An asteroid is moving directly towards the centre of the earth. When at a distance of 10R(R is the radius of the earth) from the earth’s centre, it has a speed of 12 km/s. Neglecting the effect of earth’s 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-------

Moderate

Solution

$\begin{array}{l} A p p l y i n g C o n s e r v a t i o n o f E n e r g y, \\ U_{i} + K_{i} = U_{f} + K_{f} \\ \Rightarrow - \frac{G M m}{10 R} + \frac{1}{2} m {(12 \times 10^{3})}^{2} = \frac{- G M m}{R} + \frac{1}{2} m {υ_{f}}^{2} \\ \Rightarrow \frac{1}{2} m {υ_{f}}^{2} = \frac{9 G M m}{10 R} + \frac{1}{2} m {(12 \times 10^{3})}^{2} \\ \Rightarrow {υ_{f}}^{2} = \frac{18 G M}{10 R} + {(12 \times 10^{3})}^{2} \\ \Rightarrow {υ_{f}}^{2} = \frac{9 \times 2 G M}{10 R} + {(12 \times 10^{3})}^{2} \\ \Rightarrow υ_{f} = \sqrt{\frac{9 {(V_{e})}^{2}}{10} + {(12 \times 10^{3})}^{2}} \\ \Rightarrow υ_{f} = \sqrt{\frac{9 {(11.2 \times 10^{3})}^{2}}{10} + {(12 \times 10^{3})}^{2}} [E s c a p e V e l o c i t y = V_{e} = \sqrt{\frac{2 G M}{R}} = 11.2 k m / s] \\ \Rightarrow υ_{f} = \sqrt{112 + 144} \times 10^{3} m / \sec \\ \Rightarrow υ_{f} = 16 k m / \sec \end{array}$

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