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

Total energy of rocket = K.E. + P.E.
`E = (1)/(2) mv^(2) - (GMm)/(R )`
Due to loss of 20% of energy, energy remained = 80% of E
`= (80)/(100)((1)/(2)mv^(2) - (GMm)/(R ))`
At the highest point h from the surface of mars, K.E. will be zero and total energy will be wholly potential.
Therefore we can write
`(4)/(5)((1)/(2)mv^(2) - (GMm)/(R )) = -(GMm)/(R + h)`
`(4)/(2 xx 5)(mv^(2) - (2GMm)/(R )) = -(GMm)/(R + h)`
`(2)/(5) (v^(2) - 2(GM)/(R )) = - (GM)/(R + h)`
`(2)/(5) (v^(2) R - 2GM) = - (GMR)/(R + h)`
Solving, we get `R + h = (5)/(2)(-(GMR)/(v^(2)R-2GM))`
`R + h = (5)/(2)((-GMR)/(v^(2)R - 2GM))`
`R + h = (-5 xx 6.67 xx 10^(-11) xx 6.4 xx 10^(23) xx 3,395 xx 1,000)/(2(2 xx 10^(3) xx 2 xx 10^(3) xx 3,395 xx 10^(3) - 2 xx 6.67 xx 10^(-11) xx 6.4 xx 10^(23)))`
`= (-72,46,28,800xx 10^(12))/(2(13,580 xx 10^(12) - 85.376 xx 10^(12)))`
`= 5.046 xx 10^(6) m`
`:. h = 5.046 xx 10^(6) - 3.395 xx 10^(3) = 1,651 km`