Question

The escape speed of a body on the earths surface is $11.2km{s}^{-1}$. A body is projected with thrice of this speed. The speed of the body when it escapes the gravitational pull of earth is

• A. $11.2km{s}^{-1}$
• B. $22.4\sqrt{2}km{s}^{-1}$
• C. $\frac{22.4}{\sqrt{2}}km{s}^{-1}$
• D. $22.4\sqrt{3}km{s}^{-1}$

Answer 1

Answer: (B)

Let $v$ be the speed of the body when it escapes the gravitational pull of the earth and $u$ be speed of projection of the body from the earths surface.
According to law of conservation of mechanical energy,
$\frac{1}{2}m{u}^2-\frac{G{M}_{E}m}{R_E}=\frac{1}{2}m{v}^2-0$
where $m$ and $M_E$ be masses of the body and earth respectively and $R_E$ is the radius of the earth.
or $v^2=u^2-\frac{2G{M}_E}{R_E}$
$v^2=u^2-v_e^2$
$v=\sqrt{u^2-v_e^2}=\sqrt{{\left(3v_e\right)}^2-v_e^2}\left(\because v_e=\sqrt{\frac{2G{M}_E}{R_E}}\right)$
$=\sqrt{8}{v}_e=11.2\sqrt{8}km{s}^{-1}$
$=22.4\sqrt{2}km{s}^{-1}$