 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/4c/ii/solution

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

Now

E = u^{2} /2 - k / R < 0

, so the orbit is bound. Since

L = u b

, the initial energy is

E = \frac{L ^{2}}{2 b ^{2}} - \frac{k}{R} .

q

, the radial speed vanishes and the transverse speed is

L / q

E = \frac{L ^{2}}{2 q ^{2}} - \frac{k}{q} .

Equating these expressions and multiplying by

2 q^{2}

yields

(\frac{L ^{2}}{b ^{2}} - \frac{2 k}{R}) q^{2} + 2 k q - L^{2} = 0 .

Thus $A = L^{2} / b^{2} - 2 k / R$ , $B = 2 k$ , and $C = - L^{2}$ .

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