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Q.
If $v_0$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_e$ is the escape velocity from the earth, then relation between the two is
Orbital velocity of satellite $=v_{0}$
and escape velocity from the earth $=v_{e}$.
We have orbital velocity $\left(v_{\circ}\right)=\sqrt{g R}$
and escape velocity $\left( v _{ e }\right)=\sqrt{2 g R}$
Therefore ratio of orbital velocity to escape velocity
$\frac{v_{o}}{v_{e}}=\sqrt{\frac{g R}{2 g R}}=\frac{1}{\sqrt{2}}$ or
$v_{e}=\sqrt{2} v_{o} .$