Kepler #39;s second law (law of areas) i

Considering a small area swept by a planet in time dt is, $d A=\frac{1}{2} r \cdot(r d \theta)$

$\therefore$ So, rate of swept of area.

$\frac{d A}{d t}=\frac{1}{2} r\left(r \frac{d \theta}{d t}\right)$

We know angular momentum, $\vec{L}=m r^2 \frac{d \vec{\theta}}{d t}$

So, $\frac{d A}{d t}=\frac{1}{2 m}\left(m r^2 \frac{d \theta}{d t}\right)=\frac{L}{2 m}$

Now, a planet sweeps equal amount of area in same time interval of its revohution. (from 2 nd law) So, $\frac{d A}{d t}=$ constant; Hence $L$ is also constant

$\therefore$ The $2^{nd}$ law is nothing but a shatement of conservation of angular momentum.