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	\begin{document}

      \begin{center}
      {\bf \large The Kepler Problem\\}
      \vspace{0.5cm}
      {\small Math 241 Homework \\}  
      {\small John Baez}  
      \end{center}

The goal of this problem is to see why two particles interacting via
gravity move along nice curves like ellipses, parabolas and hyperbolas.
This is called the {\bf Kepler problem} since it was Kepler who discovered
that the orbits of planets were elliptical, and explaining this was the
first major triumph of Newtonian mechanics.  
However, let's start quite generally by studying an arbitrary central 
force, and specialize to the $1/r^2$ force law of gravity only when
that becomes necessary.

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 the center-of-mass frame.  This 
means we can assume
\be          m_1 \q_1(t) + m_2 \q_2(t) = 0 \label{com} \ee
for all times $t$.   Using conservation of angular momentum and symmetry
under rotations we can assume both particles lie in the $xy$ plane at
all  times.  Thus we may assume the $z$ component of $\q_1(t)$ and
$\q_2(t)$ equal zero at all times.   In short, we have reduced the problem to
a 2-dimensional problem!

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}
\[              m = {m_1 m_2 \over m_1 + m_2}   .\]

\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 
\be     E =  {1\over 2}m \left|\dot \q\right|^2 + V(|\q|) \label{E} \ee
\vskip1em

3.   Let $\J$ be the total angular momentum of the 2-particle system.
Show that 
\be   \J  = m \q \times \dot \q \label{J}  \ee 
Show that the $x$ and $y$ components of $\J$ are zero.   Let's call the
$z$ component $j$.  
\vskip1em

Now let's work in polar coordinates: the point $\q$ lies in the
$xy$ plane so write it in polar coordinates as $(r, \theta)$.
As usual, let's write time derivatives with dots:
\[      \dot r = {dr\over dt}, \qquad \dot \theta = {d\theta\over dt} .\]

\vskip1em

4.  Use equations (\ref{E}) and (\ref{J}) to show that
\be    E = {1\over 2} m(r^2 \dot \theta^2 + \dot r^2) + V(r)  \label{E2} \ee
and
\be    j = m r^2 \dot \theta    \label{J2}  \ee

\vskip1em
5.  We can use equation (\ref{J2}) to solve for $\dot \theta$ in terms of $r$:
\be       \dot \theta = {j\over mr^2}   \label{dottheta} \ee
Use this and equation (\ref{E2}) to express $E$ in terms of $r$:
\[         E = {1\over 2}m \dot r^2 + U(r)   \]
where 
\[         U(r) = V(r) + {j^2 \over 2mr^2}  \]
Thus the energy looks just like the energy of a particle of mass $m$
in a potential $U$ on the half-line $\{0 < r < \infty\}$.  We have reduced
the problem to a 1-dimensional problem!   $U$ is called the {\bf
effective potential}.  Note that the second term creates the effect
of a repulsive force equal to $j^2/mr^3$, called the {\bf centrifugal force}.

\vskip1em
6.   Show that 
\be      \dot r = \sqrt{{2\over m}(E - U(r))} . \label{dotr} \ee
We could solve this differential equation to find $r$ as a function
of $t$, but it's nicer to find $r$ as a function of $\theta$, since
this allows us to see the shape of the particles' orbits.  In fact
it turns out to be easier to first find $\theta$ as a function of $r$
and then solve for $r$ in terms of $\theta$ --- so that's what we'll do.

\vskip1em
7.   Using equations (\ref{dottheta}) and (\ref{dotr}) show that
\[      {d\theta\over dr} = {\dot \theta \over \dot r}    = 
       {j/mr^2 \over \sqrt{{2\over m}(E - U(r))}}    \]
Conclude that
\be         \theta = \theta_0 + 
\int  {(j/mr^2)\;dr \over \sqrt{{2\over m}(E - U(r))}} \label{int} \ee

\vskip1em

{\textit{\textbf{
Now let's specialize to the case of gravity, where $V(r) = -Gm_1m_2/r$.}}}

\vskip1em
8.  To reduce the clutter a little bit, write
\[             V(r) = -k/r  .\]
Sketch a graph of the effective potential $U(r)$ in this case, and
say what a particle moving in this potential
would do, depending on its energy $E$.

\vskip1em

9.  Show using equation (\ref{int}) that
\[       \theta = \theta_0 + 
\arccos {{j\over mr} - {k\over j}\over
\sqrt{{2E\over m} + {k^2 \over j^2}}}  .\]
This is the only part of this homework where you really need to
{\it sweat}.  However, some ways to do it are
easier than others, so think a bit before you plunge into an 
enormous masochistic calculation --- if you do it intelligently,
you will only need a {\it medium-sized} masochistic calculation!
For example, you may want to derive a general formula for
\[         \int {dx \over \sqrt{ax^2 + bx + c}}  \]
and then use it to do the integral in equation (\ref{int}).

\vskip1em
10.  Reduce the clutter a bit more by defining
\[           p = j^2/km, \qquad e = \sqrt{1 + {2Ej^2\over mk^2}}  .\]
Show that in terms of these variables we have
\[         \theta = \theta_0 + \arccos\left( {p/r - 1\over e}\right)  \]
and thus
\be       r = {p \over 1 + e \cos(\theta - \theta_0)}.   \label{ellipse} \ee
Note that when $\theta = \theta_0$ the denominator is maximized,
so $r$ is minimized.  We call this point the {\it perihelion}
of the orbit, since in Newton's original application to the earth
going around this sun, this is the point on the earth's orbit where
its distance to the sun is minimized.

\vskip1em
11.  Show that equation (\ref{ellipse}) describes an ellipse, parabola
or hyperbola  in polar coordinates, depending on the value of the parameter
$e$, which we call the {\bf eccentricity}.  To do this, first simplify 
things by rotating the coordinate system so that $\theta_0 = 0$.   Then 
express the variables $r,\theta$ in terms of $x,y$ and show that equation
(\ref{ellipse}) becomes the equation
\[          (1 - e^2)x^2 + 2epx + y^2 = p^2  .\]
Show that for $e = 0$ this describes a circle of radius $p$.  
Show also that for $0 < e < 1$
it describes an ellipse, for $e = 1$ it describes a parabola, and
for $e > 1$ it describes a hyperbola.  Newton used the elliptic
case to predict when the comet discovered by Edmund Halley would
return!  However, he didn't give Halley much credit for obtaining the 
necessary data --- so they wound up bitter enemies.


\end{document}

