Two uniform solid spheres of equal radii R, but mass M and 4M have a centre to centre separation is 6R as shown in fig.
The spheres are held fixed. A projectile of mass m projected with minimum velocity from the surface of sphere directly towards the centre of the second sphere so that reaches the surface of the second sphere. If the speed with which projectile reaches the surface of the sphere of mass 4M is $3 \sqrt{\frac{xGM}{5 R}} find x$

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answer is 3.

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Detailed Solution

null point between the two masses is $\frac{6 R}{\sqrt{\frac{4 M}{M} + 1}} = 2 R$
From null point to the surface of heavy sphere apply law of conservation of energy
$\frac{- GMm}{2 R} - \frac{G 4 Mm}{4 R} = \frac{- GMm}{5 R} - \frac{G 4 Mm}{R} + \frac{1}{2} {mv}^{2}$
$\Rightarrow \frac{- 3 GM}{2 R} = \frac{- 21 GM}{5 R} + \frac{1}{2} V^{2} \Rightarrow \frac{V^{2}}{2} = \frac{27 GM}{10 R} \Rightarrow V = 3 \sqrt{\frac{3 GM}{5 R}}$