JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Conservation of Momentum

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) $\frac{m}{20} (u - \frac{2 GM}{3 R})^{2}$
(B) $\frac{3 m}{8} (u + \frac{5 GM}{6 R})^{2}$
(C) $\frac{m}{20} (u^{2} - \frac{113}{200} \frac{GM}{R})$
(D) $5 m (u^{2} - \frac{119}{200} \frac{GM}{R})$ ✓ Correct

Answer: (D)

Solution

Launched radially with speed

u

, the satellite reaches height

R

, a distance

2 R

from the centre. Energy conservation,

\frac{1}{2} u^{2} - \frac{GM}{R} = \frac{1}{2} v^{2} - \frac{GM}{2 R}

, gives its radial speed there as

v^{2} = u^{2} - \frac{GM}{R}

. To leave the remaining body in a circular orbit at radius

2 R

it must end with the tangential orbital speed

v_{c} = \frac{GM}{2 R}

and no radial motion. Conserving momentum when the piece of mass

\frac{m}{10}

is ejected: radially

m v = \frac{m}{10} V_{r}

V_{r} = 10 v

, and tangentially

0 = \frac{9 m}{10} v_{c} + \frac{m}{10} V_{t}

V_{t} = - 9 v_{c}

. The kinetic energy carried off by the rocket is $\frac12\cdot\frac{m}{10}\left(V_r^2+V_t^2\right)=\frac{m}{20}\left(100v^2+81v_c^2\right)=\frac{m}{20}\left(100\Big(u^2-\frac{GM}{R}\Big)+\frac{81GM}{2R}\right)=5m\left(u^2-\frac{119}{200}\frac{GM}{R}\right).

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Conservation of Momentum

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