If the angular momentum of a planet of mass $$m$$, moving around the Sun in a circular orbit is $$L$$, about the center of the Sun, its areal velocity is:

We consider a planet of mass $$m$$ describing a circular orbit of radius $$r$$ around the Sun. For such a motion the angular velocity about the Sun is denoted by $$\omega$$ (this is the rate of change of the polar angle $$\theta$$).

First we recall the definition of angular momentum about the centre of the Sun for a particle of mass $$m$$:

$$\displaystyle L = m\,r^{2}\,\omega.$$

Next we determine the areal velocity, i.e. the rate at which area is swept out by the radius vector. In a short time interval $$dt$$, the radius vector turns through a small angle $$d\theta$$, and the infinitesimal sector (of a circle) swept out has area

$$\displaystyle dA = \tfrac12\,r^{2}\,d\theta.$$

Dividing both sides by $$dt$$ gives the areal velocity:

$$\displaystyle \frac{dA}{dt} = \tfrac12\,r^{2}\,\frac{d\theta}{dt}.$$

But $$\frac{d\theta}{dt} = \omega$$, so

$$\displaystyle \frac{dA}{dt} = \tfrac12\,r^{2}\,\omega.$$

Now we substitute for $$r^{2}\,\omega$$ from the angular momentum expression $$L = m\,r^{2}\,\omega$$. Solving that equation for $$r^{2}\,\omega$$, we have

$$\displaystyle r^{2}\,\omega = \frac{L}{m}.$$

Placing this into the formula for the areal velocity:

$$\displaystyle \frac{dA}{dt} = \tfrac12\left(\frac{L}{m}\right).$$

Simplifying, we obtain

$$\displaystyle \frac{dA}{dt} = \frac{L}{2m}.$$

Hence, the correct answer is Option C.