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WW

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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 EQN:R^2+D^2=(R+h)^2. (by Pythagoras)

We can simplify that to EQN:D^2=2hR+h^2, and hence

EQN:D=\sqrt(2hR+h^2). Using the circumference of the Earth as

EQN:D=\sqrt{2hR+h^2}. Using the circumference of the Earth as

40 million metres (the original definition of the metre), and

realising that EQN:h^2 is negligable compared with EQN:2hR,

this simplifies to EQN:D=3568.25sqrt(h) metres, or about

~realising that EQN:h^2 is negligable compared with EQN:2hR,

this simplifies to EQN: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 EQN:1.9267\sqrt(h)

metres (accurate to 4% - it's actually EQN:1.9267\sqrt{h}

Nautical Miles).

The exact same calculation can be used to compute orbital velocity.

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CategoryMaths