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LaTeX/trajectory_design.tex
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@@ -81,69 +81,68 @@
\ddot{\vec{r}} = - \frac{\mu}{r^2} \hat{r}
\end{equation}
\subsubsection{Kepler's Laws and Equations}
\subsection{Kepler's Laws}
Now that we've fully qualified the forces acting within the Two Body Problem, we can concern
ourselves with more practical applications of it as a force model. It should be noted,
firstly, that the spacecraft's position and velocity (given an initial position and velocity
and of course the $\mu$ value of the primary body) is actually analytically solvable for all
future points in time. This can be easily observed by noting that there are three
one-dimensional equations (one for each component of the three-dimensional space) and
three unknowns (the three components of the second derivative of the position).
Now that we've fully qualified the forces acting within the Two Body Problem, we can concern
ourselves with more practical applications of it as a force model. It should be noted,
firstly, that the spacecraft's position and velocity (given an initial position and velocity
and of course the $\mu$ value of the primary body) is actually analytically solvable for all
future points in time. This can be easily observed by noting that there are three
one-dimensional equations (one for each component of the three-dimensional space) and
three unknowns (the three components of the second derivative of the position).
In the early 1600s, Johannes Kepler produced just such a solution, by taking advantages of
what is also known as ``Kepler's Laws'' which are\cite{murray1999solar}:
In the early 1600s, Johannes Kepler produced just such a solution, by taking advantages of
what is also known as ``Kepler's Laws'' which are\cite{murray1999solar}:
\begin{enumerate}
\item Each planet's orbit is an ellipse with the Sun at one of the foci. This can be
expanded to any orbit by re-wording as ``all orbital paths follow a conic section
(circle, ellipse, parabola, or hyperbola) with a primary mass at one of the foci''.
\begin{enumerate}
\item Each planet's orbit is an ellipse with the Sun at one of the foci. This can be
expanded to any orbit by re-wording as ``all orbital paths follow a conic section
(circle, ellipse, parabola, or hyperbola) with a primary mass at one of the foci''.
Specifically the path of the orbit follows the trajectory equation:
Specifically the path of the orbit follows the trajectory equation:
\begin{equation}
r = \frac{\sfrac{h^2}{\mu}}{1 + e \cos(\theta)}
\end{equation}
\begin{equation}
r = \frac{\sfrac{h^2}{\mu}}{1 + e \cos(\theta)}
\end{equation}
Where $h$ is the angular momentum of the satellite, $e$ is the
eccentricity of the orbit, and $\theta$ is the true anomaly, or simply
the angular distance the satellite has traversed along the orbit path.
Where $h$ is the angular momentum of the satellite, $e$ is the
eccentricity of the orbit, and $\theta$ is the true anomaly, or simply
the angular distance the satellite has traversed along the orbit path.
\item The area swept out by the imaginary line connecting the primary and secondary
bodies increases linearly with respect to time. This implies that the magnitude of the
orbital speed is not constant. For the moment, we'll just take this
value to be a constant:
\item The area swept out by the imaginary line connecting the primary and secondary
bodies increases linearly with respect to time. This implies that the magnitude of the
orbital speed is not constant. For the moment, we'll just take this
value to be a constant:
\begin{equation}\label{swept}
\frac{\Delta t}{T} = \frac{k}{\pi a b}
\end{equation}
\begin{equation}\label{swept}
\frac{\Delta t}{T} = \frac{k}{\pi a b}
\end{equation}
Where $k$ is the constant value, $a$ and $b$ are the semi-major and
semi-minor axis of the conic section, and $T$ is the period. In the
following section, we'll derive the value for $k$.
Where $k$ is the constant value, $a$ and $b$ are the semi-major and
semi-minor axis of the conic section, and $T$ is the period. In the
following section, we'll derive the value for $k$.
\item The square of the orbital period is proportional to the cube of the semi-major
axis of the orbit, regardless of eccentricity. Specifically, the relationship is:
\item The square of the orbital period is proportional to the cube of the semi-major
axis of the orbit, regardless of eccentricity. Specifically, the relationship is:
\begin{equation}
T = 2 \pi \sqrt{\frac{a^3}{\mu}}
\end{equation}
\begin{equation}
T = 2 \pi \sqrt{\frac{a^3}{\mu}}
\end{equation}
Where $T$ is the period and $a$ is the semi-major axis.
\end{enumerate}
Where $T$ is the period and $a$ is the semi-major axis.
\end{enumerate}
\subsection{Analytical Solutions to Kepler's Equations}
\subsection{Kepler's Equation}
Kepler was able to produce an equation to represent the angular displacement of an
orbiting body around a primary body as a function of time, which we'll derive now
for the elliptical case\cite{vallado2001fundamentals}. Since the total area of an
ellipse is the product of $\pi$, the semi-major axis, and the semi-minor axis ($\pi
a b$), we can relate (by Kepler's second law) the area swept out by an orbit as a
function of time, as we did in Equation~\ref{swept}.
