"The line that joins any planet to the sun sweeps equal areas in equal intervals of time". It is shown in figure.

The planet $P$ moves around the sun in an elliptical orbit. The shaded area is the area $\Delta \mathrm{A}$ swept out in a small interval of time $\Delta t$.

This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

The law of areas can be understood as a consequence of conservation of angular momentum which is valid for any central force.

The line of action of force which is acting on planet is passing through the sun, hence the torque of any planet $\tau=r \mathrm{~F} \sin \pi=0$. So the conservation of angular momentum of any planet remains constant.

Let the sun be at the origin and let the position and momentum of the planet be $\vec{r}$ and $\vec{p}$.

The area swept out by the planet of mass $m$ in time interval $\Delta t$ is $\Delta \mathrm{A}$.

From figure area of right triangle $\Delta S P P^{\prime}$

(Note : For $\Delta t \rightarrow 0$ )

$\Delta \mathrm{A}=\frac{1}{2} \times$ base $\times$ perpendicular

$\Delta \mathrm{A}=\frac{1}{2}(\vec{r} \times \vec{v} \Delta t)$

(Here, $\mathrm{PP}^{\prime}=\Delta \mathrm{S}=\nu \Delta t$ )

Dividing $\Delta t$ on both side of equation $(1)$,

$\frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta t}=\frac{1}{2}(\vec{r} \times \vec{v})$

but momentum of planet $\vec{p}=m \vec{v}$

$\therefore \vec{v}=\frac{\vec{p}}{m}$