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### Gravitation NCERT Solutions Class 11 Physics - Solved Exercise Question 8.25

Question 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.
Solution:
Initial velocity of the rocket, v = 2 km/s = 2 × 103 m/s
Mass of Mars, M = 6.4 × 1023 kg
Radius of Mars, R = 3395 km = 3.395 × 106 m
Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2
Mass of the rocket = m
Initial kinetic energy of the rocket = (1/2)mv2
Initial potential energy of the rocket = -GMm / R
Total initial energy = (1/2)mv2- GMm / R
If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available = (80/100) X (1/2) mv2 - GMm / R  =  0.4mv2 - GMm / R
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = -GMm / (R + h)

Applying the law of conservation of energy for the rocket, we can write:
0.4mv2 - GMm / R  =  -GMm / (R + h)
0.4v2 = GM / R  -  GM / (R + h)
= GMh / R(R + h)
(R + h) / h  =  GM / 0.4v2R
R / h  =  ( GM / 0.4v2R )  -  1
h = R / [ (GM / 0.4v2R) - 1 ]
= 0.4R2v2 / (GM - 0.4v2R)
= 0.4 X (3.395 X 106)2 X (2 X 103)2 / [ 6.67 X 10-11 X 6.4 X 1023  0.4 X (2 X 103)2 X (3.395 X 106) ]
= 18.442 X 1018 / [ 42.688 X 1012  -  5.432 X 1012 ]
= 18.442 X 106 / 37.256
= 495 X 103 m = 495 km