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.
Answers
Answer:
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) × (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 × (3.395 × 106)2 × (2 × 103)2 / [ 6.67 × 10-11 × 6.4 × 1023 - 0.4 × (2 × 103)2 × (3.395 × 106) ]
= 18.442 × 1018 / [ 42.688 × 1012 - 5.432 × 1012 ]
= 18.442 × 106 / 37.256
= 495 × 103 m = 495 km.
Answer:
Refer to the Attachment.
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