A rocket is fired vertically with a speed of 5 km `s^(-1)` from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = `6.0 × 10^(24)` kg, mean radius of the earth = `6.4 xx 10^(6) m , G = 6.67 xx 10^(-11) " N "m^(2) kg^(-2)`.

Explanation

`implies` Mass of planet mars
`M = 6.4 xx 10^(23) kg`
Radius of planet mars
`R = 3395 km = 3.395 xx 10^(6) m`
Velocity of rocket `v = 2 kms^(-1) = 2 xx 10^(3) ms^(-1)`
Maximum height gained by the rocket = h
{Change in potential energy of rocket} = {Final potential energy - Initial potential energy}
`=-G(Mm)/((R+h))-(-G(Mm)/R)`
`=- G(Mm)/(R) + (GMm)/(R) `
`= GMm [1/R-(1)/(R+H)]`
`Delta.p E = GMm (h)/((R+h)R) .....(1)`
Since 20% potential energy is used in atmo - sphere .
`:. 80%` kinetic energy `(Deltap.E) = 80%` kinetic energ .
`= 1/2 mv^(2) xx(80)/100`
`= 0.4 mv^(2) " "...(2)`
From law of conservation of energy .
`Deltap . E = Delta K.E `
`(GMmh)/(R(R+H)) = 0.4 mv^(2)`
`" [ " :. ` From equ (1) and (2)]
`(Gmh)/(R^2+Rh)= 0.4 v^2`
`:. GMh = 0.4 v^(2) R^(2) + 0.4 v^(2) Rh`
`:. GMh - 0.4 v^(2) Rh = 0.4 v^(2) R^(2)`
`h[GM-0.4v^(2)R] = 0.4 v^(2)R^(2)`
`:. h = (0.4 v^(2) R^(2))/(GM - 0.4 v^(2)R)`
`:. h = R^(2) / ((GM)/(0.4 v^(2))-R)`
`:. h = ((3.395 xx10^(6))^2)/((6.67xx10^(-11) xx 6.4 xx10^(3))/(0.4 xx2 xx 10^(3))-3.395 xx10^(6))`
`:. h = (11.5 26 xx10^(12))/(26.68 xx10^(-6)-3.395 xx10^6)`
`= (11.526 xx10^(12))/(23.68 xx10^(6) - 3.395 xx10^6)`
`= (11.526 xx10^(12))/(23.285 xx10^6)`
`= 0.49499 xx 10 ^6`
`= 495 xx 10^3 m`
`:. h = 495 km`