Question

In an orbit if the time period of revolution of a satellite is $T$, then potential energy $\left(\text{PE}\right)$ is proportional to $\frac{1}{T^x}$ (Answer upto a maximum of 3 decimal places is only allowed)

Answer 1

According to Kepler's third law,
$$T^2\propto r^3...\left(1\right)$$ Where, $r=$ radius of orbit.

For a satellite of mass $m$ orbiting in an oribit of radius $r$ around a planet of mass $\text{M}$, the$$\text{Potential energy}\text{(PE)}=\frac{-GMm}{r}...\left(2\right)$$From equation $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow \text{PE}\propto \frac{GMm}{T^{\frac{2}{3}}}$

$\Rightarrow \text{PE}\propto T^{-\frac{2}{3}}$

Accepted answer : $0.6,0.66,0.666$

Key Concept - The potential energy of a planet - satellite system is given by $$U=-\frac{GMm}{r}$$Where $r$ is the distance between the centers of the planet and the satellite.