 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-4/3a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For an attractive central force, the transverse acceleration vanishes. In polar coordinates,

2 \overset{r}{˙} \dot{θ} + r \ddot{θ} = 0 ⟹ \frac{d}{d t} (r^{2} \dot{θ}) = 0.

h = r^{2} \dot{θ}

is constant. Assume

h \neq = 0

, so angle can parameterize the orbit, and put

u = 1/ r

. Primes denote angle derivatives. The chain rule gives

\dot{θ} = h u^{2}, \overset{r}{˙} = - h u^{'}, \overset{r}{¨} = - h^{2} u^{2} u^{''}, r \dot{θ}^{2} = h^{2} u^{3} .

The outward radial component of the attractive force is

- f (u)

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

u^{''} + u = \frac{f ( u )}{m h ^{2} u ^{2}} .

The sign reflects

f

being the positive inward force magnitude. Zero angular momentum produces radial motion and is outside this angular parametrization.

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