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    $\begingroup$ This is a good question. I don't know the answer for sure, but I'd guess that, at high enough rotation rates, the most stable configuration is a dumbbell shape, which breaks the rotational symmetry. That is, the shape will be a nontrivial function of the longitude angle $\phi$, as well as $\theta$. One group of people who probably know a lot about this are nuclear physicists: I think that rapidly-rotating nuclei suffer all kinds of strange deformations. Moreover, at least in some circumstances it's considered appropriate to model nuclei with a "liquid drop model." $\endgroup$ Commented Jun 1, 2011 at 17:28
  • $\begingroup$ Discussed in a fictional context in Hal Clement's old war horse Mission of Gravity. $\endgroup$ Commented Jun 1, 2011 at 17:38
  • $\begingroup$ @Ted, such deformations are dicussed for some heavy nuclei, and vibrations of such ellipsoids as a model for fission. But this is always for surface tension models (liquid). For soap bubbles such transformations are easily shown experimentally. For a gravity "drop" things will be different, I think. $\endgroup$ Commented Jun 1, 2011 at 17:54
  • $\begingroup$ You're probably right that the gravitational case is quite different, so the nuclear case may not be very relevant to teh question at hand (although it's interesting in its own right, of course). $\endgroup$ Commented Jun 1, 2011 at 18:07
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    $\begingroup$ It looks to me like the key buzzwords to look for here are the Maclaurin and Jacobi sequences. These seem to be equilibrium shapes for rotating self-gravitating bodies. At slow rotation rates, the stable equilibrium seems to be an oblate spheroid (as you'd expect), but at higher rates it switches to a prolate spheroid, breaking the azimuthal symmetry. At least, that's the impression I get from a quick scan of various web pages, but I haven't tried to understand the details. There seem to be other sequences too (e.g., ptp.ipap.jp/link?PTP/67/844). $\endgroup$ Commented Jun 1, 2011 at 22:40