A particle of mass is projected with a velocity from the surface of the earth. ( = escape velocity) The maximum height above the surface reached by the particle is [2021]

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Solution

Q.
A particle of mass $' m'$ is projected with a velocity $v = k V_{e} (k < 1)$ from the surface of the earth. ( $V_{e}$ = escape velocity)
The maximum height above the surface reached by the particle is [2021]

1 $\frac{R k^{2}}{1 - k^{2}}$

2 $R {(\frac{k}{1 - k})}^{2}$

3 $R {(\frac{k}{1 + k})}^{2}$

4 $\frac{R^{2} k}{1 + k}$

Ans.
(1)

The particle is fired vertically upwards from the Earth’s surface with a velocity $v$ and reaches a height $h$ .

Energy of the particle 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 particle at a height $h$

$E_{f} = - \frac{G M_{E} m}{R_{E} + h} (∵at height h, velocity of the particle is zero)$

According to law of conservation of energy,

   $E_{i} = E_{f}$

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

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

$\frac{1}{2} m v^{2} = \frac{m g h R_{E}}{R_{E} + h}$ ...(i) $(∵ g = \frac{G M_{E}}{R_{E}^{2}})$

As per question,

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

Using (i) and (ii), we get

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

If $R_{E} = R$ , then $h = \frac{R k^{2}}{1 - k^{2}}$

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