A rocket starts vertically upward with speed $v_{0}$ . Show that its speed $v$ at height $h$ is given by $v_{0}^{2} - v^{2} = \frac{2 g h}{1 + \frac{h}{R}}$ , hence $R$ is the radius of the Earth.

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Solution

By principle of conservation of energy, we have
$(M . E .)_{i n i t i a l} = (M . E .)_{f i n a l}$
$(M . E .)_{i n i t i a l} = T o t a l e n e r g y o f t h e r o c k e t a t t h e s u r f a c e o f t h e e a r t h = K . E ._{i n i t i a l} + P . E ._{i n i t i a l}$
$(M . E .)_{f i n a l} = T o t a l e n e r g y o f t h e r o c k e t a t a h e i g h t h = K . E ._{f i n a l} + P . E ._{f i n a l}$

Given, the rocket starts vertically upwards with speed $v_{0}$
Let the speed of the rocket at height $h$ be $v$ and mass of earth and rocket be $M$ and $m$ respectively and radius of earth be $R$ .
Then,
$(M . E .)_{i n i t i a l} = \frac{1}{2} m v_{0}^{2} - \frac{G M m}{R}$

$(M . E .)_{f i n a l} = \frac{1}{2} m v^{2} - \frac{G M m}{R + h}$

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

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

$\frac{1}{2} m (v_{0}^{2} - v^{2}) = G M m (\frac{1}{R} - \frac{1}{R + h})$

$\frac{1}{2} m (v_{0}^{2} - v^{2}) = G M m (\frac{h}{R (R + h)})$

We know, $G M = g R^{2}$

$\frac{1}{2} m (v_{0}^{2} - v^{2}) = g R^{2} m (\frac{h}{R (R + h)})$

$\frac{1}{2} m (v_{0}^{2} - v^{2}) = g R^{2} m ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{h}{R^{2} (1 + \frac{h}{R})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$

$\frac{1}{2} m (v_{0}^{2} - v^{2}) = \frac{m g h}{(1 + \frac{h}{R})}$

$(v_{0}^{2} - v^{2}) = \frac{2 g h}{(1 + \frac{h}{R})}$

Hence, proved.

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