Question

Three points masses each of mass m rotate in a circle of radius r with constant angular velocity $\omega$ due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side a, then the value of $\omega$ is

• A. $\sqrt{\frac{Gm}{a^3}}$
• B. $\sqrt{\frac{3Gm}{a^3}}$
• C. $\sqrt{\frac{Gm}{3a^3}}$
• D. $\sqrt{\frac{Gm}{2a^3}}$

Answer 1

The correct option is B $\sqrt{\frac{3Gm}{a^3}}$

$F_0=\frac{G{m}^2}{a^2},F=2F_{0}cos{30}^{\circ }=\frac{2G{m}^2}{a^2}\frac{\sqrt{3}}{2}$
or $mr{\omega }^2=\frac{\sqrt{3}G{m}^2}{a^2},\omega =\sqrt{\frac{3Gm}{a^3}}[\because 2rcor{30}^{\circ }=a]$