Question

If the time period of a satellite is T, its kinetic energy is proportional to

• A. $KE\alpha T^{-1/2}$
• B. $KE\alpha T^{-2/3}$
• C. $KE\alpha T^{3/2}$
• D. None of these

Answer 1

The correct option is B $KE\alpha T^{-2/3}$

Velocity of satellite in orbit $V=\sqrt{\frac{GM}{R}}$,

Kinetic Energy, $KE=\frac{1}{2}m{V}^2=\frac{GMm}{2R}$ $\Rightarrow KE\propto \frac{1}{R}$ ....1

From the kepler's laws of periods,

$T^2=\left(\frac{4{\pi }^2}{GM}\right)R^3$ $\Rightarrow T^2\propto R^3$ ...2

From the equations 1 and 2,

$KE\propto T^{-2/3}$