Treat the just-bound orbit as a <parabolic Kepler orbit> with $q\simeq a_p$. Its specific angular momentum is $h=\sqrt{2GM_*q}$, so its instantaneous angular speed is $h/r^2$. Equating this to the planet's mean motion $n_p=\sqrt{GM_*/a_p^3}$ gives
$$
\boxed{r=(2qa_p^3)^{1/4}\simeq2^{1/4}a_p}.
$$
The parabolic orbit equation $r=2q/(1+\cos f)$ then gives $f\simeq47^\circ$, or $50^\circ$ to the nearest ten degrees.

If the bodies were in conjunction at periapsis, <Barker equation> gives
$$
n_pt=\sqrt2\left[\tan\frac f2+\frac13\tan^3\frac f2\right]
\simeq37^\circ.
$$
The planetesimal is therefore at its maximum lead, approximately
$$
\boxed{f-n_pt\simeq10^\circ}
$$
ahead of the planet as seen from the star.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 4 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/4/i/solution