A satellite of mass M is launched vertically upwards with 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):

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$5 M (u^{2} - \frac{119}{200} \frac{G M_{e}}{R})$

$\frac{3 M}{8} {[u + \sqrt{\frac{5 G M_{e}}{6 R}}]}^{2}$

$\frac{M}{20} {[u - \sqrt{\frac{2 G M_{e}}{3 R}}]}^{2}$

$\frac{M}{20} (u^{2} + \frac{113}{200} \frac{G M_{e}}{R})$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} A p p l y i n g e n e r g y c o n s e r v a t i o n, t o f i n d v e l o c i t y o f s a t e l l i t e w h e n i t r e a c h e s a t h e i g h t R . \\ \frac{- G M_{e} M}{R} + \frac{1}{2} M u^{2} = \frac{- G M_{e} M}{2 R} + \frac{1}{2} M v^{2} \\ \Rightarrow \frac{- G M_{e}}{R} + \frac{1}{2} u^{2} = \frac{- G M_{e}}{2 R} + \frac{1}{2} v^{2} \\ \Rightarrow \frac{- G M_{e}}{R} + \frac{G M_{e}}{2 R} + \frac{1}{2} u^{2} = + \frac{1}{2} v^{2} \\ \Rightarrow \frac{- 2 G M_{e} + G M_{e}}{2 R} + \frac{1}{2} u^{2} = + \frac{1}{2} v^{2} \\ \Rightarrow \frac{- G M_{e}}{R} + u^{2} = v^{2} \\ \Rightarrow v = \sqrt{u^{2} - \frac{G M_{e}}{R}} \end{array}$

$\begin{array}{l} A s t h e s a t e l l i t e r e a c h a t h e i g h t R, i t h a s o n l y r a d i a l v e l o c i t y v, n o \tan g e n t i a l v e l o c i t y . \\ A f t e r e j e c t i n g r o c k e t, s a t e l l i t e h a s o n l y \tan g e n t i a l v e l o c i t y a n d m o v e s w i t h o r b i t a l s p e e d = \sqrt{\frac{G M_{e}}{2 R}} \\ L e t t h e r o c k e t h a s v e l o c i t y V_{r} i n r a d i a l d i r e c t i o n a n d V_{T} i n \tan g e n t i a l d i r e c t i o n . \\ A p p l y i n g m o m e n t u m c o n s e r v a t i o n i n \tan g e n t i a l d i r e c t i o n, \\ \frac{M}{10} V_{T} = \frac{9 M}{10} \sqrt{\frac{G M_{e}}{2 R}} \\ \Rightarrow T a n g e n t i a l v e l o c i t y, V_{T} = 9 \sqrt{\frac{G M_{e}}{2 R}} = \sqrt{\frac{81 G M_{e}}{2 R}} - - - - - - (1) \\ N o w a p p l y i n g m o m e n t u m c o n s e r v a t i o n i n r a d i a l d i r e c t i o n, \\ \frac{M}{10} V_{r} = M (\sqrt{u^{2} - \frac{G M_{e}}{R}}) \\ \Rightarrow V_{r} = 10 (\sqrt{u^{2} - \frac{G M_{e}}{R}}) - - - - - - (2) \\ T o t a l K i n e t i c e n e r g y o f r o c k e t = \frac{1}{2} \frac{M}{10} (V_{T}^{2} + V_{r}^{2}) \\ = \frac{M}{20} (81 \frac{G M_{e}}{2 R} + 100 u^{2} - 100 \frac{G M_{e}}{R}) \\ F r o m (1) & (2) \\ = \frac{M}{20} (100 u^{2} - \frac{119 G M_{e}}{2 R}) \\ = \frac{100 M}{20} (u^{2} - \frac{119 G M_{e}}{200 R}) \\ = 5 M (u^{2} - \frac{119 G M_{e}}{200 R}) \end{array}$