An asteroid of mass m is approaching earth, initially at a distance $10 R_{E}$ with speed $v_{i}$ . It hits earth with a speed $v_{f}$ ( $R_{E}$ and $M_{E}$ are radius and mass of the earth), then

v_{f}^{2} = v_{i}^{2} + \frac{2 G m}{R_{E}} (1 + \frac{1}{10})

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v_{f}^{2} = v_{i}^{2} + \frac{2 G m_{E}}{R_{E}} (1 + \frac{1}{10})

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v_{f}^{2} = v_{i}^{2} + \frac{2 G m_{E}}{R_{E}} (1 - \frac{1}{10})

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v_{f}^{2} = v_{i}^{2} + \frac{2 G m}{R_{E}} (1 - \frac{1}{10})

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Solution

The correct option is C $v_{f}^{2} = v_{i}^{2} + \frac{2 G m_{E}}{R_{E}} (1 - \frac{1}{10})$
Initial energy of the asteroid is $E_{i} = K_{i} + U i = \frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{10 R_{E}}$
Final energy of the asteroid is $E_{f} = \frac{1}{2} m v_{f}^{2} - \frac{G M_{E} m}{R_{E}}$
According to law of conservation of energy, $E_{i} = E f$

$\frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{10 R_{E}} = \frac{1}{2} m v_{f}^{2} - \frac{G M_{E} m}{R_{E}}$

$v_{f}^{2} - \frac{2 G M_{E}}{R_{E}} = v_{i}^{2} - \frac{2 G M_{E}}{10 R_{E}}$

$v_{f}^{2} = v_{i}^{2} + \frac{2 G m_{E}}{R_{E}} (1 - \frac{1}{10})$

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