In planetary motion the areal velocity of position vector of a planet depends on angular velocity (ω) and the distance of the planet from sun (r). If so the correct relation for areal velocity is

The correct relation for areal velocity is:
(dA/dt) ∝ ωr²

Step-by-Step Derivation:

1. Let us consider a planet moving around the Sun. Let the position vector of the planet relative to the Sun be $r$.

2. In a very small time interval $dt$, the position vector sweeps out a small triangular area $dA$.

3. The area of a triangle formed by two adjacent vectors, the position vector $r$ and the small displacement vector $ds$, is given by:
$dA=\frac{1}{2}rds$

4. Since the displacement is along the arc of the orbit, we can relate the linear displacement $ds$ to the angular displacement $d\theta$ as:
$ds=rd\theta$

5. Substituting this expression for $ds$ into the area equation gives:
$dA=\frac{1}{2}r\left(rd\theta \right)=\frac{1}{2}{r}^{2}d\theta$

6. To find the areal velocity, which is the rate at which area is swept out per unit time, we divide both sides by $dt$:
$\frac{dA}{dt}=\frac{1}{2}{r}^2\frac{d\theta }{dt}$

7. We know that the rate of change of angular displacement is defined as the angular velocity $\omega$:
$\omega =\frac{d\theta }{dt}$

8. Substituting $\omega$ into our areal velocity equation, we obtain:
$\frac{dA}{dt}=\frac{1}{2}\omega r^2$

9. Since $\frac{1}{2}$ is a constant, the relationship can be written in terms of proportionality as:
$\frac{dA}{dt}\propto \omega r^2$