Ncert Exercises & Solutions

8.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.

$T o t a l e n e r g y = P o t e n t i a l e n e r g y + K i n e t i c e n e r g y$
$= (\frac{- G m M_{e}}{R_{e}}) + \frac{1}{2} m v^{2}$
$A t h i g h e s t p o i n t v e l o c i t y = 0$
$S o, t o t a l e n e r g y o f r o c k e t = (\frac{- G m M_{e}}{R_{e}}) + 0$
$A p p l y i n g c o n s e r v a t i o n o f e n r g y,$
$T o t a l e n e r g y i n i t i a l l y (a t e a r t h' s s u r f a c e) = T o t a l e n e r g y a t h e i g h t h$
$\Rightarrow (\frac{- G m M_{e}}{R_{e}}) + \frac{1}{2} m v^{2} = (\frac{- G m M_{e}}{R_{e}})$
$\Rightarrow \frac{1}{2} v^{2} = \frac{g R_{e} h}{R_{e} + h}$
$\Rightarrow h = \frac{R_{e} v^{2}}{2 g R_{e} - v^{2}}$
$\Rightarrow h = \frac{6.4 \times 10^{6} \times {(5 \times 10^{3})}^{2}}{2 \times 9.8 \times 6.4 \times 10^{6} - {(5 \times 10^{3})}^{2}} = 1.6 \times 10^{6} m$
$T h u s t h e h e i g h t a c h i e v e d b y t h e r o c k e t f r o m e a r t h' s c e n t r e = R_{e} + h$
$= 6.4 \times 10^{6} + 1.6 \times 10^{6}$
$= 8 \times 10^{6} m$

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