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Two particles of equal mass 'm' go aroun...

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 centre of mass is :

A

`sqrt((Gm)/(4R))`

B

`sqrt((Gm)/(3R))`

C

`sqrt((Gm)/(2R))`

D

`sqrt((Gm)/(R ))`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `.omega.` be the angular velocity. The mutual force of gravitational attraction provides the necessary centripetal force of rotation. Thus

`(Gm.m)/((R+R)^(2))=m.R.omega^(2)`
or `omega^(2)=(Gm)/(4R^(3))`
or `omega=sqrt((Gm)/(4R^(3))`
Now, `v=Romega=Rsqrt((Gm)/(4R^(3)))=sqrt((Gm)/(4R))`
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Knowledge Check

  • Three identical objects each of mass M move along circle of radius R under the action of their mutual gravitational attraction, the speed of each is :

    A
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    B
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