<p>Well, I didn't vote you down but it's a strange question in that there are some uncertainties, but it's also not hard to do the math.</p>

<p>For example, which Galaxy? - they vary quite a bit, and from what angle?  10,000 light years from the side or from the "top" for lack of a better word, above the brighter galactic core or 10,000 light years from the tip of a spiral arm?</p>

<p>Andromeda is 10 times brighter than the Milky way, and galaxies get a good deal larger than Andromeda and a good deal smaller than the Milky way.</p>

<p>But, the short answer to your question is, from 10,000 light years, a galaxy wouldn't appear that bright . . . er, mostly but it would appear quite large.</p>

<p>We can see the Milky way as kind of a white smudge across the southern hemisphere, but much of that bright smudge that we see is stars that are closer than 10,000 light years, so that's not really apples to apples, as much of that smudge is alot closer to us than your example.</p>

<p>Lets take our sun, from 10,000 light years away, lets, just for fun, say you had 10 billion of our suns, each 10,000 light years away.   10 billion suns is a lot, but light decreases by the square of the distance and there's 63,000 AUs in a light year so each sun at that distance would be 1/(397 million billion) less bright than the sun is from the earth.   10 billion of them, it's still 1/40 million times less bright than the sun is form the earth, roughly 1/100th the brightness of the full moon - that's what 10 billion of our suns would look like from that distance.   All together, that's still 10 times brighter than Venus, so if they were close together they'd shine clear as day, but spread out, like a galaxy would be at 10,000 light years away - not so much.</p>

<p>Now a galaxy is more complicated cause some of the stars light would be blocked by dust and some stars would be much brighter than the sun, others much dimmer and a single supernova can outshine an entire galaxy for a couple of weeks.   There's no precise answer, but in general, a galaxy would look very large but not very bright from 10,000 light years away.</p>

<p>Another, and perhaps better way to look at this is to take a look at Andromeda, which is about 10 times the brightness of the Milky way and we have a pretty clear view of it, but it's also about 2.5 million light years away from us.    If it was, in your scenario, 250 times closer it would be 62,500 times brighter and much larger.   It would spread out over most of the sky, though, like the Milky way, we'd see a smudge of brightness, or we might be able to make out the spiral arms depending on the angle.</p>

<p>Andromeda's apparent brightness or apparent magnitude is 3.4, and each number in the brightness scale is 2.5 times brighter, so 62,500 times as bright, would be between 12 and 12.1 scales brighter, or -8.6 to -8.7, which for an object in the sky, is fairly bright, (yeah, negative numbers = brighter), but it's still quite a bit less bright than the full moon, and at the same time, spread out over a much larger area - so, if Andromeda was that close, the galactic center would probably be a fair bit more visible than Venus, brighter in total but also more spread out, and much less bright than the full moon.</p>

<p>Short answer - the photos are probably brighter by a healthy amount.</p>

<p>Here's a very cool and loosely related article.   <a href="http://www.slate.com/blogs/bad_astronomy/2014/01/01/moon_and_andromeda_relative_size_in_the_sky.html" rel="nofollow">http://www.slate.com/blogs/bad_astronomy/2014/01/01/moon_and_andromeda_relative_size_in_the_sky.html</a></p>

CC BY-SA 3.0
2015-07-09T16:02:20.173Z
Very informative, this was a great answer! Thanks a lot!
Petter
CC BY-SA 3.0
2016-01-05T15:33:41.063Z
There is also the Magellanic clouds. Smaller, but closer, than Andromeda, and clearly visible on the Southern sky.
Thriveth
CC BY-SA 3.0
2015-07-09T12:56:48.523Z
2015-07-09T12:56:48.523Z
userLTK
<p>The surface brightness of an extended source in independent of distance. So for the surface brightness of a large spiral go look at M31 (The Great Galaxy in Andromeda) . What will change with distance is the apparent size, but the surface brightness will remain the same.</p>

<p>The surface brightness of a near edge-on galaxy will be greater than one face-on due to the greater number of stars close to any sight-line, and the integrated brightness will of course vary with distance.</p>

CC BY-SA 3.0
2015-07-09T10:58:08.220Z
2015-07-09T13:08:29.787Z
2015-07-09T13:08:29.787Z
Conrad Turner
<p>In images of other galaxies they look very bright and vivid against the blackness of space (<a href="https://upload.wikimedia.org/wikipedia/commons/9/98/Andromeda_Galaxy_%28with_h-alpha%29.jpg" rel="noreferrer">example</a>). If I was located a few tens of thousands of lightyears away from a galaxy, would I see something similar or would it be more like a slightly brighter smudge in space?</p>

<p>I'm wondering since it seems we can barely see the features of our own galaxy with the naked eye from earth.</p>

<p>Edit: I see that someone has downvoted the question, it would have been good to have some feedback! I realise the question might be a bit vague, but I thought it was interesting to know if galaxies would look as bright as we see them in the pictures.</p>

CC BY-SA 3.0
2015-07-05T19:10:17.550Z
The Milky Way is about 1,000,000,000,000,000,000 km across. From 1,000,000 km away from the edge, it will almost stretch 180°. I would say that most of the nearby stars would be distinguishable and due to the density near the edge, they would overlap.
LDC3
CC BY-SA 3.0
2015-07-06T06:50:51.577Z
If you're close enough to resolve individual starts I understand that you would be able to tell quite clearly, I was thinking more like if the galaxy took ~90 degrees of your field of vision. Would it look very bright from there or would it be quite dim?
Petter
CC BY-SA 3.0
2015-07-07T02:13:56.557Z
OK, for 90° field of view $tan(45°)=\frac {500,000,000,000,000,000}{d}$ or $d=\frac {500,000,000,000,000,000}{tan(45°)}$, $d=500,000,000,000,000,000\ km$. I don't know what sort of resolution you would get, nor how bright the galaxy would be.
LDC3
<galaxy>
How clearly would I be able to see a galaxy with the naked eye if viewed from a "close" distance?
Petter
# How clearly would I be able to see a galaxy with the naked eye if viewed from a "close" distance?

