A projectile is fired vertically upwards from the surface of Earth with a velocity of $k v_{e}$ where $v_{e}$ is the escape velocity and $k < 1$ . Neglecting air resistance, the maximum height to which it will rise, measured from the centre of the Earth, is ( $R_{E}$ = radius Earth)

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$\frac{1 - k^{2}}{R_{E}}$

$\frac{R_{E}}{1 - k^{2}}$

$\frac{R_{E}}{k^{2}}$

$\frac{k^{2}}{R_{E}}$

answer is A.

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Detailed Solution

Let a projectile be fired vertically upwards from the Earth's surface with a veloclty $v$ and reaches a height $h$ . Energy of the projectile at the surface of the Earth is

$E_{i} = \frac{1}{2} m v^{2} - \frac{G M_{E} m}{R_{E}}$

Energy of the projectile at a height $h$

$E_{f} = - \frac{G M_{E} m}{R_{E} + h}$ ( $∴$ At height $h$ velocity of the projectile is zero.)

According to law of conservation of energy,

$E_{i} = E_{f} \frac{1}{2} m v^{2} - \frac{G M_{E} m}{R_{E}} = - \frac{G M_{E} m}{R_{E} + h} \frac{1}{2} m v^{2} = G M_{E} m [\frac{1}{R_{E}} - \frac{1}{R_{E} + h}] = \frac{G M_{E} m h}{(R_{E}) (R_{E} + h)} \frac{1}{2} m v^{2} = \frac{m g h R_{E}}{R_{E} + h} (∴ g = \frac{G M_{E}}{R_{E}^{2}})$

As per question,

$v = k v_{e} = k \sqrt{2 g R_{E}} (v_{e} = \sqrt{2 g R_{E}})$

and $h = r - R_{E}$

$∴ \frac{1}{2} m k^{2} (2 g R_{E}) = \frac{m g (r - R_{E}) R_{E}}{r} k^{2} = \frac{r - R_{E}}{r} \Rightarrow (1 - k^{2}) r = R_{E} \Rightarrow r = \frac{R_{E}}{1 - k^{2}}$