Editing ThreeDimensionalConfigurations/ChallengesPt6 (section)

====Are Orbits Exact Circles====
After plotting <math>(y')_{z_0}</math> as a function of <math>x'</math> (between the just-derived limits) for several different values of <math>z_0</math>, we noticed that each Lagrangian trajectory appears to be a circle.  If this is exactly the case &hellip;

<font color="red">1.) RADIUS OF CIRCLE, MEASURED PERPENDICULAR TO x'-AXIS:</font> &nbsp; Given simply by this last expression, namely,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math> \biggl[ \frac{x'}{a} \biggr]_\mathrm{radius}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 -\frac{z_0^2 \cos^2\theta}{(c^2  \cos^2\theta + b^2 \sin^2\theta )} \biggr]^{1 / 2}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math> \Rightarrow ~~~ (c^2  \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}\biggl[ \frac{x'}{a} \biggr]_\mathrm{radius}  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) - z_0^2 \cos^2\theta \biggr]^{1 / 2}  
</math>
  </td>
</tr>
</table>


<font color="red">2.) CENTER OF CIRCLE:</font> &nbsp; The y'-coordinate of the center of the circle is the value of <math>(y')_{z_0}</math> obtained when the argument of the square root goes to zero.  That is,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>( c^2 \cos^2\theta + b^2 \sin^2\theta )[(y')_{z_0}]_0 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- z_0 b^2\sin\theta 
\pm 
bc \cancelto{0}{\biggl[ \epsilon^2 (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ [(y')_{z_0}]_0 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-  \frac{z_0 b^2\sin\theta }{ ( c^2 \cos^2\theta + b^2 \sin^2\theta )} \, .
</math>
  </td>
</tr>
</table>
And, along the <math>x'</math>-axis, the inner and outer edges of the circle are identified by the positions at which <math>x'/a = 0 ~\Rightarrow ~ \epsilon = 1</math>.  That is,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>( c^2 \cos^2\theta + b^2 \sin^2\theta )[(y')_{z_0}]_\mathrm{limits} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- z_0 b^2\sin\theta 
\pm 
bc \biggl[ \cancelto{1}{\epsilon^2} (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ [(y')_{z_0}]_\mathrm{limits} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl\{ z_0 b^2\sin\theta 
\mp 
bc \biggl[ (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2}\biggr\}( c^2 \cos^2\theta + b^2 \sin^2\theta )^{-1} \, .
 </math>
  </td>
</tr>
</table>

<font color="red">3.) RADIUS OF CIRCLE, MEASURED PERPENDICULAR TO x'-AXIS:</font> &nbsp; The radius of the "circle" along the <math>x'</math>-axis is,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>[(y')_{z_0}]_\mathrm{radius} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
bc \biggl[ (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{-1}
 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}[(y')_{z_0}]_\mathrm{radius} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
bc \biggl[ 1  -  \frac{z_0^2 \cos^2\theta }{( c^2 \cos^2\theta + b^2 \sin^2\theta )}  \biggr]^{1 / 2}  \, .
 </math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center"><math>\frac{z_0}{z_\mathrm{max}}</math></td>
  <td align="center"><math>(x')_\mathrm{radius}</math></td>
  <td align="center"><math>(y')_0</math></td>
  <td align="center"><math>(y')_\mathrm{radius}</math></td>
  <td align="center"><math>\biggl[ \frac{y'}{x'}  \biggr]_\mathrm{radius}</math></td>
</tr>
<tr>
  <td align="right">0.000</td>
  <td align="right">1.000</td>
  <td align="right">0.000</td>
  <td align="right">0.974797</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
<tr>
  <td align="right">- 0.700</td>
  <td align="right">0.714143</td>
  <td align="right">- 0.625325</td>
  <td align="right">0.696144</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
<tr>
  <td align="right">- 0.975</td>
  <td align="right">0.222205</td>
  <td align="right">- 0.870989</td>
  <td align="right">0.216605</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
</table>

How do the two radii compare?  As the (immediately above) table illustrates, each trajectory's x'-radius is slightly larger than that trajectory's y'-radius.  Hence, the orbits are not circular!  However, as the last column (bgcolor="yellow") tabulates, the degree of flattening is very slight and, surprisingly, the ratio of radii is ''identical'' in every case.

Let's examine the analytic expression for the ratio of radii:

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[ \frac{x'/a}{[(y')_{z_0}]} \biggr]_\mathrm{radius} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 -\frac{z_0^2 \cos^2\theta}{(c^2  \cos^2\theta + b^2 \sin^2\theta )} \biggr]^{1 / 2}
(bc )^{-1} \biggl[ (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{- 1 / 2} ( c^2 \cos^2\theta + b^2 \sin^2\theta )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}}{bc} \, .
</math>
  </td>
</tr>
</table>

From this last expression, we see that the two radii will be the same &#8212; thereby making the LHS unity &#8212; only if,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
b^2 c^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 \sin^2\theta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 (1 - \cos^2\theta) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ b^2 (c^2 - 1)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(c^2 - b^2) \cos^2\theta  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \cos^2\theta
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{b^2 (c^2 - 1)}{(c^2 - b^2) } = 0.907244
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow \theta
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm 0.30948 \, .
</math>
  </td>
</tr>
</table>
This is not the case for our example model; its tilt angle is, instead, <math>\theta = -0.332029</math>.