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%TCIDATA{Created=Mon Sep 01 15:10:12 1997}
%TCIDATA{LastRevised=Tue Sep 09 13:30:26 1997}

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\SetTitle{Review of mechanics from PHY205}
\SetAuthor{T. Curtright}
\Setdate{29 August 1997}
\TitlePage{}

In this first lecture, we will review a few ideas from mechanics, as you
learned them in PHY205, using gravity as an example.

Kepler's first law, based on Brahe's observations, is that planetary orbits
are ellipses, with the sun at one focus of the ellipse.

Kepler's second law is that the rate of change of the area swept out by the
planet as it follows it's orbit is a constant: $dA/dt=constant.$ Now, in
PHY205, you learned that the second law is just the conservation of angular
momentum, $L$, with $dA/dt=L/2m$. This allows us to deduce that the force
between the sun and the planet is directed along the line that joins them.

There is also Kepler's third law, which relates the period of the orbit to
the length of the semi-major axis $a$ of the ellipse: $T^{2}=4\pi
^{2}\,a^{3}/GM.$ In PHY205 you discussed how this followed from an ``inverse
square law'' for the force, but only for the special case of circular orbits
for which $a=r$, the radius of the circle.

I think it is important to go beyond the simple case of circular orbits, and
to explain how the first and second laws may actually be used to deduce that
the force obeys an inverse square law. Very little mathematics is required
to do this. Just the basic equations for the ellipse, the concept of a
derivative, and some linear algebra. The task is further simplified if we
use some physics. We assume the force is conservative, with a corresponding
potential energy. We then work out the position dependence of this potential
energy for a planet moving along an elliptical orbit. It really isn't so
hard to do. All we need do is work out the kinetic energy at each point on
the orbit.

First, let's recall some facts about ellipses. These you just look up in a
math handbook if you don't know them from memory. Points on an ellipse
centered on the origin, with $x$ and $y$ axes oriented conveniently along
the axes of the ellipse, satisfy the equation 
\begin{equation}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\;.
\end{equation}
For such an ellipse, the parameters $a$ and $b$ are constants known as the
``semi-major'' and ``semi-minor'' axes. The ``foci'' of this ellipse are two
points located on the $x$ axis at distances $x=\pm c$, where $%
c^{2}=a^{2}-b^{2}$. Sometimes, to be very erudite, the ratio $\epsilon =c/a$
is called the ``eccentricity'' of the ellipse.\eject

As an example, let's plot the case $\frac{x^{2}}{2}+y^{2}=1$, or $y=\pm 
\sqrt{1-x^{2}/2}$. This is an ellipse, centered on the origin, with
semi-major axis $a=\sqrt{2}$, semi-minor axis $b=1$, and foci on the $x$
axis at $x=\pm 1$. I have also indicated on the figure below the situation
with general parameters.

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Returning now to the orbit problem, if we assume that the mass $M$ is at the
left-hand focus of the ellipse, and that this mass is so much heavier than $m
$ that it essentially does not move (a reasonable assumption for the sun and
any of the planets), then for the mass $m$ the (z-component of) angular
momentum about $M$ is 
\begin{equation}
L=m(x+c)v_{y}-myv_{x}\;,
\end{equation}
where $v_{x}=dx/dt$ and $v_{y}=dy/dt$ are the $x$ and $y$ components,
respectively, of the velocity for the mass $m$. (In class, I could not
resist calling these by their somewhat archaic names, ${\dot{x}}=dx/dt$ and $%
{\dot{y}}$$=dy/dt$. This notation was used originally by Newton.) The
equation for $L$ is one equation which we shall need. The other is obtained
by taking one time derivative of the equation for the ellipse. This gives 
\begin{equation}
2xv_{x}/a^{2}+2yv_{y}/b^{2}=0\;.
\end{equation}
Regarding these two equations as simultaneous linear equations for $v_{x}$
and $v_{y}$, we solve them to find 
\begin{eqnarray}
v_{x} &=&\frac{-yL/b^{2}}{m\left( 1+cx/a^{2}\right) }\;,\, \\
v_{y} &=&\frac{xL/a^{2}}{m\left( 1+cx/a^{2}\right) }\;.
\end{eqnarray}
It is easy to check that these results give reasonable values at special
points. For example, $v_{x}=0$ at $y=0$, and $v_{y}=0$ at $x=0$. Etc.

Now let us compute the kinetic energy $K$ of the mass $m$. 
\begin{equation}
K=\frac{1}{2}\;m\left( v_{x}^{2}+v_{y}^{2}\right) =\frac{L^{2}}{2m\left(
1+cx/a^{2}\right) ^{2}}\;\left( x^{2}/a^{4}+y^{2}/b^{4}\right) \;.
\end{equation}
But, the mass $m$ is moving along the ellipse, so we may simplify this by
replacing $y^{2}=b^{2}\left( 1-x^{2}/a^{2}\right) $. After some algebra the
result is not so complicated as a function of $x$. 
\begin{eqnarray}
K &=&\frac{L^{2}}{mb^{2}}\;\left( \frac{1-cx/a^{2}}{1+cx/a^{2}}\right)  \\
&=&\frac{L^{2}}{mb^{2}}\;\left( -1+\frac{2}{1+cx/a^{2}}\right) \;.
\end{eqnarray}
We have written the last equality (which is easily seen to be identical to
the previous expression after combining the two terms over the same
denominator) so that it exhibits the dependence on $x$ in a form we
recognize. Namely, the distance from $M\,$to $m$ is 
\begin{equation}
d=a+cx/a.
\end{equation}
Therefore, the kinetic energy is 
\begin{equation}
K=\frac{L^{2}}{mb^{2}}\;\left( -1+\frac{2a}{d}\right) \;.
\end{equation}
Now we can finally see what the potential $U$ is which leads to this
expression. This is because the total energy $E$ is the sum 
\begin{equation}
E=K+U\;,
\end{equation}
or alternatively, 
\begin{equation}
U=E-K\;.
\end{equation}
Since $E$ is a constant, $L$ is a constant, and $a$, $b$, and $c$ are
constants for a given elliptical orbit, the only position dependence is in
the $1/d$ term in $K.$ That is, 
\begin{equation}
U=E+\frac{L^{2}}{mb^{2}}-\frac{L^{2}}{mb^{2}}\;\frac{2a}{d}\text{\ }.
\end{equation}
Therefore, we have deduced from the first and second laws of Kepler that the
potential energy goes like the inverse of the distance, up to an overall
irrelevant additive constant. Accordingly, the force will go like the
inverse square of the distance.

If we choose for $U$ to vanish as $d\rightarrow \infty $, which is
customary, then we have a relation between $E$, $L$, and $b$. Namely 
\begin{equation}
E=-\frac{L^{2}}{mb^{2}}\;.
\end{equation}
Further identifying the parameters in the potential with those of Newton, $%
U=-GMm/d$, as you should recall from the PHY205 discussion of circular
orbits, we have the additional relation 
\begin{equation}
GMm=\frac{2aL^{2}}{mb^{2}}\;.
\end{equation}
Thus, another way to express the total energy of the mass $m$ is 
\begin{equation}
E=-\frac{GMm}{2a}\;.
\end{equation}
The total energy is negative here, as you may recall, because the mass $m$
is in a bounded orbit and cannot escape to infinity.

\end{document}