are equally valid.
Note well: Just because inertial frames are special in some sense does not mean that non-inertial frames are invalid.
Answer 3 (score 12)
If you start with the idea that the planets, the sun, the moon and the earth are all bodies that all move through space, exclude the apparently fixed stars, and then see what evidence there is as to how they move relative to each other, then in that context there is some evidence to be found in naked-eye astronomy aided by navigational instruments available even to the ancients.
The patterns of observed movement of the planets is evidence of heliocentric orbit. The visible planets follow certain patterns. First, Mercury and Venus:
- They are always seen in the vicinity of the sun.
- The observed angular separations of both Mercury and Venus from the sun have a regular pattern.
- Mercury has a much closer maximum separation than Venus, and its angular separation changes at a much faster rate.
- Both planets stay close to the ecliptic, and never oscillate normal to it.
- Both planets’ orbits around the sun can be documented and predicted with relative ease. This can be done imprecisely even without a telescope, though it is much harder for Mercury, being so close to the sun.
Beginning with the premise of bodies moving through the heavens, I believe the evidence is there for Mercury and Venus having a heliocentric orbit. Kepler described it precisely, but the ancient Greeks were able to model their motion very well without telescopes in the Antikythera Mechanism in geocentric terms.
If an ancient Greek astronomer had wanted to precisely model the motion of the inner planets in heliocentric terms, he could have. The way to do it is to assume the fixed stars are rigidly fixed, and measure the angular distances between them all, and then plot the motions of the moving planets among them. Sextants and other devices were used by ancient mariners who were highly skilled even with primitive ones. So this could have been done to realize the “simple experiement or calculation” you are asking for. Whether it ever was done, with that question in mind, is a bit different issue.
Now for the earth itself. Even in the ancient world the relationship between the sidereal day and the solar day has been well understood. The precession of the sun around the ecliptic plane is evidence of a heliocentric orbit. One just has to model it to make this clear. Ancient calculations relating to sidereal time and the Metonic cycle reveal that the earth’s heliocentric motion could have been mathematically modeled, if conceived of and desired.
As for the outer planets, to my mind this is the least intuitive, but there is evidence for a heliocentric orbit for them too, but only by building on the idea that earth and the inner planets orbit the sun. This comes from observing their retrograde motion. These planets will move retrograde against the “fixed background stars” at certain times, and those times can be correlated to their angular separation from the sun. Also the different planets move through the zodiac at different speeds, which also correlate with the amplitude of retrograde motion.
If you simulate all this with a heliocentric orrery, it is very plainly evident that we on an inner, faster planet observe an outer, slower planet in its orbit. The ancient Greeks had enough skill to model the motions of Mars, Jupiter and Saturn in their Antikythera Mechanism in geocentric terms. So it follows that a precise, mathematical model of heliocentric motion for the outer planets was within their reach, if they ever reached for it.
There is also some evidence that at least some ancient thinkers were able to decode all this into a heliocentric model. The ancient Greek Aristarchus of Samos had a heliocentric model. However, Plato and others seemed to disfavor it, and this reconstruction of the Antikythera Mechanism which is believed to come well after Aristarchus’ day features a geocentric orrery which models planetary retrograde motion. And heliocentric thinking stayed within the minority in the west until the modern age. Perhaps the obvious geocentric orbit of the moon, or the question of the stars (whether they should be included in any correct model or not), or the lack of a universal theory of gravity, sufficiently obscured for them what to us is clear.
96: Gravitation - Pulling or Pushing force? (score 11135 in 2015)
Question
According to Newton, gravity is the pulling - or in fact the attracting - force of any heavenly body towards any object to its center. But, to the contrary, Einstein once said that the four dimensions of space and time push the object downwards.
So can which one of them is correct?
Answer accepted (score 3)
I’m guessing that this misunderstanding is a result of the oft-used rubber sheet analogy. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes other bits of matter/energy move in different ways. I’m guessing that this is your confusion.
