Question 23 of 223

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 $\dfrac{M}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( $G$ is the gravitational constant; ${M}_{e}$ is the mass of the earth):

A$\dfrac{M}{20}\left({u}^{2}+\dfrac{113}{200}\dfrac{G{M}_{e}}{R}\right)$B$5M\left({u}^{2}-\dfrac{119}{200}\dfrac{G{M}_{e}}{R}\right)$C$\dfrac{3M}{8}{\left(u+\sqrt{\dfrac{5G{M}_{e}}{6R}}\right)}^{2}$D$\dfrac{M}{20}{\left(u-\sqrt{\dfrac{2G{M}_{e}}{3R}}\right)}^{2}$

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Answer: (B) $5M\left({u}^{2}-\dfrac{119}{200}\dfrac{G{M}_{e}}{R}\right)$

$\dfrac{-{GM}_{e}M}{R}+\dfrac{1}{2}{Mu}^{2}=\dfrac{-{GM}_{e}M}{2R}+\dfrac{1}{2}{Mv}^{2}$

$v=\sqrt{{u}^{2}-\dfrac{G{M}_{e}}{R}}$
${V}_{\tau }\rightarrow$ Transverse velocity of rocket
${V}_{R}\rightarrow$ Radial velocity of rocket
$V=\sqrt{\dfrac{G{M}_{e}}{2R}}$
$\dfrac{M}{10}{V}_{T}=\dfrac{9M}{10}\sqrt{\dfrac{G{M}_{e}}{2R}}$
$\dfrac{M}{10}{V}_{r}=M\sqrt{{u}^{2}-\dfrac{{GM}_{e}}{R}}$
Kinetic energy $=\dfrac{1}{2}\dfrac{M}{10}\left({V}_{T}^{2}+{V}_{r}^{2}\right)=\dfrac{M}{20}\left(81\dfrac{{GM}_{e}}{2R}+100{u}^{2}-100\dfrac{{GM}_{e}}{R}\right)$
$=\dfrac{M}{20}\left(100{u}^{2}-\dfrac{119G{M}_{e}}{2R}\right)$
$=5M\left({u}^{2}-\dfrac{119{GM}_{e}}{200R}\right)$