Update after @M. Enns's answer Consider a circular orbit of radius $a$ passing through the centre of a central force is given by the equation $r=2a\cos\theta$. Then using the orbit equation one can show that the force varies as $$\textbf{F}(|\textbf{r}|)=-\frac{k}{r^5}\hat{\textbf{r}}.\tag{1}$$

But an orbit passing through the centre of force O will have zero angular momentum when it passes through O, and nonzero when it doesn't pass through O. This shows that the angular momentum is changing throughout the motion. I wished to attach the figure but I couldn't. This problem is solved in the book Theoretical Mechanics by Spiegel.

But here is the fallacy. The angular momentum should be conserved for motion under any central force. However, this doesn't seem to work in this case. But the proof of conservation of angular momentum is based on the simple observation that $$\frac{d\textbf{L}}{dt}=\textbf{r}\times\textbf{F}=0\tag{2}$$ since $\textbf{F}=f(r)\hat{\textbf{r}}$ for any central force.

Is the proof (2) reliable at $r=0$ where the force (1) become singular?

  • 1
    $\begingroup$ If it passes through the origin, is it not the case that particle's motion will be in the $\hat{r}$ direction? Then, the position vector and the velocity vector will be always parallel, so that $r \times F$ still vanishes. $\endgroup$ Jan 8, 2018 at 18:24
  • $\begingroup$ @ZachMcDargh That's true. But what about the fact that $f(r)$ blows up at r=0? $\endgroup$
    – SRS
    Jan 8, 2018 at 21:41
  • $\begingroup$ Of course the situation only occurs when $L=0$ identically. In that case, the theory is pathological at the point when the two objects have the same position, and so it cannot hold in that regime. What is your question? $\endgroup$ Jan 8, 2018 at 23:24
  • $\begingroup$ The force is not defined at the origin. I believe that an answer becomes ill posed at the origin. For example, the derivative is not defined. $\endgroup$ Sep 28, 2022 at 18:48

1 Answer 1


I've looked at this problem, and the preceding pages in Spiegel's book, and it would seem to me that any particle travelling in this orbit would have an infinite velocity at O, at least if the velocity was non-zero anywhere else in the orbit since the potential is infinitely negative when r=0 so the kinetic energy would have to be infinite. Wouldn't it be the same situation as two massive particles moving towards each other until they are zero distance apart?

If that's right than your angular momentum at O becomes a zero times infinity situation...

Here's a Q&A from Quora that addresses a particle moving through the origin in central force motion.

  • $\begingroup$ In other words, the orbit posited is not physical? It's seems certainly true that the orbit is practically impossible. Some structure would be needed to generate the force, and that structure would occupy some volume that would exclude the orbiting object. $\endgroup$
    – garyp
    Jan 8, 2018 at 18:16
  • $\begingroup$ @garyp You mean $r=0$ is not physically accessible due to the finite size of the force providing agency? $\endgroup$
    – SRS
    Oct 24, 2018 at 7:36
  • $\begingroup$ Yes, that's what I was saying. $\endgroup$
    – garyp
    Oct 24, 2018 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.