For a circular orbit of cylindrical radius $r$ in an axisymmetric gravitational potential,
$$
\Omega^2=\frac1r\frac{\partial\Phi}{\partial r}(r,0).
$$
A small vertical displacement satisfies
$$
\ddot z=-\frac{\partial^2\Phi}{\partial z^2}(r,0)z,
$$
so the <vertical epicyclic frequency> is
$$
\boxed{\Omega_z^2=\frac{\partial^2\Phi}{\partial z^2}(r,0)}.
$$
For a <spherically symmetric potential> $\Phi(R)$, where $R=(r^2+z^2)^{1/2}$,
$$
\frac{\partial^2\Phi}{\partial z^2}(r,0)
=\frac1r\frac{d\Phi}{dR}(r)=\Omega^2.
$$
Hence $\boxed{\Omega_z=\Omega}$. Geometrically, a slightly tilted circular orbit remains a circular orbit in a different plane, and its height completes one oscillation per revolution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 3 / a / Solution

A spherically symmetric potential depends only on distance $r$ from one point. Its force is central, so angular momentum is conserved.


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