For the <Paczyński-Wiita potential>, circular balance gives
$$
r\Omega^2=\frac{d\Phi}{dr}=\frac{GM}{(r-r_S)^2}.
$$
Therefore
$$
\boxed{\Omega(r)=\frac{\sqrt{GM}}{r^{1/2}(r-r_S)}},
\qquad
\boxed{h(r)=r^2\Omega=\frac{\sqrt{GMr^3}}{r-r_S}}.
$$
The squared <radial epicyclic frequency> is
$$
\kappa_r^2=\frac1{r^3}\frac{d(h^2)}{dr}
=\frac{GM(r-3r_S)}{r(r-r_S)^3}.
$$
Circular orbits change from stable to unstable where $\kappa_r^2$ vanishes, so
$$
\boxed{r_{\rm ISCO}=3r_S}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / a / Solution

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{title2=$\Phi=-GM/(r-r_S)$}
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The Paczyński-Wiita potential is a Newtonian pseudo-potential that reproduces the Schwarzschild innermost stable circular orbit at $r=3r_S$.


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