In planetary motion the areal velocity of position vector of a planet depends on angular velocity $\omega$…
In planetary motion the areal velocity of position vector of a planet depends on angular velocity $\omega$ and the distance of the planet from sun $r$. The correct relation for areal velocity is
$\frac{d A}{d t} \propto \omega r$
$\frac{d A}{d t} \propto \omega^2 r$
$\frac{d A}{d t} \propto \omega r^2$
$\frac{d A}{d t} \propto \sqrt{\omega r}$
Solution
Areal velocity $\frac{d A}{d t}$ depends on angular velocity $(\omega)$ and distance of the planet from sun.
$\frac{d A}{d t} \propto w^a r^b \Rightarrow \frac{d A}{d t}=K \omega^a r^b$
Writing the dimensional formula on both sides,
$\left[\mathrm{L}^2 \mathrm{~T}^{-1}\right]=k\left[\mathrm{~T}^{-1}\right]^a[\mathrm{~L}]^b$
On comparing, $-a=-1 \Rightarrow a=1$
$b=2$
$\therefore \quad \frac{d A}{d t} \propto \omega r^2$