Three particles each of mass ‘m’ are located at the three vertices of an equilateral triangle of side ‘a’. If the system is revolving in a uniform circle due to their mutual force of attraction, then the angular velocity of each particle is

The force of attraction on body C due to bodies at A and B are

${\text{F}}_{\text{1}}=\frac{{\text{Gm}}^{\text{2}}}{{\text{a}}^{\text{2}}}{\text{ F}}_{\text{2}}=\frac{{\text{Gm}}^{\text{2}}}{{\text{a}}^{\text{2}}}$

$\text{F}=\sqrt{{\text{F}}_1^2+{\text{F}}_{\text{2}}^{\text{2}}+2{\text{F}}_{\text{1}}{\text{F}}_{\text{2}}\cos 60°}$

$=\sqrt{3}{\text{F}}_{\text{1}}$

$\text{F}=\sqrt{3}\frac{{\text{Gm}}^{\text{2}}}{{\text{a}}^{\text{2}}}={\text{F}}_{\text{c}}=\text{mr}{\omega }^2$                 $\text{CO}=\frac{2}{3}\text{CD}$               

$\text{m}\left(\frac{1}{\sqrt{3}}\text{a}\right){\omega }^2=\sqrt{3}\frac{{\text{Gm}}^{\text{2}}}{{\text{a}}^{\text{2}}}$                    $\text{In }{\Delta }^{\text{le}}\text{ CD}=\text{h}=\frac{\sqrt{3}}{2}\text{a}$

${\omega }^2=\frac{3\text{Gm}}{{\text{a}}^3}\Rightarrow \omega =\sqrt{\frac{3\text{Gm}}{{\text{a}}^3}}$                 $\text{CO}=\text{r}=\frac{2}{3}\left(\frac{\sqrt{3}}{2}\text{a}\right)=\frac{\sqrt{3}}{3}\text{a}=\frac{\text{a}}{\sqrt{3}}$

$\omega =\sqrt{\frac{3\text{Gm}}{{\text{a}}^3}}$