The distance to the horizon is calculated as follows:

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 $D=3568.25sqrt(h)$ 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.

CategoryMaths
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