First slide

Gravitational potential and potential energy

Question

A body is released from a point of distance R' from the centre of earth. Its velocity at the time of striking the earth will be $(R^{'} > R_{e})$

Moderate

A

$\sqrt{2 {gR}_{e}}$
B

$\sqrt{R_{e} g}$
C

$\sqrt{2 g (R^{'} - R_{e})}$
D

$\sqrt{2 {gR}_{e} (1 - \frac{R_{e}}{R^{'}})}$

Solution

Let the mass of the particle be m
PE at a distance of $R^{'} = - (GMm) / R^{'}$
PE at a distance of $R_{e} = - (GMm) / R_{e}$
Decrease in PE = Increase in KE
$\begin{array}{l} \Rightarrow - \frac{GMm}{R^{'}} + \frac{GMm}{R_{e}} = \frac{1}{2} {mv}^{2} \\ v^{2} = 2 GM [\frac{1}{R_{e}} - \frac{1}{R^{'}}] \Rightarrow v^{2} = \frac{2 GM}{R_{e}} [1 - \frac{R_{e}}{R^{'}}] \\ ∴ v^{2} = \frac{2 GM}{R_{e}} (1 - \frac{R_{e}}{R^{'}}) \Rightarrow v = \sqrt{\frac{2 {GMR}_{e}}{R_{e}^{2}}} (1 - \frac{R_{e}}{R^{'}}) \\ ∴ v = \sqrt{2 {gR}_{e} (1 - \frac{R_{e}}{R^{'}})} \end{array}$

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