= Solution
Set $\theta=0$ initially and choose its positive sense to agree with the initial tangential motion. The inward radial component is $\dot r(0)=-4\cos(\pi/3)=-2$, while the tangential <speed> is $4\sin(\pi/3)=2\sqrt3$. Hence
$$
\boxed{l=2\sqrt3,\quad u(0)=1,\quad u'(0)=1/\sqrt3.}
$$
Substitution in the <Binet equation> gives $12(u''+u)=3+9u$, or $u''+u/4=1/4$. This is an <orbit equation for combined inverse-square and inverse-cube attraction>. Solving the constant-coefficient equation and applying both initial conditions gives
$$
\boxed{u(\theta)=1+\frac2{\sqrt3}\sin(\theta/2).}
$$
On $0\leq\theta\leq2\pi$, the sine is nonnegative, so $r$ stays finite and positive. At $\theta=2\pi$, $u=1$, hence $r=1$ and the particle is back at its original spatial position after one revolution. But $u'(2\pi)=-1/\sqrt3$, so its radial <velocity> is now $+2$, rather than the original $-2$. \b[The return is not a periodic return in position and <velocity>.]
After that revolution, the first zero of $u$ is at $\theta=8\pi/3$, where $\sin(\theta/2)=-\sqrt3/2$. The physical branch therefore has $r\to\infty$ as $\theta\uparrow8\pi/3$; it must not be continued into negative $u$. Since $dt/d\theta=1/(lu^2)$ and $u$ has a simple zero, reaching infinity takes infinite time. As a check, the specific <potential energy> is $\Phi(r)=-3/r-9/(2r^2)$ and the conserved energy per unit <mass> is $4^2/2+\Phi(1)=1/2$, giving escape <speed> $1$ at infinity, also obtained from $-lu'(8\pi/3)=1$.
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