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`,

Explanation

From the law conservation of energy `(-GMm)/R+(80/100)1/2mV^(2)=(-GMm)/(R+h)+0`
`GMm(1/R-1/(R+h))=0.4m (2xx10^(3))^(2)`
`rArr (GM)/R(1-R/(R+h))=1.6xx10^(6)`
`=1-(3.395xx10^(6)xx1.6xx10^(6))/(6.67xx10^(-11)xx6.4xx10^(23))`
`R+h=R/0.873=3,395/0.873=3,888.9 km`
`:.` The required height up to which the rocket will go`3,888.9-3,395=493.9 km`