A rocket starts vertically upward with speed $v_{0}$ . Show that its speed v at height h is given by $v_{0}^{2} - v^{2} = \frac{2 h g}{1 + \frac{h}{R}}$ where R is the radius of the earth and g is acceleration due to gravity at earth's surface. Deduce an expression for maximum height reached by a rocket fired with speed $0.9$ times the escape velocity.

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Solution

The gravitational potential energy of a mass m on earth's surface and at a height h is given by
$U (R) = - \frac{G M m}{R}$
and $U (R + h) = - \frac{G M m}{R + h}$
$∴ U (R + h) - U (R) = - G M m (\frac{1}{R + h} - \frac{1}{R})$
$= \frac{G M m h}{(R + h) R} = \frac{m h g}{1 + \frac{h}{R}}$ $[∵ G M = g R^{2}]$
This increase in potential energy occurs at the cost of kinetic energy which correspondingly decreases. If v is the velocity of the rocket at height h, then the decrease in kinetic energy is $1 / 2$ $m v_{0}^{2} - 1 / 2 m v^{2}$ . Thus,
$\frac{1}{2} m v_{0}^{2} - \frac{1}{2} m v^{2} = \frac{m h g}{1 + \frac{h}{R}}$ or $v_{0}^{2} - v^{2} = \frac{2 g h}{1 + \frac{h}{R}}$
Let $h_{m a x}$ be the maximum height reached by the rocket, at which its velocity has been reduced to zero. Thus, substituting $v = 0$ and $h = h_{m a x}$ in the last expression, we have
$v_{0}^{2} = \frac{2 g h_{m a x}}{1 + h_{m a x} / R}$
or $v_{0}^{2} (1 + \frac{h_{m a x}}{R}) = 2 g h_{m a x}$
or $v_{0}^{2} = h_{m a x} [2 g - \frac{v_{0}^{2}}{R}]$
or $h_{m a x} = \frac{v_{0}^{2}}{2 g - \frac{v_{0}^{2}}{R}}$
Now, it is given that
$v_{0} = 0.9 \times$ escape velocity
$= 0.9 \times \sqrt{(2 g R)}$
$∴ h_{m a x} = \frac{(09 \times 0.9) 2 g R}{2 g - \frac{(09 \times 0.9) 2 g R}{R}}$
$= \frac{1.62 g R}{2 g - 1.62 R}$
$= \frac{1.62 R}{0.38} = 4.26 R$ .

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