# Theory: Gravitational restricted three-body system

## Equation of motion

The motion of the planets can computed from Newton's equations of motion. We consider two dimensional motion. For a three-body system, a set of coupled differential equations can be used to express the system.They can be expressed using vector notation. Let ${\bf{r}} = x \hat{i} + y \hat{j} $. Where, $\hat{i}$ and $\hat{j}$ are unit vectors along x and y axis. So, the differential equation of motion of a planet is:

\begin{equation}
	\label{e1}
	 \frac{d^2 \bf{r}}{d t^2} = \frac{\bf{F}}{M_E}                   \hspace{5cm}     (1)
\end{equation}

## Dimensionless form of the differential equation

We also define the following dimensionless variables

\begin{eqnarray}
	\nonumber
	\hat{t} &=& \frac{t}{T}  \\ \nonumber
	\hat{\bf{r}} &=& \frac{\bf{r}}{R}  \\ \nonumber
	\hat{M_E} &=& \frac{M_E}{M}  \\ \nonumber
	\hat{M_S} &=& \frac{M_S}{M}  \\ \nonumber
	\hat{\bf{F}} &=& \frac{\bf{F}}{F^*}  
\end{eqnarray}

We note that the derivatives can be written in terms of the dimensionless quantities as:

\begin{eqnarray}
    \nonumber
	\frac{d \bf{r}}{d t} &=& \frac{R}{T}\frac{d \bf{\hat{r}}}{d \hat{t}}     \hspace{6cm} (2) \\  \nonumber
    \frac{d^2 \bf{r}}{d t^2} &=& \frac{R}{T^2} \frac{d^2 \bf{\hat{r}}}{d \hat{t}^2} \hspace{5.5cm} (3)
\end{eqnarray}

Substituting these values in Eq. (1), we get:

\begin{equation}
    \nonumber
	\frac{R}{T^2} \frac{d^2 \bf{\hat{r}}}{d \hat{t}^2} = \frac{F^*\bf{\hat{F}}}{\hat{M_E} M}  	\hspace{5cm} (4) \\ 
\end{equation}

\begin{equation}
    \nonumber
	\frac{d^2 \bf{\hat{r}}}{d \hat{t}^2} = \frac{\bf{\hat{F}}}{\hat{M_E} } \frac{F^*T^2}{RM}         \hspace{5cm} (5)
\end{equation}





Let,

\begin{equation}
     \nonumber
	\frac{F^*T^2}{RM} = 1 
\end{equation}

\begin{equation}
	F^* = \frac{RM}{T^2}
\end{equation}

So, Eq. (5) becomes:

\begin{equation}
	\frac{d^2 \bf{\hat{r}}}{d \hat{t}^2} = \frac{\bf{\hat{F}}}{\hat{M_E} } \hspace{5cm} (6)
\end{equation}

Eq. (6) is dimensionless. It can be broken down into two coupled first order equations:

\begin{equation}
    \nonumber
	\frac{d \bf{\hat{r}}}{d \hat{t}} = \hat{\bf{v}} \hspace{5cm} (7) 
\end{equation}
    
\begin{equation}
	\frac{d \bf{\hat{v}}}{d \hat{t}} = \frac{\bf{\hat{F}}}{\hat{M_E} } \hspace{5cm} (8)
\end{equation}

Eq. (7) and Eq. (8) are the required dimensionless coupled first order ODEs.



The force is given by Newton's law:
\begin{equation}
     \nonumber
	|{\bf{F}}| = \frac{GM_EM_S}{|\bf{r}|^2} = \frac{\hat{M}_E \hat{M}_S}{|\hat{\bf{r}}|^2}\frac{GM^2}{R^2} 
\end{equation}

So,



\begin{equation}
    \nonumber
	|{\bf{\hat{F}}}| = \frac{\hat{M}_E \hat{M}_S}{|\hat{\bf{r}}|^2}\frac{GM^2}{R^2} \frac{1}{F^*} 	=\frac{\hat{M}_E \hat{M}_S}{|\hat{\bf{r}}|^2}\frac{GM^2}{R^2} \frac{T^2}{RM} =\frac{\hat{M}_E \hat{M}_S}{|\hat{\bf{r}}|^2}\frac{GMT^2}{R^3} =\frac{\hat{G} \hat{M}_E \hat{M}_S}{|\hat{\bf{r}}|^2}	
\end{equation}
 




where,
\begin{equation}
	\hat{G} = \frac{GMT^2}{R^3}
\end{equation}


Finally, considering the direction of the force:

\begin{equation}
    \nonumber
	\bf{F} = |\bf{F}|\cos(\theta) \hat{\bf{i}}  + |\bf{F}|\sin(\theta) \hat{\bf{j}}
\end{equation}

Here,
\begin{equation}
	\theta = \tan^{-1}\frac{y}{x}
\end{equation} 



The value of the normalizing constants are taken in astronomical units:

\begin{equation}
	\nonumber
	T = 365 \times 24\times 60 \times 60 \,seconds = 1.0 \, year 
\end{equation}
\begin{equation}
	R = 1.5 \times 10^{11} \,m = 1.0 \, AU 
\end{equation}
\begin{equation}
	M = 6 \times 10^{24} \,kg = 1.0 \, M_E
\end{equation}

The sun is positioned at the origin and assumed to have no motion. Earth's initial position is taken as x = 1 AU and y = 0. The initial velocity is taken to be 30 km/s.




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MIT
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