This leaves just one unknown variable $k$, which we can determine through use of the
geometric auxiliary circle, which is a circle with radius equal to the ellipse's semi-major
axis and center directly between the two foci, as in Figure~\ref{aux_circ}.
orbiting body around a primary body as a function of time, which we'll derive now for
the elliptical case\cite{vallado2001fundamentals}. Since the total area of an ellipse is
the product of $\pi$, the semi-major axis, and the semi-minor axis ($\pi a b$), we can
relate (by Kepler's second law) the area swept out by an orbit as a function of time, as
we did in Equation~\ref{swept}. This leaves just one unknown variable $k$, which we can
determine through use of the geometric auxiliary circle, which is a circle with radius
equal to the ellipse's semi-major axis and center directly between the two foci, as in
Figure~\ref{aux_circ}.
\begin{figure}[H]
\centering
@@ -151,13 +150,15 @@
\caption{Geometric Representation of Auxiliary Circle}\label{aux_circ}
\end{figure}
In order to find the area swept by the spacecraft, $k$, we can take advantage of the fact
that that area is the triangle $k_1$ subtracted from the elliptical segment $PCB$:
In order to find the area swept by the spacecraft\cite{vallado2001fundamentals}, $k$, we
can take advantage of the fact that that area is the triangle $k_1$ subtracted from the
elliptical segment $PCB$:
\begin{equation}\label{areas_eq}
k = area(seg_{PCB}) - area(k_1)
\end{equation}
\noindent
Where the area of the triangle $k_1$ can be found easily using geometric formulae:
\begin{align}
@@ -165,8 +166,11 @@
&= \frac{ab}{2} \left(e \sin E - \cos E \sin E \right)
\end{align}
Now we can find the area for the elliptical segment $PCB$ by first finding the circular
segment $POB'$, subtracting the triangle $COB'$, then applying the fact that an ellipse is
\noindent
Where $E$, notably, is the eccentric anomaly, or the angular distance from the
periapsis to the vertical projection of the spacecraft on the auxiliary circle. Now we
can find the area for the elliptical segment $PCB$ by first finding the circular segment
$POB'$, subtracting the triangle $COB'$, then applying the fact that an ellipse is
merely a vertical scaling of a circle by the amount $\frac{b}{a}$.
\begin{align}
@@ -185,10 +189,10 @@
\end{equation}
Which we can then substitute back into the equation for the swept area as a function of
time (Equation~\ref{swept}):
time (Equation~\ref{swept}) for period of time since the spacecraft left periapsis:
\begin{equation}
\frac{\Delta t}{T} = \frac{E - e \sin E}{2 \pi}
\frac{\Delta t}{T} = \frac{t_2 - t_{peri}}{T} = \frac{E - e \sin E}{2 \pi}
\end{equation}
Which is, effectively, Kepler's equation. It is commonly known by a different form:
@@ -200,18 +204,81 @@
Where we've defined the mean anomaly as $M$ and used the fact that $T =
\sqrt{\frac{a^3}{\mu}}$. This provides us a useful relationship between Eccentric Anomaly
($E$) which can be related to spacecraft position, and time, but we still need a useful
algorithm for solving this equation.
algorithm for solving this equation in order to use this equation to propagate a
spacecraft.
\subsubsection{LaGuerre-Conway Algorithm}\label{laguerre}
\subsection{LaGuerre-Conway Algorithm}\label{laguerre}
For this application, I used an algorithm known as the LaGuerre-Conway
algorithm\cite{laguerre_conway}, which was presented in 1986 as a faster and more
robust algorithm for directly solving Kepler's equation and has been in use in many
applications since. This algorithm is known for its convergence robustness and also
its speed of convergence when compared to higher order Newton methods.
For this thesis, the algorithm used to solve Kepler's equation was the general numeric
root-finding scheme first developed by LaGuerre in the 1800s and first applied to
Kepler's equation by Bruce Conway in 1985\cite{laguerre_conway}. In his paper, Conway
makes a compelling argument for utilizing the less common LaGuerre method over higher
order Newton or Newton-Raphson methods.
This thesis will omit a step-through of the algorithm itself, but psuedo-code for
the algorithm will be discussed briefly in Section~\ref{conway_pseudocode}.
The Newton-Raphson methods, while found to generally have quite impressive convergence
rates (generally successfully solving Kepler's equation correctly within 5 iterations),
were prone to failures in convergence given certain specific initial conditions.
Therefore LaGuerre's algorithm is proposed as an alternative.
The algorithm can be relatively easily derived by examining the polynomial equation with
$m$ roots:
\begin{equation}
g(x) = (x - x_1) (x - x_2) ... ( x - x_m)
\end{equation}
\noindent
We can then generate some useful convenience functions as:
\begin{align}
\ln|g(x)| &= \ln|(x - x_1)| + \ln|(x - x_2)| + ... + \ln|( x - x_m)| \\
\frac{d\ln|g(x)|}{dx} &= \frac{1}{x - x_1} + \frac{1}{x - x_2} + ... + \frac{1}{x -
x_m} = G_1(x)
\end{align}
and
\begin{align}
\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}
x - x_i = b, i=2,3,...,m
\end{equation}
\noindent
We can then rewrite $G_1$ and $G_2$ as:
\begin{align}
G_1 &= \frac{1}{a} + \frac{n-1}{b} \\
G_2 &= \frac{1}{a^2} + \frac{n-1}{b^2}
\end{align}
\noindent
Which may be solved for $a$ in terms of $G_1$, $G_2$:
\begin{equation}
a = \frac{n}{G_1 \pm \sqrt{(n-1)(nG_2 - G_1^2)}}
\end{equation}
\noindent
With corresponding iteration function:
\begin{equation}
x_{i+1} = x_i - \frac{n g(x_i)}{g'(x_i) \pm \sqrt{(n-1)^2 f'(x_i)^2 - n (n-1) f(x_i)
f''(x_i)}}
\end{equation}
This iteration scheme can be shown to be globally convergent, regardless of the initial
guess. More relevantly, Conway also showed that the application of this method to
Kepler's equation was shown to converge with similar speed to many of the best common
higher order Newton-Raphson solvers. However, LaGuerre's method was also found to be
incredibly robust, converging to the correct value for every one of Conway's 500,000
tests. Because of this robustness, it is very useful for propagating spacecraft states.
\section{Interplanetary Considerations}\label{interplanetary}