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Step-by-Step Solution
Step 1: Recognize that Angular Momentum is Constant
The force acting on the planet is the gravitational pull from the Sun, which is always directed toward the Sun. Hence, the torque about the Sun is zero, making the angular momentum
$ \overrightarrow{L} $ of the planet constant.
Step 2: Express Angular Momentum in Terms of Perpendicular Velocity and Radius
Let the planet have mass $M$, and at some instant, let its velocity perpendicular to the radius vector be $v_{\perp}$. If the distance between the planet and the Sun is $r$, the angular momentum is given by:
$ L = M \, r \, v_{\perp} $
Step 3: Relate Angular Momentum to Areal Velocity
The areal velocity (rate of sweeping out area) for the planet is given by:
$ \frac{dA}{dt} = \frac{1}{2} \times (\text{base}) \times (\text{height}) \, / \, (\text{time interval}) $
In this context, the "base" is $r$ and the "height" is $v_{\perp} \, dt$. Hence,
$ \frac{dA}{dt} = \frac{1}{2} \,\bigl(r \times v_{\perp}\bigr) = \frac{1}{2} \, r \, v_{\perp} $
Step 4: Substitute the Expression for $v_{\perp}$ in Terms of $L$
Since $ L = M \, r \, v_{\perp} $, we can solve for $v_{\perp}$:
$ v_{\perp} = \frac{L}{M \, r} $
Substitute this back into the expression for areal velocity:
$ \frac{dA}{dt} = \frac{1}{2} \, r \, \frac{L}{M \, r} = \frac{L}{2M} $
Step 5: Conclude the Correct Answer
The magnitude of the areal velocity of the planet is therefore:
$ \frac{L}{2M} $
