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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small January 24, 2008 \\}
{\small John Baez\\}
{\small homework by: Scot Childress}
\end{center}
\section*{The Kepler Problem}
\newcommand{\eff}{\text{eff}}
(\textbf{Background}) 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 $\ddot q(t)$ vanish for all $t$.
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
\noindent 1. \emph{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
\vskip 1em
Recall that the energy of such a particle is given by
\begin{equation} E = {1\over 2} m \dot q(t)^2 + V(|q(t)|).\label{energy}\end{equation}
Noting that in polar coordinates
\begin{equation}\label{qdot}\dot q = (\dot r \cos\theta - r\dot\theta\sin\theta, \dot r \sin\theta + r\dot\theta\cos\theta),\end{equation}
we see that
\[\dot q(t)^2 = \dot r^2 \cos^2\theta -r\dot r\dot\theta \sin 2\theta + r^2\dot\theta^2\sin\theta + \dot r^2\sin^2\theta + r\dot r\dot\theta \sin 2\theta + r^2\dot\theta^2\cos^2\theta\]
which reduces nicely to $\dot q(t)^2 = \dot r^2 + r^2\dot\theta^2.$ Substitution of this last expression for $\dot q(t)^2$ into \eqref{energy} and noting that $r = |q(t)|$ yields \eqref{E2}.
Now we will show that the angular momentum $J$ is a vector with vanishing $x$ and $y$ components with the $z$ component given by \eqref{J2}. The angular momentum is
\[J = mq\times \dot q\]
and if we use the expression for $\dot q$ obtained in \eqref{qdot}, we have
\begin{align*}
q\times \dot q &= (r\cos\theta \hat\imath + r\sin\theta \hat\jmath)\times [(\dot r \cos\theta - r\dot\theta\sin\theta)\hat\imath + (\dot r \sin\theta + r\dot\theta\cos\theta)\hat\jmath]\\
&= [r^2\dot\theta\cos^2\theta + r\dot r \cos\theta\sin\theta - r\sin\theta(\dot r\cos\theta - r\dot\theta\sin\theta)]\hat k\\
&= r^2\dot\theta \hat k,
\end{align*}
so that $J = mr^2\dot \theta \hat k$.
\vskip 1em
\noindent 2. \emph{We use equation (\ref{J2}) to solve for $\dot \theta$ in terms of $r$:
\be \dot \theta = {j\over mr^2} \label{dottheta} \ee
Combining this and equation (\ref{E2}) we express $E$ in terms of} $r$:
\be\label{energy2} E = {1\over 2}m \dot r^2 + V_{\eff}(r) \ee
\emph{where}
\[ V_{\eff}(r) = V(r) + {j^2 \over 2mr^2}. \]
The only thing to note here is that $\dot\theta^2 = j^2/m^2r^4.$
\vskip1em
\noindent 3. We solve \eqref{energy2} for $\dot r$ to obtain
\be \label{rdot} \dot r = \sqrt{{2\over m}(E-V_{\eff}(r))}.\ee
It should be noted that in our use of the symbol for the positive square root \emph{we are not asserting that $\dot r$ is positive}! It is entirely possible that the above root is negative! This, as we will discuss below (in \# 5) will not effect the form of our solution for $r$ in terms of $\theta.$
\vskip1em
\noindent 4. \emph{Using \eqref{dottheta} and \eqref{rdot} show that}
\[{d\theta \over dr} = \frac{\dot \theta}{\sqrt{{2\over m}(E-V_{\eff}(r))}} = \frac{j/mr^2}{\sqrt{{2\over m}(E-V_{\eff}(r))}}.\]
By the chain rule, we have that
\[\dot\theta = {d\theta \over dr}\dot r,\]
which when combined with \eqref{rdot} (and subsequently \eqref{dottheta}) gives:
\[{d\theta \over dr} = \frac{\dot \theta}{\sqrt{{2\over m}(E-V_{\eff}(r))}} = \frac{j/mr^2}{\sqrt{{2\over m}(E-V_{\eff}(r))}}.\]
Upon integration we arrive at
\be \theta = \theta_0 + \int \frac{(j/mr^2)dr}{\sqrt{{2\over m}(E-V_{\eff}(r))}} \label{thetaint1}.\ee
\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
\noindent 5. \emph{Sketch a graph of the effective potential } $V_{\eff}(r)$ \emph{in this case, and say what a particle moving in this potential would do, depending on its energy} $E$.
\vskip1em
\begin{center}
\includegraphics[height=5in, width=5in]{energy001}
\end{center}
\vspace{- .5 in}
Figure $I$ shows a sketch of $V_{\eff}$ in the case that $|j| > m$ (this is the case where $V_{\eff}'(r) < 0$ for $r < j^2/2mk$) and $II$ shows $V_{\eff}$ where $|j|0$ for $r < j^2/2mk.$) In both sketches, the zero is at $r = j^2/2mk$ and the $r$-axis is a horizontal asymptote as $r\to\infty.$
Let us briefly discuss the behavior of a particle with energy $E < 0$ with $|j|>m$. Such a particle is shown in $III$. As was discussed in the example in class, the particles radius $r$ would oscillate within the classically allowed region (the $r$ values lying between the intersection points of $E$ and $V_{\eff}(r)$). The particle would be moving fastest at the minimum value of $V_{\eff}$ and would change from moving away from the origin to moving towards it (or vice a versa) at the intersection points.
