Distance To The Horizon |
|
Suppose you are at height h and the radius of the Earth is R. Then we have a right-angled triangle, with the hypotenuse being from your location to the centre of the Earth, and the right-angle at the "Horizon Point", the point of maximum visibility. Calling the distance from you to that point D, we have $R^2+D^2=(R+h)^2.$ (by Pythagoras)
We can simplify that to $D^2=2hR+h^2,$ and hence $D=\sqrt{2hR+h^2}.$ Using the circumference of the Earth as 40 million metres (the original definition of the metre), and realising that $h^2$ is negligable compared with $2hR,$ this simplifies to $D=3568.25\sqrt{h}$ metres, or about 2 Nautical Miles times the square root of the height in metres (accurate to 4% - it's actually $1.9267\sqrt{h}$ Nautical Miles).
The exact same calculation can be used to compute orbital velocity.