﻿ Example: Elliptical Orbit & Hodograph | Primer on Geometric Algebra | David Hestenes

Example: Elliptical Orbit & Hodograph

Remember k>0 for an attractive force. We already have position, velocity, and acceleration as functions of direction:

$\Large{\begin{array}{l}{\bf{r}}\,\,\, = \,\,\,r{\bf{\hat r}}\,\,\, = \,\,\,\left( {\frac{\ell }{{1 + \varepsilon \cos \theta }}} \right)\,\,\hat {\boldsymbol{\varepsilon}} \,{e^{{\bf{\hat L}}\theta }}\,\,\,\,\,~~~~~~~~~~\,\,\,\,\,\left\{ {\ell = \frac{{{L^2}}}{{mk}}} \right\}\\{\bf{v}}\,\,\, = \,\,\,{\bf{\dot r}}\,\,\, = \,\,\,\frac{k}{L}\,\,\left( {{\bf{\hat r}}\,\, + \,\,{\boldsymbol{\varepsilon}} } \right)\,\,{\bf{\hat L}}\\{\bf{\dot v}}\,\,\, = \,\,\,{\bf{\ddot r}}\,\,\, = \,\,\, - \,\,\frac{k}{{m{r^2}}}\,\,{\bf{\hat r}}\end{array}}$

To end up with elliptical motion, the values of m, k, rinit, and vinit will result in 0<ε<1.

Over hundreds of years many, many ellipse parameter relationships and theorems have emerged. The point of the orbit closest to the center of force is called the pericenter in the general case (perigee for a satellite around Earth, perihelion for a planet around the sun). The center of force is at one focus of the ellipse. The point furthest from the focus is generally called the apocenter (apogee for an Earth satellite, aphelion for a solar planet). The distance from the center of the ellipse to the focus is called the focal length, f. Here are some common ellipse parameters and relationships derivable from the orbit and velocity equations above:

$\Large{\begin{array}{l}{r_{\min }}\,\,\, = \,\,\,r\left( {\theta = 0} \right)\,\,\, = \,\,\,\frac{\ell }{{1 + \varepsilon }}~~~~\,\,\,\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,\,{r_{\max }}\,\,\, = \,\,\,r\left( {\theta = \pi } \right)\,\,\, = \,\,\,\frac{\ell }{{1 - \varepsilon }}\\\\a\,\,\, = \,\,\,{\rm{semimajor ~ axis}}\,\,\,{\rm{ = }}\,\,\,\frac{{{r_{\min }}\, + \,\,{r_{\max }}}}{2}\,\,\, = \,\,\,\frac{\ell }{{1 - {\varepsilon ^2}}}\\\\f\,\,\, = \,\,\,{\rm{focal ~ length}}\,\,\,{\rm{ = }}\,\,\,\frac{{{r_{\max }}\, - \,\,{r_{\min }}}}{2}\,\,\, = \,\,\,a\, - \,{r_{\min }}\,\,\, = \,\,\,a\varepsilon \\\\b\,\,\, = \,\,\,{\rm{semiminor ~ axis}}\end{array}}$

The semiminor axis intersects the ellipse where the particle is midway between pericenter and apocenter. At that moment,
v is parallel to ε. We can use that property to find b:

$\Large{\begin{array}{l}0\,\,\, = \,\,\,{\boldsymbol{\varepsilon}} \wedge {\bf{v}}\,\,\, = \,\,\,{\left\langle {{\boldsymbol{\varepsilon}} \,{\bf{v}}} \right\rangle _2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,0\,\,\, = \,\,\,{\boldsymbol{\varepsilon}} \cdot \left( {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right)\,\,\, = \,\,\,{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}\,\, + \,\,{\varepsilon ^2}\\{r_{mid}}\,\,\, = \,\,\,\frac{\ell }{{1\,\, + \,\,{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}}}\,\,\, = \,\,\,\frac{\ell }{{1\,\, - \,\,{\varepsilon ^2}}}\,\,\, = \,\,\,a\\\\\\{\rm{Right}}\,\,{\rm{triangle:}}\,\,\,\,\,\,{a^2}\,\,\, = \,\,\,{b^2}\, + \,\,{f^2}\,\,\, = \,\,\,{b^2}\, + \,\,{a^2}{\varepsilon ^2}\\\\~~~~~~~~~~~~b\,\,\, = \,\,\,a\,\,\sqrt {1\, - \,{\varepsilon ^2}} \end{array}}$

Specific Example:

Initial conditions: (We need to specify mk, and initial values of v and r.  All other values can be calculated.)

