For a bound orbit let $E<0$ be the specific energy and $L$ the specific angular momentum. Introduce the dimensionless radius suggested in the question,
$$
x=-\frac{\mu}{b\Phi(r)}
=\frac{b+\sqrt{b^2+r^2}}b,
\qquad
r^2=b^2x(x-2).
$$
The radial energy equation becomes
$$
\dot x^2
=\frac{-2E(x-x_-)(x_+-x)}{b^2(x-1)^2},
$$
where $x_-$ and $x_+$ are the two <turning points>. Comparing coefficients gives
$$
x_-+x_+=2-\frac{\mu}{bE}.
$$
The time from periapsis to apoapsis is
$$
\begin{aligned}
\frac{T_r}{2}
&=\frac b{\sqrt{-2E}}
\int_{x_-}^{x_+}
\frac{x-1}{\sqrt{(x-x_-)(x_+-x)}}\,dx\\
&=\frac{\pi b}{\sqrt{-2E}}
\left(\frac{x_-+x_+}{2}-1\right)
=\frac{\pi\mu}{(-2E)^{3/2}}.
\end{aligned}
$$
Thus the radial <orbital period> is
$$
\boxed{T_r=\frac{2\pi GM}{(-2E)^{3/2}}}.
$$
It depends on $E$ but not on $L$, which is the defining isochrone property.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-320/2/b/solution