 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/10c/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

V = - km / r

, one has

d V / d r = km u^{2}

, so

u^{''} + u = K, K = \frac{k m ^{2}}{ℓ ^{2}} .

After choosing the angular origin, the solution is

u = K (1 + e cos θ), r = \frac{p}{1 + e c o s θ}, p = \frac{1}{K} .

For

0 \leq e < 1

, using

r = x^{2} + y^{2}

and

r + e x = p

gives

(1 - e^{2}) (x + \frac{e p}{1 - e ^{2}})^{2} + y^{2} = \frac{p ^{2}}{1 - e ^{2}} .

This is an ellipse, with the force centre at one focus, as described by a Kepler orbit.

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