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\subsection{Potential in 3D}

Suppose $\bf{F} = \bf{F}(\bf{x})$.

\begin{claim*}
    There exists a conserved energy if and only if
    the force is of the form $\bf{F} = -\nabla V$,
    i.e. $F_i = -\pfrac{V(\bf{x})}{x^i}$. Then,
    \[ E = \half m \dot{\bf{x}} \cdot \dot{\bf{x}}
    + V(\bf{x}) \]
    is conserved.
\end{claim*}

\begin{proof}
    (To prove $E$ is conserved if $\bf{F} =
    -\nabla V$):
    \begin{align*}
        \dfrac{E}{t}
        &= m \ddot{\bf{x}} \cdot \dot{\bf{x}} +
        \pfrac{V}{x^i} \dfrac{x^i}{t} \\
        &= \dot{\bf{x}} \cdot (m \ddot{\bf{x}} +
        \nabla V) \\
        &= 0
    \end{align*}
    so $E$ is conserved. \myskip
    To prove $\bf{F} = -\nabla V$ if $E$ is
    conserved, we introduce the \emph{work done}
    $W$ on a particle, which moves from
    $\bf{x}(t_1)$ to $\bf{x}(t_2)$ along a
    trajectory $C$.
    \begin{align*}
        W
        := \int_C \bf{F} \cdot \dd \bf{x} \\
        &= \int_{t_1}^{t_2} \bf{F} \cdot
        \dfrac{\bf{x}}{t} \dd t \\
        &= m \int_{t_1}^{t_2} \ddot{\bf{x}} \cdot
        \dot{\bf{x}} \dd t \\
        &= m \int_{t_1}^{t_2} \half \dfrac{}{t}
        (\dot{\bf{x}} \cdot \dot{\bf{x}}) \dd t \\
        &= T(t_2) - T(t_1)
    \end{align*}
    where $T := \half m \dot{\bf{x}} \cdot
    \dot{\bf{x}}$ is the kinetic energy. If we
    want a conserved energy $E = T + V$
    \[ \implies W = \int_C \bf{F} \cdot \dd \bf{x}
    = V(\bf{x}(t_1)) - V(\bf{x}(t_2)) .\]
    depends only on the end-points, not on the
    trajectory $C$, i.e. for a closed path $C$,
    \[ \oint_C \bf{F} \cdot \dd \bf{x} = 0 \]
    A result from vector calculus states that this
    can only hold if $\bf{F} = -\nabla V$. 
\end{proof}

\myskip
Forces of this form are called
\emph{conservative}.
\begin{note*}
    $\bf{F} = -\nabla V \iff \nabla \times \bf{F}
    = 0$.
\end{note*}

\subsubsection*{Central Forces}

A special class of force has potential $V(r)$,
with $r = |\bf{x}|$
\[ \implies \bf{F}(r) = -\nabla V = -\dfrac{v}{r}
\hat{\bf{r}} .\]
i.e. points towards/away from the origin. These
are \emph{central forces}. Central forces have an
additional conserved quantity called \emph{angular
momentum}.
\[ \bf{L} = m \bf{x} \times \dot{\bf{x}} .\]
Note that it depends on choice of origin.
\[ \dfrac{\bf{L}}{t} = m \dot{\bf{x}} \times
\dot{\bf{x}} + m \bf{x} \times \ddot{\bf{x}} =
\bf{x} \times \bf{F} = 0 \]
because $\bf{F} = \bf{x}$. For non-central forces,
\[ \dfrac{\bf{L}}{t} = \bf{\tau} \]
where $\bf{\tau} = \bf{x} \times \bf{F}$ is the
\emph{torque}.

\subsection{Gravity}

Gravity is a conservative force. Fix a particle of
mass $M$ at the origin. A particle of mass $m$ in
its presence experiences a potential energy.
\[ V(r) = -\frac{GMm}{r} \]
where $G = 6.67 \times 10^{-11}
\mathsf{m^3kg^{-1}s^{-2}}$ is Newton's constant.
\[ \implies \bf{F} = -\nabla V = -\frac{GMm}{r^2}
\hat{\bf{r}} \]
It's common to define the \emph{gravitational
field} $\phi(r) = -\frac{GM}{r}$, a property of
the mass $M$ alone. \\
The gravitational field due to many fixed masses
$M_i$ at position $\bf{r}_i$ is
\[ \phi(\bf{r}) = -G \sum_i \left(
\frac{M_i}{|\bf{r} - \bf{r}_i} \right) .\]
The potential energy of mass $m$ is then $V =
m\phi$.

\begin{claim*}
    The external gravitational field of a
    spherically symmetric object of mass $M$ is
    \[ \phi(r) = -\frac{GM}{r} \]
    i.e. as if all the mass were at $r = 0$.
\end{claim*}

\begin{proof}
    Volume integral (see vector calculus course).
\end{proof}

\subsubsection*{Some 1d Problems}

Sit on a spherical planet of radius $R$. If one
only moves a distance $x \ll R$ above the surface,
then
\[ V(R + z) = -\frac{GMm}{R + z} = -\frac{GMm}{R}
\left(1 - \frac{z}{R} + \cdots \right) \approx
\text{constant} + \frac{GM}{R^2} m \cdot z \]
\[ V \approx \text{constant} + mgz \]
How fast must we jump to escape the gravitational
pull of the earth?
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/3a2fd1a8829211ec.png}
\end{center}
