Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 316 / 2 / v

Table of contents
- Solution v

Solution

Eliminate

a

using the Poynting–Robertson drag invariant,

a = C e^{4/5} / (1 - e^{2})

. The eccentricity rate becomes

\overset{e}{˙} = - \frac{5 A}{2 C ^{2}} e^{- 3/5} (1 - e^{2})^{3/2} .

Since the point-star limit corresponds to

a \to 0

and

e \to 0

, the inspiral time under Poynting–Robertson drag is

t (e_{0}) = \frac{2 C ^{2}}{5 A} \int_{0}^{e_{0}} \frac{u ^{3/5}}{( 1 - u ^{2} ) ^{3/2}} d u, C = a_{0} (1 - e_{0}^{2}) e_{0}^{- 4/5} .

For

e_{0} \to 0

, the integral is

(5/8) e_{0}^{8/5} + o (e_{0}^{8/5})

, recovering

a_{0}^{2} / (4 A)

despite the apparent singularity of

C

.

For

e_{0} \to 1

, use

C ≃ 2 q_{0}

and

1 - e_{0} = 2 q_{0} / (Q_{0} + q_{0}) ≃ 2 q_{0} / Q_{0}

. The endpoint asymptotic integral is

[2 (1 - e_{0})]^{- 1/2}

t ≃ \frac{4}{5 A} Q_{0}^{1/2} q_{0}^{3/2} .

Most of this long lifetime is spent near the initially large apocentre distance, before substantial circularization. The result assumes weak drag and a valid orbital average; near-parabolic release can violate that assumption.

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