For a circular orbit, the <derivative> of
$$
V_{\rm eff}(r)=-\frac mr+\frac{h^2}{2r^2}-\frac{mh^2}{r^3}
$$
vanishes. Thus
$$
\frac m{r^2}-\frac{h^2}{r^3}+\frac{3mh^2}{r^4}=0,
\qquad
\boxed{h^2=\frac{mr^2}{r-3m}.}
$$
A real finite <angular momentum> therefore requires
$$
\boxed{r>3m.}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 38B / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/38b/b/i/solution