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

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,
$\therefore 2\frac{G M^{2}}{\left(\sqrt{2} \, R\right)^{2}}cos45^\circ +\frac{G M^{2}}{\left(2 R\right) \left( \, \right)^{2}}=\frac{M v^{2}}{R}$
$\Rightarrow v=\sqrt{\frac{G M}{R} \left(\frac{2 \sqrt{2} + 1}{4}\right)}$