The rubber sheet analogy fails massively in one area, any demonstration of it involves gravity on Earth. If I use a bowling ball to deform a sheet, and then role a golf ball along the sheet nearby, the golf ball will move a bit towards the bowling ball because of the force of gravity around me - not “gravity” in the simulation. It thus makes it seem like gravity pulls the golf ball “down” because the bowling ball pulls the rubber sheet “down”. This is the result of using a two-dimensional analogy of a three-dimensional universe.
The point is, there is no “pushing” going on in the general relativistic model of gravity. Gravity is attractive (so long as the strong energy conditions holds for the object in question), just like Newton postulated.
Answer 2 (score 3)
I’m guessing that this misunderstanding is a result of the oft-used rubber sheet analogy. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes other bits of matter/energy move in different ways. I’m guessing that this is your confusion.
The rubber sheet analogy fails massively in one area, any demonstration of it involves gravity on Earth. If I use a bowling ball to deform a sheet, and then role a golf ball along the sheet nearby, the golf ball will move a bit towards the bowling ball because of the force of gravity around me - not “gravity” in the simulation. It thus makes it seem like gravity pulls the golf ball “down” because the bowling ball pulls the rubber sheet “down”. This is the result of using a two-dimensional analogy of a three-dimensional universe.
The point is, there is no “pushing” going on in the general relativistic model of gravity. Gravity is attractive (so long as the strong energy conditions holds for the object in question), just like Newton postulated.
Answer 3 (score 1)
Earlier answers pointed out that this question is based on a mistake about general relativity, in which gravity neither pushes nor pulls: but it is also based on a mistake about Newton’s view and about what he wrote. Newton was emphatic in refusing to commit himself to identify gravitational attraction either as a push or as a pull. He said he was concerned only with the mathematical quantities and relations of the forces, etc. He wrote that he used the words ‘attraction’ and ‘impulse’ indifferently, ‘one for another’ in respect of the general propensity of massive bodies to approach each other, and he expressly abstained from speculation about their physical nature or cause (see quotations below).
Accordingly, just as the other answers point out that GRT does not specify gravitation as either a push or a pull, neither did Newton’s physics. Whether in GRT or in Newtonian physics, the answer ‘push’ or ‘pull’ is unnecessary to an account of the physics. The question in effect poses a false antithesis.
Here is the basis so far as what Newton wrote. In the opening parts of the ‘Principia’, in Definition VIII, Newton wrote (quoted here from the 1729 English translation of his original Latin, available online at https://books.google.com/books?id=Tm0FAAAAQAAJ):
“I … use the words Attraction, Impulse or Propensity of any sort towards a centre, promiscuously, and indifferently, one for another; considering those forces not Physically but Mathematically : Wherefore, the reader is not to imagine, that by those words, I any where take upon me to define the kind, or the manner of any Action, the causes or the physical reason thereof, or that I attribute Forces, in a true and Physical sense, to certain centres (which are only Mathematical points) …”
In the same vein, his Definition V stated that “a centripetal force is that by which bodies are drawn or impelled, or any way tend, towards a point as to a centre”.
Later on he emphasised yet again the same abstention from speculation (Principia, Book 1, Section XI, Scholium following Proposition 69):
“I here use the word attraction in general for any endeavour, of what kind soever, made by bodies to approach to each other; whether that endeavour arise from the action of the bodies themselves as tending mutually to, or agitating each other by spirits emitted; or whether it arises from the action of the aether or of the air, or of any medium whatsoever, whether corporeal or incorporeal, any how impelling bodies placed therein towards each other. In the same general sense I use the word impulse, not defining in this treatise the species or physical qualities of forces, but investigating the quantities and mathematical proportions of them; as I observed before in the definitions.”
97: Why is the discovery of merging neutron stars important? (score 10994 in 2017)
Question
I’m fairly certain people here will have heard about it, already, but apparently, two supernova leftovers clashed some 130 million years ago and some billion billion kilometres away …
What I haven’t heard yet, however, is why we should care.
I mean sure, it’s an interesting phenomenon and measuring it can’t have been easy.
But now that we’ve heard it … what changes?
I’ll admit it, I don’t know particularly much about astronomy, but I’m curious:
What’s the significance of having achieved this? Why does it matter whether or not we know?
