Editing Appendix/Ramblings/BordeauxSequences (section)

=====Surrounding Torus=====
We'll assume that the surrounding ''2<sup>nd</sup>'' object is a thin torus (1) with the same density as the central object, (2) with a major axis, <math>~a</math>, which ensures that the torus is spinning with the Keplerian frequency prescribed by the mass of the central object, (3) and with a minor cross-sectional radius, <math>~b</math>.  The second of these constraints means that,
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<math>~\alpha_t \equiv \frac{a}{R_\mathrm{eq}} </math>
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<math>~=</math>
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<math>~\biggl[\frac{1}{3\Omega^2} (1 - e^2)^{1 / 2} \biggr]^{1 / 3} \, .</math>
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The mass of the torus is given by the expression,
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<math>~M_t</math>
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<math>~=</math>
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<math>~2\pi a (\pi b^2)\rho 
=
2\pi^2 R_\mathrm{eq}^3 \rho \biggl[ \alpha_t \beta_t^2 \biggr]
\, ,</math>
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where,
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<math>~\beta_t</math>
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<math>~\equiv</math>
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<math>~\frac{b}{R_\mathrm{eq}} \, .</math>
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Given that <math>~\alpha_t</math> is known once the eccentricity of the central Maclaurin spheroid has been selected and, given that the density of the torus must match the density of the central object, the mass of the torus will only depend on the choice of <math>~0 < \beta_t \le \beta_\mathrm{max}</math>.  The maximum allowed value, <math>~\beta_\mathrm{max}</math>, is set by ensuring that equatorial-plane location of the inner edge of the torus is no smaller than the equatorial radius of the central spheroid, <math>~R_\mathrm{eq}</math>.  This means,
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<math>~\beta_\mathrm{max}</math>
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<math>~=</math>
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<math>~\alpha_t - 1 \, .</math>
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So, the maximum torus mass is,
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<math>~M_\mathrm{max}</math>
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<math>~=</math>
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<math>~
2\pi^2 R_\mathrm{eq}^3 \rho \biggl[ \alpha_t \beta_\mathrm{max}^2 \biggr]
=
2\pi^2 R_\mathrm{eq}^3 \rho ~\alpha_t (\alpha_t - 1)^2 \, .
</math>
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The moment of inertia of the torus is,
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<math>~I_t</math>
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<math>~=</math>
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<math>~M_t R_\mathrm{eq}^2\biggl[ \biggl(\frac{a}{R_\mathrm{eq}} \biggr)^2 + \frac{3}{4} \biggl( \frac{b}{R_\mathrm{eq}} \biggr)^2 \biggr]</math>
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&nbsp;
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<math>~=</math>
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<math>~2\pi^2 R_\mathrm{eq}^5 \rho \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) \, .</math>
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Hence, the (square of the) angular momentum of the torus is,
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<math>~L_t^2 = I_t^2 \omega_0^2</math>
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<math>~=</math>
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<math>~
\biggl[ 2\pi^2 R_\mathrm{eq}^5 \rho \alpha_t \beta_t^2 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr) \biggr]^2 \Omega^2 (4\pi G\rho)
</math>
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&nbsp;
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<math>~=</math>
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<math>~
2^4 \pi^5 \alpha_t^2 \beta_t^4 \biggl( \alpha_t^2 + \frac{3}{4} \beta_t^2 \biggr)^2 G R_\mathrm{eq}^{10} \rho^3  \Omega^2 \, .
</math>
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