For $e=0$, regard
$$
T_p(x)=\frac1x+2\cos I\sqrt x.
$$
Its minimum occurs at $x=(\cos I)^{-2/3}$ and equals $3(\cos I)^{2/3}$. A circular orbit is therefore possible for $T_p=5/2$ only if
$$
3(\cos I)^{2/3}\leq\frac52.
$$
The smallest required <orbital inclination> is consequently
$$
\boxed{I_{\min}=\arccos\left(\frac56\right)^{3/2}\simeq40.5^\circ}.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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