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% 08/02/2022 10AM

\subsubsection*{The Kepler Problem}

\[ V(r) = -\frac{km}{r} \implies F(r) =
-\frac{km}{r^2} \]
$k = GM$ for gravity, $k = -\frac{qQ}{4\pi\eps_0}$
for Coulomb force. \\
The orbit equation is
\[ \dfrac[2]{u}{\theta} + u = \frac{k}{l^2} \]
This is a harmonic oscillator with a displaced
centre:
\[ u = A\cos(\theta - \theta_0) + \frac{k}{l^2} \]
Pick $A > 0$. \\
We choose our polar coordinates such that the
closest point (periapsis) occurs at $\theta = 0
\implies \theta_0 = 0$.
\[ \implies r = \frac{r_0}{e\cos\theta + 1} \]
where
\[ r_0 := \frac{l^2}{k} \qquad \text{and} \qquad e
:+ \frac{A l^2}{k} \]
This is the equation of a \emph{conic section}.
$e$ is the \emph{eccentricity}. First look at
attractive forces ($k > 0$): \myskip
\ul{$e < 1$ ellipses}:
\[ \frac{r}{r_0} \in \left[ \frac{1}{1 + e},
\frac{1}{1 - e} \right] \]
\[ \implies r \text{ is positive and bounded.} \]
\[ r = r_0 - re \cos\theta \]
\[ \implies x^2 + y^2 = (r_0 - ex)^2 \]
\[ \implies \cdots \]
\[ \implies \frac{(x - x_0)^2}{a^2} +
\frac{y^2}{b^2} = 1 \]
which is the formula of an ellipse where
\[ x_0 := -\frac{er_0}{1 - e^2} \]
\[ a^2 := \frac{r_0^2}{(1 - e^2)^2} \]
\[ b^2 := \frac{r_0^2}{1 - e^2} < a^2 \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/88ed0a1e8b8311ec.png}
\end{center}
$O$ is the origin and sits at a focus of the
ellipse, a distance $|x_0| = ea$ from the centre.
When $e = 0$, $a = b$ and ellipse $\to$ circle.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c55275208b8311ec.png}
\end{center}
\myskip
\ul{$e > 1$ hyperbolae}: $r \to \infty$ as
$\cos\theta \to -\frac{1}{e}$. Repeat steps above:
\[ \implies \frac{1}{a^2} \left( x - \frac{r_0
e}{e^2 - 1} \right)^2 - \frac{y^2}{b^2} = 1 \]
with
\[ a^2 = \frac{r_0^2}{(e^2 - 1)^2}, \qquad b^2 =
\frac{r_0^2}{e^2 - 1} .\]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/6e25a8708b8411ec.png}
\end{center}

\subsubsection*{The Energy of the Orbit}

\begin{align*}
    E
    &= \half m \dot{r}^2 + \frac{ml^2}{2r^2} -
    \frac{km}{r} \\
    &= \half m \left( \dfrac{r}{\theta} \right)^2
    \frac{l^2}{r^4} + \frac{ml^2}{2r^2} -
    \frac{km}{r} \\
    &= \cdots \\
    &= \frac{1}{2l^2} mk^2 (e^2 - 1)
\end{align*}
\[ \left( \dot{r} = -\frac{l}{r^2}
\dfrac{r}{\theta} \right) \]
\begin{itemize}
    \item $e < 1 \implies E < 0$, bound orbits.
    \item $e > 1 \implies E > 0$, unbound orbit.
\end{itemize}
A repulsive force:
\[ V = -\frac{km}{r} \]
$k > 0$ is attractive, $k < 0$ is repulsive. We
get
\[ r = \frac{|r_0|}{e \cos\theta - 1} \]
with
\[ |r_0| = \frac{l^2}{|k|} \]
\[ e = \frac{Al^2}{|k|} \]
but $r > 0 \implies e > 1$ for solutions to make
sense. Solutions now look like
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e4773b9c8b8411ec.png}
\end{center}
$r \to \infty$ at
$\cos\theta = +\frac{1}{e}$ i.e. $\theta <
\frac{\pi}{2}$.

\subsubsection*{Kepler's Laws of Planetary Motion}

In 1605, Kepler stated three laws obeyed by all
planets (in the solar system).
\begin{enumerate}[K1]
    \item Each planet moves on an ellipse with the
        sun at one focus.
    \item The line between the planet and the sun
        sweeps out equal areas in equal times.
    \item The period of the orbit is proportional
        to $(\text{radius})^{3/2}$.
\end{enumerate}
Let's see how they follow from Newton's laws:
\begin{enumerate}[(K1)]
    \item[K2] follows from conservation of angular
        momentum.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/5fb546788b8511ec.png}
        \end{center}
        \[ \delta A = \half r^2 \delta \theta \]
        \[ \implies \dfrac{A}{t} = \half r^2
        \dfrac{\theta}{t} = \half l \]
        is constant. This then is true for any
        central force.
    \item[K3] holds on dimensional grounds if we
        assume the inverse square law:
        \[ F = -\frac{GMm}{r^2} \implies [GM] =
        L^3 T^{-2} \]
        so if $R$ is some definition of radius,
        the period $T$ must obey $T^2 \propto
        \frac{R^3}{GM}$.
    \item[K1] requires the full solution to
        inverse square law.
\end{enumerate}