<p>The Andromeda galaxy is not too different from ours, and yet you can see it with your naked eyes at 2.5 million light years distance. You do need to be under a very dark sky, far from cities, and it does look like a mere white smudge, but it is visible.</p>

<p>Now compare that with the fact that the Milky Way itself is quite visible to the naked eye if you're far from cities...</p>

<p><a href="https://astronomy.stackexchange.com/questions/22012/how-do-i-see-the-milky-way">How do I see the milky way?</a></p>

<p>...and basically interpolate somewhere in between "little white smudge" and "river of stars filling up half the sky" to get an idea of how our galaxy would look like from its vicinity.</p>

<p>It would not be bright anyway, and you would have to adapt your eyes to complete darkness first. But you would definitely see its outline. Whether you would distinguish anything of the shape of its arms or not, is harder to say, and would probably depend a lot on the actual distance.</p>

CC BY-SA 3.0
2017-08-11T23:51:58.340Z
2017-08-11T23:51:58.340Z
Florin Andrei
<p>Just to add some bad mathematics to this problem, lets use Andromeda.</p>

<p>Short answer.   The numbers agree what Florin Andrei said.   </p>

<p>Longer answer:</p>

<p><a href="https://en.wikipedia.org/wiki/Andromeda_Galaxy" rel="nofollow noreferrer">Andromeda has</a> an apparent magnitude of 3.44 from about 2.54 million light years distance.</p>

<p>The tricky part of the question is, how close would you be, above Andromeda's center to see the galaxy as a whole, but not so close that individual stars stand out.    The width of the center of galaxies is tricky and I had no success looking up specific numbers.  Galaxies have fuzzy boundaries anyway.   </p>

<p>The average width of the Milky way is given as about 1,000 light years, but it's thickest in the middle.  Ballpark estimates, I would guess the thickness of the Milky way to be maybe 3,000 - 6,000 light-years at the center (<a href="https://upload.wikimedia.org/wikipedia/commons/thumb/a/a1/ESO_-_The_Milky_Way_panorama_%28by%29.jpg/825px-ESO_-_The_Milky_Way_panorama_%28by%29.jpg" rel="nofollow noreferrer">see picture</a>).   The thickness of Andromeda, a larger galaxy, perhaps 7,000-10,000.</p>

<p>And as for how close you could be where individual stars wouldn't stand out and you'd see the galaxy as a whole.  From Earth, the most distant star that can be seen by the naked eye is about 4,000 light years away.   <a href="https://ww2.kqed.org/quest/2007/09/28/the-unaided-eye/" rel="nofollow noreferrer">Source.</a>  Given that Andromeda would appear like a white smudge, similar to the Milky way does from the Southern hemisphere, individual stars would stand out much less over the white smudge, so an additional 1,000 light years or so from the fuzzy edge should be enough to make the galaxy appear as a galaxy without visible individual stars.  </p>

<p>If we give a distance of 8,000 light years from Andromeda, or, about 310-320 times closer than we are now.  That would make it about a 100,000 times brighter and give it a brightness of about about -9.1, which might be a tricky number to conceptualize.   It's significantly less bright than a full moon but many times brighter than Venus, but it would be spread out over a large area of the sky.</p>

<p>If you moved a bit further away, the spiral arms of Andromeda might be more visible, say 15,000 or 20,000 light years above it's center, but the brightness would drop to a -7.6 or -6.8 respectively.</p>

<p>This is pretty ballpark cause it's comparing brightness from our view at a significant angle to the view from directly above.  The brightness might be less obstructed from directly above so my estimates might be low.</p>

<p>But to give an idea of what a -9.1 over much of the sky would mean, it's probably easier to look at that in terms of number of zero magnitude stars.   See <a href="http://www.skyandtelescope.com/astronomy-resources/astronomy-questions-answers/if-you-combine-the-magnitudes-of-all-visible-stars-how-bright-a-star-will-you-come-up-with/" rel="nofollow noreferrer">this article</a>.</p>

<p>The entire night sky (360 degrees, all directions) has a brightness of about 500-550 zero magnitude stars.   A -9.1 would be the equivalent of about 4,200 zero magnitude stars, so at about 8,000 light years, Andromeda would look like a not very bright white smudge over perhaps nearly half of the sky, with dimmer regions outside the center.  But the overall brightness would be several times brighter than our moonless night sky on a clear night, but still about 20-25 times dimmer than the full moon.</p>

<p>At greater distances, the brightness drops with the square of the distance, but from far enugh away to see the spiral arms, they would probably be very faint, but perhaps discernable.  Hard to say. </p>

CC BY-SA 3.0
2017-08-12T16:02:21.550Z
2017-08-12T16:02:21.550Z
userLTK
<p>If someone was far enough above the milky way to see the entire diameter, how much can he actually see with naked eye? I'm aware that on Earth, we only see stars within 1,000 light-years (roughly) by naked eyes. I found myself wondering if someone is more than 1,000 light years away from milky way then we won't be able to see it?</p>

CC BY-SA 3.0
2017-08-15T22:05:51.110Z
Related question (not a duplicate). https://astronomy.stackexchange.com/q/19756/10117
iMerchant
<milky-way>
If someone was above the galaxy, how much of it can he see?
Eric Palmer-Smith
# If someone was above the galaxy, how much of it can he see?

