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Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

A

`v=1/(2R)sqrt((1/(Gm)))`

B

`v=sqrt(((GM)/(2R)))`

C

`v=1/2sqrt(((Gm)/R))`

D

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

Text Solution

Verified by Experts

The correct Answer is:
C

`F_(G)=(mv^(2))/R=(GM^(2))/(4R^(2))`
`v=1/2sqrt((GM)/R)`
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