Two uniform soild spheres of equal radii `R` but mass `M` and `4M` have a centre to centre separation `6 R`, as shows in Fig. (a) The two spheres are held fixed. A projectile of mass `m` is projected from the surface of the sphere of mass `M` directly towards the centre of ten second. Obtain an expression for the minimum speed `upsilon` of the projectile so that it reaches the surface of second sphere.

Explanation

Refer to Fig. (b) let the projectile fired with minimum velocity `upsilon` from the surface of sphere `A` reach the surface of sphere `B`. Let `C` be the neutral point at a distance `x` from centre of sphere `A` where the gravitational pull on body due to sphere `A` is equal and opposite to the gravitational pull on body due to sphere `B`. Then

`(GM m)/(x^(2)) = (G 4M m)/((6 R - x)^(2))` or `(6R - x)^(2) = 4 x^(2)`
or `6R - x = +- 6 R` is not of any concern in this problem. Thus, `AC = x = 2 R`. Therefore, it is sufficient to project the particle with a minimum speed `upsilon` which enable it to each `C` Beyond that the greater gravitational pull of `4 M` will be suffficiently large to attract the projectile.
The mechanical energy of projectile at the surface of `M` is
`E_(i) = (1)/(2) m upsilon^(2) + ((-GM m)/(R )) + ((-G 4M m)/(5R))`
At the natural point `C`, the speed apporaches zero. The mechanical energy at `C` is purely potential . So
`E_(C) = (-GMm)/(2R) - (G 4 M m)/(4R)`
Using principle of conservation of mechanical energy
`(1)/(2) m upsilon^(2) - (-GM m)/(R ) - (4 GM m)/(5R)`
`= (-GM m)/(2R ) - (4 GM m)/(4R)`
or `upsilon^(2) = (2GM)/(R ) [(4)/(5) - (1)/(2)] = (3 GM)/(5R)`
or `upsilon ((3GM)/(5R))^(1//2)`