A satellite is orbiting in circular path near about earth surface. How much times its kinetic energy make to escape it from its orbit:

Kinetic energy of a satellite in the orbit is given as $KE=\frac{1}{2}m{v}_o^2$ , where $v_O$ is the orbital velocity of satellite.
We know that, escape velocity, $v_e=\sqrt{2}{v}_o$ 

Kinetic energy of satellite to escape,

${\mathrm{KE}}^{\mathrm{\prime }}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}^2=\frac{1}{2}\mathrm{m}{\left(\sqrt{2}{\mathrm{v}}_{\mathrm{o}}\right)}^2=2$

$\times \frac{1}{2}{\mathrm{mv}}_{\mathrm{o}}^2=2\mathrm{KE}$

Hence, kinetic energy required to escape is two times the kinetic energy in orbit.