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
$K_i+U_i=K_f+U_f$
$\frac{1}{2}m{u}^2+\left(-\frac{GMm}{R}\right)=\frac{1}{2}m{v}^2-\frac{GMm}{2R}$
$v=\sqrt{u^2-\frac{GM}{R}}$ ....(i)
By momentum conservation, we have
$\frac{m}{10}{v}_T=\frac{9m}{10}\sqrt{\frac{GM}{2R}}$
$\&\frac{m}{10}{v}_r=mv$
$\Rightarrow \frac{m}{10}{v}_r=m\sqrt{u^2-\frac{GM}{R}}$
Kinetic energy of rocket
$=\frac{1}{2}m\left(v_T^2+v_r^2\right)$
$=\frac{m}{20}\left(81\frac{GM}{2R}+100u^2-100\frac{GM}{R}\right)$
$=\frac{m}{20}\left(100u^2-\frac{119GM}{2R}\right)$
$=5m\left(u^2-\frac{119GM}{200R}\right).$