Question

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

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

Answer 1

Answer: (B)

Key Idea : Applying conservation of energy. By law of conservation of energy $(U+K{)}_{\text{surface }}=(U+K{)}_{\infty }$
$\Rightarrow -\frac{GMm}{R}+\frac{1}{2}m{\left(3v_e\right)}^2=0+\frac{1}{2}m{v}^2$
$\Rightarrow -\frac{GM}{R}+\frac{9v_e^2}{2}=\frac{1}{2}{v}^2$
Since, ${\nu }_e^2=\frac{2GM}{R}$
$\therefore -\frac{v_e^2}{2},+\frac{9v_e^2}{2}=\frac{1}{2}{v}^2$
$\Rightarrow v^2=8v_e^2$
$\therefore v=2\sqrt{2}{v}_e$
$=2\sqrt{2}\times 11.2$
$=31.7km{s}^{-1}$