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.

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Initial kinetic energy of the rocket = 12mv2

    Initial potential energy of the rocket = -GMmR

    Total initial energy = 12mv2-GMmR

    If 20% of initial kinetic energy is lost due to Martian atmosphere resistance, then only 80% of its kinetic energy helps in reaching a height

    Total initial energy available = 0.8 *12mv2 -GMmR

    Maximum height reached by the rocket = h

    At this height, the velocity and hence the kinetic energy of the rocket becomes zero.

    Total energy of the rocket at height h = -GMmR+h

    Applying the law of conservation of energy for the rocket, we can write:

    0.4 *mv2 -GMmR = -GMmR+h

    0.4 v2 = GM( 1R -1R+h)

    0.4 v2 = GM( R+h-RR(R+h) )

    ...more

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