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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small January 17, 2008 \\}
{\small John Baez}
\end{center}
\section*{The Kepler Problem}
The goal of this problem is to see why particles moving in an inverse
square force law --- for example, {\it 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 particle moving in a central force.
Its position is a function of time, say $\q \maps \R \to \R^3$,
satisfying Newton's law:
\[ m\ddot \q = f(|\q|) {\q \over |\q|} \]
Here $m$ is its 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 angular momentum
we can choose coordinates where the particle lies in the $xy$ plane at
all times. Thus we may assume the $z$ component of $\q(t)$ and
In short, we have reduced the problem to a 2-dimensional problem!
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:
\[ \dot r = {dr\over dt}, \qquad \dot \theta = {d\theta\over dt} .\]
\vskip 1em
\textit{\textbf{Now here's where you come in!}}
\vskip 1em
1. Show that the energy $E$ of the particle is given by
\be E = {1\over 2} m(r^2 \dot \theta^2 + \dot r^2) + V(r) \label{E2} \ee
and the angular momentum $J$ is a vector with vanishing $x$ and $y$
components, and $z$ component given by
\be j = m r^2 \dot \theta . \label{J2} \ee
\vskip1em
2. 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 $r$:
\[ E = {1\over 2}m \dot r^2 + V_{\eff}(r) \]
where
\[ V_{\eff}(r) = V(r) + {j^2 \over 2mr^2}. \]
Thus the energy looks just like the energy of a particle of mass $m$
in a potential $V_{\eff}$ on the half-line $\{0 < r < \infty\}$. We have
reduced the problem to a 1-dimensional problem! $V_{\eff}$ 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
3. Show that
\be \dot r = \sqrt{{2\over m}(E - V_{\eff}(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
4. Using equations (\ref{dottheta}) and (\ref{dotr}) show that
\[ {d\theta\over dr} =
{j/mr^2 \over \sqrt{{2\over m}(E - V_{\eff}(r))}} . \]
Conclude that
\be \theta = \theta_0 +
\int {(j/mr^2)\;dr \over \sqrt{{2\over m}(E - V_{\eff}(r))}} \label{int} \ee
\vskip1em
\textit{\textbf{
Now let's specialize to the case of gravity, where $f(r) = -k/r^2$
and thus $V(r) = -k/r$ for some constant $k$.}}
\vskip1em
5. Sketch a graph of the effective potential $V_{\eff}(r)$ in this case, and
say what a particle moving in this potential
would do, depending on its energy $E$.
\vskip1em
6. 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
7. 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
8. 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.
\vskip1em
9. How are the three kinds of orbits --- ellipse, parabola
or hyperbola --- related to the energy $E$?
\end{document}