Answer accepted (score 159)
Reasons why this is important:
It is the first simultaneous detection of a gravitational wave and electromagnetic signal, and the strongest GW signal yet in terms of signal to noise (Abbott et al. 2017a). It spectacularly corroborates the reality of the GW detection technology and analysis. The progenitor has been unambiguously located in a (relatively) nearby galaxy (Soares-Santos et al. 2017), allowing a host of other telescopes to obtain detailed measurements.
It shows that GWs travel at the speed of light, a further verification of Einstein’s General Relativity (Abbott et al. 2017b).
It shows that most of the very heavy elements such as gold, platinum, osmium etc. are plausibly produced by merging neutron stars and constrains the rate of such mergers in the local universe (e.g. Chornock et al. 2017; Tanvir et al. 2017).
It shows that short gamma ray bursts — some of the most energetic explosions in the universe — can be caused by neutron star mergers (e.g. Savchenko et al. 2017; Goldstein et al. 2017).
It is the closest detected short gamma ray burst (with a known distance). That the progenitor has also been characterised allows a closer investigation of the interesting physics underlying the ejection and jet mechanisms thought to be responsible for the gamma rays and later X-ray and radio emission (e.g. Margutti et al. 2017; Alexander et al. 2017).
It provides observational constraints on how matter behaves at extremely high densities, testing our understanding of fundamental physics to its limits — for example, the details of the gravitational wave signal moments before merger are diagnostic of the interior conditions of neutron stars at densities of \(\sim 10^{18}\) kg/m\(^3\) (Hinderer et al. 2010; Postnikov et al. 2010).
It provides an independent way of measuring the expansion of the universe. Merging binary gravitational wave sources are known as “standard sirens”, because the distance to the GW source pops straight out of the analysis and can be compared with the redshift of the identified host galaxy (Abbott et al. 2017c). The result agrees with measurements made using the cosmic microwave background and the distance-redshift relation calibrated by other means, verifying our estimation of distances, at least in the local universe.
Finally, this event will turn out to be important because it was lucky; in the sense that the source was detected well-inside the sensitivity horizon of LIGO (Abbott et al. 2017a). The detection itself, was not unexpected given the rates predicted based on studying the neutron star binary systems in our own Galaxy (e.g. Kim et al. 2015), but the fact that it was so close — within the closest 5% of the sensitive survey volume where it could have been detected — is fortunate.
In the end, if someone thinks none of the above is interesting or important, then nothing I can write will convince them otherwise. The vast majority of people I speak to are curious and fascinated to find out about our cosmic origins and how the universe works.
Answer 2 (score 39)
Because its awesome (SMBC)
So this guy called Copernicus suggested that the Earth orbits the Sun (not the other way round) - What changes?
This guy Newton had a theory for how a mass responds to force, and how gravity works - So what?
Another guy called Maxwell had this idea of how light could actually be waves of electromagnetic fields - does this matter?
A guy called Monet decided to paint some pictures of some waterlilies. Who cares?
Last February, some guys from Denver carried a ball over a line more often than some guys from Carolina carried the ball over another line. So what?
It is worth finding things out because it means finding things out. It is worth understanding our world because it is there to be understood. Discovery is its own reward. It is not measured in by £ or $.
The observations of GW170817 show that heavy elements are created by the merger of neutron stars. Heavy elements like gold, platinum on Earth were likely created in a neutron star merger in The Milky Way, billions of years ago, that enriched the interstellar dust.
If it doesn’t matter to you that’s fine. If Monet’s Waterlilies or the Superbowl leave you cold, that’s no problem either. But not everything of value is of use.
98: Do asteroids have a gravitational field? (score 10973 in )
Question
I know that asteroids are huge chunks of rock, orbiting a solar system. Do asteroids have a gravitational field and do they gravitationally attract each other to form planets?
Answer accepted (score 8)
By definition, gravity is a result of mass. Any body with a non-zero mass (even atoms) will have a gravitational field associated with it. The higher the mass the stronger will be the field. This is basic of classical mechanics. Until we reach quantum scale where the gravitational force is dominated by other 3 forces and the gravitational field becomes irrelevent.
