Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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 {\dfrac{{Gm}}{{4R}}} $
(B) $\sqrt {\dfrac{{Gm}}{{3R}}} $
(C) $\sqrt {\dfrac{{Gm}}{{2R}}} $
(D) $\sqrt {\dfrac{{Gm}}{R}} $

seo-qna
Last updated date: 19th Jun 2024
Total views: 53.4k
Views today: 0.53k
Answer
VerifiedVerified
53.4k+ views
Hint In this question, we need to find the value of the velocity at which both the particles are revolving about their center of mass. This value of force should be equal to the centripetal force which is directed towards the center of the circle i.e. the center of mass.
Formula used ${F_{gravitation}} = \dfrac{{G{M_1}{M_2}}}{{{R^2}}}$, which is the general form of the gravitational force of attraction. Here $G$ is the gravitational constant with a value of $G = 6.6743 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{\operatorname{s} ^{ - 2}}$ and $R$is the distance between the two bodies.
We will also use ${F_{centripetal}} = \dfrac{{m{v^2}}}{R}$.

Complete step by step solution:
Given two particles are moving in a circular path of radius $R$. The center of mass of the particles is at the center of the circle whose radius is $R$, which they are tracing.
The particles are of equal mass $m$, which means that the center of mass is at the midpoint of the two particles. The particles revolve around their center of mass in a circular path as said in the question. The radius of the circular path is $R$, thus the diameter of the circle is $2R$.
The gravitational attraction force on the particles is $F = \dfrac{{G \times m \times m}}{{{{(2R)}^2}}}$.
This is equal to the centripetal force since no other force is acting on any particle.
The centripetal force is the force which is directed towards the center during a rotational motion. This force is equal to the net force towards the center which is the gravitational attraction force.
Thus, ${F_{gravitation}} = {F_{centripetal}}$
$ \Rightarrow \dfrac{{G{m^2}}}{{{{(2R)}^2}}} = \dfrac{{m{v^2}}}{R}$
$ \Rightarrow v = \sqrt {\dfrac{{Gm}}{{4R}}} $

The correct answer is option $(A).$

Note: We did not calculate the exact point of the center of mass in this problem as it is understood that the center of mass is at the center of the circle. In case the masses involved were distinct, then the position of the center of mass would be vital to calculate the centripetal force with respect to different radius of the circular paths of different bodies with different masses.