Question

Two particles of equal masses $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

• A. $\frac{1}{2R}\sqrt{\frac{1}{Gm}}$
• B. $\sqrt{\frac{Gm}{2R}}$
• C. $\frac{1}{2}\sqrt{\frac{Gm}{R}}$
• D. $\sqrt{\frac{4Gm}{R}}$

Answer 1

The correct option is C $\frac{1}{2}\sqrt{\frac{Gm}{R}}$

Given,
Masses of the two particles are $m_1=m_2=m$

Radius of circular motion of each particle, $r_1=r_2=R$

Distance between the particles $=2R$




So ,the gravitational force of attraction between the particles

$F_G=\frac{G\left(m\right)\left(m\right)}{(2R{)}^2}$

And, if speed is $u$ then, centripetal force

$F_C=\frac{m{u}^2}{R}$

Both the particles perform circular motion under the influence of gravitational force of attraction between them. Hence,

$F_G=F_C$

$\Rightarrow \frac{G{m}^2}{(2R{)}^2}=\frac{m{u}^2}{R}$

$\Rightarrow \frac{G{m}^2}{4R^2}=\frac{m{u}^2}{R}$

$\Rightarrow u=\frac{1}{2}\sqrt{\frac{Gm}{R}}$

Hence, option (c) is the correct answer.