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