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Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. The speed v of earth particle is

Answer»

`(1)/(3)SQRT((Gm)/(R ))`
`(1)/(2)sqrt((Gm)/(R ))`
`(1)/(2)sqrt((R )/(Gm))`
`sqrt((Gm)/(R ))`

Solution :The two masses, separated by a distance 2R are going round their common CENTRE of mass, the centre of the circle.
Attractive force`=-G(mm)/(4R^(2))`. But the two masses are going round the centre of mass or the reduced mass `mu=(mm)/(m+m)` is going round a circle of radius= distance of separation
`:.` Centrifugal force `=(m)/(2)omega^(2).2 R=(m)/(2)v^(2).(1)/(2R )`
`rArr (m)/(2)xx(v^(2))/(2R)=(Gm^(2))/(4R^(2)) rArr v=sqrt((Gm)/(R ))`


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