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    \begin{document}

       \begin{center}
      {\large Classical Mechanics Homework\\}
      {\small January 24, 2008 \\}
      {\small John Baez}
      {\small homework by Brian Rolle}
      \end{center}

\section*{The Kepler Problem}


\textbf{1. }Show that the energy of a particle is given by
$\displaystyle E=\frac{1}{2}m(r^2\dot{\theta}^2+\dot{r}^2)+V(r)$ and
the angular momentum $J$ has $z$ coordinate
$j=mr^2\dot{\theta}$ and vanishing $x$ and $y$ coordinates.\\

Solution: We know $E=T+V$ where $T$ is the kinetic energy and $V$ is
the potential energy.  Kinetic energy is $\displaystyle
T=\frac{1}{2}m\dot{q}(t)\cdot \dot{q}(t)$ where
$q(t)=(r(t)\cos\theta(t),r(t)\sin\theta(t),0)$.  Since
\\$\dot{q}(t)=(\dot{r}\cos\theta-r\dot{\theta}\sin\theta,\dot{r}\sin\theta+r\dot{\theta}\cos\theta,0)$,
$\dot{q}\cdot\dot{q}=\dot{r}^2+r^2\dot{\theta}^2$ and $\displaystyle
T=\frac{1}{2}m(r^2\dot{\theta}^2+\dot{r}^2)$ .  By assumption,
$V$ depends on $r$, since the force depends only on $r$.\\
The angular momentum is given $j=q(t)\times m\dot{q}(t)$. Since $q$
and $\dot{q}$ lie in the $xy$ plane, $q(t)\times
\dot{q}(t)=(0,0,r^2\dot{\theta})$ and so the $z$ coordinate of the
angular momentum is $mr^2\dot{\theta}$.

\medskip

\textbf{2. } Use the angular momentum to solve for $\dot{\theta}$
and write $E$ as $\displaystyle
E=\frac{1}{2}m\dot{r}^2+V_{\operatorname{eff}}(r)$ where
$\displaystyle
V_{\operatorname{eff}}(r)=V(r)+\frac{j^2}{2mr^2}$.\\

Solution: Since $j=mr^2\dot{\theta}$, we have $\displaystyle
\dot{\theta}=\frac{j}{mr^2}$. The equation for $T$ gives
$\displaystyle T=
\frac{1}{2}m\left(r^2\left(\frac{j}{mr^2}\right)^2+\dot{r}^2\right)$
and $\displaystyle
E=\frac{1}{2}m\dot{r}^2+\frac{j^2}{2mr^2}+V(r)=\frac{1}{2}m\dot{r}^2+V_{\operatorname{eff}}(r)$,
where $\displaystyle
V_{\operatorname{eff}}(r)=\frac{j^2}{2mr^2}+V(r)$.

\medskip

\textbf{3. } Show that $\displaystyle
\dot{r}=\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}$.\\

Solution: Solving $\displaystyle
E=\frac{1}{2}m\dot{r}^2+V_{\operatorname{eff}}(r)$ for $\dot{r}^2$
gives $\displaystyle
\dot{r}^2=\frac{2}{m}(E-V_{\operatorname{eff}}(r))$.  Taking the
square root gives $\displaystyle
\dot{r}=\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}$.

\medskip{4. } Show that $\displaystyle
\frac{d\theta}{dr}=\frac{j/mr^2}{\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}}$.\\

Solution: By the chain rule $\displaystyle
\frac{d\theta}{dr}=\frac{d\theta}{dt}\frac{dt}{dr}$.  Implicit
differentiation (like in 9A) gives $\displaystyle
\frac{dt}{dr}=\frac{1}{\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}}$,
provided $\dot{r}>0$. Since $\displaystyle
\frac{d\theta}{dt}=\frac{j}{mr^2}$, we have $\displaystyle
\frac{d\theta}{dr}=\frac{j/mr^2}{\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}}$.

Integrating this gives $\displaystyle \theta=\theta_0+\int
\frac{(j/mr^2)dr}{\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}}$.

\newpage

\textbf{5. } Sketch the graph of $V_{\operatorname{eff}}(r)$ when
$\displaystyle V=-\frac{k}{r}$ and describe what a particle in this
potential would do, depending on it's energy $E$.\\

Solution: We have $\displaystyle
V_{\operatorname{eff}}(r)=\frac{j^2}{2mr^2}-\frac{k}{r}=k\left(\frac{\frac{j^2}{2mk}-r}{r^2}\right)$.
This is minimized when $\displaystyle r=\frac{j^2}{mk}$, and then
$\displaystyle V_{\operatorname{eff}}=-\frac{k^2m}{2j^2}$.  Since
$\displaystyle \frac{1}{2}m\dot{r}^2$ must be nonnegative, we know
that a particles total energy can never be less than $\displaystyle
-\frac{k^2m}{2j^2}$. If a particles total energy is positive, then
$r$ can go to infinity.  If $E<0$, then the particle will be stay in
between to values of $r$, determined by when the energy equals
$V_{\operatorname{eff}}$.

