Question

Four particles of equal mass are moving round a circle of radius r due to their mutual gravitational attraction. Find the angular velocity of each particle?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{all}\mathrm{particles}\mathrm{will}\mathrm{remain}\mathrm{diametrically}\mathrm{opposite}\mathrm{so}\mathrm{that}\mathrm{the}\mathrm{force}\mathrm{on}\mathrm{each}\mathrm{particle}\mathrm{will}\mathrm{be}\mathrm{directed}\mathrm{along}\mathrm{the}\mathrm{radius}. \\ \mathrm{Suppose}\mathrm{we}\mathrm{consider}\mathrm{the}\mathrm{motion}\mathrm{of}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{particles}. \\ \mathrm{The}\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{particle}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{F}=\frac{{\mathrm{Gm}}^2}{4{\mathrm{r}}^2} \\ \mathrm{If}\mathrm{the}\mathrm{speed}\mathrm{is}\mathrm{v},\mathrm{then} \\ \mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}} \\ \mathrm{According}\mathrm{to}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{law}, \\ \frac{{\mathrm{Gm}}^2}{4{\mathrm{r}}^2}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{4\mathrm{r}}} \\ \mathrm{Angular}\mathrm{velocity},\mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{GM}}{4\mathrm{r}}}\times \frac{1}{\mathrm{r}} \\ =\sqrt{\frac{\mathrm{GM}}{4{\mathrm{r}}^3}} \end{gathered}$$