An asteroid of mass m is approaching earth, initially at a distance of 10 R_e with speed v_i. It hits the earth with a speed v_f (R_e and M_e are radius and mass of earth), then :

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$v_{f}^{2} = v_{i}^{2} + \frac{2 Gm}{R_{e}} (1 - \frac{1}{10})$

$v_{f}^{2} = v_{i}^{2} + \frac{2 {GM}_{e}}{R_{e}} (1 + \frac{1}{10})$

$v_{f}^{2} = v_{i}^{2} + \frac{2 Gm}{M_{e} R} (1 - \frac{1}{10})$

$v_{f}^{2} = v_{i}^{2} + \frac{2 {GM}_{e}}{R_{e}} (1 - \frac{1}{10})$

answer is C.

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Detailed Solution

$\begin{array}{l} - \frac{{GM}_{e} m}{10 R_{e}} + \frac{1}{2} {mv}_{i}^{2} = - \frac{{GM}_{e} m}{R_{e}} + \frac{1}{2} {mv}_{f}^{2} \\ \Rightarrow v_{f}^{2} = v_{i}^{2} + \frac{2 {GM}_{e}}{R_{e}} (1 - \frac{1}{10}) \end{array}$

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