I'm working on a central force problem in which the potential is $$ U(r)=-\frac{\alpha}{r}\left( 1+ \frac{\beta}{r} \right) $$ I'm asked what condition has to be satisfied for the orbit to be closed.

I'm aware that Bertrand's theorem suggests that the form of the potential allows closed orbits and in other books like Marion's they have a condition that goes like: $$ \Delta{\phi}=2\int_{ r_{min}}^{r_{max}}\frac{\frac{L}{r^{2}}\text{d}r}{\sqrt{2\mu \left[E-\frac{L^{2}}{2\mu r^{2}}+\frac{\alpha}{r}\left(1+\frac{\beta}{r}\right) \right]}}=2\pi\frac{a}{b} $$ with $a$ and $b$ being natural numbers.

I already have the condition I'm asked (It's not a homework). But I don't understand where that $\frac{a}{b}$ having to be rational comes from. Is that a geometrical reasoning? It has to do with Bertrand's theorem? It looks somewhat like Lissajous curves and it may be something simple that I don't know.

  • $\begingroup$ Your question is not clear. Is the issue the interpretation of the ratio $a/b$? $\endgroup$ Jul 8, 2017 at 2:52
  • $\begingroup$ Yes, exactly. Let me edit the question $\endgroup$ Jul 8, 2017 at 2:53

1 Answer 1


The orbit is in 2d and "oscillates" between a minimum and a maximum $r$. The position in the plane if given by $(r(t),\phi(t))$ but here $t$ has been eliminated and you have $r(\phi)$.

As you go once from $r_{min}$ to $r_{max}(\phi)$, the body will advance along the orbit by an angular distance $\Delta \phi$. As you go from $r_{min}$ to $r_{max}$ and back to $r_{min}$, you advance by an angle $2\Delta \phi$.

To get a closed orbit, you must eventually come back to your starting point, meaning you must make an integer number $b$ of trips between $r_{min}$ and $r_{max}$ while advancing by an integer multiple $a$ of $2\pi$. This is the geometrical origin of the $2\pi a/b$ factor.

Edit: In answer to a comment, two situations are illustrated below. In both cases $r_{min}=1$ and $r_{max}=3$, and these values are shows as red thick lines. These values restrict the orbits to a ring of inner radius $1$ and outer radius $3$. The radius oscillates between $1$ and $3$ with some frequency $\omega_r$, as can be seen by the black lines in the figures.

enter image description here enter image description here

The parametric equations for the figures on the left and the right are, respectively, $$ r(\phi)=2+\cos\left(\sqrt{3}\phi\right)\, ,\qquad \hbox{and}\qquad r(\phi)=2+\cos\left(\phi\right) $$
In the first case, the ratio $\omega_\phi/\omega_r$ is not commensurate since $\sqrt{3}$ is irrational, and the orbit does not close. The best way to see this is to note that the start of the parametric curve is at $r=3,\phi=0$ but, at the end of the curve, $r\ne 3$. Because the ratio $\omega_\phi/\omega_r$ is irrational, the orbit would eventually densely fill the ring.

In the second case, on the other hand, the ratio $\omega_\phi/\omega_r$ is commensurate, and one can show (if we follow the curve through its $\phi$ evolution) that in fact it goes from $r_{min}\to r_{max}\to r_{min}$ exactly once when $\phi$ goes from $0\to 2\pi$.

  • $\begingroup$ @DavidLeonardoRamos, there is a simpler requirement for a closed orbit. Based on the way that gravitational potential energy is defined, a closed orbit occurs when $U(r)<0$. $\endgroup$ Jul 8, 2017 at 3:55
  • $\begingroup$ How's that? You mean always being negative? I don't see it clear. $\endgroup$ Jul 8, 2017 at 4:36
  • 1
    $\begingroup$ @DavidWhite I think you may be misunderstanding the question: closed as in closing on itself in 2d. $\endgroup$ Jul 8, 2017 at 6:33
  • $\begingroup$ @ZeroTheHero, can you elaborate? $\endgroup$ Jul 8, 2017 at 13:40
  • $\begingroup$ I think he means that is not sufficient that the orbit is bounded by $r_{min}, r_{max}$ $\endgroup$ Jul 8, 2017 at 13:45

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