1.

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 :

Answer»

`sqrt((Gm)/(4R))`
`sqrt((Gm)/(3R))`
`sqrt((Gm)/(2R))`
`sqrt((Gm)/(R ))`

Solution :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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