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 projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is defined and the position where the two forces cancel each other exactly if ON=r. we have
`(GMm)/(r^(2))=(4GMm)/((6R-r)^(2))`
`(6R-r)^(2)=4r^(2)`
`6R-r=+-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_(1)=(1)/(2)mv^(2)-(GMm)/(R)-(4GMm)/(5R)`
At the neutral point N, the speed approches 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 taht the speed of the projectile is zero at N. but is nonzero when it strikes the heavier spehre 4 M. the calculation of this speed is left as an exercise to the students.