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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 10C / iv / Solution

To first order in

r_{0}

,

\frac{K}{( 1 - r _{0} u ) ^{2}} = K (1 + 2 r_{0} u) + O (r_{0}^{2}) .

The expanded orbit equation is therefore

u^{''} + ω^{2} u = K, ω^{2} = 1 - 2 K r_{0} .

Its required family of solutions is

u (θ) = X [1 + A cos (ω θ)], X = \frac{K}{1 - 2 K r _{0}}, ω = 1 - 2 K r_{0},

with any

0 < A < 1

. Successive periapses occur when

ω θ

changes by

2 π

, so their angular separation is

Δ θ = \frac{2 π}{1 - 2 K r _{0}} = 2 π (1 + K r_{0}) + O (r_{0}^{2}) .

The periapsis advance per revolution is consequently

2 π K r_{0} + O (r_{0}^{2})

.

Solved by gpt-5.6-sol high.

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