The masses and radii of the earth an moon are `M_(1) and R_(1) and M_(2), R_(2)` respectively. Their centres are at a distacne r apart. Find the minimum speed with which the particle of mass m should be projected from a point mid-way between the two centres so as to escape to infinity.

Options

A. `2[(G(M_(1)+M_(2)))/(md)]^(1//2)`
B. `2[(G(M_(1)+M_(2)))/(d)]^(1//2)`
C. `2[(2(M_(1)-M_(2)))/(md)]^(1//2)`
D. `2[(G(M_(1)-M_(2)))/(d)]^(1//2)`

Answer

b

Explanation

The potential energy of particle P due to earth is
`-("GmM"_(1))/(r)=-("GmM"_(1))/(d//2)=-("2GM"_(1)m)/(d)`
`"P.E. due to moon at p is "-("2GM"_(2)m)/(d)`

`therefore" Total potential energy "=-(2Gm)/(d)(M_(1)+M_(2))`
If the particle P is projected with a velocity v, its kinetic energy `=1//2mv^(2)`.
Therefore, the total initial energy of the particle is
`E_(1)=-("2Gm")/(d)(M_(1)+M_(2))+(1)/(2)mv^(2)`.
If the particle is to escape to infinity, its final potential and kinetic energy will be zero. Thus the total energy `E_(f)=0`.
From the principle of conservation of energy,
`E_(1)=E_(f)=-("2Gm")/(d)(M_(1)+M_(2))+(1)/(2)mv^(2)=0`
Which gives.
`v=2[(G(M_(1)+M_(2)))/(d)]^(1//2)`