Explanation

(d) From Kepler’s second law of planetary motion, a line joining any planet to the sun sweeps out equal areas in equal times, that is, the aerial velocities of the planet remains constant $dA$= area of the curved triangle $SAB$
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$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \approx \frac{1}{2}\ (AB \times SA)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \approx \frac{1}{2}(rd\theta \times r)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \approx \frac{1}{2}r^{2}d\theta$
Thus, the areal (instantaneous) velocity of the planet is
$\ \ \ \ \ \ \ \ \ \ \ \ \frac{dA}{dt} = \frac{1}{2}r^{2}\frac{d\theta}{dt} = \frac{1}{2}r^{2}\omega = constant$
where $\omega$ is angular speed of planet and $r$ its radius.