Question

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):

• A. $\frac{3m}{8} \left(u+\sqrt{\frac{5GM}{6R}}\right)^{2}$
• B. $\frac{m}{20} \left(u^{2}+\frac{113}{200} \frac{GM}{R}\right)$
• C. $5m\left(u^{2}+\frac{119}{200} \frac{GM}{R}\right)$
• D. $\frac{m}{20} \left(u+\sqrt{\frac{2GM}{3R}}\right)^{2}$

Answer 1

Answer: (C)

Applying energy conservation
$K _{ i }+ U _{ i }= K _{ f }+ U _{ f }$
$\frac{1}{2} mu ^{2}+\left(-\frac{ GMm }{ R }\right)=\frac{1}{2} mv ^{2}-\frac{ GMm }{2 R }$
$v=\sqrt{u^{2}-\frac{G M}{R}}$ ....(i)
By momentum conservation, we have
$\frac{ m }{10} v _{ T }=\frac{9 m }{10} \sqrt{\frac{ GM }{2 R }}$
$\& \frac{ m }{10} v _{ r }= mv$
$\Rightarrow \frac{ m }{10} v _{ r }= m \sqrt{ u ^{2}-\frac{ GM }{ R }}$
Kinetic energy of rocket
$=\frac{1}{2} m \left( v _{ T }^{2}+ v _{ r }^{2}\right)$
$=\frac{ m }{20}\left(81 \frac{ GM }{2 R }+100 u ^{2}-100 \frac{ GM }{ R }\right)$
$=\frac{ m }{20}\left(100 u ^{2}-\frac{119 GM }{2 R }\right)$
$=5 m \left( u ^{2}-\frac{119 GM }{200 R }\right) .$