Question

Four particles of equal mass $M$ move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

• A. $\frac{GM}{R}$
• B. $\sqrt{\left[2\sqrt{2}\frac{GM}{R}\right]}$
• C. $\sqrt{\left[\frac{GM}{R}\left(2\sqrt{2}+1\right)\right]}$
• D. $\sqrt{\left[\frac{GM}{R}\left(\frac{2\sqrt{2}+1}{4}\right)\right]}$

Answer 1

Answer: (D)

Gravitational force on each due to other three particles provides the necessary centripetal force.

Net force on topmost mass towards centre O provides necessary centripetal force. Therefore,
$\therefore 2\frac{G{M}^2}{{\left(\sqrt{2}R\right)}^2}cos45^{\circ }+\frac{G{M}^2}{\left(2R\right){\left(\right)}^2}=\frac{M{v}^2}{R}$
$\Rightarrow v=\sqrt{\frac{GM}{R}\left(\frac{2\sqrt{2}+1}{4}\right)}$