Newton's theorem of revolving orbits

Remarks

Proof idea: the new motion has the same radius at each time and angular momentum $r^2 \theta_2' = k h$, which is constant. Its radial equation $r'' - r \theta_2'^2 = f_2(r)$ follows from $r'' - r \theta_1'^2 = f(r)$, because $r \theta_2'^2 - r \theta_1'^2 = (k^2 - 1) \frac{h^2}{r^3}$. Equivalently, in the Binet equation with $u = 1/r$, the new orbit is $u(\theta_2 / k)$.

Example: start with a circular orbit of radius $\rho$ under any attractive force; then $\theta_2 = k \theta_1$ is again a circle, now traversed $k$ times as fast. Start with a Kepler ellipse under $f(r) = - \mu / r^2$, and take $k$ close to $1$. During one radial period the angle $\theta_2$ grows by $2 \pi k$, so the pericentre turns by $2 \pi (k - 1)$. So the new orbit is a turning ellipse, turning forward when $k > 1$ and backward when $k < 1$. The added force depends on the orbit through $h$. So the theorem gives one central force for each value of the angular momentum, and not one force that rotates all orbits at once. The added term is an inverse-cube force, attractive when $k > 1$ and repulsive when $k < 1$.

Going deeper

For a student

The theorem appears in Book I of Newton's Principia. Newton used it to explain the slow turning, or apsidal precession, of orbits such as that of the Moon, by an extra force. In terms of the effective potential, multiplying the angle by $k$ changes the centrifugal term $\frac{h^2}{2 r^2}$ to $\frac{k^2 h^2}{2 r^2}$. The added inverse-cube force compensates exactly, so the radial motion does not change. Together with Bertrand's theorem it shows how special the closed orbits of the Kepler problem are.