\nopagenumbers
\def\pt#1{\hbox{#1) }}
\input amssym.def
\input amssym.tex
\def\R{{\bf R}}
\def\u{{\bf u}}
\def\v{{\bf v}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\a{{\bf a}}
\font\srm=cmr8
\magnification=\magstep1

\headline={ESP Math 408D - AP\hfil Fall 1996}
\footline={\ifnum\pageno>1 \hfil Worksheet 33\hskip.3in Page \folio \fi}
\footnote{}{Mike Oehrtman}

\centerline
{\bf WORKSHEET 33} 
\bigskip
\bigskip 

\itemitem{1.\hskip12pt a)}  State Newton's second law of motion (the
one relating force and acceleration) and Newton's law of gravity.

\medskip
\itemitem{b)}  State Kepler's three laws of planetary motion.

\bigskip 
\medskip 
\item{2.}  Let $\R$ be the position vector of a planet moving about a
stationary sun at the origin.  Show that $\R$ stays within a fixed
plane.
$$
\hbox{Hint: consider } {d\over dt}(\R\times\R').\hskip3.5in
$$

\item{3.}  Because of Problem 2, we can now work in a plane.  Let's
establish a coordinate system so that this is the $x,y$-plane (or
$r,\theta$-plane), and the planet is somewhere on the $x$-axis (or
$\theta=0$ ray) at time $t=0$.  Define two pairs of unit vectors
$$
\u_r=(\cos\theta,\sin\theta)\quad\u_\theta=(-\sin\theta,\cos\theta)\qquad
\qquad\i=(1,0)\quad\j=(0,1)
$$
\itemitem{a)}  Write the position vector $\R$ in terms of $\i$ and $\j$,
then in terms of $\u_r$ and $\u_\theta$.

\medskip
\itemitem{b)}  Write the velocity vector $\v$ in terms of $\i$ and $\j$,
then in terms of $\u_r$ and $\u_\theta$.

\medskip
\itemitem{c)}  Write the acceleration vector $\a$ in terms of $\i$ and
$\j$ then, in terms of $\u_r$ and $\u_\theta$.

\bigskip 
\medskip
\item{4.}  Give the formula for Newton's law of gravity in terms of
the vecors $\i$ and $\j$, then give the formula in terms of $\u_r$ and
$\u_\theta$.

\bigskip 
\medskip
\item{5.}  Using whichever is more convenient, the $\i,\j$-system or
the $\u_r,\u_\theta$-system, write down the differential equations
governing planetary motion by combining Newton's laws and the formula
for acceleration from Problem 3c).

\bigskip 
\medskip
\item{6.}  Show that the area swept out by the vector $\R$ from time
$0$ to time $t$ is
$$
A(t)=\int_0^t{1\over2}r^2\theta'\,dt.
$$
Use this and one of your differential equations to prove Kepler's
second law: 

\item{} {\it The radius vector sweeps out equal area in equal time}. 

\eject
\item{7.}  Ok.  A quick reality check here.  The two equations you
should have for Problem 5 are $r''-r(\theta')^2=-GM/r^2$ and
$2r'\theta'+r\theta''=0$.  You did choose polar coordinates didn't
you?  Our goal in this problem is going to be to show that these
equations imply that the planet must follow a path of the form
$$
r={c\over1+e\cos\theta}
\eqno(*)$$
proving Kepler's first law:  {\it The orbit of a planet is an
ellipse}.  

\item{}  To be honest we're going to pull some TRICKS out of nowhere.
We'll use them simply because they work.  How they were thought of is
not so important right now (wait until next semester).  Here, I'll
tell you what tricks to use.

\medskip
\itemitem{a)}  From the second equation (and what you did in Problem
6), show that $r^2\theta'=r_0v_0$.  Here $r_0$ and $v_0$ are the
radius and speed, respectivly, at time $t=0$

\medskip
\itemitem{b)}  Now we want to find a solution to the differential
equation $r''-r(\theta')^2=-GM/r^2$.  First get rid of all
references to $\theta$ by substituting in the reult from part a).

\medskip
\itemitem{c)}  Now you should have an equation involving only constants,
$r$ and derivatives of $r$.  Let's simplify this by making it an
equation involving only a first derivative.  We'll do that with a
trick.  Set $p=r'$.  Show that
$$
r''=p{dp\over dr}.
$$
Hint:  Think of $p$ as $p(r(t))$, then use the chain rule.

\medskip
\itemitem{d)} Rewrite the differential equation using only constants,
$r$, $p$, and the derivative $dp/dr$.

\medskip
\itemitem{e)}  Move all the refernces to $p$ to one side of the
equation and all references to $r$ to the other side.  When this is
possible we say the equation is {\it separable}.  Now integrate both
sides and conclude that
$$
(r')^2=-{r_0^2v_0^2\over r^2}+{2GM\over r}+C
$$
for some constant $C$.  Find $C$ by evaluating everything at time
$t=0$.

\medskip
\itemitem{f)}  Divide the left and right sides of this equation by the
left and right sides of the equation $r^4(\theta')^2=r_0^2v_0^2$,
respectively.  Remember this guy from part a)?

\medskip
\itemitem{g)}  Now make the substitution $u=1/r$.  (So what is
$du/d\theta$?)  Also, to simplify the constants, set $h={GM\over
r_0^2v_0^2}$.  After some algebra, show that
$$
{du\over d\theta}=-\sqrt{(u_o-h)^2-(u-h)^2}.
$$
(Why the negative root instead of the positive one?)

\itemitem{h)}  Once again, we have a separable equation.  Move
everything involving $u$ to one side and everything involving $\theta$
to the other.  Then integrate both sides.

\itemitem{}  Hint:  use the trig substitution $u-h=(u_0-h)\cos x$.

\medskip
\itemitem{i)}  Part h) gave you $u$ as a function of $\theta$.  Now
solve for $r=1/u$.  Finally, substitute back in the value of $h$.
This should give you an equation of the form $(*)$.  What are $c$ and
$e$?  

\eject
\item{8.}  Here, we seek to discover the period of the planet's
orbit.  This is the amount of time $T$ required to complete one
revolution of the orbit. 

\medskip
\itemitem{a)}  Show that an ellipse of the form $r=c/(1+e\cos\theta)$
has semimajor axis $a=c/(1-e^2)$, and semiminor axis $b=a\sqrt{1-e^2}$.
Now write the area of an ellipse in terms of $a$ and $e$.

\medskip
\itemitem{b)}  Recall that the rate of change of the area being swept
out by the position vector is $A'=1/2r_0v_0$.  Integrate this (from
time $t=0$ to time $t=$ what?) to obtain another expression for the
area of the ellipse.

\medskip
\itemitem{c)}  Set the two area equations you have equal to each other
and solve for the period $T$.  Square both sides.  Replace $a(1-e^2)$
with the value of $c$ which you discovered in Problem 7, and deduce
Kepler's third law: {\it The square of the period is proportional to the
cube of the mean distance to the sun.  Specifically,}
$$
T^2={4\pi^2a^3\over GM}{\atop.}
$$

\medskip
\item{9.}  Only Kepler's 3rd law actually assumed that the orbit was
an ellipse.  Actually, the equation $(*)$ from problem 7 only gives an
ellipse when $0\leq e<1$.  This equation can also describe a parabola
when $e=1$ or a hyperbola when $e>1$.  For what values of $v_0$ (the
initial velocity) would the orbit be

\medskip
\itemitem{a)} an ellipse,

\medskip
\itemitem{b)} a parabola, and

\medskip
\itemitem{c)} a hyperbola?



\bye