Maybe nearing the final commit

LaTeX/trajectory_design.tex

@@ -34,7 +34,6 @@
 			the end, we'll also assume that the mass of the spacecraft ($m_2$) is much much smaller
 			than the mass of the planetary body ($m_1$) and enough so as to be considered
 			negligible.
-
 			\begin{figure}[H]
 				\centering
 				\includegraphics[width=0.65\textwidth]{LaTeX/fig/2bp}
@@ -42,21 +41,18 @@
 				the center of mass in the two body problem}
 				\label{2bp_fig}
 			\end{figure}
-
 			Under these assumptions, the force acting on the body due to the law of universal
 			gravitation is:
 			\begin{align}
 				F_2 &= - \frac{G m_1 m_2}{r^2} \frac{\vec{r}}{\left| r \right|} \\
 				F_1 &= \frac{G m_2 m_1}{r^2} \frac{\vec{r}}{\left| r \right|}
 			\end{align}
-
 			And by Newton's second law (force is the product of mass and acceleration), we can
 			derive the following differential equations for $r_1$ and $r_2$:
 			\begin{align}
 				m_2 \ddot{\vec{r}}_2 &= - \frac{G m_1 m_2}{r^2} \frac{\vec{r}}{\left| r \right|} \\
 				m_1 \ddot{\vec{r}}_1 &= \frac{G m_2 m_1}{r^2} \frac{\vec{r}}{\left| r \right|}
 			\end{align}
-
 			Where $\vec{r}$ is the position of the spacecraft relative to the primary body,
 			$\vec{r}_1$ is the position of the primary body relative to the origin of the inertial
 			frame, and $\vec{r}_2$ is the position of the spacecraft relative to the center of the
@@ -177,7 +173,6 @@
 				&= \frac{abE}{2} - \frac{ab}{2} \left( \cos E \sin E \right) \\
 				&= \frac{ab}{2} \left( E - \cos E \sin E \right)
 			\end{align}
-
 			By substituting the two areas back into Equation~\ref{areas_eq} we can get the $k$ area
 			swept out by the spacecraft:
 			\begin{equation}
@@ -224,7 +219,6 @@
 			\frac{-d^2\ln|g(x)|}{dx^2} &= \frac{1}{(x - x_1)^2} + \frac{1}{(x - x_2)^2} + ... +
 			\frac{1}{(x - x_m)^2} = G_2(x)
 		\end{align}
-
 		Now we define the targeted root as $x_1$ and make the approximation that all of the
 		other roots are equidistant from the targeted root, which means:
 		\begin{equation}
@@ -333,7 +327,6 @@
 			\begin{equation}
 				\xi = \frac{v^2}{2} - \frac{\mu}{r} = \frac{v_\infty^2}{2}
 			\end{equation}
-
 			We can then leverage the conservation of energy to determine the velocity at a
 			particular point, $r_{ins}$:
 			\begin{align}
@@ -482,14 +475,12 @@
 					\cos (\Delta \theta) &= \frac{\vec{r}_1 \cdot \vec{r}_2}{|\vec{r}_1| |\vec{r}_2|} \\
 					\Delta \theta &= \arctan(y_2/x_2) - \arctan(y_1/x_1)
 				\end{align}
-
 				The direction of motion is then chosen such that counter-clockwise orbits are
 				considered, as travelling in the same direction as the planets is generally more
 				efficient. Next, the variable $A$ is defined:
 				\begin{equation}
 					A = DM \sqrt{|r_1| |r_2| (1 - \cos(\Delta \theta))}
 				\end{equation}
-
 				A is independent of $\psi$, and therefore won't need updating as the iteration
 				proceeds. Then $\psi$ is initialized to any number within its bounds
 				($[-4\pi,4\pi^2]$), arbitrarily set to 0, representing a parabolic arc as a starting
@@ -520,7 +511,6 @@
 				\begin{equation}
 					y = |r_1| + |r_2| + \frac{A (c_3 \psi - 1)}{\sqrt{c_2}}
 				\end{equation}
-
 				We can then finally calculate the variable $\chi$, and from that, the time of
 				flight:
 				\begin{equation}