Put $u=1/r$. Since $h=r^2\dot\theta$, the radial equation becomes the <Binet equation>
$$
u''+u=\frac\mu{h^2},
$$
where primes denote derivatives with respect to the polar angle. Choosing the angular origin at <periapsis> gives
$$
u=\frac\mu{h^2}(1+e\cos f).
$$
Writing the semi-latus rectum as $h^2/\mu=a(1-e^2)$ yields the <Kepler orbit>
$$
\boxed{r=\frac{a(1-e^2)}{1+e\cos f}}.
$$
Hence the <specific angular momentum> and <specific orbital energy> are
$$
\boxed{h=\sqrt{\mu a(1-e^2)}},
\qquad
\boxed{C=-\frac\mu{2a}}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-316/1/b/solution