<p>The Earth's surface which is visible when you look at the planet from a certain distance is a <a href="https://en.wikipedia.org/wiki/Spherical_cap" rel="nofollow noreferrer">spherical cap</a> in terms of geometry. Here it is, in blue:</p>
<p><a href="https://i.sstatic.net/lpe1d.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lpe1d.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$A$</span> is the position of the observer, <br/><span class="math-container">$H$</span> is the distance from the observer to the surface of the sphere, <br/><span class="math-container">$O$</span> is the center of the sphere,<br/><span class="math-container">$r$</span> is the radius of the sphere, <br/><span class="math-container">$AB$</span> is the distance to the true horizon, <br/><span class="math-container">$\angle ABO = 90°$</span>, <br/> <span class="math-container">$\angle \theta$</span> is the angle between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the disk forming the base of the cap.</p>
<p>The area of the cap <span class="math-container">$A_c$</span> can be found according to this formula:
<span class="math-container">$$A_c=2 \pi r^2(1-\cos \theta)$$</span></p>
<p><span class="math-container">$\cos \theta$</span> in the right triangle <span class="math-container">$ABO$</span> is the ratio of the adjacent cathetus <span class="math-container">$OB = r$</span> to the hypotenuse <span class="math-container">$OA = r+H$</span>, that is, <span class="math-container">$\cos \theta = \frac{r}{r+H}$</span>, so
<span class="math-container">$$A_c=2 \pi r^2(1-\frac{r}{r+H})$$</span></p>
<p>The percentage ratio <span class="math-container">$R_\%$</span> of the visible area <span class="math-container">$A_c$</span> to the full area of the sphere <span class="math-container">$A_s$</span> is <span class="math-container">$$R_\%=\frac{A_c}{A_s}\times 100\%$$</span>
Since the full area of the sphere is <span class="math-container">$A_s = 4\pi r^2$</span>, we have:
<span class="math-container">$$R_\%=\frac{2 \pi r^2(1-\frac{r}{r+H})}{4\pi r^2}\times 100\%$$</span>
<span class="math-container">$2 \pi r^2$</span> above and below the line are canceled, so the final formula is like this:
<span class="math-container">$$\bbox[7px,border:2px solid red]{R_\%=\frac{1}{2}  \times (1-\frac{r}{r+H}) \times 100\%}$$</span></p>
<hr />
<p>Answering your question, from the altitude of 3,000 mi (4,800 km) one can see <span class="math-container">$21.4842\% \approx 21.5\%$</span> of the Earth's surface.</p>
<p>I have typed it in the Google calculator <a href="https://www.google.com/search?hl=uk&amp;sxsrf=ALeKk00J1r1Ne69vFyUpwt2koydEXPUhhg%3A1595378921793&amp;ei=6YwXX7iCMPGvrgTE67NA&amp;q=%281%20-%20%286371%20%2F%20%286371%20%2B%204800%29%29%29%20%2F%202%20*%20100%20%3D&amp;oq=%281%20-%20%286371%20%2F%20%286371%20%2B%204800%29%29%29%20%2F%202%20*%20100%20%3D&amp;gs_lcp=CgZwc3ktYWIQDDoHCAAQRxCwA1DxQ1ihU2CDwwFoAXAAeACAAWeIAewCkgEDMy4xmAEAoAEBqgEHZ3dzLXdpesABAQ&amp;sclient=psy-ab&amp;ved=0ahUKEwj4mqnP0d_qAhXxl4sKHcT1DAgQ4dUDCAw" rel="nofollow noreferrer">here</a>, so you can use it, just substitute the Earth radius of 6371 km and the altitude 4800 km for the numbers you would like.</p>
<p>Also, <a href="https://www.google.com.ua/search?bih=578&amp;biw=1229&amp;hl=uk&amp;sxsrf=ALeKk01aNnUprPs8sUec1B7NqMYdtP85Ow%3A1595387845941&amp;ei=xa8XX-OCOePKrgSA_aSABA&amp;q=f%28x%29%20%3D%20%281%20-%20%286371%20%2F%20%286371%20%2B%20x%29%29%29%20%2F%202%20*%20100&amp;oq=f%28x%29%20%3D%20%281%20-%20%286371%20%2F%20%286371%20%2B%20x%29%29%29%20%2F%202%20*%20100&amp;gs_lcp=CgZwc3ktYWIQAzoHCAAQRxCwA1D1Slj1SmCyU2gBcAB4AIABW4gBW5IBATGYAQCgAQGqAQdnd3Mtd2l6wAEB&amp;sclient=psy-ab&amp;ved=0ahUKEwij-dfu8t_qAhVjpYsKHYA-CUAQ4dUDCAw&amp;uact=5" rel="nofollow noreferrer">here is a graph</a> of how the visible area of the Earth (<span class="math-container">$r=6371 km$</span>) changes with the altitude, scale and drag the graph with the mouse.</p>
<p>As for finding the altitude from which the given percentage of the Earth's surface is visible, here it is, it is just a transformation my <em>final formula</em> so that <span class="math-container">$H$</span> is on one side of the <span class="math-container">$=$</span> and all the rest is on the other, only <span class="math-container">$R$</span> here is just a ratio, no percent, so you will put 0.38, not 38% into the formula:
<span class="math-container">$$H=2r/(\frac{1}{R} - 2)$$</span>
<a href="https://www.google.com/search?hl=uk&amp;sxsrf=ALeKk016sSVLgXb78aRF71KvXYCWrqkyYA%3A1595392165810&amp;ei=pcAXX4SBMeegrgTDqJ7YDg&amp;q=2*6371%2F%28%281%2F0.38%29%20-%202%29%20%3D&amp;oq=2*6371%2F%28%281%2F0.38%29%20-%202%29%20%3D&amp;gs_lcp=CgZwc3ktYWIQDDoHCAAQRxCwA1DtmAlY7ZgJYMGoCWgBcAB4AIABhAGIAYQBkgEDMC4xmAEAoAEBqgEHZ3dzLXdpesABAQ&amp;sclient=psy-ab&amp;ved=0ahUKEwjE58f6guDqAhVnkIsKHUOUB-sQ4dUDCAw" rel="nofollow noreferrer">The calculator</a> says that to see 38% of the Earth's surface you need to be 20,174.83 km (12,536 mi) above the surface.</p>

