Ncert Exercises & Solutions

8.25 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67×10-11 N m2 kg–2.

T h e i n i t i 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} m v^{2}

T h e i n i t i a l p o t e n t i a l e n e r g y = \frac{- G M m}{R}

T o t a l i n i t i a l e n e r g y = \frac{1}{2} m v^{2} + \frac{- G M m}{R}

I f 20 % o f k i n e t i c e n e r g y i s l o s t .

T h e n n e t i n i t i a l e n e r g y = 0.4 m v^{2} - \frac{G M m}{R}

T h e f i n a l e n e r y = \frac{G M m}{R + h}

A p p l y i n g c o n s e r v a t i o n o f e n e r g y :

0.4 m v^{2} - \frac{G M m}{R} = \frac{G M m}{R + h}

\Rightarrow h = \frac{R}{\frac{G M}{0.4 v^{2} R} - 1}

\Rightarrow h = \frac{0.4 R^{2} v^{2}}{G M - 0.4 v^{2} R}

\Rightarrow h = \frac{18.442 \times 10^{18}}{42.688 \times 10^{12} - 5.432 \times 10^{12}}

= 495 k m .

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