deletions | additions       diff --git a/PH1 stable?.tex b/PH1 stable?.tex  index 8a60a61..a29cf7d 100644  --- a/PH1 stable?.tex  +++ b/PH1 stable?.tex 
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\item use Holliday and Resnick's simulation results to compare PH1b's inclination and eccentricity with that of the expected stability zones.  \end{itemize}  \subsection{Semimajor Axis Inner Region}  The planet's semimajor axis is approximately 0.634 AU. The eccentricity is $e \approx 0.0539$, and the mass ratio is $\mu=m_2 /(m_1 +m_2 )=0.408/(1.528+.408)=.211$. The binary semimajor axis is $a_b =0.1744$. So, using Holman & \&  Wiegert's best fit equation: \begin{equation}  a_c =[0.464+(-0.380)\mu +(-.631)e+(.586)\mu e +.15e^2 +(-0.198)\mu e^2 ]a_b ,  \end{equation} 
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This means for an inner region orbit the planet would not be stable. However, the planet is orbiting both of the binary stars which means it is an outer orbit so we will next determine its stability with outer region calculations.  \subsection{Semimajor Axis Outer Region}  The planet's semimajor axis, eccentricity, the stars' mass ratio, and semimajor axis are all the same as in the last equation. Using Holman & \&  Wiegert's outer region best fit line: \begin{equation}  a_c =1.6+5.1e+(-2.22)e^2 +4.12 \mu +(-4.27)e \mu +(-5.09)\mu ^2 +4.61e^2 \mu ^2 ,  \end{equation} 
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\end{tabular}  \end{table}  We will define $\mu _1 =0.4366$ as the mass ratio between Binary 1 & \&  Binary 2 and $\mu _2 =0.000262$ as the mass ratio between the Planet and Binary 1. Using equation $(7)$ again, we get $a_c =[0.464+(-0.380)0.4366 +(-.631)0.539+(.586)0.4366 (0.539) +$ $.15(0.539^2 ) +(-0.198)(0.4366) (0.539^2 ) ]*1000 = 253.192 $  For inner orbits we again assume that semimajor axes smaller than the critical semimajor axis is stable and $.634 \stackrel{\checkmark}{<} 253.000$ which means the planet is stable!  Next, we will calculate the critical semimajor axis of the outer binary using equation $(8)$ to get: $a_c =1.6+5.10.5+(-2.22)0.5^2 +4.12 (0.000262) +(-4.27)(0.5) (0.000262)+$ 
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\subsection{Inclination}  Finally, for one last test of the systems stability, we will compare its inclination with Halliday & \&  Resnick's model. The planet's inclination is 87.4$^{\circ}$ and the mass ratio is $m_\star /M_\star = .000507/1.936 < 0.05$. As Halliday & \&  Resnick showed, the smaller the $m_\star /M_\star $ ratio is, the less impact the inclination has on stability. At a mass ratio of 0.05, every inclination was found to be stable in their integration so anything less than 0.05 mass ratio can be assumed stable. Our ratio is much less than 0.05; therefore, the planet is stable! Unfortunately, Schwamb et al. were unable to report an inclination for the outer binary so we can not make an approximation of its stability based on Halliday & \&  Resnick's model. We can, however, find the range of inclinations in which the binary \textit{would} be stable. The mass ratio for the outer binary to the inner binary is $m_\star /M_\star = 1.500/1.936 \approx .77$. For large mass ratios, inclination must be very small for the system to be stable. This ratio of 0.77 is very large so an inclination of $i_0 \approx 0^{\circ}$ would be required for the outer binary to be considered stable.