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: \[ T^{-2/3}. \]

Answer 2

In satellite motion, the time period \( T \) is related to the orbital radius \( r \) and gravitational forces. For a satellite of mass \( m \) orbiting a planet of mass \( M \), the gravitational force provides the necessary centripetal force:

\( \frac{G Mm}{r^2} = \frac{mv^2}{r} \)

Simplifying for the orbital velocity \( v \), we find:

\( v = \sqrt{\frac{GM}{r}} \)

The time period \( T \) of the satellite is given by the orbital circumference divided by the orbital speed:

\( T = \frac{2\pi r}{v} \)

Substituting \( v \) from the earlier equation:

\( T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} = 2\pi \sqrt{\frac{r^3}{GM}} \)

The kinetic energy \( K \) of the satellite is:

\( K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{2\pi r}{T}\right)^2 \)

Now, substitute the expression for \( v \):

\( K = \frac{1}{2}m \left(\sqrt{\frac{GM}{r}}\right)^2 = \frac{1}{2}m\frac{GM}{r} = \frac{GMm}{2r} \)

From \( T^2 \propto r^3 \) (Kepler's Third Law), we deduce:

\( r^3 \propto T^2 \Rightarrow r \propto T^{2/3} \)

Substitute \( r \propto T^{2/3} \) into the expression for kinetic energy:

\( K \propto \frac{1}{r} \propto \frac{1}{T^{2/3}} = T^{-2/3} \)

Thus, the kinetic energy is proportional to \( T^{-2/3} \).