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\subsection{Back to Central Forces}

Since $V = V(r) \implies \nabla V = \dfrac{V}{r}
\hat{\bf{r}}$. From Newton's second law and (4.2)
we have
\[ m (\ddot{r} - r \dot{\theta}) \hat{\bf{r}} +
m(r \ddot{\theta} + 2\dot{r} \dot{\theta})
\hat{\bf{\theta}} = -\dfrac{V}{r} \hat{\bf{r}} \]
equate $\hat{\bf{\theta}}$ components
\[ \implies r \ddot{\theta} + 2 \dot{r}
\dot{\theta} = 0 \]
\[ \iff \frac{1}{r} \dfrac{}{t} (r^2 \dot{\theta})
= 0 \]
\[ \iff l = r^2\theta \]
is conserved. This is actually the magnitude
(perhaps with a sign) of the angular momentum per
unit mass. To see this, look at
\[ \bf{L} = m \bf{x} \times \dot{\bf{x}}
\stackrel{(4.1)}{=} mr \hat{\bf{r}} \times
(\cancel{\dot{r} \hat{\bf{r}}} + r \dot{\theta}
\hat{\bf{\theta}}) = mr^2 \dot{\theta}
(\hat{\bf{r}} \times \hat{\bf{\theta}}) .\]
Equate $\hat{\bf{r}}$ components: $m(\ddot{r} - r
\dot{\theta}^2) = -\dfrac{V}{r}$. \\
Replace $\dot{\theta}$ with $l$:
\[ \implies m \ddot{r} = -\dfrac{V}{r} =
\frac{ml^2}{r^3} :+ -\dfrac{V_{\text{eff}}}{r} \]
where $V_{\text{eff}}$ is the ``\emph{effective
potential}'':
\[ V_{\text{eff}}(R) = V(r) + \frac{ml^2}{2r^2} \]
($\frac{ml^2}{2r^2}$ is called the \emph{angular
momentum barrier}.) \myskip
This problem has been reduced to the kind of
one-dimensional problem we looked at before. The
energy of the particle is
\begin{align*}
    E
    &= \half m \dot{\bf{x}} \cdot \dot{\bf{x}} +
    V(r) \\
    &= \half m \dot{r}^2 + \half mr^2
    \dot{\theta}^2 + V(r) \\
    &= \half m \dot{r}^2 + \frac{ml^2}{2r^2} +
    V(r) \\
    &= \half m \dot{r}^2 + V_{\text{eff}} (r)
\end{align*}

\begin{example*}[Inverse Square Law]
    \[ V = -\frac{k}{r} \implies V_{\text{eff}} =
    -\frac{k}{r} + \frac{ml^2}{2r^2} .\]
    ($k > 0$)
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/eee9e89686ab11ec.png}
    \end{center}
    The minimum is at $r_* = \frac{ml^2}{k}$
    \[ \implies E_{\text{min}} = V_{\text{eff}}
    (r_*) = -\frac{k^2}{2ml^2} .\]
    Kinds of motion:
    \begin{itemize}
        \item $E = E_{\text{min}} \implies$
            particle sits at $r = r_*$. But
            \[ \dot{\theta} = \frac{l}{r^2}
            \implies \text{particle moves in a
            circle at constant angular speed.} \]
        \item $E_{\text{min}} < E < 0 \implies$
            particle oscillates back and forth.
            Meanwhile, $\dot{\theta} =
            \frac{l}{r^2}$ also changes. This is a
            non-circular orbit. (Definitions: The
            smallest value of $r$ reached by the
            particle is the \emph{periapsis}
            (\emph{perehelion} for orbiting the
            sun). The furthest distance is the
            \emph{apoapsis} (\emph{aphelion} for
            the sun).)
        \item $E > 0 \implies$ this motion is not
            an orbit. The particle escapes to
            $\infty$.
    \end{itemize}
\end{example*}

\subsubsection*{Stability of Circular Orbits}

Circular orbits are equilibrium points, $r_*$, of
$V_{\text{eff}}$:
\[ V_{\text{eff}}' (r_*) = 0 .\]
They are stable if it's a minimum, i.e.
$V_{\text{eff}}'' (r_*) > 0$. \myskip
For $V(r) = -\frac{k}{r^n}$ for $n \ge 1$
\[ \implies V_{\text{eff}} = -\frac{k}{r^n} +
\frac{ml^2}{2r^2} .\]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c3753d7c86ac11ec.png}
\end{center}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d7c8c1d686ac11ec.png}
\end{center}
Can check that circular orbits are stable for $(n
+ 1) - 3 < 0 \implies n < 2$. (In a univserse with
$d$ space dimensions, potential energy of gravity
is $V \sim \frac{1}{r^{d - 2}} \implies$ circular
orbits not stable in $d > 3$).

\subsubsection*{The Orbit Equations}

If we want to compute $r(t)$, we use
\[ E = \half m \dot{r}^2 + V_{\text{eff}}(r) .\]
and integrate
\[ \implies t = \pm \sqrt{\frac{m}{2}} \int
\frac{\dd r}{\sqrt{E - V_{\text{eff}}(r)}} \]
Here, we'll instead try to understand the shape of
the trajectory by computing $r(\theta)$ (can then
use $l = r^2 \dot{\theta}$ to get $\theta(t)$ and
$r(t)$). \\
A trick: define $u = \frac{1}{r}$. If $r =
r(\theta)$ then
\[ \dfrac{r}{t} = \dfrac{r}{\theta} \dot{\theta} =
\dfrac{r}{\theta} \frac{l}{r^2} = -l
\dfrac{u}{\theta} \]
and
\[ \dfrac[2]{r}{t} = \dfrac{}{t} \left( -l
\dfrac{u}{\theta} \right) = -l
\dfrac[2]{u}{\theta} \dot{\theta} = -l^2
\dfrac[2]{u}{\theta} \frac{1}{r^2} = -l^2 u^2
\dfrac[2]{u}{\theta} \]
The equation of motion is
\[ m \ddot{r} - \frac{ml^2}{r^3} = -\frac{V}{r} =
F(r) \]
\[ - ml^2 u^2 \dfrac[2]{u}{\theta} - ml^2u^3 = F
\left( \frac{1}{u} \right) \]
\[ \implies \boxed{\dfrac[2]{u}{\theta} + u =
-\frac{1}{ml^2u^2} F \left( \frac{1}{u} \right)} \]
(this is known as the \emph{orbit equation}).
