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Constant areal velocity gives constant specific angular momentum $l=r^2\dot{\theta }$, hence zero transverse acceleration and a central force. For a focus-based ellipse $u=1/r=(1+\epsilon \cos \theta )/A$, one has $u_{\theta \theta }+u=1/A$. The Binet equation then gives the displayed radial acceleration, proving an attractive inverse-square law. An orbit's shape alone does not specify its acceleration without a time law.

The radius vector from the force centre to an orbiting body sweeps out equal areas in equal times. Its areal velocity is constant because a central force conserves angular momentum.

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.

The transfer ellipse has semi-major axis and semi-minor axisConservation of angular momentum gives constant areal velocity $h/2$. The satellite sweeps half the ellipse, whose area is $\pi ab/2$, soUsing part (i),This is half the orbital period from Kepler's third law.