= Solution
For <inferring an inverse-square force from Kepler laws>, first use the area swept by an infinitesimal radius sector:
$$
\dot{\mathcal A}=\frac12r^2\dot\theta.
$$
<Kepler's second law> makes the <areal velocity> constant, so $l=r^2\dot\theta$ is constant. Then $r\ddot\theta+2\dot r\dot\theta=0$, meaning that the acceleration has no transverse component. <Newton's second law> therefore gives a <central force> directed along the Sun–planet line.
For the radial dependence, set $u=1/r$. Since $\dot\theta=lu^2$,
$$
\dot r=-l u_\theta,\qquad
\ddot r=-l^2u^2u_{\theta\theta},\qquad
r\dot\theta^2=l^2u^3.
$$
The radial acceleration is the <Binet equation> expression
$$
a_r=-l^2u^2(u_{\theta\theta}+u).
$$
The focus-based <ellipse> has $u=(1+\varepsilon\cos\theta)/A$, hence $u_{\theta\theta}+u=1/A$. Thus
$$
\boxed{a_r=-\frac{l^2}{A r^2}.}
$$
For a planet of mass $m$, the <force> is $F_r=-ml^2/(Ar^2)$: it is attractive and obeys the <inverse-square law>. This deduction uses both the shape of the orbit and its constant <areal velocity>; the orbit shape alone would not determine its time-dependent acceleration.
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