Past exam of the physics B course of the University of Cambridge / 2015 / Paper 2 / Section A / A3

Interpret the two radial turning values as $r_{\min }=r_0$ and $r_{\max }=4r_0$; the printed maximum/minimum order is reversed. The potential is $V(r)=A{r}^2/2$. At either turning point,Equating these two energies gives $L^2=mA{r}_{\min }^2{r}_{\max }^2$. A radial impulse changes the radial velocity and energy but exerts no torque about the centre, so it preserves $L$. For radial-impulse circularization of a harmonic central orbit, the circular orbit satisfies $L^2/(m{r}_c^3)=A{r}_c$. HenceThe original orbit crosses this radius. Cancelling its radial velocity there leaves the tangential speed required for the circular orbit; the impulse is not applied at an original radial turning point.