Acceleration vector for an object moving in a elliptic trajectory

I am interested in deriving what the radial and tangential components of the acceleration vector should be for an object following an elliptical trajectory centered on the origin, in which the relation between the speed $v$ (module) is related to the distance of the object to the origin ($r$) by $v=kr^\beta$ where $k$ and $\beta$ are constants. I tried to find information online, but much of the information about elliptical motion is devoted to objects subject to gravitation.

I know that the components of acceleration in polar coordinates are $$a_r = \frac{d^2r}{dt^2}-r \left(\frac{d \theta}{dt}\right)^2$$ and $$a_\theta = r\frac{d^2\theta}{dt^2}+2 \frac{dr}{dt} \frac{d\theta}{dt}$$ but I do not know how to implement the constraint $v=kr^\beta$ there or whether that is the way to go.

• Can you expand your question to indicate what you've done, e.g. what approaches you've tried, so far? As it stands it's unclear what you're asking. – John Rennie Mar 19 '14 at 7:03

Take the ansatz \begin{align} \vec{r}(\theta(t)) = \begin{pmatrix} \rho_1\cos(\theta(t))\\ \rho_2\sin(\theta(t)) \end{pmatrix} \end{align} with a yet unknown scalar smooth function $\theta$.

Solve the ode $$k \bigg(\left|\vec{r}\big(\theta(t)\big)\right|\bigg)^{\beta} = |\vec{r}'(\theta(t))| \cdot\dot\theta(t)$$ for $\theta$ or explicitely $$\dot\theta(t) = \frac{k\cdot\left|\vec{r}(\theta(t))\right|^{\beta}}{|\vec{r}'(\theta(t))|}$$ Then use the found $\theta$ to calculate $$\vec{a}(t)=\frac{d}{dt}\left(\vec{r}'(\theta(t)) \dot\theta(t)\right).$$

A particle moving along a path has velocity and acceleration decomposed as

\begin{aligned} \vec{v} & = v \vec{e} \\ \vec{a} & = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} \\ \end{aligned}

where $v$ is the tangential velocity, $\rho$ is the radius of curvature, and $\vec{e}$, $\vec{n}$ are the tangent and normal vectors. For an ellipse with path coordinates $\vec{r}(t)=(a \cos t, b \sin t)$ the path properties are \begin{aligned} R(t) & = \sqrt{b^2+(a^2-b^2)\cos^2 t} \\ \vec{e}(t) & = \left( -\frac{a \sin t}{\sqrt{a^2+(b^2-a^2)\cos^2 t}} , \frac{b \cos t}{\sqrt{a^2+(b^2-a^2)\cos^2 t}} \right) \\ \vec{n}(t) & = \left( -\frac{b \cos t}{\sqrt{a^2+(b^2-a^2)\cos^2 t}} , -\frac{a \sin t}{\sqrt{a^2+(b^2-a^2)\cos^2 t}} \right) \\ \rho(t) & = \frac{ \left( (b^2-a^2) \cos^2 t + a^2 \right)^{\frac{3}{2}}} {a b} \\ v & = k \left(\sqrt{b^2+(a^2-b^2)\cos^2 t} \right)^\beta \\ \dot{v} & = \beta k \left(\sqrt{b^2+(a^2-b^2)\cos^2 t} \right)^{\beta-2} (b^2-a^2) (\sin t \cos t) \dot{t} \\ \end{aligned}

where $R(t)$ is the distance to origin.

Have fun putting it all together :-)