If the time of revolution of a satellite is T, then the potential energy will be proportional to :
Answer & explanation
Correct answer: option 1
Velocity of satellite : \(v = \sqrt{\frac{GM}{r}}\)
Kinetic Energy : \(K.E. \propto v^2 \text{ and } v^2 \propto \frac{1}{r}\)
Also, Time period square: \(T^2 = r^3\)
\(\Rightarrow K.E. \propto T^{-2/3}\)