Show that the escape velocity of a body from the earth's surface is root 2 times its velocity in circular orbit just above the earth's surface.

Dear Student ,
Orbital velocity of a satellite is given by the minimum velocity of a satellite which rotate in a orbit around the earth near the earth's surface .
So orbital velocity is , 
$$\begin{gathered} v_0=\sqrt{\raisebox{1ex}{$GM$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}=R\sqrt{\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$\left(R+h\right)$}\right.} \\ where M is the mass of the planet R is the radius of the planet and h is the height of the satellite near the earth\text{'}s surface . \\ Now , if the satellite is revolving near the earth\text{'}s surface then ,(R+h)=-R \\ Now orbital velocity , v_0=\sqrt{gR}.......\left(1\right) \\ Again escape velocity is the minimum velocity with which a body has to be projected vertically upwards from earth\text{'}s surface \\ so that it just crosses the earth\text{'}s gravitational field . \\ So , escape velocity of an object is , \\ {v}_e=\sqrt{\raisebox{1ex}{$2GM$}\!\left/ \!\raisebox{-1ex}{$R$}\right.} \\ =\sqrt{2gR} ..............\left(2\right) \\ Now from equation 1 and 2 we can write that , \\ {v}_e=\sqrt{2}{v}_0 \end{gathered}$$
Hence it is proved .
Regards