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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 uniform circular motion of radius $R$ and angular speed $\omega$, the acceleration is directed toward the centre and has magnitude ${\omega }^{2}R=v^2/R$. Writing $\mathbf{r}$ for the position relative to the centre gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. The required resultant force is $m\ddot{\mathbf{r}}$; it may be supplied by a normal force, gravity, or another physical interaction.

For the radial acceleration $-k{r}^n$ with $k>0$, a circular orbit of radius $a$ has squared specific angular momentum $h^2=k{a}^{n+3}$. At fixed $h$, its radial small oscillations have squared frequency $(n+3)k{a}^{n-1}$, so the orbit is stable for $n>-3$. This is a special case of the effective potential stability criterion.

The local vector-calculus identityseparates the convective acceleration into a kinetic-energy gradient and a vorticity cross-product contribution. Substitution into the steady Euler equations for an inviscid fluid yields the vorticity form of the Bernoulli equation.

For perturbing acceleration $R\hat{\mathbf{r}}+T\hat{\mathbf{\theta }}$ on an osculating orbital element system with fixed central parameter $\mu$,To prove it, differentiate the specific orbital energy $-\mu /(2a)$, obtaining $\dot{a}=(2a^2/\mu )(R\dot{r}+Th/r)$. Substitute $\dot{r}=\mu e\sin f/h$, $h/r=\mu (1+e\cos f)/h$ and $h^2=\mu a(1-e^2)$. Here $f$ is the true anomaly; a normal perturbing acceleration does no instantaneous work.

The Jeans swindle is the removal of the unperturbed uniform density's gravitational field when treating an infinite homogeneous Newtonian medium as static. Without such subtraction or an external supporting prescription, a nonzero constant density and zero background gravitational acceleration cannot satisfy Poisson's equation. Linear perturbations then obey ${\nabla }^2\delta \varphi =4\pi G\delta \rho$. This is a specified idealization, not an exact isolated infinite equilibrium.

The kinematic jerk is the time derivative of acceleration. For a Kepler orbit with relative position $\mathbf{r}$, relative velocity $\mathbf{v}$, $r=|\mathbf{r}|$, $\dot{r}=\mathbf{r}\cdot \mathbf{v}/r$ and total mass $M$, differentiation of Newton's law of universal gravitation gives $\mathbf{j}=-GM(\mathbf{v}-3\dot{r}\mathbf{r}/r)/r^3$.