CC BY-SA 4.0
2020-07-22T03:28:26.047Z
2020-07-22T04:54:28.437Z
2020-07-22T04:54:28.437Z
Yellow Sky
<p>My latter three questions remain unanswered but I won't wait with this one, I'd like to learn the formula and hope this question <em>gets</em> an answer.</p>
<p>As you're going higher respectively farther away from the Earth, more and more of the Earth's surface is visible. At infinite distance the percentage of the surface that is visible would be 50%. Now I'd like to know how you calculate the percentage that is visible from altitude/distance, and how you calculate the altitude/distance from how much of the surface is visible.</p>
<p>For example, how much of the Earth's surface is visible at 3,000 mi (4,800 km) altitude or at what altitude do you see 38% of the Earth's surface?</p>

CC BY-SA 4.0
2020-07-21T13:30:19.700Z
Lots and lots of formulae you need are here, enjoy! https://en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon
Yellow Sky
CC BY-SA 4.0
2020-07-21T17:09:35.360Z
How do you conclude the _percentage_ of the surface from the formulae in the other question or in the WP link?
Ioannes
CC BY-SA 4.0
2020-07-21T23:21:51.713Z
Answers to [Help computing solid angle subtended by the Earth as seen from LEO](https://space.stackexchange.com/q/26633/12102) in Space Exploration SE should get you what you need
uhoh
<distances><mathematics>
How to calculate how much of the Earth's (or another planet's) surface is visible?
Ioannes
# How to calculate how much of the Earth's (or another planet's) surface is visible?

(Asked by: username_0 on 2020-07-21)

My latter three questions remain unanswered but I won't wait with this one, I'd like to learn the formula and hope this question *gets* an answer.

As you're going higher respectively farther away from the Earth, more and more of the Earth's surface is visible. At infinite distance the percentage of the surface that is visible would be 50%. Now I'd like to know how you calculate the percentage that is visible from altitude/distance, and how you calculate the altitude/distance from how much of the surface is visible.

For example, how much of the Earth's surface is visible at 3,000 mi (4,800 km) altitude or at what altitude do you see 38% of the Earth's surface?

**Tags:** distances, mathematics

> **Comments:**
> **username_1**: Lots and lots of formulae you need are here, enjoy! https://en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon  
> **username_0**: How do you conclude the _percentage_ of the surface from the formulae in the other question or in the WP link?  
> **username_2**: Answers to [Help computing solid angle subtended by the Earth as seen from LEO](https://space.stackexchange.com/q/26633/12102) in Space Exploration SE should get you what you need  

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## 1 answers

### Answer by username_1 on 2020-07-22. Score: 3 | Accepted answer

The Earth's surface which is visible when you look at the planet from a certain distance is a [spherical cap](https://en.wikipedia.org/wiki/Spherical_cap) in terms of geometry. Here it is, in blue:

[![enter image description here](https://i.sstatic.net/lpe1d.jpg)](https://i.sstatic.net/lpe1d.jpg)

$A$ is the position of the observer,   
$H$ is the distance from the observer to the surface of the sphere,   
$O$ is the center of the sphere,  
$r$ is the radius of the sphere,   
$AB$ is the distance to the true horizon,   
$\angle ABO = 90°$,   
 $\angle \theta$ is the angle between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the disk forming the base of the cap.

