= Solution
With the Sun fixed, its gravitational force on the planet is a <central force>. In the fixed orbital plane, the transverse acceleration in <polar coordinates> therefore vanishes:
$$
r\ddot\theta+2\dot r\dot\theta=0,\qquad \frac{d}{dt}(r^2\dot\theta)=0.
$$
During $dt$, the position vector sweeps an infinitesimal sector of area $dA=\tfrac12r^2d\theta$. Hence the <areal velocity> is
$$
\boxed{\frac{dA}{dt}=\frac12r^2\dot\theta=\frac{L}{2m}=\text{constant}.}
$$
Integrating over any two equal time intervals gives equal swept areas, proving <Kepler's second law>. Here $L$ is the signed component of <angular momentum> normal to the orbital plane. The argument uses <conservation of angular momentum>; an inverse-square dependence is not needed for this particular law, only a central force.
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