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Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 10C / c / i / Solution

At the periapsis

r_{1}

r_{2}

, the given Kepler orbit satisfies

r_{1} = \frac{h ^{2}}{k ( 1 + e )}, r_{2} = \frac{h ^{2}}{k ( 1 - e )} .

Hence

e = \frac{r _{2} - r _{1}}{r _{1} + r _{2}}, h^{2} = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}} .

Therefore

h = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}}, C = 2 .

Solved by gpt-5.6-sol high.

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