When it comes to graitational field of asteroids, it exists, but is very weak. However over a course of few million years these small asteroids combine together to form large masses of bodies that we now call planets. That is one of the prominent theory of Solar system formation, where the gravity of small dust particles from first our generation disintegrated star over a course of time accumulated to give us what we now know as our Solar system. Think of it like this, every planet that you see now would once have been an asteroid at some point during its evolution.
Another proof to support this is the presence of numerous binary asteroids that orbit each other around a common center of mass, which requires gravitational attraction.
Answer 2 (score 6)
Sure. Any mass has its gravitational field. However, its size is proportional to the mass, so as most asteroids have little mass, they have little gravitational field, and therefore pull only very slightly at each other, resulting in not enough effect to get them to lump together.
Typically, their difference in momentum/speed is too large to be removed by the little pull of the gravitation between them.
Answer 3 (score 4)
You asked two questions.
Do asteroids have a gravitational field.
Of course. Even a microscopic grain of dust has a gravitational field.
Do they gravitationally attract each other to form planets?
Not any more. During the formation of the solar system, asteroid-like and comet-like objects collided to build larger objects, which in turn collided to form even larger objects, and so on, eventually building the cores of giant planets and later, the terrestrial planets. But that stage ended long ago, shortly after the solar system formed.
Asteroids do of course gravitationally attract other objects, but this attraction is so weak due to the small masses of asteroids that it is easily overwhelmed by other perturbing forces. The vast majority of the asteroids lie between Mars and Jupiter, and Jupiter is the primary culprit in explaining why no planet exists in that gap.
When two astronomical bodies collide, one of the outcomes is a purely inelastic collision that makes two bodies form a single body. This only happens with a rather mild collision. A more energetic collision will result in some mass being expelled. An even more energetic collision will result in lots of mass being expelled; the colliding bodies become many smaller bodies. With a few exceptions, the latter is what is what is happening amongst the asteroids today, and for the last four-plus billion years or so.
Jupiter is such a huge perturbing body that collisions in the asteroid belt are generally very energetic. Instead of forming ever larger bodies, the asteroid belt is gradually being broken up into smaller and smaller bodies. Some of these collisional bodies are ejected from the solar system thanks to interactions with Jupiter. The smallest results of these collisions migrates sunward thanks to the Poynting-Robertson effect.
99: Years, months, days, and … weeks? (score 10933 in 2017)
Question
Why do we divide time into weeks? Is there any celestial reason why humans do this?
- one year: earth revolution around the sun
- one month: moon revolution around the earth
- one week: 7 days = ???
- one day: earth rotation about its axis
Answer accepted (score 17)
The synodic period of the moon is \(29.53\) days, a little shorter than a calendar month, which is on average about \(30.4\) days. This is slightly longer than its orbital period, but corresponds to the periodic visual appearance of the moon as viewed from Earth. I mention this to make it clear that we should be forgiving of a little imprecision.
Conventionally, the moon’s appearance is divided into four phases: first quarter, full, last quarter, new. That means that on average, each phase lasts about \(7.4\) days. Since calendars count days in integer amounts, a \(7\)-day period seems to be a natural choice.
The social importance of the seven-day period in Western cultural probably has much more to do with its religious significance in Abrahamic religions than astronomy per se (although certainly not unique to it). But its ultimate origin probably does lie in the natural division of the moon’s appearance into four phases, which correspond to an apparent geocentric celestial latitude difference between the Moon and Sun of \(0^\circ\), \(90^\circ\), \(180^\circ\), and \(270^\circ\).
That, the explicit answer to your question is
- 1 week = 7 days = one lunar phase.
Answer 2 (score 4)
The modern 7-day week was adopted in the late Roman Empire, replacing an 8-day “nundinal week” that had previously been in use mainly for market intervals. As far as I know, there was no particular reason for the 8-day length other than that it was a convenient size to separate market days.
In the early Empire, Christians and Jews both used a 7-day week for religious purposes and as they grew to be a major part of the Empire’s population, it became convenient to link the market week with the religious week. This worked because 7 worked about as well as 8 for markets, and so by the late Empire, it was a done deal.
