Explanation

(c) Applying law of conservation of energy for asteroid at a distance $10{\ R}_{e}$ and at earth’s surface.
[image:8_image47.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{i} + U_{i} = K_{f} + U_{f}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
$Now,\ \ K_{f} = \frac{1}{2}{mv}_{i}^{2}\ \ and\ U_{i} = - \frac{GM_{e}m}{{10R}_{e}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{f} = \frac{1}{2}{mv}_{f}^{2}\ \ and\ U_{f} = - \frac{GM_{e}m}{R_{e}}$
Substituting these values in Eq. (i), we get
$\ \ \ \ \ \ \ \ \frac{1}{2}{mv}_{i}^{2} - \frac{GM_{e}m}{{10R}_{e}} = \frac{1}{2}{mv}_{f}^{2} - \frac{GM_{e}m}{R_{e}}$
$\Longrightarrow \ \ \frac{1}{2}{mv}_{f}^{2} = \frac{1}{2}{mv}_{f}^{2} + \frac{GM_{e}m}{R_{e}} - \frac{GM_{e}m}{{10R}_{e}}$
$\Longrightarrow \ v_{f}^{2} = v_{i}^{2} + \frac{2GM_{e}}{R_{e}} - \frac{2GM_{e}}{10R_{e}}$
$\therefore\ \ \ \ v_{f}^{2} = v_{i}^{2} + \frac{GM_{e}m}{R_{e}}\left( 1 - \frac{1}{10} \right)$