Q 11-07-146JEE MainJEE Main 2021 (18 Mar, Shift 2)Medium
A particle of mass $m$ moves in a circular orbit under the central potential field, $U(r) = \dfrac{-C}{r}$, where $C$ is a positive constant. The correct radius-velocity graph of the particle's motion is:
Answer: (A) see figure
The force is $F = -\dfrac{dU}{dr}$, of magnitude $\dfrac{C}{r^2}$, directed towards the centre. It supplies the centripetal force:
$$\frac{C}{r^2} = \frac{mv^2}{r} \Rightarrow r = \frac{C}{mv^2}$$
So $r \propto \dfrac{1}{v^2}$: $r$ falls steeply as $v$ increases and approaches the $v$-axis asymptotically, which is graph (1).
Solution by Sreeraj P, M.Sc Physics