<p>If the atmosphere could be redistributed so that it has sea level density <span class="math-container">$D$</span> up to height <span class="math-container">$H$</span> and <span class="math-container">$0$</span>  density above that, then we can calculate 
<span class="math-container">$$H = \frac{P}{gD} = \frac{100,000}{9.8 \times 1.27} \approxeq 8,000$$</span>
(SI units), where <span class="math-container">$P$</span> is sea level pressure.</p>

<p>So it is the same as looking through <span class="math-container">$8$</span> km of air horizontally.</p>

CC BY-SA 4.0
2019-12-08T03:03:12.303Z
2019-12-08T03:03:12.303Z
Keith McClary
<p>Extinction is proportional to air mass, and according to the Wikipedia page for air mass, for the model you are describing (the isothermal one), the air mass is around <span class="math-container">$37.2$</span> times that at the zenith at the horizon. This means that in order to achieve the same extinction you would need to observe the object <span class="math-container">$37$</span> times closer than if it were on the horizon (as far as it can be seen due to the curvature of the Earth, this is just an average value though, as it depends from person to person based on their heights).<br> Now, the distance of the horizon is around 5km at sea level (this can be calculated with trigonometry), meaning you'd have to observe it at 130 or so meters away from you. A surprising result, yet one I believe to make sense due to the exponential decay of the air's density and the fact that extinction at the horizon is many magnitudes greater than the one at zenith.</p>

CC BY-SA 4.0
2019-12-06T21:20:41.867Z
Interesting, but how is it possible? The 100 km of atmosphere above us is not enought to diminish the brightness of an object stronger than the same object located at 130 m? I feel like this is not only surprising but wrong. Maybe the value is very model-dependent?
Swike
CC BY-SA 4.0
2019-12-06T21:49:29.577Z
The magnitude being proportional might be too rough an approximation, yes ... according to this [here](http://star-www.rl.ac.uk/star/docs/sc6.htx/sc6se8.html) however, it would appear that changing a model only makes calculating air mass more difficult and the extinction is still proportional ...
Tosic
CC BY-SA 4.0
2019-12-07T08:13:19.267Z
"...just as badly..." and "...image I can get from it is still better..." are probably more about [astronomical seeing](https://en.wikipedia.org/wiki/Astronomical_seeing) than about extinction.
uhoh
CC BY-SA 4.0
2019-12-06T20:08:06.437Z
2019-12-06T20:08:06.437Z
Tosic
<p>When I see an object in space (let's say the ISS) above my head, my line of sight traverses ~<span class="math-container">$100\; km$</span> of atmosphere. The vast majority of the extinction, absorption and turbulence happens closer to the surface of Earth (since the density of air follows an exponential decay with altitude).</p>

<p>If I tried to observe another object, this time on the surface of Earth, the line of sight towards it would have to traverse huge amounts of air and the density would in principle be constant throughout the entire optical path.</p>

<p>My question is: How should I calculate the maximum distance of an object on the surface of Earth such that the image I can get from it is still better than the image I can get for any object outside the atmosphere in the zenith of an observer located on the surface of Earth? What is the distance for an object on the surface of Earth at which anything I photograph in space looks crispier than this object?</p>

<photography><space><atmospheric-effects>
What is the horizontal distance needed to observe an object just as badly as if it where in space at the zenith?
Swike
# What is the horizontal distance needed to observe an object just as badly as if it where in space at the zenith?

(Asked by: username_0 on 2019-12-06)

When I see an object in space (let's say the ISS) above my head, my line of sight traverses ~$100\; km$ of atmosphere. The vast majority of the extinction, absorption and turbulence happens closer to the surface of Earth (since the density of air follows an exponential decay with altitude).

If I tried to observe another object, this time on the surface of Earth, the line of sight towards it would have to traverse huge amounts of air and the density would in principle be constant throughout the entire optical path.

My question is: How should I calculate the maximum distance of an object on the surface of Earth such that the image I can get from it is still better than the image I can get for any object outside the atmosphere in the zenith of an observer located on the surface of Earth? What is the distance for an object on the surface of Earth at which anything I photograph in space looks crispier than this object?

**Tags:** photography, space, atmospheric-effects


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## 2 answers

### Answer by username_1 on 2019-12-08. Score: 2

If the atmosphere could be redistributed so that it has sea level density $D$ up to height $H$ and $0$ density above that, then we can calculate
$$H = \frac{P}{gD} = \frac{100,000}{9.8 \times 1.27} \approxeq 8,000$$
(SI units), where $P$ is sea level pressure.

So it is the same as looking through $8$ km of air horizontally.


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### Answer by username_2 on 2019-12-06. Score: 1

Extinction is proportional to air mass, and according to the Wikipedia page for air mass, for the model you are describing (the isothermal one), the air mass is around $37.2$ times that at the zenith at the horizon. This means that in order to achieve the same extinction you would need to observe the object $37$ times closer than if it were on the horizon (as far as it can be seen due to the curvature of the Earth, this is just an average value though, as it depends from person to person based on their heights).  
 Now, the distance of the horizon is around 5km at sea level (this can be calculated with trigonometry), meaning you'd have to observe it at 130 or so meters away from you. A surprising result, yet one I believe to make sense due to the exponential decay of the air's density and the fact that extinction at the horizon is many magnitudes greater than the one at zenith.

> **Comments:**
> **username_0**: Interesting, but how is it possible? The 100 km of atmosphere above us is not enought to diminish the brightness of an object stronger than the same object located at 130 m? I feel like this is not only surprising but wrong. Maybe the value is very model-dependent?  
> **username_2**: The magnitude being proportional might be too rough an approximation, yes ... according to this [here](http://star-www.rl.ac.uk/star/docs/sc6.htx/sc6se8.html) however, it would appear that changing a model only makes calculating air mass more difficult and the extinction is still proportional ...  
> **username_3**: "...just as badly..." and "...image I can get from it is still better..." are probably more about [astronomical seeing](https://en.wikipedia.org/wiki/Astronomical_seeing) than about extinction.  

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