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\documentclass{article}
\pagestyle{empty}
\usepackage{amsfonts,amssymb}
%\usepackage{latexsym}



	\textwidth 6in
	\textheight 8.5in	\evensidemargin .25in
	\oddsidemargin .25in
	\topmargin .25in
	\headsep 0in
	\headheight 0in
	\footskip .5in
%If you want single spaced copy, delete the next two lines.
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%	\renewcommand{\baselinestretch}{1.5}
	\begin{document}

      \begin{center}
      {\bf \large The Kepler Problem Revisited: \\
           The Runge--Lenz Vector\\ }
      \vspace{0.5cm}
      {\small Answers to Math 241 Homework \\}  
      {\small Toby Bartels}  
      \end{center}


1.  
$$ \begin{array}{ccl}
\dot \J 
&=& m \dot \q \times \dot \q + m \q \times \ddot \q   \\
\\
&=& \0 - k \q \times { \hat \q / q ^ 2 }         \\
\\
&=& \0 
\end{array}
$$
since cross products of parallel vectors are zero.

\vskip 1em

2.  
$$ \begin{array}{ccl} 
m \ddot \q \times \J 
&=& - k \hat \q \times { \J / q ^ 2 }        \\
\\
&=& - m k \hat \q \times { ( { \q \times \dot \q } ) / q ^ 2 }      \\
\\
&=& { - m k ( { \hat \q \cdot \dot \q } ) \q / q ^ 2 }     
+ { m k ( { \hat \q \cdot \q } ) \dot \q / q ^ 2 }        \\
\\
&=& ( { { m k / q ^ 3 } } )
\big ( { ( { \q \cdot \q } ) \dot \q
- ( { \q \cdot \dot \q } ) \q } \big ) 
\end{array}
$$
so \( \ddot \q \times \J
= ( { { k / q ^ 3 } } )
\big ( { ( { \q \cdot \q } ) \dot \q
- ( { \q \cdot \dot \q } ) \q } \big ) \)
Now, differentiate \( q ^ 2 = \q \cdot \q \)
to get \( 2 q \dot q = 2 \q \cdot \dot \q \),
and then differentiate \( \q = q \hat \q \)
to get \( \dot \q = \dot q \hat \q + q \dot { \hat \q } \).
Then \( q ^ 3 \dot { \hat \q } = q ^ 2 \dot \q - q \dot q \q
= ( { \q \cdot \q } ) \dot \q - ( { \q \cdot \dot \q } ) \q \),
and we see that \( \ddot \q \times \J = k \dot { \hat \q } \).


\vskip 1em
3. 
Since \( \dot \J = \0 \),
we obtain \( {d\over dt} ( \dot \q \times \J )  
= \ddot \q \times \J = k \dot { \hat \q } \),
as desired.

\vskip 1em
4.
Since \( k \) is a constant,
the above formula says that
$$ {d\over dt} ( \dot \q \times \J - k \hat \q   ) = \0 .$$
Thus if \( \x : = \dot \q \times \J - k \hat \q \),
then we have \( \dot \q \times \J = k \hat \q + \x \)
with \( \dot \x = \0 \).

\vskip 1em
5. 
If \( \A = \frac 1 k \dot \q \times \J - \hat \q \),
then 
$$ \begin{array}{ccl} 
\A \cdot \q
&=& \frac 1 k ( { \dot \q \times \J } ) \cdot \q - \hat \q \cdot \q  \\
\\
&=& \frac 1 k ( { \q \times \dot \q } ) \cdot \J - q \hat \q \cdot \hat \q  \\
\\
&=& \frac 1 { k m } \J \cdot \J - q 
\end{array}
$$
since \( \J = m \q \times \dot \q \) and \( | { \hat \q } | = 1 \).

\vskip 1em
6.  
\( A q \cos \theta = \A \cdot \q = \frac 1 { k m } \J \cdot \J - q \),
so \( q ( { A \cos \theta + 1 } ) = \frac 1 { k m } \J \cdot \J \),
resulting in the desired formula.
\end {document}


