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Q.
Radius of orbit of satellite of earth is $R$. Its Kinetic energy is proportional to:
BHUBHU 2003
Solution:
Gravitational force provides the required centripetal force.
The gravitational force provides the required centripetal force in orbit of earth.
$\therefore \frac{G M_{e} m}{R^{2}} =\frac{m v_{0}^{2}}{R} $
$ \Rightarrow v_{o} =\sqrt{\frac{G M_{e}}{R}} $
Kinetic energy $ =\frac{1}{2} m v_{o}^{2} $
$ \therefore KE =\frac{1}{2} m\left(\frac{G M_{e}}{R}\right)^{2 / 2} $
$=\frac{1}{2} \frac{m G M_{e}}{R} $
$\Rightarrow KE \propto \frac{1}{R} $
