# Describing points on Earth's surface visible from space

I'd like to describe the points on Earth's surface that are visible to a viewer who is 100 miles above the North Pole. I'm assuming a spherical Earth and radius of 3960 miles if it's needed. My thoughts are that you'll be able see a portion of the Earth, all the way to a "horizon", and this horizon can be described by a circle. Is there any way to find the equation of this circle? All I really need is the radius to do this. I just don't seem to have enough information to find the distance from the viewer to the horizon, which seems like the most direct way forward. Is there something else I can do? Thanks so much!

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Drop down a dimension and try this with a point outside of a circle first. Can you do that? Find the distance between the two points (because you won't get a circle) – BeaumontTaz Aug 10 '14 at 0:04
Thanks for the hint! I'll give that a try. – steve Aug 10 '14 at 0:14
That suggestion has two-fold benefits. It's easier because it's down a dimension which leads to insight in the higher dimension, and in this case, the distance between the two points is the same as the diameter of the circle you're interested in, so you don't actually have to go back up a dimension. – BeaumontTaz Aug 10 '14 at 0:22
The picture here will help. And the problem gets easier if you say that the radius is 1 and scale the height of the person to $100/3960$. Then just scale back up when you have an answer. Look, in particular, at the sin and exsin lines in the image, and then work with the definition of exsec in the table. – BeaumontTaz Aug 10 '14 at 0:28
See if you can put together a typed up answer and answer you own question! Best of luck! – BeaumontTaz Aug 10 '14 at 0:32

First, draw a triangle with vertices the viewer, the center of the Earth, and a point on the Earth's surface. This is a right triangle because a radius and tangent line form a right angle at the point of tangency. The hypotenuse is 4060 miles, the longer leg is 3960 miles, and the short leg is 895.545 miles (from the Pythagorean theorem). The angle at the viewer can be found by the law of cosines and is equal to $77.257^\circ$. Next the radius of the circle we are interested in is the length of the altitude drawn to the side with length 4060 miles. The length of this altitude is $895.545\sin 77.257=873.487$ miles. Then the equation for the circle we are looking for is $x^2+y^2=762979$ and it is $3862$ miles above the center of the Earth.
Answer looks great! Those are the same numbers I got with a different method, too! In general you can show that for a sphere of radius $R$ and the astronaut at a height $h$ the angle that you're interested in is $\theta=\sec^{-1}(1+h/R)$. The radius of the circle is $r=R\sin\theta$ and the distance from the center is $d=R\cos\theta$ – BeaumontTaz Aug 10 '14 at 0:54