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Two bodies of masses m(1) and m(2) are p...

Two bodies of masses `m_(1)` and `m_(2)` are placed distant d apart. Show that the position where the gravitational field due to them is zero, the potential is given by, `V = - (G)/( d) ( m_(1) +m_(2) + 2 sqrt( m_(1) m_(2)))`

A

`V=-(G)/(d)(m+M)`

B

`V=-(Gm)/(d)`

C

`V=-(GM)/(d)`

D

`V=-(G)/(d)(sqrt(m)+sqrt(M))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_("net")=0`
`:. (Gm)/(x^(2))=(GM)/((d-x)^(2)),x` is distance from `m`
`:. (x)/(d-x)=(sqrt(m))/(sqrt(M))rArrx=(sqrt(m))/(sqrt(m)+sqrt(M)),d`
and `d-x=(sqrt(M))/(sqrt(m)+sqrt(M)).d`
Since, gravitational potential between two bodies of masses M and m respectively, is given by
`V=-(Gm)/(x)-(GM)/(d-x)`
`=-(GM(sqrt(m)+sqrt(M)))/(sqrt(m).d)-(GMsqrt(m)+sqrt(M))/(sqrt(M).d)=-(G)/(d)(sqrt(m)+sqrt(M))^(2)`.
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