State and prove Kepler's second law (Law of Areas) of planetary motion.

(N/A) Kepler's second law states: "The line that joins any planet to the sun sweeps out equal areas in equal intervals of time."
Proof:
Consider a planet $P$ moving around the sun $S$ in an elliptical orbit. Let $\vec{r}$ be the position vector of the planet with respect to the sun. In a small time interval $\Delta t$, the planet moves from $P$ to $P^{\prime}$, covering a displacement $\Delta \vec{r} = \vec{v} \Delta t$.
The area $\Delta A$ swept out by the position vector in time $\Delta t$ is given by the area of the triangle $SPP^{\prime}$:
$\Delta A = \frac{1}{2} |\vec{r} \times \Delta \vec{r}| = \frac{1}{2} |\vec{r} \times (\vec{v} \Delta t)| = \frac{1}{2} |\vec{r} \times \vec{v}| \Delta t$
Dividing by $\Delta t$, we get the areal velocity:
$\frac{dA}{dt} = \frac{1}{2} |\vec{r} \times \vec{v}|$
Since the gravitational force exerted by the sun on the planet is a central force, it acts along the line joining the sun and the planet. Thus, the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Because the torque is zero, the angular momentum $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v}) = m(\vec{r} \times \vec{v})$ is conserved.
Substituting $\vec{r} \times \vec{v} = \frac{\vec{L}}{m}$ into the areal velocity equation:
$\frac{dA}{dt} = \frac{1}{2} |\frac{\vec{L}}{m}| = \frac{L}{2m}$
Since $L$ and $m$ are constants, $\frac{dA}{dt}$ is constant. This proves that the planet sweeps out equal areas in equal intervals of time.