$\Large{m\,\,\, = \,\,\,1\,\,~~~~~~~~~~\,\,k\,\,\, = \,\,\,20\,\,~~~~~~~~~~\,\,{{\bf{r}}_{{\rm{init}}}}\,\,\, = \,\,\,2{\bf{\hat a}}\,\,~~~~~~~~~~\,\,{{\bf{v}}_{{\rm{init}}}}\,\,\, = \,\,\,4{\bf{\hat b}}}$

These values were chosen to give convenient dimensions in the interactive diagram below.
Assume appropriate units are used. The constants of motion fit conditions for elliptical motion:

$\Large{\begin{array}{l}{\bf{L}}\,\,\, = \,\,\,L\,{\bf{\hat L}}\,\,\, = \,\,\,m\,{{\bf{r}}_{{\rm{init}}}} \wedge {{\bf{v}}_{{\rm{init}}}}\,\,\, = \,\,\,8{\bf{\hat a\hat b}}\,\,\, = \,\,\,8{\bf{\hat L}}\\\\E\,\,\, = \,\,\,\frac{1}{2}\,\,m\,{v_{{\rm{init}}}}^2\,\,\,\, - \,\,\,\frac{k}{{{r_{{\rm{init}}}}}}\,\,\, = \,\,\, - \,2\\\\\\{\varepsilon ^2}\,\,\, = \,\,\,1\,\, + \,\,\frac{{2E\,{L^2}}}{{m\,{k^2}}}\,\,\, = \,\,\,\frac{9}{{25}}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\varepsilon \,\,\, = \,\,\,3/5\,\,\, = \,\,\,0.6\end{array}}$

Other ellipse parameters are also easy to calculate:

$\Large{\begin{array}{l}\ell \,\,\, = \,\,\,\frac{{{L^2}}}{{mk}}\,\,\, = \,\,\,3.2\,\,\,~~~~~~~~~~\,\,\,{r_{\min }}\,\,\, = \,\,\,\frac{\ell }{{1 + \varepsilon }}\,\,\, = \,\,\,2\,\,\,~~~~~~~~~~\,\,\,{r_{\max }}\,\,\, = \,\,\,\frac{\ell }{{1 - \varepsilon }}\,\,\, = \,\,\,8\\\\a\,\,\, = \,\,\,\frac{1}{2}\,\,\left( {{r_{\min }}\, + \,\,{r_{\max }}} \right)\,\,\, = \,\,\,5\,\,\,~~~~~~~~~~\,\,\,f\,\,\, = \,\,\,a\varepsilon \,\,\, = \,\,\,3\,\,\, = \,\,\,a\,\, - \,\,{r_{\min }}\\\\\\b\,\,\, = \,\,\,a\,\,\sqrt {1 - {\varepsilon ^2}} \,\,\, = \,\,\,4\end{array}}$

In the following interacive diagram, we let the hodogram ride along with the particle.
This is not necessary (the orbit and hodogram are independent), but may aid visualization.
Also, the orbit (drawing distances) and hodogram (drawing velocities) have different units.

Finally, we can calculate some hodograph parameters:

$\Large{\begin{array}{l}{\bf{v}}({\rm{pericenter)}}\,\,\, = \,\,\,\frac{k}{L}\,\,\left( {{\bf{\hat r}}\,\, + \,\,{\boldsymbol{\varepsilon}} } \right)\,\,{\bf{\hat L}}\,\,\, = \,\,\,2.5\,\left( {1 + \varepsilon } \right)\,\hat {\boldsymbol{\varepsilon}} \,{\bf{\hat L}}\,\,\, = \,\,\,4{\bf{\hat b}}\\{\bf{v}}({\rm{apocenter)}}\,\,\, = \,\,\,2.5\,\left( {\, - \,\,1\,\, + \,\,\varepsilon } \right)\,\hat {\boldsymbol{\varepsilon}} \,{\bf{\hat L}}\,\,\, = \,\,\, - \,\,1\,\,{\bf{\hat b}}\\{\bf{u}}\,\,\, \equiv \,\,\,k\,{{\bf{L}}^{ - 1}}\,{\boldsymbol{\varepsilon}} \,\,\, = \,\,\,\frac{{k\varepsilon }}{L}\,\,\hat {\boldsymbol{\varepsilon}} \,{\bf{\hat L}}\,\,\, = \,\,\,1.5\,{\bf{\hat b}}\\\left| {{\bf{v}} - {\bf{u}}} \right|\,\,\, = \,\,\,{\rm{hodograph ~ radius}}\,\,\, = \,\,\,\frac{k}{L}\,\,\, = \,\,\,2.5\end{array}}$

