Explanation

(c) From Kepler’s second law of motion, a line joining any planet to the sun sweeps out equal areas in equal intervals of time. Let any instant $t$, the planet is in position $A$. Then area swept out by $SA$ is
[image:7_image173.jpeg]
$dA = area\ of\ the\ curved\ triangle\ SAB$
$= \frac{1}{2}(AB \times SA) = \frac{1}{2}(rd\theta \times r) = \frac{1}{2}r^{2}d\theta$
The instantaneous areal speed is
$\frac{dA}{dt} = \frac{1}{2}r^{2}\frac{d\theta}{dt} = \frac{1}{2}r^{2}\omega\ $
Let $J$ be angular momentum, $I$ the moment of inertia and $m$ the mass, then
$J = I\omega = mr^{2}\omega$
$\therefore$ $\frac{dA}{dt} = \frac{J}{2m} =$constant
Hence, angular momentum of the planet is conserved.