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{\bf PHYS 410/555 Computational Physics: The Orbiting Dumbbell}\\
{(Following Giordano, {\em Computational Physics}, Section 4.6)}
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{\em Background:} With the exception of Hyperion, which is one 
of Saturn's satellites, all of the moons in the solar system
are ``spin-locked''; a moon which is spin-locked has a rotational
frequency, $\omega$ about its own spin axes which is the same 
as its orbital frequency, $\Omega$.  The supposed mechanism
by which the spin-locking comes about is somewhat involved; however
the point is that Hyperion is somehow exceptional---study of its 
$\omega(t)$ suggests that it is tumbling {\em chaotically} in
its orbit about Saturn, which is presumably due to both to its 
peculiar shape (like that of an egg), and the fact that it 
is in an elliptical orbit about Saturn.

To investigate the effects of a non-spherical distribution 
of mass on a satellite's spin as it orbits its parental 
body, we consider the model of an ``orbiting dumbbell''.

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Consider two test masses, $m_1$, $m_2$ connected by a massless rigid rod
of length $d$, in orbit about a mass, $M >> m_1, m_2$ as shown in the 
figure above.
Let $(x_i,y_i), i = 1 , 2$ be the coordinates of the two test masses,
let $(x_c,y_c)$ be the coordinates of the dumbbell's center of mass,
and let $\theta$ be the angle the rod makes with the $x$-axis.
Defining
$$
\mu \equiv \frac{m_2}{m_1 + m_2}
$$
the distances of the masses from the center of mass are
$$
d_1 = \mu \, d \quad\quad\quad d_2 = (1 - \mu) d
$$
then
$$
x_i  =  x_c \pm d_i \cos\theta \quad\quad\quad 
y_i  =  y_c \pm d_i \sin\theta
$$
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The moment of inertia of the dumbbell about $(x_c,y_c)$ is 
$$
I = m_1 d_1{}^2 + m_2 d_2{}^2 =  
  \frac{m_1 m_2{}^2}{(m_1 + m_2)^2} d{}^2 +
  \frac{m_2 m_1{}^2}{(m_1 + m_2)^2} d{}^2 =
  \frac{m_1 m_2}{m_1 + m_2} d^2
$$
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The equations of motion for the body are 
$$
\lb m_1 + m_2 \rb {\bf a_c} = \lb m_1 + m_2 \rb {\ddot {\bf r}}_c = 
\sum {\bf F} = {\bf F}_1 + {\bf F}_2
$$
$$
I \alpha = I {\ddot \theta} = \sum \tau = 
  {\bf d}_1 \times {\bf F}_1 + {\bf d}_2 \times {\bf F}_2
$$
where 
$$
{\bf F}_1 = -\frac{G M m_1}{r_1{}^3} \lbb x_1 , y_1 \rbb
$$
$$
{\bf F}_2 = -\frac{G M m_2}{r_2{}^3} \lbb x_2 , y_2 \rbb
$$
are the gravitational forces acting on $m_1$ and $m_2$ respectively.
The translational equations yield:
$$
\lb m_1 + m_2 \rb \, {\ddot x}_c = 
   -G M \lb \frac{m_1}{r_1{}^3}x_1 + \frac{m_2}{r_2{}^3} y_2 \rb
$$
$$
\lb m_1 + m_2 \rb \, {\ddot y}_c = 
   -G M \lb \frac{m_1}{r_1{}^3}y_1 + \frac{m_2}{r_2{}^3} y_2 \rb
$$
or 
$$
{\ddot x}_c = -G M \lb \frac{1 - \mu}{r_1{}^3} x_1 + \frac{\mu}{r_2{}^3} x_2 \rb
$$
$$
{\ddot y}_c = -G M \lb \frac{1 - \mu}{r_1{}^3} y_1 + \frac{\mu}{r_2{}^3} y_2 \rb
$$
The rotational equation gives:
\begin{eqnarray*}
I {\ddot \theta} = {\bf d}_1 \times {\bf F}_1 + {\bf d}_2 \times {\bf F}_2
&=& -G M \frac{m_1}{r_1{}^3} d_1 \lb \cos\theta y_1 - \sin\theta x_1 \rb +
   G M \frac{m_2}{r_2{}^3} d_2  \lb \cos\theta y_2 - \sin\theta x_2 \rb \\
&=& -G M \frac{m_1}{r_1{}^3} d_1 \lb \cos\theta y_c - \sin\theta x_c \rb +
   G M \frac{m_2}{r_2{}^3} d_2 \lb \cos\theta y_c - \sin\theta x_c \rb \\
&=& G M \lb \frac{m_2}{r_2{}^3}d_2 - \frac{m_1}{r_1{}^3}d_1 \rb
    \lb \cos\theta y_c - \sin\theta x_c \rb \\
&=& G M \frac{m_1 m_2}{m_1 + m_2} d \lb \frac{1}{r_1{}^3} - \frac{1}{r_2{}^3}\rb
    \lb \sin\theta x_c - \cos\theta y_c \rb
\end{eqnarray*}
so
$$
{\ddot \theta} = \frac{GM}{d} \lb \frac{1}{r_1{}^3} - \frac{1}{r_2{}^3}\rb
    \lb \sin\theta x_c - \cos\theta y_c \rb
$$
Summarizing, we have:
\begin{eqnarray*}
{\ddot x}_c &=& 
 -G M \lb \frac{1 - \mu}{r_1{}^3} x_1 + \frac{\mu}{r_2{}^3} x_2 \rb \\
{\ddot y}_c &=& 
 -G M \lb \frac{1 - \mu}{r_1{}^3} y_1 + \frac{\mu}{r_2{}^3} y_2 \rb \\
{\ddot \theta} &=& \frac{GM}{d} \lb \frac{1}{r_1{}^3} - \frac{1}{r_2{}^3}\rb
    \lb \sin\theta x_c - \cos\theta y_c \rb 
\end{eqnarray*}
where
\begin{eqnarray*}
\mu &\equiv& \frac{m_2}{m_1 + m_2} \\
d_1 &=& \mu \, d \\
d_2 &=& (1 - \mu) d \\
x_i &=& x_c \pm d_i \cos\theta \\
y_i &=& y_c \pm d_i \sin\theta \\
r_i{}^3 &=& \lb x_i{}^2 + y_i{}^2 \rb^{3/2}
\end{eqnarray*}
The total (conserved) energy of the system is
$$
E_{\rm tot} = T_{\rm trans} + T_{\rm rot} + V_{\rm grav}
$$
where
\begin{eqnarray*}
T_{\rm trans} &\equiv& 
   \frac{1}{2}\lb m_1 + m_2 \rb \lb {\dot x_c}^2 + {\dot y_c}^2 \rb\\
T_{\rm rot} &\equiv&
   \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2} \, d^2 \, {\dot \theta}^2 \\
V_{\rm grav} &\equiv& -G M \lb \frac{m_1}{r_1} + \frac{m_2}{r_2} \rb
\end{eqnarray*}

