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 teh second. Obtain an expression for the minimum speed `upsilon` of the projectile so that it reaches the surface of second sphere.

Explanation

The two spheres exert gravitational forces on the projectile in mutually opposite directions. At the neutral point N, there two forces cancel each other. If ON = r, then
`(GMm)/r^2 =(G(4M)m)/((6R-r)^2)` or `(6R-r)^2 = 4r^2 rArr 6R-r=pm2r` or r=2R or -6R
The neutral point r = – 6R is inadmissible.
`therefore` ON=r=2 R
It will be sufficient to project the particle m with a minimum speed v which enables it to reach the point N.
Therefore, the particle m gets attracted by the gravitational pull of 4 M.
The total mechanical energy of m at surface of left sphere is `E_i`=KE of m+ PE due to left sphere + PE due to right sphere
`=1/2mv^2 -(GMm)/R-(4GMm)/(5R)`
At the neutral point, speed of the particle becomes zero. The energy is purely potential.
`therefore E_N=PE` due to left sphere + PE due to right sphere `=-(GMm)/(2R) - (4GMm)/(4R)`
By conservation of mechanical energy, `E_i=E_N`
Or `1/2mv^2 -(GMm)/R-(4GMm)/(5R)=-(GMm)/(2R) -(4GMm)/(4R)` or `v^2=(2GM)/R (4/5-1/2)=(3GM)/(5R) therefore v=sqrt((3GM)/(5R))`