 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/10c/a/solution

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

The torque about the origin is

τ = r \times F = - \frac{km}{r ^{3}} r \times r = 0.

Since

\dot{L} = τ

L = m r \times \overset{r}{˙}

V (r) = - \frac{km}{r},

so the conserved energy is

E = \frac{1}{2} m ∣ \overset{r}{˙} ∣^{2} - \frac{km}{r} .

Directly,

\dot{E} = \overset{r}{˙} \cdot (m \overset{r}{¨} + km r / r^{3}) = 0

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