Q. Two particles of equal mass $m$ go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their center of mass is
A) $\sqrt{\frac{Gm}{R}}$
B) $\sqrt{\frac{Gm}{4R}}$
C) $\sqrt{\frac{Gm}{3R}}$
D) $\sqrt{\frac{Gm}{2R}}$
Explanation
(b) Gravitational force provides necessary centripetal force
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$\frac{{Gm}^{2}}{(2R)^{2}} = \frac{{mv}^{2}}{R}$
$\Longrightarrow \ \ v = \sqrt{\frac{Gm}{4R}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.