 Solution

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

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

ℓ = m r^{2} \dot{θ} .

Put

u = 1/ r

. Since

\dot{θ} = ℓ u^{2} / m

\overset{r}{˙} = - \frac{ℓ}{m} u^{'}, \overset{r}{¨} = - \frac{ℓ ^{2}}{m ^{2}} u^{2} u^{''}, r \dot{θ}^{2} = \frac{ℓ ^{2}}{m ^{2}} u^{3},

where primes now denote differentiation with respect to

θ

. Substitution into the radial equation gives the Binet equation

u^{''} + u = \frac{m}{ℓ ^{2} u ^{2}} \frac{d V}{d r}_{r = 1/ u} .

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