(Asked by: username_0 on 2015-07-05)

In images of other galaxies they look very bright and vivid against the blackness of space ([example](https://upload.wikimedia.org/wikipedia/commons/9/98/Andromeda_Galaxy_%28with_h-alpha%29.jpg)). If I was located a few tens of thousands of lightyears away from a galaxy, would I see something similar or would it be more like a slightly brighter smudge in space?

I'm wondering since it seems we can barely see the features of our own galaxy with the naked eye from earth.

Edit: I see that someone has downvoted the question, it would have been good to have some feedback! I realise the question might be a bit vague, but I thought it was interesting to know if galaxies would look as bright as we see them in the pictures.

**Tags:** galaxy

> **Comments:**
> **username_1**: The Milky Way is about 1,000,000,000,000,000,000 km across. From 1,000,000 km away from the edge, it will almost stretch 180°. I would say that most of the nearby stars would be distinguishable and due to the density near the edge, they would overlap.  
> **username_0**: If you're close enough to resolve individual starts I understand that you would be able to tell quite clearly, I was thinking more like if the galaxy took ~90 degrees of your field of vision. Would it look very bright from there or would it be quite dim?  
> **username_1**: OK, for 90° field of view $tan(45°)=\frac {500,000,000,000,000,000}{d}$ or $d=\frac {500,000,000,000,000,000}{tan(45°)}$, $d=500,000,000,000,000,000\ km$. I don't know what sort of resolution you would get, nor how bright the galaxy would be.  

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## 2 answers

### Answer by username_2 on 2015-07-09. Score: 2 | Accepted answer

Well, I didn't vote you down but it's a strange question in that there are some uncertainties, but it's also not hard to do the math.

For example, which Galaxy? - they vary quite a bit, and from what angle? 10,000 light years from the side or from the "top" for lack of a better word, above the brighter galactic core or 10,000 light years from the tip of a spiral arm?

Andromeda is 10 times brighter than the Milky way, and galaxies get a good deal larger than Andromeda and a good deal smaller than the Milky way.

But, the short answer to your question is, from 10,000 light years, a galaxy wouldn't appear that bright . . . er, mostly but it would appear quite large.

We can see the Milky way as kind of a white smudge across the southern hemisphere, but much of that bright smudge that we see is stars that are closer than 10,000 light years, so that's not really apples to apples, as much of that smudge is alot closer to us than your example.

Lets take our sun, from 10,000 light years away, lets, just for fun, say you had 10 billion of our suns, each 10,000 light years away. 10 billion suns is a lot, but light decreases by the square of the distance and there's 63,000 AUs in a light year so each sun at that distance would be 1/(397 million billion) less bright than the sun is from the earth. 10 billion of them, it's still 1/40 million times less bright than the sun is form the earth, roughly 1/100th the brightness of the full moon - that's what 10 billion of our suns would look like from that distance. All together, that's still 10 times brighter than Venus, so if they were close together they'd shine clear as day, but spread out, like a galaxy would be at 10,000 light years away - not so much.

Now a galaxy is more complicated cause some of the stars light would be blocked by dust and some stars would be much brighter than the sun, others much dimmer and a single supernova can outshine an entire galaxy for a couple of weeks. There's no precise answer, but in general, a galaxy would look very large but not very bright from 10,000 light years away.

Another, and perhaps better way to look at this is to take a look at Andromeda, which is about 10 times the brightness of the Milky way and we have a pretty clear view of it, but it's also about 2.5 million light years away from us. If it was, in your scenario, 250 times closer it would be 62,500 times brighter and much larger. It would spread out over most of the sky, though, like the Milky way, we'd see a smudge of brightness, or we might be able to make out the spiral arms depending on the angle.

Andromeda's apparent brightness or apparent magnitude is 3.4, and each number in the brightness scale is 2.5 times brighter, so 62,500 times as bright, would be between 12 and 12.1 scales brighter, or -8.6 to -8.7, which for an object in the sky, is fairly bright, (yeah, negative numbers = brighter), but it's still quite a bit less bright than the full moon, and at the same time, spread out over a much larger area - so, if Andromeda was that close, the galactic center would probably be a fair bit more visible than Venus, brighter in total but also more spread out, and much less bright than the full moon.

Short answer - the photos are probably brighter by a healthy amount.

Here's a very cool and loosely related article. <http://www.slate.com/blogs/bad_astronomy/2014/01/01/moon_and_andromeda_relative_size_in_the_sky.html>

> **Comments:**
> **username_0**: Very informative, this was a great answer! Thanks a lot!  
> **username_3**: There is also the Magellanic clouds. Smaller, but closer, than Andromeda, and clearly visible on the Southern sky.  

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### Answer by username_4 on 2015-07-09. Score: 2

The surface brightness of an extended source in independent of distance. So for the surface brightness of a large spiral go look at M31 (The Great Galaxy in Andromeda) . What will change with distance is the apparent size, but the surface brightness will remain the same.

The surface brightness of a near edge-on galaxy will be greater than one face-on due to the greater number of stars close to any sight-line, and the integrated brightness will of course vary with distance.


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