 Solution

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.