Max possible line of sight on earth This question might be too simple for this site, but I'm wondering what the max distance is that I would be able to see, given that I'm on earth at no more than 100m above sea level. Assume that the object I'm trying to see is based at 0, and 100m tall above sea level, an arbitrary distance away.
Essentially, given the shape and dimensions of earth, and the 2 constraints above, how far away would I be able to see a building 100m tall (from sea level).
See is defined as, eyes with perfect eyesight being able to detect (straight line) to the building while begin themselves at 0-100m from sea level. (Probably best if they are at 100m up?)
I've tried to be precise in my question, but if there is any ambiguity, please feel free to leave a comment?
 A: 
Suppose first you are at sea level at point $C$. The building is at $B$ a height $0.1$km. You can just see the top of the building if $BC$ is perpendicular to $OC$ (where $O$ is the earth's centre). So we have $BC^2=OB^2-OC^2=(R+0.1)^2-R^2$ We have $R=6371$km approx, so $BC=36$km approx. 
If you are 100m above the surface at point $X$, then by symmetry $BX$ is twice this distance or approx 71km.
In practice the distances would be less if you wanted to be sure it was the building you were seeing.
A: Call the height of the first point $h_1$, the height of the second point $h_2$. The total height of the first point, assuming the earth spherical and its radius $R_0$, the total height of the first point is $H_1=R_0+h_1$. For the second point you have: $H_2=R_0+h_2$. Because the LOS (line of sight) is tangent to the orizon, you have: $d_1=\sqrt{(H1^2-R_0^2)}$ and $d_2=\sqrt{(H2^2-R_0^2)}$ where $d_1$ and $d_2$ are the distances of the points to the horizon. If $d_1+d_2\ge D$ the LOS is free and you can see the point $2$ from the point $1$ (and vice versa). If $d_1+d_2\lt D$, you don't. $D$ is the distance between the two points in a straight line
