Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 34D / b / Solution

At a circular orbit, $V^{\prime }=0$. Differentiating the Schwarzschild effective potential at fixed $h$ givesA timelike circular orbit requires $r>3r_s/2$. Evaluating the second derivative after this substitution givesA stable nondegenerate minimum therefore requires $r>3r_s$. The boundary $r=3r_s$ is the marginal innermost stable circular orbit, not a strict stable minimum.