A rocket of mass `m` is field vertically from the surface of Mars of mass `M`, radius `R` with a sped `upsilon`. If `20%` of its initial energy is lost due to Martain atmosheric resistance, how far will the rocket go from the surface of Mars before returing to it? Let `G` be the gravational constant.

Explanation

Let `m =` mass of the rocket, `M =` mass of the Mars and `R =` radius of Mars. Let `upsilon` be the initial velocity of rocket.
Initial `K.E. = (1)/(2) m upsilon^(2)`, Initial `P.E. = -(GM m)/(R )`, Total initial energy `= (1)/(2) m upsilon^(2) - (GM m)/(R )`
Since, `20%` of `K.E.` is lost, only `80%` is left behind to rach the height. Therefore,
Total initial energy avilable `= (80)/(100) xx (1)/(2) m upsilon^(2) - (GM m)/(R ) = 0.4 m upsilon^(2) - (G M m)/(R )`
if the rocket reaches the height point which is at a height `h` from the surface of mars, its `K.E.` is zero and
`P.E. = (-GM m)/((R + h))`
Using principle of conservation of energy, we have
`0.4 m upsilon^(2) - (GM m)/(R ) = (GM m)/((R + h))` or `(GM)/(R + h) = (GM)/(R ) 0.4 upsilon^(2) = (1)/(R )[Gm - 0.4 R upsilon^(2)]`
or `(R + h)/(R ) = (GM)/(GM - 0.4R upsilon^(2))`
or `(h)/(R ) = (GM)/(GM - 0.4 R upsilon^(2)) - 1 = (0.4 upsilon^(2)R)/(GM - 0.4 R upsilon^(2))`
or `h = (0.4 upsilon^(2) R^(2))/(GM - 0.4 R upsilon^(2))`
`= (0.4 xx (2 xx 10^(3))^(2) xx (3.395 xx 10^(6))^(2))/((6.67 xx 10^(-11)) xx (6.4 xx 10^(23)) - 0.4 xx (3.395 xx 10^(6)) xx (2 xx 10^(3))^(2)m)`
`= 495 km`