= Solution
By spherical symmetry, place the geodesic in the equatorial plane $\theta=\pi/2$. With
$$
f(r)=1-\frac{2M}{r},
$$
the <geodesic Lagrangian> is
$$
2\mathcal L
=-f\dot t^{\,2}+f^{-1}\dot r^{\,2}+r^2\dot\phi^{\,2}
=Q,
$$
where $Q$ is the constant squared norm of the tangent: $Q=-1$ for a unit-speed timelike geodesic and $Q=0$ for a null geodesic. The cyclic coordinates $t$ and $\phi$ give the conserved quantities
$$
E=f\dot t,
\qquad
L=r^2\dot\phi.
$$
Substitution into the normalization gives
$$
Q=-\frac{E^2}{f}+\frac{\dot r^{\,2}}f+\frac{L^2}{r^2}.
$$
Multiplying by $f$ and rearranging produces
$$
\boxed{
\frac12\dot r^{\,2}+V(r)=\frac12E^2,
\qquad
V(r)=\frac12\left(1-\frac{2M}{r}\right)
\left(\frac{L^2}{r^2}-Q\right)
}.
$$
This is the <timelike geodesic effective potential> form, with the same derivation applying to any affinely parametrized causal geodesic.
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