# Proof of the conservation of angular momentum in a central force field

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?

• 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. Jan 8, 2018 at 18:24
• @ZachMcDargh That's true. But what about the fact that $f(r)$ blows up at r=0?
– SRS
Jan 8, 2018 at 21:41
• 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? Jan 8, 2018 at 23:24
• 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. Sep 28, 2022 at 18:48

• @garyp You mean $r=0$ is not physically accessible due to the finite size of the force providing agency?