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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 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=3568.25\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} Nautical Miles). The exact same calculation can be used to compute orbital velocity. ---- CategoryMaths