Question

The escape velocity on earth is 11.2 km $s^{- 1}$ . If the body is projected out with twice this velocity, then the speed of the body far away from the earth, ignoring the presence of any other object in universe, will be [MP PET 2013]

Moderate

Solution

short cut : $v_{i n f} = \sqrt{v^{2} - {v^{2}}_{e}}$
Using energy conservation law, $(PE)_{i} + (KE)_{i} = (PE)_{f} + (KE)_{f}$
$\frac{- G M m}{R} + \frac{1}{2} m {(2 v_{e})}^{2} = 0 + \frac{1}{2} m {(v^{'})}^{2}$
$\Rightarrow - m (\frac{1}{2} v_{e}^{2}) + \frac{1}{2} m {(2 v_{e})}^{2} = \frac{1}{2} m {(v^{'})}^{2} (∵ \frac{G M}{R} = \frac{v_{e}^{2}}{2})$
$\Rightarrow v^{'} = \sqrt{3} v_{e} = \sqrt{3} \times 11.2 = 19.4 km / s$

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