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Question

The escape velocity of a body on the earth's surface is Ve. A body is thrown vertically up with a speed (kVe)(k<1). The maximum height reached by the body above the earth is

A
R(k21k2)
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B
R2k2(1k2)
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C
R(1k2k)
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D
Rk2
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Solution

The correct option is A R(k21k2)
Given Ve is the escape velocity of a body on the earth's surface.
Let us assume the body reaches a maximum height h above earth's surface when thrown vertically up with speed kVe.

As there are no other forces than the gravitational foce in action, we can conserve the total mechanical energy
(Kinetic energy + Gravitational Potential energy) of the system.

If mass of the particle is m, mass of earth is M, radius of earth is R.
From prior knowledge we know that
Ve=2GMR

Initially the kinetic energy is 12m(kVe)2

At the maximum height, Kinetic energy of the particle is 0.

Initial potential energy on the surface of earth is GMmR

Final potential energy at a height h from the surface of earth is GMmR+h

12m(kVe)2GMmR=0GMmR+h

Substituting Ve from above formula,

12mk2(2GMR)GMmR=GMmR+h

GMmRk2GMmR=GMmR+h

1R(1k2)=1R+h

R+h=R(11k2)

h=R(11k2)R

h=R(k21k2)

The maximum height reached by the body is R(k21k2)

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