Thank you for reporting, we will resolve it shortly
Q.
In a satellite if the time of revolution is $T$, then $P E$ is proportional to
Gravitation
Solution:
According to Kepler's third law,
$T^{2} \propto r^{3}$
$r=$ radius of orbit
For a satellite of mass $m$ orbiting in an orbit of radius $r$ around planet of mass $M$,
Potential energy $( PE )=\frac{- GMm }{r}$
$\Rightarrow PE \propto \frac{ GMm }{T^{2 / 3}} $
$\Rightarrow PE \propto T^{-2 / 3}$