JEE PYQ: Gravitation Question 29
Question 29 - 2020 (07 Jan Shift 1)
A satellite of mass $m$ is launched vertically upwards with an initial speed $u$ from the surface of the earth. After it reaches height $R$ ($R$ = radius of the earth), it ejects a rocket of mass $\frac{m}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ($G$ is the gravitational constant; $M$ is the mass of the earth):
(1) $\frac{m}{20}\left(u^2 + \frac{113}{200}\frac{GM}{R}\right)$
(2) $5m\left(u^2 - \frac{119}{200}\frac{GM}{R}\right)$
(3) $\frac{3m}{8}\left(u + \sqrt{\frac{5GM}{6R}}\right)^2$
(4) $\frac{m}{20}\left(u - \sqrt{\frac{2GM}{3R}}\right)^2$
Show Answer
Answer: (2)
Solution
At height $R$: $v_0 = \sqrt{u^2 - \frac{GM}{R} + \frac{GM}{2R}} = \sqrt{u^2 - \frac{GM}{2R}}$… After ejection, the $\frac{9m}{10}$ satellite enters circular orbit at $r = 2R$: $v_c = \sqrt{\frac{GM}{2R}}$. By momentum conservation and energy: KE of rocket $= 5m\left(u^2 - \frac{119}{200}\frac{GM}{R}\right)$.
BETA
