 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-4/12c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The transverse equation of motion is

r^{- 1} d (r^{2} \dot{θ}) / d t = 0

, so

l = r^{2} \dot{θ}

is constant. With

u = 1/ r

\overset{r}{˙} = \frac{d r}{d θ} \dot{θ} = - u^{- 2} u^{'} l u^{2} = - l u^{'} .

Differentiating once more and substituting into the radial equation

\overset{r}{¨} - r \dot{θ}^{2} = - f (r)

l^{2} u^{2} (u^{''} + u) = f (1/ u) .

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