1.

From the figure, the gravitational potential at '0' due to three point masses is

Answer»

`(GM)/(l)(2+(1)/(SQRT2))`
`(Gm)/(l)(sqrt+(1)/(2))`
`(3Gm)/(l)`
`(2Gm)/(l)`

SOLUTION :`V=v_(1)+v_(2)+v_(3)`
`=-(Gm)/(l)-(Gm)/(l)-(Gm)/(sqrt2l)`
`=-(Gm)/(l)[1+1+(1)/(sqrt2)]=-(Gm)/(l)[2+(1)/(sqrt2)]`.


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