A particle of mass `m_(0) ` is projected from the surface of body (1) with initial velocity `v_(0)` towards the body (2). As shown in figure find the minimum value of `v_(0)` so that the particle reaches the surface of body (2). Radius of body (1) is `R_(1)` and mass of body (1) is `M_(1)`. Radius body (2) is `R_(2)` and mass of the body (2) is `M_(2)` distance between the center of two bodies is d.

Explanation

On the line joining two centre, there will be a point where gravitational intensity is zero. To the right of this point attraction force due to smaller sphere will be more. So if particle crosses this point, it will automaticaly reach the surface of smaller sphere so consevation of energy is applied upto this point only.
There fore on m towards `M_(1)` is `F_(1)=(GM_(1)m)/(r^(2))`
The forece on m towards `M_(2)` is `F_(2)=(GM_(2)m)/((d-r)^(2))`
Accoring to question net force on m is zero i.e., `F_(1)=F_(2)`
`implies(GM_(1)m)/(r^(2))=(GM_(2)m)/((d-r)^(2))implies((d-r)/(r))^(2)=(M_(2))/(M_(1))implies(d)/(r)-1=(sqrt(M_(2)))/(sqrt(M_(1)))`
`impliesr=d[(sqrt(M_(1)))/(sqrt(M_(1))+sqrt(M_(2)))]` ....(i)
`E_(i)=E_(i)`
`K_(i)+U_(i)=K_(f)+U_(f)` ltbr `(1)/(2)m(v_(0)^(2))_(min)-(GM_(1)m)/(R_(1))-(GM_(2)m)/(d-R_(2))=(-GM_(1)m)/(r)-(GM_(2)m)/(d-r)` ..(ii)
Solving equation (i) and (ii) we get `(v_(0))_(min)`
`v_(0)=sqrt(2((GM_(1))/(R_(1))+(GM_(2))/(d-R_(2))-(G)/(d)(M_(1)+M_(2)+2sqrt(M_(1)M_(2)))))`