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.

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