 Solution

In Newtonian gravity, the radial equation with conserved specific angular momentum is $\ddot{r}=-GM/r^2+h^2/r^3$. The substitution $u=1/r$ therefore gives the linear differential equationChoosing $\varphi =0$ along perihelion, its elliptic orbit solution isHere $p$ is the semilatus rectum. One can check the last equality directly from $r_{\mathrm{peri}}=p/(1+e)$, $r_{\mathrm{apo}}=p/(1-e)$, and $a_{\mathrm{orb}}=(r_{\mathrm{peri}}+r_{\mathrm{apo}})/2$.

To find Schwarzschild perihelion precession, write $u=u_0+\delta u$ in the relativistic equation and retain first order in $\mu /p$. The linear ordinary differential equation for the correction isThe constant and double-frequency forcing change the orbit's shape without accumulating an angular shift. The resonant $\cos \varphi$ term producessince $(d^2/d{\varphi }^2+1)(\varphi \sin \varphi )=2\cos \varphi$. The oscillating part of the perturbed orbit is consequently the first-order expansion ofSuccessive perihelia are separated by $2\pi /(1-3\mu /p)$, slightly more than a full revolution. Their forward angular advance is thusin radians per orbit. This leading expression applies when $\mu /p\ll 1$. For a perfectly circular orbit there is no uniquely located perihelion; the same expression describes the limiting mismatch of radial and azimuthal frequencies for a small nonzero orbital eccentricity.