JEE PYQ: Gravitation Question 28
Question 28 - 2020 (06 Sep Shift 2)
Two planets have masses $M$ and $16M$ 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 the surface of smaller planet, the minimum firing speed needed is:
(1) $2\sqrt{\frac{GM}{a}}$
(2) $4\sqrt{\frac{GM}{a}}$
(3) $\sqrt{\frac{GM^2}{ma}}$
(4) $\frac{3}{2}\sqrt{\frac{5GM}{a}}$
Show Answer
Answer: (4)
Solution
The null point (where gravitational fields cancel) is at distance $x$ from smaller planet: $\frac{GM}{x^2} = \frac{G(16M)}{(10a-x)^2}$, giving $10a - x = 4x$, so $x = 2a$. Using energy conservation from surface of larger planet to null point: $\frac{1}{2}mv^2 - \frac{G(16M)m}{2a} - \frac{GMm}{8a} = -\frac{G(16M)m}{8a} - \frac{GMm}{2a}$. Solving: $v = \frac{3}{2}\sqrt{\frac{5GM}{a}}$.
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