Kinetic energy, $K=\frac{1}{2}m\upsilon_{0}^{2} $
or$\quad v_{0}=\sqrt{\frac{2k}{m}}$
When kinetic energy is doubled, then
$v'_{0}=\sqrt{\frac{2\times2K}{m}}=\sqrt{2}\sqrt{\frac{2K}{m}}=\sqrt{2v_{0}}$
But $\sqrt{2v_{0}}=v_{e}$ (escape velocity)
Hence satellite will escape out of earth’s gravitational field.