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I found it fascinating that many asteroids rotate with a period of just seconds. For this fast of a rotation on this size of object, I thought that would actually cause significant acceleration on the "equator" of the rotation. This is not the case for planets where gravitation still causes a net downward force on its equator. An entity on the surface of such as asteroid will have to hold on to the surface in order to keep from flying off.

Here is what I calculate for the surface acceleration (again, off of the surface) for the notable examples in the above link:

Asteroid    Period (s)  Radius (m)  v^2/r (m/s)
-----------------------------------------------
2010 JL88     24.5       15          0.987
2010 WA       31          3          0.123
2008 HJ       42.7       24          0.520
2000 DO8      78         30          0.195
2003 DW10    100         20          0.079
2003 EM1     111.6       33          0.105

I consider these accelerations to be quite considerable. If the surface was in any way sandy, for instance, the top layer would just fly off. Obviously some sort of cohesion is necessary.

My Question

Are the fastest rotation periods of asteroids limited by asteroid material, or is there no astrophysical way for them to be spun fast enough to matter in the first place? Given what we know about asteroids, what is the surface acceleration at which we expect them to fall apart at, and how does that compare to what we observe?

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@dotancohen: The original title seemed clearer. Apparently "UTS" means "Ultimate Tensile Strength"; that should be explained in the question. –  Keith Thompson Jan 17 '12 at 20:18
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hmm, to propose yet another alternative, I would suggest "does the structure of asteroid rock limit an asteroid's speed of rotation?" The current title does go out of its way to preserve the detail of the question, the only problem I can see is that it's just a little long and verbose. –  Alan Rominger Jan 17 '12 at 20:36

2 Answers 2

up vote 1 down vote accepted

These objects have been called "superfast rotators" lately. A google search for this term found a very nice paper which talks about tensile strength and theoretical explanations for high rotation rates.

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I inadvertently stumbled upon a relevant quote about this.

http://books.google.com/books?id=JwHTyO6IHh8C&lpg=PA471&ots=AI73Vfs-4N&dq=asteroid%20internal%20pressure&pg=PA471#v=onepage&q=asteroid%20internal%20pressure&f=false

While no asteroid larger than ~150m shows evidence for global cohesion, almost all asteroids smaller than this must be cohesive. One might suppose that every larger asteroid is a gravitational aggregate of smaller pieces, whereas every small asteroid is a fast-rotating collisional shard. But the term "monolith" for these smallest asteroids is misleading. Consider a spherical object of uniform density $\rho$ rotating with a frequency $\omega$; the mean stress accross its equator is ~$R^2 \rho \omega^2$. For the well-studied fast rotator 1998 KY26 (Ostro et al., 1999), self-gravity is not capable of holding it together; however, its ~11-min period and ~30-m diameter requires only a tensile strength of ~$300 dyn/cm^2$ (presuming $\rho \approx 1.3 g/cm^3$ for this C-type), order of magnitude weaker than the tensile strength of snow.

The question was somewhat misguided. Of course you can't determine the strength of asteroids from their rotation, largely because the rotational stress isn't very restrictive.

I find the $R^2$ term from this reference somewhat misleading, because the important to observe about such a arbitrary spherical rotating body is that the maximum period of rotation without the equator exceeding escape velocity is irrelevant of $R$. The centripetal acceleration is directly proportional to $R$ and so is gravity.

For the examples in the OP, however, there is a valid point that the necessary material strength just isn't very important due to the $R^2$ dependence. To apply the reference's logic to the first one:

$$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{24.5 s} $$

$$ \sigma \approx R^2 \rho \omega^2 = (15 m) \left(1.3 \frac{g}{cm^3} \right) \left( \frac{2 \pi}{24.5 s} \right)^2 = 19 kPa $$

The snow reference comes to $30 Pa$. Even so, a few kPa isn't much at all.

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