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

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$\frac{d A}{d t} \propto \sqrt{ω r}$

\frac{d A}{d t} \propto ω r

$\frac{d A}{d t} \propto ω r^{2}$

\frac{d A}{d t} \propto ω^{2} r

answer is C.

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Detailed Solution

ΔA = Area swept out by the position vector = Area SP₁P_2;SP₁ =r, SP₂ = r + Δr

$∴ Δ A = \frac{1}{2} (S P_{1}) (S P_{2}) \sin Δ θ = \frac{1}{2} r (r + Δ r) \cdot Δ θ$

$\Rightarrow Δ A = \frac{1}{2} r^{2} \cdot \frac{Δ θ}{Δ t}$
As Δt → 0

$\frac{d A}{d t} = \frac{1}{2} r^{2} \cdot \frac{d θ}{d t} = \frac{1}{2} r^{2} ω$

$∴ \frac{d A}{d t} \propto ω r^{2}$

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