Set $x=a/a_p$ and $c_I=\cos I$. At fixed $T_p$ and $I$,
$$
e^2=1-\frac{(T_p-x^{-1})^2}{4c_I^2x}.
$$
For a prograde orbit, the unsquared equation also requires $x\geq1/T_p$. Differentiation shows that the only stationary values occur at
$$
x=\frac1{T_p},
\qquad
x=\frac3{T_p}.
$$
The first is an endpoint maximum with $e=1$. The second is the interior minimum
$$
\boxed{e_{\min}^2=1-\frac{T_p^3}{27\cos^2I}},
\qquad
\boxed{a_{\min}=\frac{3a_p}{T_p}}.
$$
It lies on the physical locus when $T_p^3\leq27\cos^2I$. Equality makes the minimum circular. If $T_p^3>27\cos^2I$, a forbidden interval surrounds $3/T_p$ and the allowed locus splits into branches ending at $e=0$. As $a\to\infty$, $e\to1$.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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