= Solution
{c}
The initial transverse <velocity> gives $h=a(\sqrt{2k}/a)=\sqrt{2k}$. The initial <mechanical energy> is
$$
E=\frac m2\left(\frac{2k}{a^2}+\frac{2k}{a^2}\right)-\frac{mk}{a^2}
=\frac{mk}{a^2}.
$$
Since $\Phi(1/u)=-mku^2$ and $mh^2/2=mk$, the $u^2$ terms cancel in the conserved <energy> equation:
$$
mk(u')^2=\frac{mk}{a^2}.
$$
The initial radial <velocity> fixes the sign, because $\dot r=-hu'$ gives $u'(0)=1/a$. The <continuous> <derivative> $u'$ cannot change sign while its square remains $1/a^2$. As $u(0)=1/a$, this gives
$$
u=\frac{1+\theta}{a},\qquad
\boxed{r(\theta)=\frac{a}{1+\theta}\quad(\theta\ge0).}
$$
The future <orbit> is a <hyperbolic spiral>: it starts at $(a,0)$, turns counterclockwise and winds indefinitely towards the origin. Its radial <distance> decreases on every turn. The positive-$r$ continuation has $\theta>-1$, but only $\theta\ge0$ is needed after release.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv-q2-orbit.png]
{title=Future orbit in question 2, scaled by the initial radius}
To decide whether infinitely many turns require infinite time, use the conserved <angular momentum>:
$$
\frac{dt}{d\theta}=\frac{r^2}{h}
=\frac{a^2}{\sqrt{2k}(1+\theta)^2}.
$$
Integrating from $0$ gives
$$
t=\frac{a^2}{\sqrt{2k}}\frac{\theta}{1+\theta}.
$$
Consequently
$$
\boxed{t_* =\frac{a^2}{\sqrt{2k}}<\infty,\qquad
r(t)=a-\frac{\sqrt{2k}}a\,t\quad(0\le t<t_*).}
$$
\b[The particle reaches the origin in finite time], in the sense that $r\to0$ as $t\uparrow t_*$. Meanwhile $\theta=t/(t_*-t)\to\infty$. The <central force> is undefined at the origin, so these formulas describe motion up to collision. This is <finite-time collision in an inverse-cube central force>.
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