Why doesn’t horizon distance move exactly proportional to the height of the observer? For instance if someone is 8 inches above the surface of the Earth, they can see approximately 1 mile to the horizon. However, if someone is viewing the horizon at an eye level of 5’5 they can only see about 3 miles out. If the height of the observer increases by a factor of about 8, from 8 inches to 65 inches, why does the distance they can see only increase by a factor of 3?(from 1 mile to 3 miles)
 A: Consider the following image showing the earth (with radius $R$)
and an observer at height $h$ above the ground.
The distance from the observer to the horizon is $s$.

The theorem of Pythagoras applied to the right triangle gives
$$R^2+s^2=(R+h)^2$$
With a little bit of algebra we get
$$R^2+s^2=R^2+2Rh+h^2$$
$$s^2=2Rh+h^2$$
Because $h$ is much smaller than $R$
we can neglect the $h^2$ on the right side.
$$s^2 \approx 2Rh$$
$$s \approx \sqrt{2Rh}$$
Now you see that $s$ is not proportional $h$,
but proportional to $\sqrt{h}$.
A: On earth, the distance to the horizon, say $d_h$ and the height of an observer, say $h_o$ cannot have a linear relationship $$d_h=\text{constant}\cdot h_o$$ or  proportional relationship you speak of, since this would assume the earth has some geometry other than spherical.
Instead, in reality, since the earth has curvature and is a sphere, you can show with a little trigonometry, $$d_h\propto \sqrt h_o$$

The exact relationship is $$d_h=r_e\cos^{-1}(\frac{r_e}{r_e+h_o})$$ where $r_e$ is the radius of earth, and this equation is usually simplified to the approximate relationship $$d_h\approx\text{1.22}\sqrt h_o$$ where $d_h$ is in miles and $h_o$ is in feet.
