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):
Answer: (B) $5M\left(u^2 - \dfrac{119}{200}\dfrac{GM_e}{R}\right)$
**Speed at height $R$** (distance $2R$ from the centre):
$$v^2 = u^2 - 2GM_e\left(\frac1R - \frac1{2R}\right) = u^2 - \frac{GM_e}{R}$$
This velocity is radial (vertical).
**After ejection** the satellite (mass $\tfrac{9M}{10}$) must move in a circle of radius $2R$: no radial velocity and tangential speed $v_0 = \sqrt{\dfrac{GM_e}{2R}}$.
**Momentum conservation**, with rocket velocity components $v_r$ (radial) and $v_t$ (tangential):
Radial: $Mv = \dfrac{M}{10}v_r \Rightarrow v_r = 10v$
Tangential: $0 = \dfrac{9M}{10}v_0 + \dfrac{M}{10}v_t \Rightarrow v_t = -9v_0$
**Kinetic energy of the rocket:**
$$K = \frac12\cdot\frac{M}{10}\left(100v^2 + 81v_0^2\right) = \frac{M}{20}\left[100\left(u^2 - \frac{GM_e}{R}\right) + \frac{81GM_e}{2R}\right]$$
$$K = 5M\left(u^2 - \frac{GM_e}{R} + \frac{81}{200}\frac{GM_e}{R}\right) = 5M\left(u^2 - \frac{119}{200}\frac{GM_e}{R}\right)$$
Solution by Sreeraj P, M.Sc Physics