\vspace{.5 in}
\noindent 6. \emph{Carry out the integration in} \eqref{thetaint1}.
We must compute
\[\int \frac{(j/mr^2)dr}{\sqrt{{2\over m}(E + k/r - j^2/2mr^2)}}.\]
Too much has been made of this bugaboo! Let's put this ``beast'' to rest by an elementary trigonometric substitution:
\[{j\over m}\left({1\over r} - {mk\over j^2}\right) = \sqrt{{2E\over m} + {k^2\over j^2}}\cos u.\]
(The sign of the radical here is chosen to match the sign of the radical in \# 3) This substitution comes from completing the square under the radical---a simple and computationally economical process---and recalling the pythagorean identity for sine and cosine. All showboating aside, we see that
\[(j/mr^2)dr = \sin u \sqrt{{2E\over m} + {k^2\over j^2}}du,\]
and upon substitution, the integral becomes
\[\int u du = u\]
(the constant of integration already being accounted for in $\theta_0$, and any sign changes from radicals canceling). Reversing the trignometric substitution we see that $u$ and hence the sought after antiderivative is
\[\arccos \frac{{j\over mr} - {k\over j}}{\sqrt{{2E\over m} + {k^2\over j^2}}}.\]
Whence,
\be \label{theta3} \theta = \theta_0 + \arccos \frac{{j\over mr} - {k\over j}}{\sqrt{{2E\over m} + {k^2\over j^2}}}.\ee
\vskip1em
\noindent 7. \emph{Reduce the clutter in \eqref{theta3} by defining}
\[p = j^2/km, \qquad e= \sqrt{1 + {2Ej^2 \over mk^2}}.\]
Note that
\[\sqrt{{2E\over m} + {k^2\over j^2}} = {k\over j}e,\]
so that
\[\frac{{j\over mr} - {k\over j}}{\sqrt{{2E\over m} + {k^2\over j^2}}} = {j\over k}\frac{{j\over mr} - {k\over j}}{e} = {p/r-1\over e},\]
from whence it follows that
\[\theta = \theta_0 + \arccos\left({p/r-1\over e}\right).\]
Solving for $r$ yields:
\be\label{theta4} r = \frac{p}{1+e\cos(\theta-\theta_0)}.\ee
We should note that if the sign of the radical for $\dot r$
\vskip1em
\noindent 8. \emph{Show that equation \eqref{theta4} describes an ellipse, parabola, or hyperbola in polar coordinates, depending on the value of the parameter $e$, which we call the} {\bf \emph{eccentricity}}.
Begin by making a shift (a rotation) of $\theta_0$ in $\theta$. We will call the new coordinates that result from this shift $r'$ and $\theta'.$ We have that
\[r' = \frac{p}{1+e\cos\theta'}\]
by \eqref{theta4}, or equivalently
\[r' + er'\cos\theta' = p.\]
Making the standard change to cartesian coordinates, the above reads
\[\sqrt{x^2 + y^2} + ex = p.\]
Now a little algebra yields
\[x^2 + y^2 = p^2 - 2ex + e^2x^2,\]
or put a little differently,
\be\label{conic1} (1-e^2)x^2 + 2ex + y^2 = p^2; \ee
which we immediately recognize as the equation of a conic.
The particular conic that \eqref{conic1} describes will be determined by the value of $e$. If $e = 0$, for instance, then \eqref{conic1} reduces to
\[ x^2 + y^2 = p^2,\]
a circle centered at the origin with radius $p.$ If $e = 1$, then \eqref{conic1} reduces to
\[2(x- p^2/2) = y^2 ,\]
a parabola with vertex (in the original polar coordinates) $(p^2/2, \theta_0)$ opening \emph{towards} the origin.
Let's exhaust all of the cases. If $e\neq 0$ or $1$, then we may rewrite \eqref{conic1} as
\be \label{conic2} \frac{\left(x + {ep\over 1-e^2}\right)^2}{p^2 {1+e^2\over 1 - e^2}} + \frac{y^2}{p^2(1+e^2)} = 1.\ee
We see that in this case \eqref{conic2} represents either a hyperbola ( $e > 1$) with vertices (in rotated cartesian coordinates)
\[\left({-ep\over 1 - e^2}, \pm p(1-e^2)^{1/2}\right)\]
opening in the $y$ direction or an ellipse ($ 0 < e < 1$) with center (again in rotated cartesian coordinates)
\[\left({-ep\over 1 - e^2}, 0\right).\]
\vskip1em
\noindent 9. \emph{How are the three kinds of orbits related to the energy} $E$?
Recall that $e$ is given by
\[e = \sqrt{1 + {2Ej^2\over mk^2}},\]
so that
\[E = {mk^2\over 2j^2}(e^2 - 1).\]
Using this, we compile the following chart:
\begin{center}
\begin{tabular}[c]{|c|c|}
\hline
Orbit(type) & Energy\\
\hline
Circular & $E = -mk^2/2j^2$\\
\hline
Parabolic& $E = 0$\\
\hline
Hyperbolic& $E > 0$\\
\hline
Elliptic& $-mk^2/2j^2 < E < 0$\\
\hline
\end{tabular}
\end{center}
\end{document}