The wikipedia article on Alpha Centauri has an image of that system from earth. I'm hoping to make the opposite image. I realize that there are various discussion of roughly what it would look like (e.g., What would our Sun look like from other solar systems?), but presumably one or another of modern simulators can actually essentially take a simulated photo of this at this point. What simulator would you recommend that I try to do this with? Thanks!

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    $\begingroup$ SpaceEngine is a nice program for that. It simulates the universe pretty accurately, and you can get it on Steam. With it, you can not only visit other star systems, real or procedurally generated, but also other galaxies. For example, you can look at the Milky Way from the LMC or Andromeda. You can also go to highly distant objects such as TON 618, for example. It takes all factors such as position, size and luminosities into consideration. So, in this simulator, if you went to Alpha Centauri and looked at the Sun, it would be a generally accurate representation of what you would really see. $\endgroup$
    – 4NT4R3S
    Jan 22 at 19:55
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    $\begingroup$ FWIW, the angular diameter of Neptune's orbit, as seen from Alpha Centauri AB, is ~45 arcseconds. $\endgroup$
    – PM 2Ring
    Jan 22 at 20:32
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    $\begingroup$ It really depends on what you use to look. Anything beyond the largest telescopes will simply give you a comparatively bright star (similar to what alpha centauri looks from here - its brightest component is also a G-star like our Sun) $\endgroup$ Jan 22 at 22:09
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    $\begingroup$ Have a look at Where can I find the positions of the planets, stars, moons, artificial satellites, etc. and visualize them? You might really like Universe Sandbox $\endgroup$
    – uhoh
    Jan 23 at 12:24

2 Answers 2


The Sun has an absolute visual magnitude of about 4.75. Alpha Cen is 1.33 pc away. Thus the apparent magnitude of the Sun when viewed from Alpha Cen is 0.37.

The Sun would appear as a point of light, just as Alpha Cen appears as a point of light to us (although a telescope will split it into a binary system). Note that this isn't just a question of angular resolution (Jupiter would be separated from the Sun by several arcseconds viewed from Alpha Cen), it is a question of brightness contrast - Jupiter would be around 25 magnitudes fainter than the Sun.

Even a space telescope like HST or JWST would be unable to resolve the visible disk of the Sun or to see any of the surrounding planets against the glare of the Sun.

An open-source program like Celestia allows you to "fly" to Alpha Cen and look back at the Sun.

I had a quick play with Celestia and produced this image as viewed from Alpha Cen A, with a field of view of around 75 degrees. The Sun is the bright star in the centre and is viewed towards the "w" of Cassiopeia (seen to the right of the Sun).

View from Alpha Cen towards the Sun

  • $\begingroup$ Wow. This is fantastic. For a sense of scale, how wide would pluto's orbit be in this picture if we were looking down on the disc? (I realize w're not, and you wouldn't be able to see the planets! I'm just trying to get a sense of scale since it's not possible to tell from this pictures what's in front and what's behond.) $\endgroup$ Jan 23 at 23:20
  • $\begingroup$ @jackisquizzical About 50 arcseconds diameter. Roughly similar to the angular resolution of 20:20 vision and smaller than a pixel on this picture. $\endgroup$
    – ProfRob
    Jan 24 at 0:18
  • $\begingroup$ So, thanks. Now, I already know the answer to this, but let me confirm: Is there any way to estimate what it would look like through a telescope of a given magnification. My kneejerk answers is: Sure, just zoom in by x, but that's not right because it doesn't take into account various optical distortions. If I assume-away atmospheric effects (it's a space telescope :-) and then it's presumably a bit closer to just an x zoom. I'm thinking back to the 85mm picture of the Alpha Centauri system I referenced above. Sol to Jupiter should be about that same size. Or am I totally confused? $\endgroup$ Jan 24 at 0:55
  • $\begingroup$ Another way to ask this, perhaps, is how close would I need to be to Sol to get the system (say, 40au) into the field of view? I guess I could do the simple trig to figure this out. $\endgroup$ Jan 24 at 1:03
  • $\begingroup$ @jackisquizzical Your FOV would need to $40/1.33$ arcsec across to encompass 40 au at 1.33 pc. The rest of your question depends on the details of your optical device. The picture I have shown in my answer approximates to what you would see without a telescope. As I say, it doesn't matter what telescope you use, the Sun (and solar system) will appear as a point of light. $\endgroup$
    – ProfRob
    Jan 24 at 6:47

Actually, although all of the above is useful discussion, I've found a wonderful NASA asteroid tracker that happens to be able to also show the starfield form any angle, although you have to look at it through the annoying asteroid field: https://www.jpl.nasa.gov/asteroid-watch/eyes-on-asteroids

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    $\begingroup$ How is the asteroid viewer relevant to your original question? The distance from the Sun to Neptune is about 30 AU, the distance to Alpha Centauri is almost 275,000 AU, around 9136 times further. $\endgroup$
    – PM 2Ring
    Feb 3 at 0:11
  • $\begingroup$ @PM2Ring The image I refer to in the question is through a telescope. I wanted to know what our solar system would look like from A Cen also through a telescope, that is, if someone in the A Cen system had a telescope trained on the Sol system. (This wasn't stated as clearly as it could have been in the original question.) $\endgroup$ Feb 4 at 1:18
  • $\begingroup$ Ok, but that asteroid viewer shows the planets from a point in the Solar System. It's like looking at your own backyard from your back porch, compared to looking at it from 100 km away. ;) JPL have several 3D Solar System viewers. Here's one without the asteroid clutter: ssd.jpl.nasa.gov/tools/orbit_viewer.html And I have a program using accurate JPL data that can show orbits of Solar System bodies in 3D: astronomy.stackexchange.com/a/49823/16685 $\endgroup$
    – PM 2Ring
    Feb 4 at 5:48

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