Question

If $ A $ is the areal velocity of a planet of mass $ M $, then its angular momentum is:

• A. \( \frac{MA}{2} \)
• B. \( MA \)
• C. \( 2MA \)
• D. \( \frac{MA}{3} \)

Hint:

The areal velocity is related to the angular momentum of a planet by the formula \( L = 2MA \), where \( A \) is the areal velocity and \( M \) is the mass of the planet.

Answer 1

The Correct Option is C

The areal velocity of a planet is defined as the rate at which the planet sweeps out area in its orbit. For a planet of mass \( M \) and distance \( r \) from the Sun, the areal velocity \( A \) is given by: \[ A = \frac{dA}{dt} = \frac{1}{2} r^2 \omega, \] where \( r \) is the radius vector and \( \omega \) is the angular velocity. 
Step 1: Relationship between angular momentum and areal velocity
The angular momentum \( L \) of a planet is given by: \[ L = M r^2 \omega, \] where \( M \) is the mass of the planet and \( r^2 \omega \) is the angular momentum per unit mass. 
Now, using the expression for areal velocity \( A = \frac{1}{2} r^2 \omega \), we can solve for \( r^2 \omega \): \[ r^2 \omega = 2A. \] 
Step 2: Final angular momentum
Substitute \( r^2 \omega = 2A \) into the expression for angular momentum: \[ L = M \times 2A = 2MA. \] Thus, the angular momentum of the planet is \( 2MA \), which corresponds to option (C).