\vspace{3 in}

\textbf{6. }  Show $\displaystyle \theta=\theta_0+\arccos\frac{
\frac{j}{mr}-\frac{k}{j}}{ \sqrt{\frac{2E}{m}+\frac{k^2}{j^2}}}$.\\

Solution: $\displaystyle \int
\frac{(j/mr^2)dr}{\sqrt{\frac{2}{m}(E-V_{\operatorname{eff}}(r))}}=\int
\frac{(j/mr^2)dr}{\sqrt{\frac{2}{m}(E+\frac{k}{r}-\frac{j^2}{2mr^2})}}=-\int
\frac{dw}{\sqrt{\frac{2Em}{j^2}+\frac{2km}{j^2}w-w^2}}$, where
$\displaystyle w=\frac{1}{r}$.  Using the fact $\displaystyle
\int\frac{dx}{\sqrt{ax^2+bx+c}}=-\frac{1}{\sqrt{-a}}\arccos\left(\frac{-2ax-b}{\sqrt{b^2-4ac}}\right)$
when $a<0$, we have $\displaystyle -\int
\frac{dw}{\sqrt{\frac{2Em}{j^2}+\frac{2km}{j^2}w-\frac{1}{j}w^2}}=\arccos
\left(\frac{2w-\frac{2km}{j^2}}{\sqrt{\frac{4k^2m^2}{j^4}+4\frac{2Em}{j^2}}}\right)=\arccos
\left(\frac{\frac{j}{mr}-\frac{k}{j}}{\sqrt{\frac{k^2}{j^2}+\frac{2E}{m}}}\right)$.
So $\displaystyle \theta=\theta_0+\arccos\frac{
\frac{j}{mr}-\frac{k}{j}}{ \sqrt{\frac{2E}{m}+\frac{k^2}{j^2}}}$.

\newpage

\textbf{7. } Letting $\displaystyle p=\frac{j^2}{km}$ and
$\displaystyle e=\sqrt{1+\frac{2Ej^2}{mk^2}}$, show $\displaystyle
\theta=\theta_0+\arccos\left(\frac{p/r-1}{e}\right)$.\\

Solution: $\displaystyle \theta=\theta_0+\arccos\frac{
\frac{j}{mr}-\frac{k}{j}}{
\sqrt{\frac{2E}{m}+\frac{k^2}{j^2}}}=\theta_0+\arccos\frac{
\frac{j}{mr}-\frac{k}{j}}{\frac{j}{k}
\sqrt{\frac{2Ej^2}{mk^2}+1}}=\theta_0+\arccos\frac{
\frac{j}{mr}\frac{j}{k}-1}{\sqrt{\frac{2Ej^2}{mk^2}+1}}=\\
\theta_0+\arccos\left(\frac{p/r-1}{e}\right)$.  Solving for $r$
gives $\displaystyle r=\frac{p}{1+e\cos(\theta-\theta_0)}$.

\textbf{8. }  The equation $\displaystyle
r=\frac{p}{1+e\cos(\theta-\theta_0)}$ describes an ellipse, parabola
or hyperbola based on the value of $e$.\\

Solution: By a rotation, we can assume $\theta_0=0$.  So we have
$p=r+er\cos\theta$ or $p=\sqrt{x^2+y^2}+ex$.  So
$x^2+y^2=p^2-2epx+x^2$, or $(1-e^2)x^2+2epx+y^2=p^2$.

If $e=0$, we have $p^2=x^2+y^2$, the equation of a circle.

\medskip

If $ 0<e<1$, let $\displaystyle k=\frac{ep}{1-e^2}$.  Then we have
$\displaystyle
x^2+2kx+\frac{y^2}{1-e^2}=\frac{p^2}{1-e^2}=(x-k)^2+\frac{y^2}{1-e^2}=\frac{p^2}{1-e^2}+k^2$,
and ellipse.

\medskip

If $e=1$, we have $2px=-y^2+p^2$, a parabola.

\medskip

If $e>1$, let $\displaystyle k=\frac{ep}{1-e^2}$.  Then we have
$\displaystyle
x^2+2kx+\frac{y^2}{1-e^2}=\frac{p^2}{1-e^2}=(x-k)^2-\frac{y^2}{e^2-1}=\frac{p^2}{1-e^2}+k^2$,
a hyperbola.

\medskip

\textbf{9. }  How are the 3 kinds of orbits related to energy?\\

Solution:  The hyperbola corresponds to $e>1$, which requires the
energy to be positive.  The parabola correspond to $e=1$, which
requires energy to be 0.  The ellipse corresponds to $0<e<1$ which
corresponds which requires energy to be negative.





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