For $T_p=5/2$ and $I=0$, the fixed-<Tisserand parameter> curve begins at
$$
\frac a{a_p}=\frac25,
\qquad e=1.
$$
It falls smoothly to
$$
\frac a{a_p}=\frac65,
\qquad
e_{\min}=\sqrt{1-\frac{(5/2)^3}{27}}
=\sqrt{\frac{91}{216}}\simeq0.649,
$$
and then rises asymptotically back toward $e=1$ as $a/a_p\to\infty$. Since $T_p<3$, this coplanar locus never reaches a <circular Kepler orbit>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-316/2/c/solution