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

Treat the just-bound orbit as a parabolic Kepler orbit with $q\simeq a_p$. Its specific angular momentum is $h=\sqrt{2G{M}_{\ast }q}$, so its instantaneous angular speed is $h/r^2$. Equating this to the planet's mean motion $n_p=\sqrt{G{M}_{\ast }/a_p^3}$ givesThe parabolic orbit equation $r=2q/(1+\cos f)$ then gives $f\simeq 47^{\circ }$, or $50^{\circ }$ to the nearest ten degrees.

If the bodies were in conjunction at periapsis, Barker equation givesThe planetesimal is therefore at its maximum lead, approximatelyahead of the planet as seen from the star.