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Four identical particles of mass M are l...

Four identical particles of mass `M` are located at the corners of a square of side `'a'`. What should be their speed if each of them revolves under the influence of others' gravitational field in a circular orbit circumscribing the square?

A

`1.2 sqrt((GM)/a)`

B

`1.41 sqrt((GM)/a)`

C

`1.16 sqrt((GM)/a)`

D

`1.35sqrt((GM)/a)`

Text Solution

Verified by Experts

The correct Answer is:
C
radius of the circle `a/(sqrt(2))=r`
FBD of any one particle
`(GM^2)/(2a^2)+(2GM^2)/(a^2)1/(sqrt(2))=(Mv^2)/(a//sqrt(2))" " implies " " sqrt((GM)/(a))sqrt(1/(2sqrt(2))+1)=1.16 sqrt((GM)/(a))`
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