Question

In a satellite if the time of revolution is T, then its kinetic energy is proportional to

• A. $\frac{1}{T}$
• B. $\frac{1}{T^2}$
• C. $\frac{1}{T^3}$
• D. $T^{-\frac{2}{3}}$

Answer 1

The correct option is D $T^{-\frac{2}{3}}$

According to Kepler's third law which states that:
Time period T $\propto r^{\frac{3}{2}}$ where r = radius of the circular path/orbit around which the satellite moves
$\therefore r\propto T^{\frac{2}{3}}$
And kinetic energy $K.E\propto \frac{1}{r}$
$\therefore K.E\propto \frac{1}{T^{\frac{2}{3}}}or,$
$K.E\propto T^{-\frac{2}{3}}$