# View from asteroid belt?

At a distance of 2-4 AUs, (in the asteroid belt), would you be able to stare directly at the sun? What about the rest of the view from this distance? Other planets, if at all visible, would just be tiny specks, like they are when viewed from earth? Would Jupiter appear larger by any noticeable amount? Apart from the sun and stars and some planets, I imagine nothing else would be visible and the sky would be mostly black - given the distance asteroids have from each other and their smallness, you probably wouldn't be able to see any other asteroids, right?

What would light levels be like at this distance? How far away would you be able to see an object (say, a small ship or astronaut), before it disappears from view?

Simon Bridge
Homework Helper
Welcome to PF.

Light falls off by inverse square law so at 2AU you get 1/4 of the light at Earth.
But it is direct, from the Sun, rather than diffuse like on the Earth's surface. So you want to compare with illumination on the moon... look at the NASA photos.

Staring right at the Sun is probably still not a good idea. It would take longer to burn your eyes out. You should be able to work out the angular size of the sun's disk at 2AU from geometry - it's a good exercize so I leave it to you (or someone else to tell you.)

Jupiter can get brighter because you can get closer, but it can also get farther away - draw the orbits. It will still be a bright star.

You'd be able to see more stars than on the Earth - and nebulae and so forth. And you wold probably not see other asteroids, yep. You can think of the light levels as 0.25 that on the moon at midday (lunar time). How far away you can see something depends on it's albedo, and size, - it would be much further than on the earth because - no atmosphere.

So for example, if you're 2 AU from the sun and look at it, then back away to 4 AU from the sun it will be now appear 4 times dimmer even though you just moved back twice as far. If you move back four times as far, it will appear 16 times dimmer, and so on.

Hope this helps a bit, I don't have the mathematical knowledge to work out the numbers for you for specifics from those distances and objects but I imagine if you know the luminousity, the distance to object, and what the minimum luminousity the human eye can see is, then you can figure what objects will be visible or not.

I am inclined to think it would be very unlikely to see other asteroids while on an asteroid within the belt. I am certain the sun is still visible from there and I don't know about how easy to see Jupiter would be. Would be interested in seeing someone with actual numbers work this out or present the already worked out information.

Drakkith
Staff Emeritus
What about the rest of the view from this distance? Other planets, if at all visible, would just be tiny specks, like they are when viewed from earth?
Imagine yourself as the Sun. All planets are about the same brightness all the time, as most have orbits fairly close to circular. Move out to Mercury and now as Mercury and the planets orbit around the Sun they vary in brightness to a slightly greater extent, as at different points in Mercury's orbit it is further or closer to any specific planet. Jump out to Venus and the effect is even greater. Same for Earth, then Mars, and then the asteroid belt. Jupiter would be a dazzling sight at opposition (when it is closest to you and opposite of the Sun in your sky), you could probably see the larger moons with the naked eye, and would fade away to an even dimmer magnitude than it gets to here on Earth as you move away from it.

Would Jupiter appear larger by any noticeable amount?
Absolutely. Near opposition it would be several times larger in the sky as it is here on Earth due to the reduced distance. This is counteracted by a much smaller apparent size when it is on the far side of the sun.

Apart from the sun and stars and some planets, I imagine nothing else would be visible and the sky would be mostly black - given the distance asteroids have from each other and their smallness, you probably wouldn't be able to see any other asteroids, right?
I'd expect that if you had a sharp eye you could spot many of the larger asteroids, especially ones like Ceres or Vesta if they were close enough. It really depends on what's near you. But the sky would mostly appear just like it does here on Earth.

What would light levels be like at this distance? How far away would you be able to see an object (say, a small ship or astronaut), before it disappears from view?
Thanks to the reduced light levels you would only be able to see them at about 1/4 the distance as you normally would.

Janus
Staff Emeritus
Gold Member

So for example, if you're 2 AU from the sun and look at it, then back away to 4 AU from the sun it will be now appear 4 times dimmer even though you just moved back twice as far. If you move back four times as far, it will appear 16 times dimmer, and so on.

Hope this helps a bit, I don't have the mathematical knowledge to work out the numbers for you for specifics from those distances and objects but I imagine if you know the luminousity, the distance to object, and what the minimum luminousity the human eye can see is, then you can figure what objects will be visible or not.

I am inclined to think it would be very unlikely to see other asteroids while on an asteroid within the belt. I am certain the sun is still visible from there and I don't know about how easy to see Jupiter would be. Would be interested in seeing someone with actual numbers work this out or present the already worked out information.
As as how the Suns will "appear", it is a little different. Our visual sensitivity is not linear but logarithmic The Sun, while only 1/4 as luminous would, to our eyes only be ~1.5 magnitudes dimmer

Drakkith
Staff Emeritus
As as how the Suns will "appear", it is a little different. Our visual sensitivity is not linear but logarithmic The Sun, while only 1/4 as luminous would, to our eyes only be ~1.5 magnitudes dimmer
How much dimmer is something that is 1.5 magnitudes less? I know a difference of 1 magnitude is 2.512 times as dim, but not how to figure the other .5 magnitude.

Janus
Staff Emeritus
$$2.512 ^{-1.5}$$= ~0.25