Question

The escape velocity of a projectile on the earths surface is $11.2\,\,km{{s}^{-1}}$ . A body is projected out with thrice this speed. The speed of the body far away from the earth will be:

• A. $22.4\,km{{s}^{-1}}$
• B. $31.7\,km{{s}^{-1}}$
• C. $33.6\,km{{s}^{-1}}$
• D. none of these

Answer 1

Answer: (B)

$31.7\,km{{s}^{-1}}$

Answer 2

Key Idea : Applying conservation of energy. By law of conservation of energy $(U+K)_{\text {surface }}=(U+K)_{\infty}$
$\Rightarrow \quad-\frac{G M m}{R}+\frac{1}{2} m\left(3 v_{e}\right)^{2}=0+\frac{1}{2} m v^{2}$
$\Rightarrow -\frac{G M}{R}+\frac{9 v_{e}^{2}}{2}=\frac{1}{2} v^{2}$
Since, $\nu_{e}^{2}=\frac{2 G M}{R}$
$\therefore -\frac{v_{e}^{2}}{2},+\frac{9 v_{e}^{2}}{2}=\frac{1}{2} v^{2}$
$\Rightarrow v^{2} =8 \,v_{e}^{2}$
$\therefore v =2 \sqrt{2} \,v_{e}$
$=2 \sqrt{2} \times 11.2$
$=31.7 \,kms ^{-1}$