A rocket is fired with a velocity 0.6 times the escape velocity on the surface of earth.How high will it go from the surface?
We know,
Radius of earth, R = 6400000 m
Mass of object be m
Speed of the rocket, v = 0.6(2GM/R)1/2
The energy of the rocket at the surface of earth is = ½ mv2 – GMm/R
At the maximum height from the surface the velocity of the rocket is zero and PE is = -GMm/(R+h)
So, by law of conservation of energy,
½ mv2 – GMm/R = -GMm/(R+h)
=> ½ {0.6(2GM/R)1/2}2 – GM/R = -GM/(R+h)
=> ½ {0.36(2GM/R)} – GM/R = -GM/(R+h)
=> 0.36 – 1 = -R/(R+h)
=> 0.64(R+h) = R
=> h = (R – 0.64R)/0.64
=> h = 0.36R/0.64
=> h = (0.36)(6400000)/0.64
=> h = 3600000 m
This is the height reached by the rocket from the surface of the earth.