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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small January 29, 2008 \\}
{\small John Baez}
\end{center}
\section*{The 2-Body Problem}
The goal of this problem is to understand a pair of particles
interacting via a central force such as gravity. We'll reduce it
to problem you've already studied --- the case of a {\it single}
particle in a central force.
Suppose we have a system of two particles interacting by a central force.
Their positions are functions of time, say $\q_1,\q_2 \maps \R \to \R^3$,
satisfying Newton's law:
\[ m_1\ddot \q_1 = f(|\q_1 - \q_2|) {\q_1 - \q_2\over |\q_1 - \q_2|} \]
\[ m_2 \ddot \q_2 = f(|\q_2 - \q_1|) {\q_2 - \q_1\over |\q_2 - \q_1|} . \]
Here $m_1,m_2$ are their masses, and the force is described by
some smooth function $f \maps (0,\infty) \to \R$. Let's write
the force in terms of a potential as follows:
\[ f(r) = -{d V\over dr} .\]
Using conservation of momentum and symmetry under translations
and Galilei boosts we can work in coordinates where
\be m_1 \q_1(t) + m_2 \q_2(t) = 0 \label{com} \ee
for all times $t$. This coordinate system is called the
{\bf center-of-mass frame}.
We could use equation (\ref{com}) to solve for $\q_2$ in terms of
$\q_1$, or vice versa, but we can also use it to express both $\q_1$
and $\q_2$ in terms of the {\bf relative position}
\[ \q(t) = \q_1(t) - \q_2(t) . \]
This is more symmetrical, so this is what we will do. Henceforth we
only need to talk about $\q$. Thus we have reduced the problem to a
1-body problem!
\vskip2em
\noindent
{\textit{\textbf{Now here's where you come in:}}}
\vskip2em
1. Show that $\q(t)$ satisfies the equation
\[ m\ddot \q = f(|\q|) {\q\over |\q|} \]
where $m$ is the so-called {\bf reduced mass}
\[ m = {m_1 m_2 \over m_1 + m_2} .\]
Note that this looks exactly like Newton's second law
for a single particle!
\vskip1em
2. Recall that the total energy $E$ of the 2-particle system is
the sum of the kinetic energies of the particles plus the potential
energy. Express $E$ in terms of $\q$ and the reduced mass.
Show that
\[ E = {1\over 2}m \left|\dot \q\right|^2 + V(|\q|) \]
Note that this looks exactly like the energy of a single
particle!
\vskip1em
3. Let $\J$ be the total angular momentum of the 2-particle system.
Show that
\[ \J = m \q \times \dot \q \]
Note that this looks exactly like the angular momentum
of a single particle!
\vskip1em
At this point we're back to a problem you've already solved:
a {\em single} particle in a central force. The only difference
is that now $\q$ stands for the {\em relative} position and
$m$ stands for the {\em reduced} mass!
So, we instantly know that two bodies orbiting each other
due to the force of gravity will {\em both} have an orbit that's
either an ellipse, or a parabola, or a hyperbola... when viewed
in the center-of-mass frame.
\end{document}