 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-4/10b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

E = \frac{m}{2} (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - \frac{GM m}{r} = \frac{m h ^{2}}{2} ((u^{'})^{2} + u^{2}) - GM m u .

Put

γ = GM / h^{2}

and

φ = θ - θ_{0}

. From part (i),

u = γ (1 + e cos φ)

and

u^{'} = - γ e sin φ

. Consequently

(u^{'})^{2} + u^{2} = γ^{2} (1 + 2 e cos φ + e^{2}),

and the angular terms cancel against the gravitational potential energy:

E = \frac{m G ^{2} M ^{2}}{2 h ^{2}} (e^{2} - 1) .

For a nonradial Kepler orbit, negative, zero and positive mechanical energy correspond respectively to

e < 1

e = 1

and

e > 1

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