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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 4 / 10B / b / Solution

Zero energy gives

e = 1

. Since the periapsis is

d

,

r_{0} = 2 d

and

l = 2 k d

. Put

T = tan (θ /2)

; then

r = d (1 + T^{2}), \frac{d t}{d T} = \frac{2 d ^{3}}{k} (1 + T^{2}) .

Taking

t = 0

at periapsis yields

θ (t) = 2 tan^{- 1} T (t), T^{3} + 3 T = 3 t \frac{k}{2 d ^{3}} .

Solved by gpt-5.6-sol high.

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