# A geometric question about spherical planet and being obscured by the horizon.

Is there a general formula to use to determine the amount that a distant object is obscured by the horizon?

Here are a couple of examples:

I can view the mountains that are 70 km away. I am at or very near to the ground. If we assume that the intervening terrain is negligible, how much of the mountains are obscured below the horizon?

A ship that stands 30m to the tip of the mast sails out onto the sea. At what distance does its mast finally vanish completely out of sight?

Of course, we need to assume a spherical Earth with a radius of 6371km.

For an observer at height $h$, the straight line distance $d$ to the horizon on a spherical planet of radius $r$ is given by Pythagoras $$r^2+d^2=(r+h)^2=r^2+2rh+h^2$$ so that for $h\ll r$ (a condition that we can take for granted even for an observer on top of Mount Everest, say) $$\tag1 d\approx\sqrt{2rh}.$$ Now if the observer is at $h_1$ and the observed object at $h_2$, then the maximum distance for the object to be visible is tha sum of the two horizon distances, $$\tag2 d\approx\sqrt{2rh_1}+\sqrt{2rh_2}.$$ If we solve this for $h_2$, we obtain how much is obscured from an object at distance $d>\sqrt{2rH_1}$ from an observer at height $h_1$: $$\tag3 h_2\approx\frac{(d-\sqrt{2h_1r})^2}{2r}.$$
Example: With $h_1=0$ (something you will never have in real life), the formula in $(3)$ can approximately be memorized as "8 inches per mile squared". But beware: This is an approximation for an approximation for an approximation.