Speed of particles under mutual gravitation – Rankers Physics

Newton's Law of Gravitation: Practice Problem & Solution

Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed $v$ of each particle is: (1995)
$\frac{1}{2}\sqrt{\frac{Gm}{R}}$
$\sqrt{\frac{4Gm}{R}}$
$\frac{1}{2R}\sqrt{\frac{1}{Gm}}$
$\sqrt{\frac{Gm}{R}}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Newton's Law of Gravitation. Understanding the underlying formula is key to arriving at the correct answer below:

The gravitational force provides the necessary centripetal force. $\frac{mv^2}{R} = \frac{Gmm}{(2R)^2} = \frac{Gm^2}{4R^2}$. Solving for $v$, we get $v^2 = \frac{Gm}{4R}$, which means $v = \frac{1}{2}\sqrt{\frac{Gm}{R}}$.

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