Math for determining second body mass in binary pair given first body parameters

math.tex

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\begin{document}
\begin{align*}
\text{==== Centre of mass ====} \\
m_{1} a_{1} &= m_{2} a_{2} \\
a_{1} &= \frac{m_{2} a_{2}}{m_{1}} \\
\text{==== Separation ====} \\
a &= a_{1} + a_{2} \\
\text{==== Kepler's third law ====} \\
\frac{P^2}{a^3} &= \frac{4 \pi}{G (m_{1} + m_{2})} \\
\text{==== Velocities ====} \\
P v_{1} &= 2 \pi a_{1} \\
P v_{2} &= 2 \pi a_{2} \\
a &= \frac{P}{2 \pi}(v_{1} + v_{2}) \\
\frac{P^2}{[\frac{P}{2 \pi}(v_{1} + v_{2})]^3} &= \frac{4 \pi^2}{G(m_{1} + m_{2})} \\
m_{1} + m_{2} &= \frac{P(v_{1} + v_{2})^3}{2 \pi G} \\
v_{1} m_{1} = v_{2} m_{2} &\Rightarrow v_{2} = \frac{v_{1} m_{1}}{m_{2}} \\
m_{1} + m_{2} &= \frac{P(v_{1} + \frac{m_{1}}{m_{2}} v_{1})^3}{2 \pi G} \\
\frac{m_{2}^3}{(m_{1} + m_{2})^2} &= \frac{P v_{1}^3}{2 \pi G} \\
\text{==== Eliminate velocity ====} \\
v_{1} &:= \frac{2 \pi a_{1}}{P} \\
\frac{m_{2}^3}{(m_{1} + m_{2})^2} &= \frac{P (2 \pi a_{1})^3}{2 \pi G P^3} \\
 &= \frac{4 \pi^2 a_{1}^3}{G P^2} \\
\text{==== Convert to polynomial ====} \\
\frac{(m_{1} + m_{2})^2}{m_{2}^3} &= \frac{G P^2}{4 \pi^2 a_{1}^3} \\
\frac{m_{2}^2 + 2 m_{1} m_{2} + m_{1}^2}{m_{2}^3} &= \frac{G P^2}{4 \pi^2 a_{1}^3} \\
m_{2}^2 + 2 m_{1} m_{2} + m_{1}^2 &= m_{2}^3 \frac{G P^2}{4 \pi^2 a_{1}^3} \\
\frac{G P^2}{4 \pi^2 a_{1}^3} m_{2}^3 - m_{2}^2 - 2 m_{1} m_{2} - m_{1}^2 &= 0 \\
\text{==== Polynomial coefficients ====} \\
a x^3 + b x^2 + c x + d &= 0 \\
a &:= \frac{G P^2}{4 \pi^2 a_{1}^3} \\
b &:= -1 \\
c &:= -2 m_{1} \\
d &:= -m_{1}^2 \\
\text{==== Eliminate $a$ ====} \\
x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} &= 0 \\
\lambda &:= \frac{1}{a} = \frac{4 \pi^2 a_{1}^3}{G P^2} \\
a &:= 1 \\
b &:= -\lambda \\
c &:= -2 \lambda m_{1} \\
d &:= -\lambda m_{1}^2 \\
\text{==== Eliminate $x^2$ ====} \\
x &:= z - \frac{b}{3} \\
(z - \frac{b}{3})^3 + b (z - \frac{b}{3})^2 + c (z - \frac{b}{3}) + d &= 0 \\
[z^3 - 3 \cdot \frac{1}{3} b z^2 + 3 (\frac{1}{3} b)^2 z - (\frac{1}{3} b)^3] + b [z^2 - 2 \cdot \frac{1}{3} b z + (\frac{1}{3} b)^2] + c (z - \frac{1}{3} b) + d &= 0 \\
z^3 - b z^2 + \frac{1}{3} b^2 z - \frac{1}{27} b^3 + b z^2 - \frac{2}{3} b^2 z + \frac{1}{9} b^3 + c z - \frac{1}{3} b c + d &= 0 \\
z^3 + (b - b)z^2 + (\frac{1}{3} b^2 - \frac{2}{3} b^2 + c)z + \frac{1}{9} b^3 - \frac{1}{27} b^3 - \frac{1}{3} b c + d &= 0 \\
z^3 + (c - \frac{1}{3} b^2) z + \frac{2}{27} b^3 - \frac{1}{3} b c + d &= 0 \\
z^3 + (c - \frac{1}{3} b^2) z - (\frac{1}{3} b c - d - \frac{2}{27} b^3) &= 0 \\
\text{==== Polynomial Coefficients ====} \\
z^3 + p z - q &= 0 \\
p &:= c - \frac{b^2}{3} \\
  &= -2 \lambda m_{1} - \frac{\lambda^2}{3} \\
q &:= \frac{1}{3} b c - d - \frac{2}{27} b^3 \\
  &= \frac{2}{3} \lambda^2 m_{1} + \lambda m_{1}^2 + \frac{2}{27} \lambda^3 \\
\text{==== Vieta's Substitution ====} \\
z &:= w - \frac{p}{3 w} \\
(w - \frac{p}{3 w})^3 + p (w - \frac{p}{3 w}) - q &= 0 \\
[w^3 - 3 \cdot \frac{p}{3 w} w^2 + 3 (\frac{p}{3 w})^2 w - (\frac{p}{3 w})^3] + (p w - p \frac{p}{3 w}) - q &= 0 \\
(w^3 - p w + \frac{p^2}{3 w} - \frac{p^3}{27 w^3}) + (p w - \frac{p^2}{3 w}) - q &= 0 \\
w^3 + p w - p w + \frac{p^2}{3 w} - \frac{p^2}{3 w} - \frac{p^3}{27 w^3} - q &= 0 \\
w^3 - \frac{p^3}{27 w^3} - q &= 0 \\
\text{==== Convert to quadratic ====} \\
w^3 w^3 - \frac{p^3}{27 w^3} w^3 - q w^3 &= 0 \\
(w^3)^2 - q (w^3) - \frac{1}{27} p^3 &= 0 \\
\text{==== Quadratic formula ====} \\
w^3 &= \frac{1}{2} q + \sqrt{\frac{1}{4} q^2 + \frac{1}{27} p^3} \\
w &= \sqrt[3]{\frac{1}{2} q + \sqrt{\frac{1}{4} q^2 + \frac{1}{27} p^3}} \\
\text{==== Substitute to get $x$ ====} \\
x &= (w - \frac{p}{3 w}) - \frac{b}{3} \\
  &= w + \frac{1}{3 w} (2 \lambda m_{1} + \frac{1}{3} \lambda^2) + \frac{1}{3} \lambda \\
\text{==== Bringing it all together ====} \\
\lambda &= \frac{4 \pi^2 a_{1}^3}{G P^2} \\
p &= -2 \lambda m_{1} - \frac{\lambda^2}{3} \\
q &= \frac{2}{3} \lambda^2 m_{1} + \lambda m_{1}^2 + \frac{2}{27} \lambda^3 \\
w &= \sqrt[3]{\frac{1}{2} q + \sqrt{\frac{1}{4} q^2 + \frac{1}{27} p^3}} \\
m_{2} &= w + \frac{1}{3 w} (2 \lambda m_{1} + \frac{1}{3} \lambda^2) + \frac{1}{3} \lambda
\end{align*}
\end{document}