 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-316/1/a/solution

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

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:

h = ∣ r \times v ∣ = r^{2} \dot{θ} .

Taking the scalar product of

\ddot{r} = - μ r / r^{3}

with

\dot{r}

gives

\frac{d}{d t} (\frac{v ^{2}}{2} - \frac{μ}{r}) = \dot{r} \cdot \ddot{r} + \frac{μ}{r ^{3}} r \cdot \dot{r} = 0.

C = \frac{v ^{2}}{2} - \frac{μ}{r} .

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