Question

A satellite of mass \(m\) rotates round the earth in a circular orbit of radius \(R\). If the angular momentum of the satellite is \(J\), then its kinetic energy (\(K\)) and the total energy (\(E\)) of the satellite are:

• A. \(K = \frac{J^2}{mR^2}, \, E = \frac{J^2}{2mR^2}\)
• B. \(K = \frac{J^2}{2mR^2}, \, E = -\frac{J^2}{2mR^2}\)
• C. \(K = \frac{J^2}{2mR^2}, \, E = -\frac{J^2}{mR^2}\)
• D. \(K = \frac{J^2}{mR^2}, \, E = -\frac{J^2}{2mR^2}\)

Hint:

For orbital motion, the total energy is negative and is equal to the kinetic energy but with a negative sign. Use the angular momentum formula to find relationships between kinetic and potential energy.

Answer 1

The Correct Option is B

Step 1: The angular momentum \( J \) is related to the kinetic energy \( K \) and the radius \( R \) by the formula:

\[ J = mRv \]

where \( v \) is the velocity of the satellite.

Step 2: The total kinetic energy is given by:

\[ K = \frac{1}{2} mv^2 \]

From the relationship \( J = mRv \), we can solve for \( v \) and substitute it into the equation for kinetic energy to get:

\[ K = \frac{J^2}{2mR^2}. \]

Step 3: The total energy \( E \) of the satellite in orbit is the sum of its kinetic and potential energy. The potential energy is \( U = -\frac{GMm}{R} \), where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. The total energy is then:

\[ E = K + U = \frac{J^2}{2mR^2} - \frac{GMm}{R}. \]

Therefore, the correct answer is \( K = \frac{J^2}{2mR^2} \) and \( E = -\frac{J^2}{2mR^2} \).