Speed of particles under mutual gravitation – Rankers Physics
Topic: Gravitation
Subtopic: Newton's Law of Gravitation

Speed of particles under mutual gravitation

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)

$\frac{1}{2}\sqrt{\frac{Gm}{R}}$
$\sqrt{\frac{4Gm}{R}}$
$\frac{1}{2R}\sqrt{\frac{1}{Gm}}$
$\sqrt{\frac{Gm}{R}}$

Solution:

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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