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Gravitation

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Question : 25 of 25
Marks: +1, -0
A rocket is fired ‘vertically’ from the surface of Mars with a speed of 2 km s −1^{-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= 6.4 \times 10^{23} kg; radius of Mars =3395= 3395 km; G=6.67×10−11G = 6.67 \times 10^{-11} N m2^{2} kg−2^{-2}.
Solution:  
Let m = mass of the rocket, M = mass of the Mars and R = radius of Mars. Let v be the initial velocity of rocket.
Initial K.E.=12mv2;\text{K.E.} = \frac{1}{2} m v^{2} ; Initial P.E. =−GMmR= - \frac{G M m}{R}
Total initial energy =12mv2−GMmR= \frac{1}{2} m v^{2} - \frac{G M m}{R}
Since 20% of K.E. is lost, only 80% is left behind to reach the height.
Therefore,
Total initial energy available
=80100×12mv2−GMmR=0.4mv2−GMmR= \frac{80}{100} \times \frac{1}{2} m v^{2} - \frac{G M m}{R} = 0.4 m v^{2} - \frac{G M m}{R}
If the rocket reaches the highest point which is at a height h from the surface of Mars, its K.E. is zero and P.E. =−GMmR+h= \frac{-G M m}{R+h}
Using principle of conservation of energy, we have
0.4mv2−GMmR=−GMmR+h0.4 m v^{2} - \frac{G M m}{R} = - \frac{G M m}{R+h}
or GMR+h=GMR−0.4v2=1R[GM−0.4Rv2]\frac{G M}{R+h} = \frac{G M}{R} - 0.4 v^{2} = \frac{1}{R} [G M - 0.4 R v^{2}]
or R+hR=GMGM−0.4Rv2\frac{R+h}{R} = \frac{G M}{G M - 0.4 R v^{2}}
or hR=GMGM−0.4Rv2−1\frac{h}{R} = \frac{G M}{G M - 0.4 R v^{2}} - 1 =0.4Rv2GM−0.4Rv2= \frac{0.4 R v^{2}}{G M - 0.4 R v^{2}}
or h=0.4×(2×103)2×(3.395×106)2[(6.67×10−11)×(6.4×1023)−0.4×(3.395×106)×(2×103)2] mh = \frac{0.4 \times (2 \times 10^{3})^{2} \times (3.395 \times 10^{6})^{2}}{\left[ (6.67 \times 10^{-11}) \times (6.4 \times 10^{23}) - 0.4 \times (3.395 \times 10^{6}) \times (2 \times 10^{3})^{2} \right]} \text{ m}
≈495 km.\approx 495 \text{ km}.
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