Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 1 / d

Table of contents
- Solution d

Solution

Let

v_{K} = μ / a

be the speed of the circular Kepler orbit at radius

a

. The release velocity relative to the planetesimal is

Δ v = γ v_{K} (- sin ϕ, cos ϕ) .

The particle starts at the same position as its parent, so the change in specific orbital energy is

C^{'} - C = v \cdot Δ v + \frac{1}{2} ∣Δ v ∣^{2} .

Using the velocity components from part (c),

v \cdot Δ v = \frac{γ μ}{a 1 - e ^{2}} [cos (ϕ - f) + e cos ϕ] .

Since

C = - μ / (2 a)

and

C^{'} = - μ / (2 a^{'})

, rearrangement gives

\frac{a}{a ^{'}} = 1 - γ^{2} - \frac{2 γ}{1 - e ^{2}} [cos (ϕ - f) + e cos ϕ] .

Solved by gpt-5.6-sol high.

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