Elliptical motion: An object is moving at a constant speed?

[rashida564]

I am confused why the acceleration doesn't point to the center of the ellipse or one of the focus, since it moves in circular motion. Shouldn't the acceleration be just in the radial direction

 

[vanhees71]

You get an ellipse for central forces describing a harmonic oscillator, $\vec{F}=-m \omega^2 \vec{r}$, or an $1/r^2$ force like in the Kepler problem of Newtonian gravitational mechanics, $\vec{F}=-G m M \vec{r}/r^3$. In the former case the force points towards the "center" of the ellipse, in the latter to one of its foci.

 

[Haborix]

Just to add to Vanhees's answer and connect it to your constant speed supposition. In the two examples that Vanhees gives for a force that points to the center or focus of an ellipse the speed of the particle is not constant. In another words, if the speed were to be constant, then there must be some component of force (acceleration) not parallel to the radial force.

 

[vanhees71]

For any central-potential motion of course the angular momentum is conserved (Noether for rotational symmetry around the center of the force), which means that the line connecting the center and the body swipes out equal areas per unit time, i.e., if the body comes closer to the center, it's moving faster.

 

[rcgldr]

If the object is at constant speed, then all accelerations are perpendicular to the current velocity. A simple example would be a car moving at constant speed, with the driver adjusting steering inputs so that the car follows an elliptical path (imagine doing this on an elliptical shaped loop of road).

Trying to create an equation for this is complicated. It may require parametric equations defining x and y as functions of time.

 

[vanhees71]

Well
$$\vec{x}=\begin{pmatrix} a \cos \phi \\ b \sin \phi \end{pmatrix}.$$
Then
$$\vec{v}=\begin{pmatrix} -a \sin \phi \\ b \cos \phi \end{pmatrix} \dot{\phi}.$$
The equation thus is
$$\vec{v}^2=\dot{\phi}^2 (a^2 \sin^2 \phi + b^2 \cos^2 \phi)=\text{const}.$$
or
$$\mathrm{d}_t \vec{v}^2 = 0 \; \Rightarrow \; \dot{\phi} \ddot{\phi} (a^2 \sin^2 \phi + b^2 \cos^2 \phi) + \dot{\phi}^3 (a^2-b^2) \sin \phi \cos \phi =0.$$
This is indeed very difficult to solve. Mathematica gives some complicated implicit solution with some elliptic functions ;-)).

 

[rcgldr]

This is indeed very difficult to solve.

What is φ in your equations? I'm wondering if it would help to express x and y as functions of time. I did a web search and found a paper for ellipse with constant speed, but they're asking $16.00 to download it, and it's possible that the paper doesn't come up with an actual solution.

 

[vanhees71]

$\varphi$ is a generalized coordinate to parametrize the ellipse. It's of course a function of time. Which paper is it? Maybe I can download it via my university account.

 

[rcgldr]

Which paper is it? Maybe I can download it via my university account.

Here is the link (maybe you can find this elsewhere). I did a search for "ellipse parametric constant speed" to search for articles, and this seemed to be the only one that could have a solution. It's from a math magazine article, archived at this web site.

https://www.jstor.org/stable/10.4169/math.mag.86.1.003?seq=1