(Asked by: username_0 on 2017-08-11)

If someone was far enough above the milky way to see the entire diameter, how much can he actually see with naked eye? I'm aware that on Earth, we only see stars within 1,000 light-years (roughly) by naked eyes. I found myself wondering if someone is more than 1,000 light years away from milky way then we won't be able to see it?

**Tags:** milky-way

> **Comments:**
> **username_1**: Related question (not a duplicate). https://astronomy.stackexchange.com/q/19756/10117  

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## 2 answers

### Answer by username_2 on 2017-08-11. Score: 3 | Accepted answer

The Andromeda galaxy is not too different from ours, and yet you can see it with your naked eyes at 2.5 million light years distance. You do need to be under a very dark sky, far from cities, and it does look like a mere white smudge, but it is visible.

Now compare that with the fact that the Milky Way itself is quite visible to the naked eye if you're far from cities...

[How do I see the milky way?](https://astronomy.stackexchange.com/questions/22012/how-do-i-see-the-milky-way)

...and basically interpolate somewhere in between "little white smudge" and "river of stars filling up half the sky" to get an idea of how our galaxy would look like from its vicinity.

It would not be bright anyway, and you would have to adapt your eyes to complete darkness first. But you would definitely see its outline. Whether you would distinguish anything of the shape of its arms or not, is harder to say, and would probably depend a lot on the actual distance.


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### Answer by username_3 on 2017-08-12. Score: 1

Just to add some bad mathematics to this problem, lets use Andromeda.

Short answer. The numbers agree what username_2 username_2 said.

Longer answer:

[Andromeda has](https://en.wikipedia.org/wiki/Andromeda_Galaxy) an apparent magnitude of 3.44 from about 2.54 million light years distance.

The tricky part of the question is, how close would you be, above Andromeda's center to see the galaxy as a whole, but not so close that individual stars stand out. The width of the center of galaxies is tricky and I had no success looking up specific numbers. Galaxies have fuzzy boundaries anyway.

The average width of the Milky way is given as about 1,000 light years, but it's thickest in the middle. Ballpark estimates, I would guess the thickness of the Milky way to be maybe 3,000 - 6,000 light-years at the center ([see picture](https://upload.wikimedia.org/wikipedia/commons/thumb/a/a1/ESO_-_The_Milky_Way_panorama_%28by%29.jpg/825px-ESO_-_The_Milky_Way_panorama_%28by%29.jpg)). The thickness of Andromeda, a larger galaxy, perhaps 7,000-10,000.

And as for how close you could be where individual stars wouldn't stand out and you'd see the galaxy as a whole. From Earth, the most distant star that can be seen by the naked eye is about 4,000 light years away. [Source.](https://ww2.kqed.org/quest/2007/09/28/the-unaided-eye/) Given that Andromeda would appear like a white smudge, similar to the Milky way does from the Southern hemisphere, individual stars would stand out much less over the white smudge, so an additional 1,000 light years or so from the fuzzy edge should be enough to make the galaxy appear as a galaxy without visible individual stars.

If we give a distance of 8,000 light years from Andromeda, or, about 310-320 times closer than we are now. That would make it about a 100,000 times brighter and give it a brightness of about about -9.1, which might be a tricky number to conceptualize. It's significantly less bright than a full moon but many times brighter than Venus, but it would be spread out over a large area of the sky.

If you moved a bit further away, the spiral arms of Andromeda might be more visible, say 15,000 or 20,000 light years above it's center, but the brightness would drop to a -7.6 or -6.8 respectively.

This is pretty ballpark cause it's comparing brightness from our view at a significant angle to the view from directly above. The brightness might be less obstructed from directly above so my estimates might be low.

But to give an idea of what a -9.1 over much of the sky would mean, it's probably easier to look at that in terms of number of zero magnitude stars. See [this article](http://www.skyandtelescope.com/astronomy-resources/astronomy-questions-answers/if-you-combine-the-magnitudes-of-all-visible-stars-how-bright-a-star-will-you-come-up-with/).

The entire night sky (360 degrees, all directions) has a brightness of about 500-550 zero magnitude stars. A -9.1 would be the equivalent of about 4,200 zero magnitude stars, so at about 8,000 light years, Andromeda would look like a not very bright white smudge over perhaps nearly half of the sky, with dimmer regions outside the center. But the overall brightness would be several times brighter than our moonless night sky on a clear night, but still about 20-25 times dimmer than the full moon.

At greater distances, the brightness drops with the square of the distance, but from far enugh away to see the spiral arms, they would probably be very faint, but perhaps discernable. Hard to say.


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