Two planets have masses M and 16 M and their radii are a and 2a, respectively. The separation between the centres of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planer, the minimum firing speed needed is :

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$\sqrt{\frac{G M^{2}}{m a}}$

$\frac{3}{2} \sqrt{\frac{5 G M}{a}}$

$2 \sqrt{\frac{G M}{a}}$

$4 \sqrt{\frac{G M}{a}}$

answer is C.

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Detailed Solution

Let gravitational field intensity at P is zero. Then if distance of P from the center of bigger sphere is r,
$\frac{- G 16 M}{r^{2}} = \frac{- G \times M}{{(10 a - r)}^{2}} \Rightarrow r = 8 a$
The particle is projected from surface of the bigger sphere, and if it just manages to cross P then it will surely land on the smaller sphere.
So applying energy conservation.
$- \frac{G 16 M m}{2 a} - \frac{G M m}{8 a} + \frac{1}{2} m ϑ^{2}$ $= - \frac{G 16 M m}{8 a} - \frac{G M m}{2 a}$
$\Rightarrow ϑ = \frac{3}{2} \times \sqrt{\frac{5 G M}{a}}$