The area of the cap $A\_c$ can be found according to this formula:
$$A\_c=2 \pi r^2(1-\cos \theta)$$

$\cos \theta$ in the right triangle $ABO$ is the ratio of the adjacent cathetus $OB = r$ to the hypotenuse $OA = r+H$, that is, $\cos \theta = \frac{r}{r+H}$, so
$$A\_c=2 \pi r^2(1-\frac{r}{r+H})$$

The percentage ratio $R\_\%$ of the visible area $A\_c$ to the full area of the sphere $A\_s$ is $$R\_\%=\frac{A\_c}{A\_s}\times 100\%$$
Since the full area of the sphere is $A\_s = 4\pi r^2$, we have:
$$R\_\%=\frac{2 \pi r^2(1-\frac{r}{r+H})}{4\pi r^2}\times 100\%$$
$2 \pi r^2$ above and below the line are canceled, so the final formula is like this:
$$\bbox[7px,border:2px solid red]{R\_\%=\frac{1}{2} \times (1-\frac{r}{r+H}) \times 100\%}$$

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Answering your question, from the altitude of 3,000 mi (4,800 km) one can see $21.4842\% \approx 21.5\%$ of the Earth's surface.

I have typed it in the Google calculator [here](https://www.google.com/search?hl=uk&sxsrf=ALeKk00J1r1Ne69vFyUpwt2koydEXPUhhg%3A1595378921793&ei=6YwXX7iCMPGvrgTE67NA&q=%281%20-%20%286371%20%2F%20%286371%20%2B%204800%29%29%29%20%2F%202%20*%20100%20%3D&oq=%281%20-%20%286371%20%2F%20%286371%20%2B%204800%29%29%29%20%2F%202%20*%20100%20%3D&gs_lcp=CgZwc3ktYWIQDDoHCAAQRxCwA1DxQ1ihU2CDwwFoAXAAeACAAWeIAewCkgEDMy4xmAEAoAEBqgEHZ3dzLXdpesABAQ&sclient=psy-ab&ved=0ahUKEwj4mqnP0d_qAhXxl4sKHcT1DAgQ4dUDCAw), so you can use it, just substitute the Earth radius of 6371 km and the altitude 4800 km for the numbers you would like.

Also, [here is a graph](https://www.google.com.ua/search?bih=578&biw=1229&hl=uk&sxsrf=ALeKk01aNnUprPs8sUec1B7NqMYdtP85Ow%3A1595387845941&ei=xa8XX-OCOePKrgSA_aSABA&q=f%28x%29%20%3D%20%281%20-%20%286371%20%2F%20%286371%20%2B%20x%29%29%29%20%2F%202%20*%20100&oq=f%28x%29%20%3D%20%281%20-%20%286371%20%2F%20%286371%20%2B%20x%29%29%29%20%2F%202%20*%20100&gs_lcp=CgZwc3ktYWIQAzoHCAAQRxCwA1D1Slj1SmCyU2gBcAB4AIABW4gBW5IBATGYAQCgAQGqAQdnd3Mtd2l6wAEB&sclient=psy-ab&ved=0ahUKEwij-dfu8t_qAhVjpYsKHYA-CUAQ4dUDCAw&uact=5) of how the visible area of the Earth ($r=6371 km$) changes with the altitude, scale and drag the graph with the mouse.

As for finding the altitude from which the given percentage of the Earth's surface is visible, here it is, it is just a transformation my *final formula* so that $H$ is on one side of the $=$ and all the rest is on the other, only $R$ here is just a ratio, no percent, so you will put 0.38, not 38% into the formula:
$$H=2r/(\frac{1}{R} - 2)$$
[The calculator](https://www.google.com/search?hl=uk&sxsrf=ALeKk016sSVLgXb78aRF71KvXYCWrqkyYA%3A1595392165810&ei=pcAXX4SBMeegrgTDqJ7YDg&q=2*6371%2F%28%281%2F0.38%29%20-%202%29%20%3D&oq=2*6371%2F%28%281%2F0.38%29%20-%202%29%20%3D&gs_lcp=CgZwc3ktYWIQDDoHCAAQRxCwA1DtmAlY7ZgJYMGoCWgBcAB4AIABhAGIAYQBkgEDMC4xmAEAoAEBqgEHZ3dzLXdpesABAQ&sclient=psy-ab&ved=0ahUKEwjE58f6guDqAhVnkIsKHUOUB-sQ4dUDCAw) says that to see 38% of the Earth's surface you need to be 20,174.83 km (12,536 mi) above the surface.


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