Solution

First of all, we need to write angular momentum at the beginning and at the surface of the planet:
$$mvsin\alpha* r=muR$$
We have triangle with the opposite side $\rho$ and hypotenuse $r$, so:
$$rsin\alpha=\rho$$
Then we can express the speed at the surface
$$u=\frac{v\rho}{R}$$
We need to write law of conservation of energy
$$\frac{mv^2}{2}=\frac{mu^2}{2}-\frac{\gamma Mm}{R}$$
Let's substitute the speed we know at the surface
$$\frac{v^2R+2\gamma M}{2R}=\frac{v^2\rho^2}{2R^2}$$
et us express the square of the sighting parameter
$$\rho^2=v^2R^2+2\gamma MR$$
and we will get an answer
$$\boxed{\rho=R\sqrt{1+2γM/(Rv2)}}$$