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\begin {document}
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\begin{center}
{\bf The Orbiting Dumbbell}
\end  {center}
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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, $M >> m_1, m_2$.  
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 
$$
I = m_1 d_1{}^2 + m_2 d_2{}^2 =  
  \frac{m_1 m_2{}^2}{(m_1 + m_2)^2} d_1{}^2 +
  \frac{m_2 m_1{}^2}{(m_1 + m_2)^2} d_2{}^2 =
  \frac{m_1 m_2}{m_1 + m_2} d^2
$$
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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 
$$
{\bf F}_1 = -\frac{G M m}{r_1{}^3} \lbb x_1 , y_1 \rbb
$$
$$
{\bf F}_2 = -\frac{G M m}{r_2{}^3} \lbb x_2 , y_2 \rbb
$$
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_2 \rb +
   G M \frac{m_21}{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 y_c - \cos\theta x_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*}
\end {document}
