Question

Three point 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 vertices 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. Zero

Answer 1

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

$Forcebetweentwoadjacentparticles=\frac{G{m}^2}{a^2}Nowtheseforcesareat{60}^{o}degreestoeachothersoresultantofthesetwo{F}_r=2Fcos\theta /2Fr=\frac{\sqrt{3}G{m}^2}{a^2}Andrcos\theta =\frac{a}{2}\Rightarrow r=\frac{a}{\sqrt{3}}NowFractsasthecentripetalforceofsingleparticletherefore\frac{\sqrt{3}G{m}^2}{a^2}=m{\omega }^{2}r\Rightarrow \omega =\sqrt{\frac{3Gm}{a^3}}$