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Q.
If the kinetic energy of a satellite orbiting around the earth is doubled then
NTA AbhyasNTA Abhyas 2020Gravitation
Solution:
$\frac{1}{2}mv_{1}^{2}=2\left(\frac{1}{2} m v_{0}^{2}\right)$
$v_{1}=\sqrt{2}v_{0}$
this is the value of escape velocity, hence the satellite will escape out from gravitation of earth