\documentclass[12pt]{article} % BB Oct 1 2011.
\usepackage{epsfig}
\textheight 9.0in \textwidth  6.0in \topmargin -0.4in \headsep 0.4in  
\oddsidemargin  0.25in  \evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
   \vrule width.#2pt} \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
\pagestyle{empty}

\begin{document}

% \begin{figure}[tb] \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=figs/KeplerUeff.eps,width=15.0cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader} \end{figure}

\centerline{\bf Solution 17 set~4.} \bigskip

\noindent
1. Yes, the Lagrangian does not depend on the time. 
\bigskip

\noindent
2. Solving for $\vec{r}_1$:
$$ M\,\vec{R} = m_1\vec{r}_1+m_2\vec{r}_2\,,
   ~~M=m_1+m_2\,,\ \vec{r} = \vec{r}_1-\vec{r}_2\,,$$
$$ m_1\vec{r}_1 = M\,\vec{R}-m_2\vec{r}_2 = 
   M\,\vec{R}+m_2\vec{r}-m_2\vec{r}_1\,,$$
$$ \vec{r}_1 =\vec{R} + \frac{m_2}{M}\,\vec{r}\,,$$
$$ m_2\vec{r}_2 = M\,\vec{R}-m_1\vec{r}_1 = 
   M\,\vec{R}-m_1\vec{r}-m_1\vec{r}_2\,,$$
$$ \vec{r}_2 =\vec{R} - \frac{m_1}{M}\,\vec{r}\,.  $$
\bigskip

\noindent
3. In the CM frame $\vec{R}=0$ and using $\mu=m_1m_2/(m_1+m_2)$ 
we can write 
$$ \vec{r}_1 = \frac{\mu}{m_1}\,\vec{r}\,,\qquad
   \vec{r}_2 = - \frac{\mu}{m_2}\,\vec{r}\,.          $$
Therefore,
$$ T_{\rm cm} = \frac{1}{2}\left( \frac{\mu^2}{m_1} +
                \frac{\mu^2}{m_2} \right)\,\vec{v}^{\,2} =
   \frac{1}{2}\,\frac{\mu^2\,(m_1+m_2)}{m_1m_2}\,\vec{v}^{\,2} 
   = \frac{1}{2}\,\mu\,\vec{v}^{\,2} $$
\bigskip

\noindent
4. Angular momentum:
$$ m_1\,\vec{r}_1\times\vec{v}_1 + m_2\,\vec{r}_2\times\vec{v}_2 =
   m_1\,\frac{\mu}{m_1}\,\vec{r}\times\vec{v}_1 - 
   m_2\,\frac{\mu}{m_2}\,\vec{r}\times\vec{v}_2 =  
   \mu\,\vec{r}\times\left(\vec{v}_1-\vec{v}_2\right) =
   \mu\,\vec{r}\times\vec{v}\ . $$
\end{document}
