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; $M_e$ is the mass of the earth):

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

Answer 1

Answer: (B)

$\frac{-G{M}_{e}M}{R}+\frac{1}{2}M{u}^2=\frac{-G{M}_{e}M}{2R}+\frac{1}{2}M{v}^2$

$v=\sqrt{u^2-\frac{G{M}_e}{R}}$
$V_{\tau }\to$ Transverse velocity of rocket
$V_R\to$ Radial velocity of rocket
$V=\sqrt{\frac{G{M}_e}{2R}}$
$\frac{M}{10}{V}_T=\frac{9M}{10}\sqrt{\frac{G{M}_e}{2R}}$
$\frac{M}{10}{V}_r=M\sqrt{u^2-\frac{G{M}_e}{R}}$
Kinetic energy $=\frac{1}{2}\frac{M}{10}\left(V_T^2+V_r^2\right)=\frac{M}{20}\left(81\frac{{\left(GM\right)}_e}{2R}+100u^2-100\frac{{\left(GM\right)}_e}{R}\right)$
$=\frac{M}{20}\left(100u^2-\frac{119G{M}_e}{2R}\right)$
$=5M\left(u^2-\frac{119{\left(GM\right)}_e}{200R}\right)$