Question

A satellite of mass \( m \), revolving round the earth of radius \( r \), has kinetic energy \( E \). Its angular momentum is

• A. \( (mEr^2)^{1/2} \)
• B. \( (mEr^2) \)
• C. \( (2mEr^2)^{1/2} \)
• D. \( (2mEr^2) \)

Hint:

Always express velocity from kinetic energy when angular momentum is asked in terms of \( E \).

Answer 1

The Correct Option is C

Step 1: Write expression for kinetic energy.
For a satellite in circular orbit, \[ E = \frac{1}{2}mv^2 \]
Step 2: Express velocity in terms of kinetic energy.
\[ v = \sqrt{\frac{2E}{m}} \]
Step 3: Write expression for angular momentum.
\[ L = mvr \]
Step 4: Substitute the value of velocity.
\[ L = m r \sqrt{\frac{2E}{m}} = \sqrt{2mEr^2} \]
Step 5: Conclusion.
The angular momentum of the satellite is \[ L = (2mEr^2)^{1/2}. \]