Guest

two particles of equal mass m go around a circle of radius R under the influence of their mutual gravitational attraction.the speed of each particle with respect to their center of mass will be??

two particles of equal mass m go around a circle of radius R under the influence of their mutual gravitational attraction.the speed of each particle with respect to their center of mass will be??

Grade:12th Pass

7 Answers

rohit yadav
34 Points
11 years ago

with respect to the center of mass the speed of each particle is (Gm/4R)1/2 as their is no external force acting on the system so velocity of the center of mass is zero velocity of each particle is (Gm/4R)1/2.

Chetan Mandayam Nayakar
312 Points
11 years ago

mv2/R=Gm2/(2R)2

Yogita Bang
39 Points
11 years ago

The gravitational force of attraction between the two masses is

F = Gm2/(2R)2

This force provides the necessary centripetal force

mv2/R = Gm2/4R2

v2 = Gm/4R

v = √(Gm/4R)

gauhar singh
7 Points
11 years ago

The Speed will be equal to their velocities in inertial frame itself as there centre of mass will be stationary.

Ojas Suvarnakar
26 Points
7 years ago
Here center of mass will be stationary because both particles are having same mass,Therefore ,Speed of the particle will be as equal as critical velocity of the particle...Therefore answer is √GM/R
Aniket lohar
21 Points
6 years ago
Two particles of equal mass(m) go around a circle of radius R under the action of their mutual gravitational attraction.Here center of mass is stationary because both particles are having same mass.1/2√(GM/R)
Yug Dedhia
15 Points
5 years ago
Hello this is master Yug Dedhia
Here,using theory of gravitation
Therefore,mv^2/ r = Gm^2/(2R)^2
Thus the answer is v=√(Gm/4R)
Thanks

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free