Circular motion without dividing by zero?

Question

When studying e.g. central potentials, I have seen the following analysis done: \[ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \f{\dot r^2}{2}+V(r)=E\] differentating this w.r.t. $t$ gives us: \[ \dot r \ddot r +\dot r \p{V}{r}=0\] \[\ddot r=-\p{V}{r}\tag{1}\] In a circular orbit $\dot r=0$ and $\ddot r=0$ is it therefore valid to say that: $$\p{V}{r}=0$$ since the derivation of (1) relies on the division of $\dot r$ which in this case is zero. Please can you explain your answer?

Answer 1

$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} $You are right that your derivation is not valid for the special case of $\dot r=0$. However, the solution is simpler than you think.

For the special case of $\dot r=0$, simply plug that into the first equation:

$$\frac{\dot r^2}{2}+V(r)=E$$ $$0+V(r)=E$$ $$\p{V}{r}=0$$

Now it happens to be that when $\dot r=0$, and $\ddot r=0$, $\p{V}{r}=-\ddot r$ simplifies to $\p{V}{r}=0$.

So what we ended up showing was that $\p{V}{r}$ is continuous at $\dot r=0$. The hole at $\dot r=0$ from your original derivation is perfectly filled in by this special case, so we can combine them to make the statement that $\p{V}{r}=-\ddot r$ for all $\dot r$.