A rocket is fired vertically with a speed of 5 `kms^(-1)` from the earth's surface. How far from the earth does the rocket go before returning to the earth ? Mass of earth=`6.0xx10^(24)`kg, mean radius of the earth =`6.4xx10^(6)`m, G=`6.67xx10^(-11)Nm^(2)Kg^(-2).`

Explanation

Suppose the rocket is fired with velocity v from the Earth.s surface and it reaches a height h from the Earth.s surface where its velocity vanishes.
If m be the mass of rocket, then its total energy
Total energy at the surface of Earth = K.E. + P.E
`= (1)/(2) mv^(2) - (GM m)/(R )`
At the heighest point, kinetic energy is zero. Therefore, its total energy at a highest h is its potential energy
`= -(GMm)/((R + h))`
According to law of conservation of energy
`(1)/(2) m v^(2) - (GMm)/(R ) = -(GMm)/((R+h))`
`(1)/(2) mv^(2) = (GMm)/(R ) - (GMm)/((R + h))`
`(1)/(2) v^(2) = (gR^(2))/(R ) - (gR^(2))/((R + h))`
`(1)/(2)v^(2) = gR(1-(R )/(R+h))`
`(1)/(2)v^(2) = gR((h)/(R+h))`
`(R+h) v^(2) = 2gRh`
`Rv^(2) = 2gRh - h v^(2) = h(2gR - v^(2))`
`h = (R v^(2))/(2gR - v^(2))`
`h = ((6.4 xx 10^(6))(5 xx 10^(3))^(2))/(2 xx 9.8 xx 6.4 xx 10^(6) - (5 xx 10^(3))^(2))`
`= 1.6 xx 10^(6)m`.
Therefore, the rocket goes `(1.6 xx 10^(6) + 6.4 xx 10^(6))m`
`= 8.0 xx 10^(6)m` from Earth centre.