Question

A space station is at a height equal to the radius of the Earth. If $'v_E'$ is the escape velocity on the surface of the Earth, the same on the space station is ....... times $v_E$.

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

$u_s = \frac{GMm}{2R}$
K.E. = P.E.
$\frac{1}{2}mv^2_s = \frac{GMm}{2R}$
$v^2_s = \frac{GM}{R} $
$v_s = \sqrt{gR}$
$[\because \,GM=gR^2]$
But $v_{e}=\sqrt{2 g R}=\sqrt{2} \sqrt{g} R$
$v_{e}=\sqrt{2} v_{s}$
$v_{s}=\frac{v_{e}}{\sqrt{2}}$