## Most recent change of DistanceToTheHorizon

Edit made on July 27, 2015 by ColinWright at 21:20:01

Deleted text in red / Inserted text in green

WW
WM
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