Consider the Earth (mass $M$, radius $R$, rotating about its own axis at $\Omega$) and the moon (mass $m$, radius $r$, with axial rotation equal to $\omega_m$), whose centre of masses are $d$ apart. They rotate around their barycentre at $\omega_e$ and $\omega_m$ radians/second respectively (I believe $\omega_m=\omega_e$ to conserve momentum).
We now tether a point on the Earth's surface to a point on the Moon's surface using a light, inextensible (and very strong) piece of string. Assume the Earth and the Moon are rigid and indestructible.
What happens?
It would be nice to find a solution of the general type, but for a first analysis it may be easier to assume that $M \gg m$, so that the Earth's rotation around the barycentre is negligible and the barycentre $\approx$ the Earth's centre of mass. Another (perhaps reasonable) assumption is that all rotation takes place in one plane and that the moon's orbit is circular, and that both bodies have uniform density.
Here are some of my thoughts based solely on intuition, feel free to ignore in answering the question:
- The moon is immediately pulled in (as $\Omega_e>\omega_m$) and sticks to the Earth like a barnacle. I think this may occur if $\Omega_e \gg \omega_m$ and $M\gg m$.
- If $\omega_m$ is large enough, the moon as before will initially be pulled in, then be thrown outwards again as $\omega_m(t)$ increases.
- If $m\approx M$, the Earth and Moon will form a dumbbell-type system (i.e. the string wouldn't have had much of an effect, as that is the case anyway in 'binary' systems where $M=m$).
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