An asteroid is moving directly towards the centre of the earth. When at a distance of 10 R (R is the radius of the earth) from the earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s)Give your answer to the nearest integer in kilometer/s ________

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answer is 16.

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Detailed Solution

$\begin{array}{l} U_{1} + K_{1} = U_{2} + K_{2} \\ - \frac{G M_{e} m}{10 R} + \frac{1}{2} m v_{0}^{2} = - \frac{G M_{e} m}{R} + \frac{1}{2} m v^{2} \\ + \frac{9}{10} \times \frac{G M_{e} m}{R} + \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} \\ \frac{9}{10} \times \frac{1}{2} M \times v_{e}^{2} + \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{-} \\ v^{2} = \frac{9}{10} v_{e}^{2} + v_{0}^{2} \\ = \frac{9}{10} \times {(11.2)}^{2} + {(12)}^{2} \\ v^{2} = 112.896 + 144 \Rightarrow v = 16.027 = 16 km/s \end{array}$

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