Editing Appendix/Ramblings/BordeauxSequences (section)

=====Central Object=====
Assume that the central object is exactly a Maclaurin spheroid.  Then from [[Apps/MaclaurinSpheroidSequence#Equilibrium_Angular_Velocity|Figure 1 (and Table 1) of our review of equilibrium models along the Maclaurin spheroid sequence]], we appreciate that all we have to do is specify the eccentricity, <math>~0 \le e \le 1</math>, and <math>~\Omega^2 \equiv \omega_0^2/(4\pi G\rho)</math> is known.  For example, if we choose <math>~e = 0.60</math>, then from that Table 1, <math>~\omega_0^2/(2\pi G\rho) = 0.1007~~\Rightarrow~~\Omega^2 \approx 0.05</math>.  Other properties of this "central" spheroid &#8212; such as its mass, moment of inertia, and angular momentum &#8212; are given by the following expressions:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \rho R_\mathrm{eq}^2 Z = \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~I_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5} M_c R_\mathrm{eq}^2 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~L_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~I_c \omega_0 = \frac{2}{5} M_c R_\mathrm{eq}^2 \omega_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{5} \biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] R_\mathrm{eq}^2 \omega_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^3\pi}{3\cdot 5} (1 - e^2)^{1 / 2} R_\mathrm{eq}^5 \rho \omega_0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~L^2_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4\pi G\biggl( \frac{2^3\pi}{3\cdot 5}\biggr)^2 (1 - e^2) R_\mathrm{eq}^{10} \rho^3 \Omega^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{2^8 \pi^3}{3^2 \cdot 5^2}\biggr) (1 - e^2) G R_\mathrm{eq}^{10} \rho^3 \Omega^2 \, . </math>
  </td>
</tr>
</table>

We note as well that the (square of the) Keplerian frequency for a massless particle orbiting in the equatorial plane at a distance, <math>~r</math>, from the center of this central object will be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\omega_K^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM_c}{r^3}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{G}{r^3} \biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] \, .</math>
  </td>
</tr>
</table>

So, if we force this orbital frequency to also equal the spin-frequency of the Maclaurin spheroid, the radius of the orbit must be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Omega^2 (4\pi G\rho)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{G}{r^3} \biggl[ \frac{4\pi}{3} \rho R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\Omega^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{r^3} \biggl[ \frac{1}{3} R_\mathrm{eq}^3 (1 - e^2)^{1 / 2} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{r}{R_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1}{3\Omega^2} (1 - e^2)^{1 / 2} \biggr]^{1 / 3} \, .</math>
  </td>
</tr>
</table>
For example, when <math>~(e, \Omega^2) = (0.6, 0.05)</math>, we have, <math>~r/R_\mathrm{eq} = (5.3333)^{1 / 3} = 1.747</math>.