A particle moves in a circular orbit of radius ' $r$ ' under a central attractive force,…

A particle moves in a circular orbit of radius ' $r$ ' under a central attractive force, $\mathrm{F}=-\frac{\mathrm{k}}{\mathrm{r}}$, where $\mathrm{k}$ is a constant. The periodic time of tis motion is
  1. $\mathrm{r}^{\frac{1}{2}}$
  2. $\mathrm{r}^{\frac{2}{3}}$
  3. $\mathrm{r}$
  4. $\mathrm{r}^{\frac{3}{2}}$

Solution

$\operatorname{mr} \omega^2=\frac{\mathrm{k}}{\mathrm{r}}$ $\omega^2=\frac{\mathrm{k}}{\mathrm{mr}}$ $\omega^2 \propto \frac{1}{r^2}$ $\omega \propto \frac{1}{\mathrm{r}}$ $\therefore \mathrm{T} \propto \mathrm{r}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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