Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 2 / c

Table of contents
- Solution c

Solution

For

T_{p} = 5/2

and

I = 0

, the fixed-Tisserand parameter curve begins at

\frac{a}{a _{p}} = \frac{2}{5}, e = 1.

It falls smoothly to

\frac{a}{a _{p}} = \frac{6}{5}, e_{m i n} = 1 - \frac{( 5/2 ) ^{3}}{27} = \frac{91}{216} ≃ 0.649,

and then rises asymptotically back toward

e = 1

a / a_{p} \to \infty

. Since

T_{p} < 3

, this coplanar locus never reaches a circular Kepler orbit.

Solved by gpt-5.6-sol high.

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