Solution
= Solution
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.