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We are all familiar with a body traversing uniformly ( or non uniformly ) along the circumference of a circle. We get nice relations because of the symmetry of the figure and the distance between every point on the trajector and the centre of the circle remains constant.

The ellipse is also quite a symmetric figure however the body traversing does not move equi distantly from the centre of the ellipse. Due to this I am unable to derive any kind of a function that can define the motion of a body moving along an elliptical path.

Planetary motion is elliptical and I am familiar with how the displacement is derived in that case, but that takes help from the Newton's law of gravitation, which gives us an extra relation to attempt the problem. However how must one approach the problem in a general case?

Can someone please help me derive the displacement and the velocity functions for a body moving uniformly on an elliptical trajectory? What properties of the ellipse can be exploited the derive the same?

Any kind of help or guidance would be appreciated. Thank you in advance!

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the equation of ellipse where the axes at (0,0) is

$$x=r(\varphi)\,\cos(\varphi)\\ y=r(\varphi)\,\sin(\varphi)$$

where

$$r(\varphi)=\frac{a\,\sqrt{1-e^2}}{\sqrt{1-e^2\,\cos^2(\varphi)}}$$

and

$$e=\sqrt{1-\frac{b^2}{a^2}}$$

a,b are the half-axis of the ellipse

enter image description here

The velocity

$$\vec v=v_r\,\vec{e}_r+v_\varphi\,\vec{e}_\varphi$$

with $$v_r=\frac{dr}{dt}=\frac{dr}{d\varphi}\,\dot{\varphi}=\\ -a\sqrt {{\frac {{b}^{2}}{{a}^{2}}}}{e}^{2}\cos \left( \varphi \right) \sin \left( \varphi \right) \left( 1-{e}^{2} \left( \cos \left( \varphi \right) \right) ^{2} \right) ^{-3/2} \dot{\varphi} $$

and

$$v_\varphi=r(\varphi)\,\dot{\varphi}$$

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  • $\begingroup$ Note that this is the equation for a centered ellipse, and planetary motion differs in that the body moves around one of the focii. $\endgroup$ – John Alexiou May 1 at 20:59
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This would more suitably be a comment, but do you mean a nice equation that describes ellipses? Perhaps you can try polar coordinates - they might be easier to work with.

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The path of a planet around a star is indeed an ellipse with the star on one of the focal points. For an ellipse with semi-major axis $a$, and semi-minor axis $b$ the focus point is $f=\sqrt{a^2-b^2}$ distance from the center of the ellipse.

With polar coordinates $(r \cos \varphi, r \sin \varphi)$ from the focal point the ellipse tracs the following curve

$$ r(\varphi) = a \left( \frac{1-\epsilon^2}{1-\epsilon \cos \varphi} \right) $$

where $\epsilon = \sqrt{ 1 - \left( \frac{b}{a}\right)^2 }$ is the orbital eccentricity.

From that you apply some differential geometry to get expressions for the tangential speed, tangential acceleration and curvature

quantity expression
radial distance $$r = a \left( \frac{1-\epsilon^2}{1-\epsilon \cos \varphi} \right)$$
tangential speed $$v = a \frac{1-\epsilon^2}{\left(1-\epsilon \cos\varphi \right)^2} \sqrt{1 + \epsilon^2 - 2 \epsilon \cos \varphi}$$
tangential acceleration $$ \dot{v} = a \frac{\epsilon (1-\epsilon^2) \sin \varphi (1-3\epsilon \cos\varphi +2\epsilon^2)}{(1-\epsilon \cos \varphi)^3 \sqrt{1-2\epsilon \cos\varphi +\epsilon^2}}$$
curvature $$ \frac{a}{\rho} = \frac{1-\epsilon\cos\varphi \left( \epsilon \cos \varphi \left( \epsilon \cos \varphi-3 \right)+3 \right)}{(1-\epsilon^2) \left( 1 -2\epsilon \cos\varphi + \epsilon^2 \right)^{3/2}} $$

Maximum radius of curvature $\rho_{\rm max} = \frac{a}{\sqrt{1-\epsilon^2}}$ at $\varphi = \frac{\pi}{2} - \sin^{-1} \epsilon$, and minim radiuis of curvature $\rho_{\rm min} = a (1-\epsilon^2)$ at $\varphi = 0, \pi, 2\pi, \ldots$.

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