Question

Two uniform solid spheres of equal radii R, but mass M and 4M have a centre to centre separation 6R, as shown in figure. A projectile of mass M is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. The minimum speed of the projectile so that it reaches the surface of the second sphere is

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Answer 2

Let the projectile of mass m be fired with minimum velocity, v from the surface of sphere of mass M to reach the surface of sphere of mass 4M. Let N be neutral point at a distance r from the centre of the sphere of mass M.

At neutral point N,

$( 6 R - r ) ^ { 2 } = 4 r ^ { 2 }$

$6 R - r = \pm 2 r$ or r = 2R or -6R

The point r = - 6R does not concern us.

Thus, ON = r = 2R.

It is sufficient to project the projectile 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

At the neutral point N, the speed approaches zero.

$\therefore$The mechanical energy at N is

$E _ { N } = - \frac { G M m } { 2 R } - \frac { G ( 4 M ) m } { 4 R } = \frac { G M m } { 2 R } - \frac { G M m } { R }$

According to law of conservations of mechanical energy,

$\frac { 1 } { 2 } \mathrm { mv } ^ { 2 } - \frac { \mathrm { GMm } } { \mathrm { R } } - \frac { 4 \mathrm { GMm } } { 5 \mathrm { R } } = - \frac { \mathrm { GMm } } { 2 \mathrm { R } } - \frac { \mathrm { GMm } } { \mathrm { R } }$

$\mathrm { v } ^ { 2 } = \frac { 2 \mathrm { GM } } { \mathrm { R } } \left[ \frac { 4 } { 5 } - \frac { 1 } { 2 } \right] = \frac { 3 } { 5 } \frac { \mathrm { GM } } { \mathrm { R } }$ or $\mathrm { v } = \left( \frac { 3 } { 5 } \frac { \mathrm { GM } } { \mathrm { R } } \right) ^ { 1 / 2 }$