Question

If the time period of revolution of a satellite is $T$, then its kinetic energy is proportional to:

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

Hint:

For a satellite in orbit, Kepler’s Third Law states that $T^2 \propto R^3$. Using this, we derive that kinetic energy varies as $KE \propto T^{-2/3}$.

Answer 1

The Correct Option is D

Step 1: Using Kepler’s Third Law
From Kepler’s Third Law, the time period $T$ of a satellite in orbit is related to the radius $R$ of its orbit as: $$T^2 \propto R^3.$$ This gives: $$R \propto T^{2/3}.$$ Step 2: Expressing Kinetic Energy
The kinetic energy of a satellite in orbit is given by: $$KE = \frac{1}{2} m v^2.$$ For circular motion, the orbital velocity is: $$v = \sqrt{\frac{GM}{R}}.$$ So the kinetic energy becomes: $$KE \propto \frac{1}{R}.$$ Step 3: Relating $KE$ to $T$
Since $R \propto T^{2/3}$, we substitute: $$KE \propto \frac{1}{T^{2/3}}.$$ Step 4: Conclusion
Thus, the kinetic energy is proportional to: $$T^{-2/3}.$$