<p>In short, it is depended on the height of the meteor and the separation of observers.</p>
<p><a href="https://i.sstatic.net/b37HRRUr.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/b37HRRUr.jpg" alt="" /></a></p>
<p>This is a rough sketch of the situation you are describing. There are 2 observers on Earth with a separation of <span class="math-container">$\theta$</span> relative to the center of Earth, <span class="math-container">$h$</span> is the height of the meteor to ground, <span class="math-container">$R_$</span> is the radius of Earth, <span class="math-container">$z$</span> is the zenith distance of the meteor observed by the second observer. Your goal is the compute the value of <span class="math-container">$z$</span>.</p>
<p>Now it is the problem of solving the triangle, there are more than one way to do it:</p>
<p>First find the length of opposite side of <span class="math-container">$\theta$</span>, denote it as <span class="math-container">$x$</span>. This can be given by cosine law:</p>
<p><span class="math-container">$x^2 = {R_}^2 + {(R_+h)}^2 - 2{R_}{(R_+h)}cos\theta$</span></p>
<p>By sine law:</p>
<p><span class="math-container">$\frac{sin(180^{\circ} - z)}{R_+h} = \frac{sin \theta}{x}$</span></p>
<p>(the reason of <span class="math-container">$180^{\circ}$</span> being here is because <span class="math-container">$z$</span> is not included in the triangle, but it's supplementary angle is included.)</p>
<p>So now you can establish a relation between <span class="math-container">$z$</span> and <span class="math-container">$h$</span> and <span class="math-container">$\theta$</span>.</p>
<p>However this is an ideal situation where we neglect the influence of atmosphere, adding atmosphere correction will cause some slightly shift to <span class="math-container">$z$</span>, but generally it can be neglected.</p>

CC BY-SA 4.0
2024-06-11T17:01:14.200Z
2024-06-11T17:01:14.200Z
Polaris5744
<p>If I observe a meteor entering the atmosphere at the zenith from my location. How far does a second observer have to be from my position to see the meteor 10 degrees above the horizon. Or what is the location limit to just not see the meteor?</p>

CC BY-SA 4.0
2024-06-11T16:12:33.427Z
Really tough to say.  How bright a meteor is, and local sky conditions cause the visibility of the meteor to occurr at different altitudes.  Meteor trajectories are usually measured using the start and end points, and the observer's location to define a plane.  Then, with multiple observers, they compute where the planes intersect.  It's totally feasible that the same meteor was visible at two locations, but the visible portions of the observed paths don't overlap.
Greg Miller
<planetary-atmosphere><meteor>
Meteor visibility from two locations
Robert
# Meteor visibility from two locations

(Asked by: username_0 on 2024-06-11)

If I observe a meteor entering the atmosphere at the zenith from my location. How far does a second observer have to be from my position to see the meteor 10 degrees above the horizon. Or what is the location limit to just not see the meteor?

**Tags:** planetary-atmosphere, meteor
> **Comments:**
> **username_1**: Really tough to say.  How bright a meteor is, and local sky conditions cause the visibility of the meteor to occurr at different altitudes.  Meteor trajectories are usually measured using the start and end points, and the observer's location to define a plane.  Then, with multiple observers, they compute where the planes intersect.  It's totally feasible that the same meteor was visible at two locations, but the visible portions of the observed paths don't overlap.  

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## 1 answers

### Answer by username_2 on 2024-06-11. Score: 3 | Accepted answer

In short, it is depended on the height of the meteor and the separation of observers.

[![](https://i.sstatic.net/b37HRRUr.jpg)](https://i.sstatic.net/b37HRRUr.jpg)

This is a rough sketch of the situation you are describing. There are 2 observers on Earth with a separation of $\theta$ relative to the center of Earth, $h$ is the height of the meteor to ground, $R\_$ is the radius of Earth, $z$ is the zenith distance of the meteor observed by the second observer. Your goal is the compute the value of $z$.

Now it is the problem of solving the triangle, there are more than one way to do it:

First find the length of opposite side of $\theta$, denote it as $x$. This can be given by cosine law:

$x^2 = {R\_}^2 + {(R\_+h)}^2 - 2{R\_}{(R\_+h)}cos\theta$

By sine law:

$\frac{sin(180^{\circ} - z)}{R\_+h} = \frac{sin \theta}{x}$

(the reason of $180^{\circ}$ being here is because $z$ is not included in the triangle, but it's supplementary angle is included.)

So now you can establish a relation between $z$ and $h$ and $\theta$.

However this is an ideal situation where we neglect the influence of atmosphere, adding atmosphere correction will cause some slightly shift to $z$, but generally it can be neglected.


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