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The distance between to stars of masses M and 16M is 10a. Their radii a and 2a respectively. A body of mass m is fired stright from the surface of larger star towards the smaller star. If the speed of projection to reach the surface of smaller star is $\frac{x}{2} \sqrt{\frac{y G M}{a}}$ then find the value of $x + y .$

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answer is 8.

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Detailed Solution

The Null point is at a distance of x from the smaller star.

$x = \frac{10 a}{\sqrt{\frac{16 M}{M}} + 1} = 2 a$

Apply law of conservation of energy at P and at N

$\frac{1}{2} {mv}^{2} - \frac{G 16 Mm}{2 a} - \frac{GMm}{8 a}$

$= - \frac{GMm}{2 a} - \frac{G 16 Mm}{8 a} \Rightarrow \frac{v^{2}}{2} - \frac{65 GM}{8 a} = \frac{- 5 GM}{2 a}$

$\Rightarrow \frac{v^{2}}{2} = \frac{65 GM}{8 a} - \frac{5 GM}{2 a} \Rightarrow v^{2} = \frac{45 GM}{4 a}$

$\Rightarrow v = \frac{3}{2} \sqrt{\frac{5 GM}{a}} ∴ x + y = 8$

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