Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown . 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 the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Explanation

The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is defined as the position where the two forces cancel each other exactly. If ON = r, we have
`(GMm)/(r^(2)) = (4GMm)/((6R - r)^(2))`
`(6 R - r)^(2) = 4 r^(2)`
6 R - r = `pm` 2r
r = 2R or - 6R.
The neutral point r = – 6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
`E_(i) = (1)/(2) m v^(2) - (GMm)/(R) - (4GMm)/(5R)` .
At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.
`E_(N) = - (GMm)/(2R) - (4GMm)/(4R) . `
From the principle of conservation of mechanical energy
`(1)/(2) v^(2) - (GM)/(R) - (4GM)/(5R) = - (GM)/(2R) - (GM)/(R)`
or
`v^(2) = (2GM)/(R) ((4)/(5) - (1)/(2))`
` v = ((3GM)/(5R))^(1//2)`
A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the heavier sphere 4 M. The calculation of this speed is left as an exercise to the students.