Question

A satellite of mass $m$ is launched vertically upward with an initial speed $u$ from the surface of the earth. After it reaches height $R$ ($R$ = radius of earth), it ejects a rocket of mass $m/10$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G = gravitational constant; M is the mass of earth)

• A. $5m[u^2-\frac{119}{200}\frac{GM}{R}]$
• B. $\frac{m}{20}{[u-\sqrt{\frac{2GM}{3R}}]}^2$
• C. $\frac{3m}{8}{[u+\sqrt{\frac{5GM}{6R}}]}^2$
• D. $\frac{m}{20}[u^2+\frac{113}{200}\frac{GM}{R}]$

Answer 1

The correct option is A $5m[u^2-\frac{119}{200}\frac{GM}{R}]$

As we know,
$$\begin{array}{cc}& T.E_{ground}=T.E_R\\ & \frac{1}{2}m{u}^2+\left(\frac{-GMm}{R}\right)=\frac{1}{2}m{v}^2+\left(\frac{-GMm}{2R}\right)\\ & \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+\left(\frac{-GMm}{2R}\right)\\ & v^2=u^2+\left(\frac{-GM}{R}\right)& & ⟹v=\sqrt{u^2+\left(\frac{-GM}{R}\right)}\\ \text{(1)}\end{array}$$
The rocket splits at height $R$. Since, seperation of rocket is impulsive therefore conservation of momentum in both radial and tangential direction can be applied.
$$\begin{array}{cc}& \frac{m}{10}{V}_T=\frac{9m}{10}\sqrt{\frac{GM}{2R}}\\ & \frac{m}{10}{V}_r=m\sqrt{u^2-\frac{GM}{R}}\end{array}$$
Kinetic energy of rocket $=\frac{1}{2}\times \frac{m}{10}(V_T^2+V_R^2)=\frac{m}{20}(81\frac{GM}{2R}+100u^2-100\frac{GM}{R})$

$$=\frac{m}{20}(100u^2-\frac{119GM}{2R})$$
$$=5m(u^2-\frac{119GM}{200R})$$