In the following interactive diagram, use your mouse (hold down the left mouse button and drag) to pull the yellow dot (our particle) around in a CCW direction. You should be able to check whether all distances and velocities seem correct around the orbit. Check values at the ellipse extrema. Can you estimate the value of the semi-latus rectum using the grid?

Interesting tidbit:

This is a central attractive force, so the acceleration is always towards the ellipse focus, but only at pericenter and apocenter is that direction also perpendicular to the orbit path. At those two points, the acceleration is all centripetal, and therefore can be written in the well known form of v2/r. Thus, looking at magnitudes in the equation of motion, we may be tempted (even knowing that temptations are often dangerous!) to write:

$\Large{m\,\,\frac{{{v^2}}}{r}\,\,\, = \,\,\,\frac{k}{{{r^2}}}}$

for either the pericenter or apocenter.
Yet, this equation is incorrect! You can try a quick check with the numbers in our example above (v=4, r=2 or v=1, r=8) to see it doesn't work. What's wrong? Well, the r in v2/r is supposed to be the radius of curvature at that point in the particle's orbit. The radius of curvature obviously varies along the ellipse. What is its value at the two positions of interest? We could use this formula properly and work backwards. Call the radius of curvature at either point ρ. By symmetry, we had better get the same value at either pericenter or apocenter.

$\Large{\begin{array}{l}m\,\,\frac{{{v^2}}}{\rho }\,\,\, = \,\,\,\frac{k}{{{r^2}}}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\rho \,\,\, = \,\,\,m{v^2}{r^2}/k\\\\{v^2}\left( {{\bf{\hat r}} = \pm \hat {\boldsymbol{\varepsilon}} } \right)\,\,\, = \,\,\,\frac{{{k^2}}}{{{L^2}}}{\left( {\varepsilon \pm 1} \right)^2}\,\,\,\,\,\,\,\,\,\,~~~~~~\,\,\,\,\,\,\,\,\,\,{r^2}\left( {{\bf{\hat r}} = \pm \hat {\boldsymbol{\varepsilon}} } \right)\,\,\, = \,\,\,\frac{{{\ell ^2}}}{{{{\left( {1 \pm \varepsilon } \right)}^2}}}\\\rho \left( {{\bf{\hat r}} = \pm \hat {\boldsymbol{\varepsilon}} } \right)\,\,\, = \,\,\,\frac{m}{k}\,\,\frac{{{k^2}}}{{{L^2}}}\,\,{\ell ^2}\,\,\, = \,\,\,\ell \end{array}}$

That's a neat result. What if we wanted the general expression for the radius of curvature at any point around an ellipse? Well, we could go on perhaps forever discovering tidbits about ellipses. We'll try for a little restraint and settle for answering this final question on the subject. Here is a formula for finding the radius of curvature for any curve in polar form, such as our r(θ):

$\Large{\rho ~~~ = ~~~ \frac{{{{\left[ {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \right]}^{3/2}}}}{{\left| {{r^2} + 2{{\left( {\frac{{dr}}{{d\theta }}} \right)}^2} - r\frac{{{d^2}r}}{{d{\theta ^2}}}} \right|}} ~~~ = ~~~ \ell {\left[ {1 + {{\left( {\frac{{\varepsilon \sin \theta }}{{1 + \varepsilon \cos \theta }}} \right)}^2}} \right]^{3/2}}}$

Getting to that last expression from the formula may be even more painful than one imagines by staring at the formula. But, there it is, and it agrees with our previous discovery for θ=0 or θ=π

Here's a better method, applicable to a general space curve. The formula involves velocity and acceleration, for which we have expressions at the top of this page, and is much easier to understand, to derive, and at least in our case, to use. We start with expressions for the unit tangent vector and the unit normal vector to any curve, expressions involving rates of change with respect to arc length and the definition of radius of curvature.

