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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 x 1023 kg; radius of Mars = 3395 km; G = 6.67 x 10-11 N m2 kg-2
Asked by Abhay Last Modified
Answer 
The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
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The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
read less
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