Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 12C (Dynamics and Relativity)

The transverse equation of motion is $r^{-1}d(r^2\dot{\theta })/dt=0$, sois constant. With $u=1/r$,Differentiating once more and substituting into the radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ gives the Binet equation

Take $\theta =0$ initially. The velocity components areso $l=2\sqrt{3}$. Since $f(1/u)=3u^2+9u^3$, the orbit equation becomesThusThe data $u(0)=1$ and $\dot{r}=-l{u}^{\prime }$ give $A=0$, $B=2/\sqrt{3}$, henceAt $\theta =2\pi$, $u=1$ again, so the particle returns to its initial position after one revolution; now $u^{\prime }=-1/\sqrt{3}$, so it is moving outward. Subsequently $u$ first reaches zero at $\theta =8\pi /3$, where $r=1/u\to \infty$. It therefore flies off to infinity.