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 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. $zero$

Answer 1

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

Since there are $3$ masses, each mass is subjected to forces of attraction by the two other masses.
The force of attraction between two masses is $F=\frac{Gmm}{a^2}$
Now the resultant force on a mass due to the other two is $F_r=2Fcos{30}^0$
$\Rightarrow F_r=2\frac{Gmm}{a^2}cos{30}^0$
The centripetal force for revolution is given by this resultant force.
$\therefore 2\frac{Gmm}{a^2}cos{30}^0=mr{\omega }^2$

$\Rightarrow \frac{Gm}{a^2}\sqrt{3}=r{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{Gm\sqrt{3}}{a^2\frac{a}{\sqrt{3}}}}$ ;

$($ $r=\frac{a}{\sqrt{3}}$ for equilateral triangle $)$
$\Rightarrow \omega =\sqrt{\frac{3Gm}{a^3}}$