Question

Three identical stars of mass $m$ each are located at the corner of equilateral triangle of side $L$ and are orbiting under mutual gravitational force. Orbital speed of such 3 star system is $V_0=\sqrt{\frac{pGm}{L}}$ then $p$ is

Answer 1

$r=L/\sqrt{3}F=\frac{G{m}^2}{L^2}\cos 30\times 2$& $F=\frac{m{V}_0^2}{r}\Rightarrow \frac{G{m}^{\text{/}2}}{L^2}\sqrt{3}=\frac{\text{/}m{V}_0^2}{r}\Rightarrow V_0^2=\frac{\sqrt{3}Gmr}{L^2}{V}_0=\sqrt{\frac{Gm}{L}}\Rightarrow p=1$