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Differentiating $\mathbf{r}=r\hat{\mathbf{r}}$ twice and using $\dot{\hat{\mathbf{r}}}=\dot{\theta }\hat{\mathbf{\theta }}$ and $\dot{\hat{\mathbf{\theta }}}=-\dot{\theta }\hat{\mathbf{r}}$ gives the displayed acceleration. For a central force, its angular component vanishes, so $r^2\dot{\theta }$ is constant and expresses conservation of angular momentum.

For unit mass in a central force with potential $\Phi (r)$ and conserved angular momentum $L$, a radial turning point $r_{\ast }$ has zero radial speed and therefore satisfies

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 potential $V(r)=-k/r$ generates the attractive inverse-square central force $F=-kr/r^3$.

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.

A central force has no angular component, so the supplied polar acceleration formula and Newton's second law give $d(r^2\dot{\theta })/dt=0$. Thus $l=r^2\dot{\theta }$ is the conserved specific angular momentum. For a nonradial orbit, $l\ne 0$ allows $\theta$ to be used as a parameter. With $u=1/r$ and a prime denoting $d/d\theta$,Also $r{\dot{\theta }}^2=l^2{u}^3$. The radial equation $\ddot{r}-r{\dot{\theta }}^2=-f(r)$ therefore becomes the Binet equationFor a purely radial trajectory $l=0$, the angular parameter is unavailable, and one instead uses the direct radial equation. The orbit in part (b) is nonradial.