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In planetary motion, the aerial 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 aerial velocity is

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$\frac{dA}{dt} \propto ωr$

$\frac{dA}{dt} \propto ω^{2} r$

$\frac{dA}{dt} \propto {ωr}^{2}$

$\frac{dA}{dt} \propto \sqrt{ωr}$

answer is C.

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Detailed Solution

According to Kepler's law, the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time. Therefore,
$\frac{dA}{dt} \propto {ωr}^{2} (∵ A = {πr}^{2})$

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