= Solution
Put $\mu=GM$. Balancing <Newton's law of universal gravitation> with the required <centripetal force> gives $mv_0^2/r_0=\mu m/r_0^2$, hence
$$
\boxed{v_0=\sqrt{\frac\mu{r_0}},\qquad E_0=\frac12mv_0^2-\frac{\mu m}{r_0}=-\frac{\mu m}{2r_0}.}
$$
The <circular Kepler orbit> is stable against small radial perturbations: its effective radial potential is $U_{\mathrm{eff}}=J^2/(2mr^2)-\mu m/r$, and at the circular value $J^2=\mu m^2r_0$, $U_{\mathrm{eff}}''(r_0)=\mu m/r_0^3>0$. Here $J$ is the conserved <angular momentum>.
Back to article page
