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

Solution

q ≃ a_{p}

h = 2 G M_{*} q

, so its instantaneous angular speed is

h / r^{2}

. Equating this to the planet's mean motion

n_{p} = G M_{*} / a_{p}^{3}

gives

r = (2 q a_{p}^{3})^{1/4} ≃ 2^{1/4} a_{p} .

The parabolic orbit equation

r = 2 q / (1 + cos f)

then gives

f ≃ 4 7^{\circ}

, or

5 0^{\circ}

to the nearest ten degrees.

If the bodies were in conjunction at periapsis, Barker equation gives

n_{p} t = 2 [tan \frac{f}{2} + \frac{1}{3} tan^{3} \frac{f}{2}] ≃ 3 7^{\circ} .

The planetesimal is therefore at its maximum lead, approximately

f - n_{p} t ≃ 1 0^{\circ}

ahead of the planet as seen from the star.

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