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For inferring an inverse-square force from Kepler laws, first use the area swept by an infinitesimal radius sector: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 $\dot{\theta }=l{u}^2$,The radial acceleration is the Binet equation expressionThe focus-based ellipse has $u=(1+\epsilon \cos \theta )/A$, hence $u_{\theta \theta }+u=1/A$. ThusFor a planet of mass $m$, the force is $F_r=-m{l}^2/(A{r}^2)$: 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.

A central force has the form $\mathbf{F}=f(r)\hat{\mathbf{r}}$, so it always points along the line from a fixed centre to the particle. Its torque about the centre vanishes:Thus angular momentum is conserved. Since $\mathbf{r}\cdot \mathbf{L}=0$, the trajectory lies in the fixed plane through the origin perpendicular to $\mathbf{L}$.

The area swept in time $dt$ is $dA=\frac{1}{2}|\mathbf{r}\times d\mathbf{r}|$, sowhich is constant. This proves Kepler's second law.