Distance To The Horizon 

Suppose you are at height h and the radius of the Earth is R. Then we have a rightangled triangle, with the hypotenuse being from your location to the centre of the Earth, and the rightangle 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.