= Solution
Substitute $r=1/u$ and $\dot r=-hu'$ into the conserved radial energy. <Differentiation> with respect to $\phi$ gives
$$
h^2u'u''-mu'+h^2uu'-3mh^2u^2u'=0.
$$
After division by $h^2u'$ and continuity through turning points,
$$
\boxed{u''+u=\frac m{h^2}+3mu^2.}
$$
Put $q=m/h^2$ and substitute
$$
u=q(1+\alpha)+A\cos[(1+\beta)\phi].
$$
Keeping the leading nontrivial powers of $m/h$ and terms through first order in $A$, the constant and oscillatory terms give
$$
q(1+\alpha)=q+3mq^2,
\qquad
-2\beta A=6mqA.
$$
Therefore
$$
\boxed{
\alpha=\frac{3m^2}{h^2},
\qquad
\beta=-\frac{3m^2}{h^2}.}
$$
The radial oscillation has angular period $2\pi/(1+\beta)$, so successive periapses advance by
$$
\boxed{\Delta\phi=-2\pi\beta
=\frac{6\pi m^2}{h^2}}
$$
to leading order. This is the <Schwarzschild perihelion precession>.
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