I'm working on a homework question in which I am supposed to discuss the types of motion that orbits may have under a certain potential. What I've done so far is draw the energy diagram for such a situation, and looked at the different types of motion depending on the initial energy.

What I'm wondering though is if it's possible to figure out if the elliptical orbits that the potential allows are precessing or not. Furthermore, I want to know if I can tell whether the elliptical orbits are closed (i.e., $r(t) = r(t+\Delta t)$ for some $t$) from my energy diagram and/or the potential given?

I'm trying to answer this question without being too quantitative. I'm also just looking for pointers in the right direction, not necessarily a full answer.

I've also graphed the corresponding energy diagram, and I get something that I'm not quite sure how to interpret. Essentially, there seems to be a certain tuning of the parameters where there can be a bound elliptical orbit with $E \gt 0$. I was under the impression that a bound orbit requires negative energy (in the sense that we define zero energy to be at infinity). Am I missing something here?

Update: I think my question about closing orbits is mostly answered. I was just wondering if something about my energy diagram could point to the fact that a certain elliptical orbit could precess. In particular, I've added two points on my graph that shows an initial given energy, which is positive. If I understand the diagram correctly, the particle should oscillate between the radii of these two points on the graph. However, unlike with regular $1/r$ potentials, these orbits seem to be bound, yet still have positive energy. Is this allowed? I was under the impression that the energy needs to be negative in order to stay bound, but it seems like it's possible to have a positive energy and still have a bound elliptical orbit.

Yukawa potential energy with centrifugal contribution

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    $\begingroup$ An elliptic(al) orbits is by definition closed. What you are describing is a bounded orbit, which may or may not be closed. $\endgroup$
    – Qmechanic
    Commented Oct 8, 2017 at 21:33
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    $\begingroup$ You might also be interested in Bertrand's theorem. $\endgroup$
    – diracula
    Commented Oct 8, 2017 at 21:47
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    $\begingroup$ @Qmechanic There's a good chance I'm mixing up my terms, but I thought that if an orbit is a precessing ellipse with an irrational fraction of $2\pi$, it won't close? $\endgroup$
    – Germ
    Commented Oct 8, 2017 at 22:15
  • $\begingroup$ @Qmechanic is right, but in reality we do call approximately-elliptical orbits "elliptical" all the time. Since gravity is long-ranged and unshielded, no real orbit can be perfectly described with a finite string of adjectives. Stating unambiguously that Mercury's orbit is elliptical. for example seems to be quite acceptable here for higher-rep users (although that answer is problematic itself). $\endgroup$
    – uhoh
    Commented Oct 9, 2017 at 2:18

1 Answer 1


It is not enough to have $r(t)=r(t+\Delta t)$: you must simultaneously have $\phi(t)=\phi(t+\Delta t)$ where $\phi$ is the angular degree of freedom. To put it differently, your need the radial and angular periods $T_r$ and $T_\phi$ to be commensurate, i.e. their ratio must be a rational number $n/m$.

This is because you need the orbit, which is described by the coordinates $(r,\phi)$, to exactly retrace itself. If $T_r/T_\phi$ is irrational, then the orbit will never close exactly on itself, and will eventually fill every point of $(r,\phi)$ space between $r_{min}$ and $r_{max}$.

Not every type of potential can produce closed orbits. This is discussed, for instance, in Marion and Thornton's Classical Dynamics and in multiple textbooks at the same level.

Edit: Simply knowing the energy is not enough.

For instance, the potential $$ V(r)=-\frac{k}{r}-\frac{\lambda}{2r^2} $$ leads to the effective potential $V_{eff}=\frac{\ell^2-\lambda m}{2 m r^2}-\frac{k}{r}$.

If $\lambda<\ell^2/\mu$ and $-mk^2/(2(\ell^2-m\lambda))\le E\le 0$ so we have bound states, the motion is described by a precessing ellipse, with $$ \phi=\phi_0-\frac{1}{\beta}\arccos\left(\frac{k-u(\beta^2\ell^2/{m})}{\sqrt{k^2+(2\beta^2\ell^2E/m)}}\right) $$ with $u=1/r$ and $\beta^2=1-m\lambda/\ell^2>0$ or, in terms of $r$: $$ r(\phi)=\frac{r_0}{1-e\cos\beta(\phi-\phi_0)} $$ for some $r_0$ and $e$. $r_0$ and $e$ can be expressed in terms of the other quantities in the problem. If $\beta$ is rational then the ratio of angular to radial period is given by $T_\phi/T_r=\beta$ and is also rational.

For instance, using $r_0=1$, the first two figures below show closed orbits, with $(e,\beta)=(\frac{4}{5},\frac{4}{5})$ and $(\frac{4}{5},\frac{1}{3})$ respectively.

enter image description here enter image description here

The figure below shows an orbit that is not closed, with $(e,\beta)=(\frac{4}{5},\frac{2+\pi}{2\pi})$.

enter image description here

Thus, even if a potential can in principle accommodate closed orbits, the orbits need not always be closed. By Bertrand's theorem, only some potential can accommodate closed orbits.

The energy diagram will guarantee that you have bounded motion, but not necessarily closed orbits.

  • $\begingroup$ Thanks for the answer. You're right, I did forget about the angular degree of freedom, so thanks for pointing that out. I imagine then that it won't be enough to look at just the energy diagram then, correct? I'd have to find the radial and angular trajectory before being able to compute this ratio. $\endgroup$
    – Germ
    Commented Oct 8, 2017 at 22:13
  • $\begingroup$ @Jeremy You are correct. Of course the energy of your particle must be such that it has two turning points in $r$ but also you need to look at $\phi$. The orbit equation is often expressed as a DE in $dr/d\phi$ and that can be a good starting point. $\endgroup$ Commented Oct 8, 2017 at 22:49
  • $\begingroup$ Okay, fantastic. I'll look into analyzing it in this manner. By the way, is there anything to be said for the fact that there seems to be a bound ellipse where $E \gt 0$? When I compare this to the usual $1/r$ potential, the energy for bound orbits is always negative and asymptotes to zero from below, while this function asymptotes to zero from above. I'm trying to figure out if there's something to glean from this? $\endgroup$
    – Germ
    Commented Oct 8, 2017 at 22:59
  • $\begingroup$ This is helpful to define the term "closed", although if it simply passes through the same point twice but with a different velocity, I'm not sure that would satisfy everyone. But this is not yet an answer to the question "How to tell if an orbit is closed given the potential?" and just mentioning Marion without including the information here is a link-only answer sans link. Can you add some actual discussion of what one would do when looking at a potential to help decide if it can have closed orbits? $\endgroup$
    – uhoh
    Commented Oct 9, 2017 at 2:08
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    $\begingroup$ @uhoh yeah I figured we were talking at cross-purposes. $\endgroup$ Commented Oct 9, 2017 at 13:59

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