Question

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

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

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,
$\frac{1}{2}mu^{2} - \frac{GM_{E}m}{R_{E}}= \frac{1}{2}mv^{2}-0$
where $m$ and $M_{E}$ be masses of the body and earth respectively and $R_{E}$ is the radius of the earth.
or $v^{2} = u^{2}-\frac{2GM_{E}}{R_{E}}$
$v^{2} = u^{2}-v_{e}^{2}$
$v = \sqrt{u^{2}-v^{2}_{e}}=\sqrt{\left(3v_{e}\right)^{2}-v_{e}^{2}} \left(\because v_{e} = \sqrt{\frac{2GM_{E}}{R_{E}}}\right)$
$= \sqrt{8}v_{e} = 11.2\sqrt{8}\,kms^{-1}$
$= 22.4\sqrt{2}\,kms^{-1}$