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):

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

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

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

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

answer is C.

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Detailed Solution

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

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