There point masses each of mass m are at the corners of an equivalent triangle of side r. Their separations do not change when the system rotates about the centre of the triangle. For this, the time period of rotation must be proportional to

Refer Fig. O is the centre of mass of the system of 3 particles present A, B and C respectively

Here, $\mathrm{AD}=\mathrm{ABsin}{60}^{\circ }=\mathrm{r}\sqrt{3}/2$

$\mathrm{OD}=\frac{1}{3}\mathrm{AD}=\frac{1}{3}\times \frac{\mathrm{r}\sqrt{3}}{2}=\frac{\mathrm{r}}{2\sqrt{3}}$

$\therefore \mathrm{OB}={\mathrm{r}}^{\mathrm{\prime }}=\frac{\mathrm{OD}}{\sin {30}^{\circ }}=\frac{\mathrm{r}/2\sqrt{3}}{1/2}=\frac{\mathrm{r}}{\sqrt{3}}$

Gravitational force on particle at B due to particle

At C is ${\mathrm{F}}_{\mathrm{BA}}=\frac{\mathrm{Gm}\times \mathrm{m}}{2}$ along BC

Gravitational force on particle at B due to particle

At A is ${\mathrm{F}}_{\mathrm{BA}}=\frac{\mathrm{Gm}\times \mathrm{m}}{{\mathrm{r}}^2}$ along BA

Resultant force on particle at B will act along BO is F

$F_{r=}\sqrt{{\left(\frac{Gmm}{r^2}\right)}^2+{\left(\frac{Gmm}{r^2}\right)}^2+\frac{2Gmm}{r^2}\times \frac{Gmm}{r^2}\cos {60}^{\circ }}$

$=\frac{{\mathrm{Gm}}^2}{{\mathrm{r}}^2}\sqrt{3}$

This force is providing the centripetal force, so

$\frac{{\mathrm{Gm}}^2}{{\mathrm{r}}^2}\sqrt{3}=\mathrm{m}\frac{\mathrm{r}}{\sqrt{3}}\frac{4{\pi }^2}{{\mathrm{T}}^2}$

$\mathrm{T}=\sqrt{\frac{4{\pi }^2{r}^3}{3\mathrm{Gm}}}\text{(or)}\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^3}{\mathrm{m}}}$ $\mathrm{and} \mathrm{T}\propto {\mathrm{m}}^{-\frac{1}{2}}$