A body is projected in vertically upward direction from the surface of the earth of radius ' R ' into space with velocity ' n V e ' ( n

Question

A body is projected in vertically upward direction from the surface of the earth of radius ' R ' into space with velocity ' n V e ​ ' ( n < 1 ) . The maximum height from the surface of earth to which a body can reach is

TestHub · Accepted Answer

By the law of conservation of mechanical energy, the total energy at the surface of the earth is equal to the total energy at the maximum height $h$ .

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

$\frac{1}{2} m v^{2} - \frac{GM m}{R} = 0 - \frac{GM m}{R + h}$

The escape velocity is given by $V_{e} = \frac{2 GM}{R}$ .

Given that the initial velocity is $v = n V_{e} = n \frac{2 GM}{R}$ , substituting this into the energy equation gives:

$\frac{1}{2} m (n^{2} \frac{2 GM}{R}) - \frac{GM m}{R} = - \frac{GM m}{R + h}$

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

Dividing both sides by $\frac{GM m}{R}$ :

$n^{2} - 1 = - \frac{R}{R + h}$

$1 - n^{2} = \frac{R}{R + h}$

$R + h = \frac{R}{1 - n ^{2}}$

$h = \frac{R}{1 - n ^{2}} - R$

$h = \frac{R - R ( 1 - n ^{2} )}{1 - n ^{2}}$

$h = \frac{n ^{2} R}{1 - n ^{2}}$

Answer: $\frac{n ^{2} R}{( 1 - n ^{2} )}$

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