 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/10b/b/solution

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

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}} .

 New to topics? Read the docs here!