$\Large{\begin{array}{l}{\bf{\hat T}}\,\,\, = \,\,\,\frac{{d{\bf{r}}}}{{ds}}\,\,\, = \,\,\,\frac{{d{\bf{r}}}}{{dt}}\,\,\frac{{dt}}{{ds}}\,\,\, = \,\,\,{\bf{v}}/v\,\,\, = \,\,\,{\bf{\hat v}}\\\\\\\frac{{d{\bf{\hat T}}}}{{ds}}\,\,\, = \,\,\,\frac{1}{\rho }\,\,{\bf{\hat N}}\,\,\, = \,\,\,\frac{1}{v}\,\,\frac{d}{{dt}}\,\left( {\frac{{\bf{v}}}{v}} \right)\,\,\, = \,\,\,\frac{{v{\bf{\dot v}} - {\bf{v}}\dot v}}{{{v^3}}}\\\\\\\frac{{{\bf{\hat T}} \wedge {\bf{\hat N}}}}{\rho }\,\,\, = \,\,\,{\bf{\hat v}} \wedge \frac{{v{\bf{\dot v}} - {\bf{v}}\dot v}}{{{v^3}}}\,\,\, = \,\,\,\frac{{{\bf{v}} \wedge {\bf{\dot v}}}}{{{v^3}}}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,~~~~~~\,\,\,\,\,\,\,\, \bbox[8px, border:2px solid red] {\rho \,\,\, = \,\,\,\frac{{{v^3}}}{{\left| {{\bf{v}} \wedge {\bf{\dot v}}} \right|}}}\end{array}}$

Now we can plug in our expressions for velocity and acceleration:

$\Large{\begin{array}{l}\rho \,\,\, = \,\,\,\frac{{\frac{{{k^3}}}{{{L^3}}}\,\,{{\left| {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right|}^3}}}{{\frac{{{k^2}}}{{Lm{r^2}}}\,\,\left| {{{\left\langle {\left( {1 + {\boldsymbol{\varepsilon}} {\bf{\hat r}}} \right){\bf{\hat L}}} \right\rangle }_2}} \right|}}\,\,\, = \,\,\,\frac{{km{r^2}\,{{\left| {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right|}^3}}}{{{L^2}\,\,\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}}\,\,\, = \,\,\,\ell \,\,\frac{{{{\left| {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right|}^3}}}{{{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}^3}}}\\\\\\\\\,\,\,\,\,~\, = \,\,\,\ell \,\,{\left[ {\frac{{\left( {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right){\bf{\hat r\hat r}}\left( {{\bf{\hat r}} + {\boldsymbol{\varepsilon}} } \right)}}{{{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}^2}}}} \right]^{3/2}}\,\,\, = \,\,\,\ell \,\,{\left[ {\frac{{\left( {1 + {\boldsymbol{\varepsilon}} {\bf{\hat r}}} \right)\left( {1 + {\bf{\hat r}}{\boldsymbol{\varepsilon}} } \right)}}{{{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}^2}}}} \right]^{3/2}}\\\\\\\\\,\,\,\,\,~\, = \,\,\,\ell \,\,{\left[ {\frac{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}} + {\boldsymbol{\varepsilon}} \wedge {\bf{\hat r}}} \right)\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}} - {\boldsymbol{\varepsilon}} \wedge {\bf{\hat r}}} \right)}}{{{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}^2}}}} \right]^{3/2}}\\\\\\\\\rho \,\,\, = \,\,\,\ell \,\,{\left[ {1\,\, - \,\,\,\frac{{{{\left( {{\boldsymbol{\varepsilon}} \wedge {\bf{\hat r}}} \right)}^2}}}{{{{\left( {1 + {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)}^2}}}} \right]^{3/2}}\end{array}}$

This is the same formula as given above in polar coordinate form. It may at first seem like a more complicated procedure, but consider the differences.

In the polar case we gave a very complicated formula (without the very complex proof) and presented the final answer without showing the agonizing trig derivatives, expanding, grouping of terms and factoring involved. Even then, it was a formula valid only for planar motion.

In the second method, we derived the
much simpler expression for radius of curvature, a formula valid for 3D space curves. Then we filled in the steps needed to apply the formula to our ellipse. Finally, notice that to apply the formula, we took NO derivatives! The only derivatives needed were finished earlier by getting the velocity and acceleration formulas, and to get those we performed only a few very simple derivatives. Finally, our orbit parameter was not the arc length, nor did we know any functions of time. We can use this formula for radius of curvature if we know velocity and acceleration as functions of any parameter at all.