A beam of light, traveling around a planet Some time ago, I happened to have found one problem:

The refractive index of the planet's atmosphere decreases with height above the surface  under the following law:
$$n=n_0-\alpha h$$
with condition $\frac{\alpha h}{n_0}\ll1$
Find the altitude $H$ above the surface of the planet, where a ray, emitted horizontally, will bypass the planet, remaining at this elevation. Radius of the planet is $R$.

Since optics is not my field I decided to ask here. What would be the height $H$ for the Earth? (approximately)
 A: If I understand the question correctly, you are asking at what height light would be able to travel in a circle around Earth due to refraction. This would be the height at which the optical path length ($OPL$) of this circular trajectory as a function of height has a local minimum. This optical path length would be given by this formula:
$$
OPL=2\pi r \cdot n=2\pi (R+h)(n_0 - \alpha h)
$$
where $R$ is the radius of Earth. If we differentiate this with respect to $h$ and set to $0$, we get:
$$
0=\frac{dOPL}{dh}=2\pi(n_0-\alpha(R+2h)) \Rightarrow h=\frac{n_0-\alpha R}{2\alpha}
$$
The problem now is that we don't have a value for $\alpha$. Finding one would be difficult bcause your formula for $n$ isn't very realistic. A more realistic (although very simplified) formula for $n$ is this:
$$
n=1+Ae^{-Bh} \Rightarrow OPL=2\pi (R+h)(1+Ae^{-Bh})
$$
which would give:
$$
0=\frac{dOPL}{dh}=2\pi(1+(1-B(R+h))Ae^{-Bh})\\
\Rightarrow e^{Bh}=A(B(R+h)-1)
$$
If we make the assumption that $h\ll R$ and use the values $A=0.00029$ and $B=0.000143$ we get:
$$
h\approx\frac{\ln{(A(BR-1))}}{B}=\frac{\ln{(0.00029(0.000143\cdot6.371\cdot 10^6-1))}}{0.000143}\approx -9300 \,\mathrm{m}
$$
The negative number means that this cannot happen on Earth. We would need a significantly thicker atmosphere, alternatively a greater planet radius.
