= Torque criterion for constant areal velocity
{title2=$\ddot A=\tau_n/(2m)$}
Assume a constant positive <mass> $m$, a twice differentiable trajectory in the fixed <plane>, a fixed origin in an <inertial frame>, and <Newton's second law> for the net <force>. The <product rule> gives the short proof
$$
2\ddot A=\mathbf n\cdot\frac{d}{dt}(\mathbf r\times\mathbf v)
=\mathbf n\cdot(\mathbf r\times\mathbf a)
=\frac{\tau_n}{m},\qquad
\boldsymbol\tau=\mathbf r\times\mathbf F_{\mathrm{net}}.
$$
The term $\mathbf v\times\mathbf v$ vanishes. This is the planar <angular momentum> balance written in <area> language.<University Physics Volume 1>[Section 11.2, equations 11.5-11.6: particle angular momentum and its torque balance]{ref}
On a connected <time> interval, <areal velocity> is constant if and only if $\tau_n=0$ throughout. Since both <position> and <acceleration> lie in the fixed <plane>, the net <torque> has only that normal component. For $r\ne0$, zero net <torque> is therefore equivalent to net <force> parallel to $\mathbf r$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net <force> along this motion, not its dependence on <position> or <time> elsewhere. Opposing transverse <forces> may cancel.
A <central force> about a fixed origin gives <conservation of angular momentum>, irrespective of its radial dependence: an <inverse-square law> is sufficient but unnecessary. With nonzero conserved <angular momentum>, $\mathbf r\cdot\mathbf L=0$ fixes the <orbital plane>. Zero <angular momentum> gives radial motion and zero <areal velocity>; it does not select a unique <orbital plane>. The converse needs the fixed-plane assumption: a constant projected <areal velocity> in three dimensions only constrains one <torque> component. Also, $r=0$ makes the instantaneous <torque> vanish regardless of <force> and makes <plane polar coordinates> singular. A moving origin or changing plane requires additional terms.
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