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A rocket is fired vertically from the surface of Mars with a speed of `2kms^(-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 `=6.4xx10^(23)kg`, radius of Mars `=3395 km`,

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From the law of conservation of energy
`(-GMm)/R+(80/100)1/2 mV^(2)=(-GMm)/(R+h)=0`
`GMm(1/R-1/(R+h))=1.6xx10^(6)`
`=1-(3.395xx10^(6)xx1.6xx10^(6))/(6.67xx10^(-11)xx6.4xx10^(23))`
`R+h=R/0.873=3395/0.873=3888.9km`
`:.` The required height up to which the rocket will go
`=9888.9-3395=493.9km`
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