A body is projected vertically upwards from earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence then the

Question

A body is projected vertically upwards from earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence then the height up to which it rises is ( R = radius of earth)

TestHub · Accepted Answer

The minimum kinetic energy required for a body to escape earth's gravitational field is given by:

$K_{e} = \frac{1}{2} m v_{e}^{2} = \frac{1}{2} m (\frac{2 GM}{R})^{2} = \frac{GM m}{R}$

Given that the kinetic energy of projection is half of the escape kinetic energy:

$K_{i} = \frac{1}{2} K_{e} = \frac{GM m}{2 R}$

Let the body rise to a height $h$ above the surface of the earth. At the maximum height, its velocity becomes zero, so the final kinetic energy $K_{f} = 0$ .

Applying the principle of conservation of mechanical energy:

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{GM m}{2 R} - \frac{GM m}{R} = 0 - \frac{GM m}{R + h}$

$- \frac{GM m}{2 R} = - \frac{GM m}{R + h}$

$\frac{1}{2 R} = \frac{1}{R + h}$

$R + h = 2 R$

$h = R$

Answer: $R$

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