See https://en.wikipedia.org/wiki/Nundinae for a good discussion.
Where did the Jewish 7-day cycle come from? It’s not certain. A practical market cycle can be anywhere in the 5-10 day range – it doesn’t matter a lot – so if early Jews preferred a 7-day religious week (from Genesis, Chapter 1, probably) then the market cycle would have synched up just like in the Roman Empire. It’s also possible they inherited it from some previous mid-Eastern culture. I doubt anyone knows for sure any more, so it must remain to some degree speculation.
Answer 3 (score 1)
The 7 days of the week certainly fit somewhat as a quarter of a month. But there was another argument for the ancients to use the number seven, because they knew about seven “planets”, seven objects which wander across the sky relative to the thousands of fixed stars. The Sun, the Moon, Mercury, Venus, Mars, Jupiter, Saturn. Earth wasn’t a planet according to the IAU definition at the time. Earth was instead the center of the universe.
The weekly calendar of the ancients was pretty busy:
Sunday - Sun you’re free, sober up and get a suntan
Monday - Moon the day after being free
Tuesday - Jupiter Zeus day, be clever and figure something out
Wednesday - Mercury go somewhere and talk about it
Thursday - Mars fight
Friday - Venus make love
Saturday - Saturn please take a bath!
100: How can ‘HD 140283’ be older than the universe? (score 10903 in 2017)
Question
Scientists have known about the star HD 140283, informally nicknamed the Methuselah star, for more than 100 years, since it cruises across the sky at a relatively rapid clip. The star moves at about 800,000 mph (1.3 million km/h) and travels the width of the full moon in the sky every 1,500 years or so, researchers said. Its apparent magnitude is 7.223.
Previous research had estimated that the Milky Way galaxy’s so-called “Methuselah star” is up to 16 billion years old. That’s a problem, since most researchers agree that the Big Bang that created the universe occurred about 13.8 billion years ago.
Later estimates show that the star could be as old as 14.5 billion years (± 0.8 billion years), which is still it older than the universe’s calculated age of about 13.8 billion years. This is an obvious dilemma.
How can a star be older than the universe itself?
Answer accepted (score 39)
It’s not - the current age is an estimate. The discrepancy comes from astronomer’s attempts to fully and correctly measure aspects of the star, particularly its distance. From the NASA page:
The new Hubble age estimates reduce the range of measurement uncertainty, so that the star’s age overlaps with the universe’s age — as independently determined by the rate of expansion of space, an analysis of the microwave background from the big bang, and measurements of radioactive decay.
Currently, if you take the uncertainty into account, the current estimate for the star’s age is 14.5 +/- 0.8 billion years or between 15.3 and 13.7 billion years - the latter end of the estimate overlaps with the current model of the age of the Universe - so a more correct way of describing this age is that HD 140283 is at least 13.7 billion years old by this estimation.
Another paper based on this star in Nature - “Nearby star is almost as old as the Universe”, a team has narrowed down the age to 13.9 +/- 0.7 billion years, stating that:
Taking into account that experimental error, the age does not conflict with the age of the Universe, 13.77 billion years.
more concisely stating that
The star’s age is therefore at least 13.2 billion years
No doubt further work is being performed to further refine the age of the star.
Answer 2 (score 15)
Later estimates shows that the star could be as old as 14.5 billion years (± 0.8 billion years), which is still it older than the universe’s calculated age of about 13.8 billion years, an obvious dilemma.
There is no dilemma. That ± 0.8 billion years is important. Subtract 0.8 billion years from that 14.5 billion year figure (later revised to 14.27 billion years) and you get 13.7 billion years, less than the age of the universe. Also note that that 0.8 billion years uncertainty is the standard error, or roughly one standard deviation. That revised age of 14.27 billion years makes the dilemma essentially non-existent. That’s older than the estimate age of the universe by 0.4 to 0.5 million years, and this is not significant even at the one sigma level.
Particle physicists generally use the five sigma level as distinguishing between “almost certain” and “maybe” (and particle physicists are reluctant to publish “maybes”). Astronomers generally use a three sigma level. Even social scientists would balk at a deviation that is less than one sigma as being significant.