Solution
= Solution
Zero energy gives $e=1$. Since the periapsis is $d$, $r_0=2d$ and $l=\sqrt{2kd}$. Put $T=\tan(\theta/2)$; then
$$
r=d(1+T^2),\qquad \frac{dt}{dT}=\sqrt{\frac{2d^3}{k}}(1+T^2).
$$
Taking $t=0$ at periapsis yields
$$
\boxed{\theta(t)=2\tan^{-1}T(t),\qquad T^3+3T=3t\sqrt{\frac{k}{2d^3}}.}
$$