<p>It varies.</p>

<p>The best astronomical sites have a visual band extinction of 0.1 mag, which means that only $\sim 10$ per cent of light is absorbed/scattered in the atmosphere.</p>

<p>In dusty, smoggy or polluted sites, this can easily reach one magnitude of extinction, which means that 60 per cent of the light is scattered.</p>

<p>These numbers are <em><a href="https://en.wikipedia.org/wiki/Air_mass_(astronomy)">per airmass</a></em> - which means looking straight up at zenith. At lower altitudes, the baseline extinction figure would be multiplied by $\sec z$ where $z$ is the angle from the zenith.</p>

<p>This amount of absorption doesn't sound like a huge difference, but what it means is that a large number of extra stars become visible, because roughly-speaking, you get a factor of two increase in the number of stars for every 0.5 magnitudes deeper you can see, and I think that is what is most striking about going to a clear, dark astronomical site.</p>

<p>Thus there wouldn't be <em>much</em> of an improvement going into space over the best astronomical sites (<em>in the visible band</em>), at least not due to the decrease in extinction. You do also escape light pollution, which is a different effect. With light pollution it is the fact that the background sky becomes too bright for your eye to pick up the fainter stars. [As Carl Witthoft notes, it is a different ball game at other wavelengths because the atmosphere is significantly more opaque to UV, X-ray and IR wavelengths.]</p>

<p>Another thing to consider is what you are breathing! Lack of oxygen can hinder visual acuity. <a href="http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm">http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm</a></p>

CC BY-SA 3.0
2016-10-13T16:16:11.140Z
Plus from space you can use IR and UV scopes, as those wavelengths don't make it thru the upper atmosphere very well.
Carl Witthoft
CC BY-SA 3.0
2016-10-13T16:57:35.887Z
Perfect. Thank you for the in-depth answer! :)
Paul
CC BY-SA 3.0
2016-10-14T05:34:03.570Z
Would not sec z only be valid on a flat Earth?
Loren Pechtel
CC BY-SA 3.0
2016-10-14T06:40:29.843Z
@LorenPechtel You mean for a plane, parallel atmosphere. Yes, but you could check the reference for how good an approximation that is and for more complicated expressions at very high airmass ($z>80$ degrees) if it matters.
ProfRob
CC BY-SA 3.0
2016-10-13T15:05:24.807Z
2016-10-13T16:26:13.810Z
2016-10-13T16:26:13.810Z
ProfRob
<p>As a child I remember my parents taking camping in California, and upon stepping out of the car I was awestruck at how thick the Milky Way galaxy was and at the number of stars everywhere. I've always wondered how much clearer it could get. What percentage of visible light from stars are absorbed by all of earths atmosphere at the clearest locations on earth? Are we taking about a small percentage or perhaps a significant amount?</p>

CC BY-SA 3.0
2016-10-13T21:17:43.167Z
that is actually a totally superb question
Fattie
<night-sky><atmospheric-effects>
How much clearer are stars in earths orbit?
Paul
# How much clearer are stars in earths orbit?

(Asked by: username_0 on 2016-10-13)

As a child I remember my parents taking camping in California, and upon stepping out of the car I was awestruck at how thick the Milky Way galaxy was and at the number of stars everywhere. I've always wondered how much clearer it could get. What percentage of visible light from stars are absorbed by all of earths atmosphere at the clearest locations on earth? Are we taking about a small percentage or perhaps a significant amount?

**Tags:** night-sky, atmospheric-effects

> **Comments:**
> **username_1**: that is actually a totally superb question  

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## 1 answers

### Answer by username_2 on 2016-10-13. Score: 22 | Accepted answer

It varies.

The best astronomical sites have a visual band extinction of 0.1 mag, which means that only $\sim 10$ per cent of light is absorbed/scattered in the atmosphere.

In dusty, smoggy or polluted sites, this can easily reach one magnitude of extinction, which means that 60 per cent of the light is scattered.

These numbers are *[per airmass](https://en.wikipedia.org/wiki/Air_mass_(astronomy))* - which means looking straight up at zenith. At lower altitudes, the baseline extinction figure would be multiplied by $\sec z$ where $z$ is the angle from the zenith.

This amount of absorption doesn't sound like a huge difference, but what it means is that a large number of extra stars become visible, because roughly-speaking, you get a factor of two increase in the number of stars for every 0.5 magnitudes deeper you can see, and I think that is what is most striking about going to a clear, dark astronomical site.

Thus there wouldn't be *much* of an improvement going into space over the best astronomical sites (*in the visible band*), at least not due to the decrease in extinction. You do also escape light pollution, which is a different effect. With light pollution it is the fact that the background sky becomes too bright for your eye to pick up the fainter stars. [As username_3 username_3 notes, it is a different ball game at other wavelengths because the atmosphere is significantly more opaque to UV, X-ray and IR wavelengths.]

Another thing to consider is what you are breathing! Lack of oxygen can hinder visual acuity. <http://navyaviation.tpub.com/14020/css/Effects-Of-Hypoxia-141.htm>

> **Comments:**
> **username_3**: Plus from space you can use IR and UV scopes, as those wavelengths don't make it thru the upper atmosphere very well.  
> **username_0**: Perfect. Thank you for the in-depth answer! :)  
> **username_4**: Would not sec z only be valid on a flat Earth?  
> **username_2**: @username_4 You mean for a plane, parallel atmosphere. Yes, but you could check the reference for how good an approximation that is and for more complicated expressions at very high airmass ($z>80$ degrees) if it matters.  

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