2011-06-01 16:19:00 8 Comments

It is common knowledge among the educated that the Earth is not exactly spherical, and some of this comes from tidal forces and inhomogeneities but some of it comes from the rotation of the planet itself. The deformation from the rotational effect makes it more oblate spheroid-like, or as I would prefer, "like a pancake". Here is one site illustrating the behavior, and image:

Literature exists detailing the mathematical expectations for a rotating planet using just hydrostatic forces, for example, see Hydrostatic theory of the earth and its mechanical implications. I like to imagine a ball of water in space held together by its own gravity. I also don't want to deviate from consideration of only hydrostatic (and gravitational) forces because I think it is sufficient for this discussion.

It would seem that the solution of the described problem is in terms of a small change in radius as a function of the azimuth angle, or z-coordinate if you take the axis of rotation to be the z-axis. This is using rotational symmetry. In other words, Earth's deformation due to rotation does not depend on longitude.

I want to ask about the extreme case. Imagine a planet rotating so fast that it is a very thin pancake. What will occur in this case? I am curious:

- Will the center hollow out, creating a donut shape?
- Will it break up into a multi-body system?

It seems to me that it would be logical for the high-rotation case to break up into 2 or more separate bodies. The reason is that a 2 body system is stable an can host a very large angular momentum. But would it be an instability that leads to this case? When would such an instability occur and could a rotating planetary body deform in a different kind of shape from the beginning, such as a dumbbell-like shape, which would transition into a 2-body system more logically than the pancake shape?

To sum up, how would a pancake shape transition into a dumbbell shape? Or would it? What are the possibilities for the described system?

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## 3 comments

## @Howard A. Landman 2014-02-07 10:57:57

@AlanSE: My reading of the paper is that, beyond a critical speed of rotation, the Maclaurin shape (C) becomes unstable to small perturbations. For the $\epsilon_1$ perturbation given in Table 2 and Figure 6, if the perturbation goes one way (thicker in the middle), you hit the Mass Shedding Limit (A), develop a sharp edge at the equator, and fling excess material off into space. If it goes the other way (thinner in the middle), then the pancake turns into a torus (I,J,K,L).

The other perturbations are essentially higher (axisymmetric) harmonics. A real perturbation could be a mix of several harmonics, or even violate axisymmetry completely. But if you stay axisymmetric, then the $\epsilon_k$ probably form an orthogonal basis set that spans the whole space of axisymmetric perturbations.

## @Alan Rominger 2014-02-06 00:24:43

I have stumbled upon a paper that presents a solution to this problem. First, I arrived at it via the following (recent) online summary of these shapes:

http://www.aleph.se/andart/archives/2014/02/torusearth.html

The paper is at Arvix:

Uniformly rotating axisymmetric fluid configurations bifurcating from highly flattened Maclaurin spheroids. Feb 2008.

Critically, the bifurcation referred to seems to be exactly what I referenced in this question. They always start with a

Maclaurin spheroids, which is what I referred to as "pancake". Then they go through a process, and in the end wind up with a torus or multiple objects. Here is the image that illustrates the real meat of their discoveries:You can see in the leftmost image, they go from a pancake to a simple torus. The rightmost process demonstrates one of the other types of processes that are physical. However, many configurations also hit upon a

mass shedding limit. In that situation, adding more rotation causes the apparent gravity on the edge to go negative. Obviously, this doesn't work, so the material "flies off into space". But that's not entirely true, it's just subtracted out of mathematical necessity because it doesn't actually have escape velocity.Moving on... I am surprised. I did not expect to see the "pinch" in the middle shown above. This still doesn't make sense to me, and I can not come up with a good argument for why it happens. In order for it to slope in, I need to be able to postulate a non-equilibrium configuration where the center is either flat or slopes out, and in that configuration, the forces/gravity pushes material

awayfrom the center. This is a very difficult proposition to accept. I see obvious in either the gravitational or hydrostatic physics that would do this. Nonetheless, the authors seem to have done an excellent and thorough job with full computer simulations backing them up with the full problem complexity included. So it appears that I'm wrong about that.If Earth were spun fast enough from its present state, the south pole and north pole would dip

inward. This is very strange, but it seems to be the correct answer, as per this paper.## @Curious George 2011-06-12 10:24:41

For an experimental test see: Liquid marbles http://adsabs.harvard.edu/abs/2001Natur.411..924A

(paywall http://www.nature.com/nature/journal/v411/n6840/full/411924a0.html)

For a paper in General Relativity see: Accurate simulations of the dynamical bar-mode instability in full general relativity http://adsabs.harvard.edu/abs/2007PhRvD..75d4023B

Basically once you start increasing rotation the pancake will become unstable and will make a transition to a rotating bar (dumbbell).

Anyway I think that nobody ever saw the bar actually breaking into 2 pieces. Typically you loose matter from the external regions, redistribute angular momentum and go back to axisymmetry.

Cheers

## @Alan Rominger 2011-06-13 03:37:46

The contribution of bar-mode instability in itself in enough for me to pick this as the answer, but that doesn't mean I'm not still thinking about this problem. I was aware of "pancake to bar" evolution in galaxies prior to this, but direct use of the polytropic index in your referenced papers indicates that this exact problem is within the scope of some prior work. Btw, this contradicts my (misplaced?) expectation of the bar shape being unstable. But I can't shake some suspicion that "bar" solutions are only possible with internal currents, i.e. transport and not hydrostatic.

## @Curious George 2011-06-13 07:00:32

I don't know if this helps but: the bar IS unstable, in the sense that simulations seem to indicate that the two arms will start becoming more and more asymmetric (one grows and the other shrink) so you go from m=2 to an m=1 mode and finally m=0 (axisymmetry).

## @Curious George 2011-06-13 07:12:41

For the internal currents: as far as I know you cannot reach the threshold for the instability unless you have differential rotation (the core rotating faster then the surface). With rigid rotation the external layer will become unbound before the instability sets in, unless you have something that keeps thing together (surface tension for droplets, weird EOS for strange stars, not sure about very compact objects in GR, if there is a windows in which it works with rigid rotation before the object collapse to a black hole, assuming that rigid rotation make sense in GR).

## @Curious George 2011-06-14 07:10:28

More recent papers on droplets (paywall, these people don't like the arxiv for some reason): Nonaxisymmetric Shapes of a Magnetically Levitated and Spinning Water Droplet; Generation and Stability of Toroidal Droplets in a Viscous Liquid. The second one seems interesting because of the generation of multiple droplets, but still I think here surface tension is playing a major role.

## @Kevin Kostlan 2020-04-24 05:32:04

Rigid rotation makes sense in GR for axisymmetric things, it means string anchored between two points will not change it's (proper) length. But planets are newtonian.