Celestia Forums

Some slightly fiddly questions for the physics-heads among you...

One of the old theories for the origin of the moon was that the early Earth was spinning very rapidly about its axis - so fast that the equatorial bulge broke off to form the moon. Now it seems to be obsolete, but I'm curious about a couple of things.

1) Is this actually physically possible for a rocky planet, or does it have to be largely molten for this to happen?! I just can't picture a few million cubic km of rock suddenly breaking off into the sky like this... or would it take a long time to actually 'fly off'?

2) For a world building project, I'd like to know how to calculate how fast a planet of any given size and mass must be rotating to split off like this. And if at all possible, how much would break off. I couldn't find any formulae for this online though, does anyone know how to figure this out?

As usual, any help would be appreciated!

Hmm....

Regarding question 1, I had thought that the spin of a planet would diminish with time due to tidal effects with its star. If this is the case, then the rapidly spinning protoplanet would either never coalesce into a single body. Of course, if an encounter with another massive object imparted more spin to the object, the result of the encounter may be that the centripetal force at the surface exceeded the gravity of the planet. If the spin were vastly increased in a short period of time, and the outward force exceeded the binding properties of the material of the mantle and crust, then a huge quantity of material could be ejected in big messy chunks.

If the Earth rotated roughly every 84 and a half minutes, its gravity would be almost exactly equal to the centripetal force at the equator. (I used a spin-habitat formula for this)

I suppose it all depends on the tensile strength of the crust and mantle material.

As for question 2, I have no idea, but you all knew that anyways

Cormoran

The splitting theory is now discredited - it's difficult to know what sort of influence (such as a giant impact) could spin a planet up to that sort of rotation rate without itself disrupting the planet. (There have been suggestions that Venus could be spun up by a tailored series of "small" tangential impacts over thousands of years, but that's not the sort of thing that can occur naturally ...)
But the considerations are exactly the same as those used to calculate the the Roche limit - when does the outward force exceed the gravitational force, since for all but the smallest bodies the tensile strength of the material is trivial compared to gravity. So finding the circular orbit period at the surface does the trick approximately, as Comoran suggested - you can imagine that the Earth would lose its oceans and its atmosphere, and then begin to isostatically heave up chunks of rock which would break loose, allowing more isostatic uplift, and so on.
In practice disruption would occur at a slightly longer rotation period, because the equilibrium ellipsoid would heave the equatorial radius outwards - decreasing gravity and increasing centrifugal "force". But the shape of the equilibrium ellipsoid depends on the density as a function of radius, so the maths begin to get a little hairy.
Somewhere I have some information on the shape of rotating equilibium ellipsoids of uniform density, which might be useful in this context, since there's obviously a crisis point in the maths when they start to fall apart. Also, some information on maximum theoretical rotation rates for "monolithic" asteroids - small unitary bodies that are effectively held together by tensile strength. I'll get back to you on those, since I'm distant from my filing cabinet at present.

Grant

Hmmm, I seem to have mislaid the article on rotation rates in monolith asteroids, but monolithic construction is considered unlikely in anything over 200m radius, so the information doesn't apply to your scenario.

From Lang's Astrophysical Formulae, there is a theoretical maximum angular rotation velocity for self-gravitating fluid bodies given by:

omega = sqrt(0.449331*pi*G*rho)

where G is the gravitational constant, rho is the mean density, and omega is in radians per second. However, such configurations are unstable, and the stable maximum is at:

omega = sqrt(0.37423*pi*G*rho)

at which point the polar radius is only 0.58 of the equatorial radius.

Plugging an Earth-like density into the above gives a minimum rotation period of around 2.5 hours. Notice the value is independent of planetary size - radius seems to have cancelled out in the algebra somewhere.
(For a water-density object, the minimum theoretical rotation period is 6.2 hours, which accords with the very marked flattening of Saturn at 10 hours.)

Grant

Hmm. Interesting that the radius doesn't matter there.

Thanks for digging that up, Grant!

Evil Dr Ganymede wrote:Hmm. Interesting that the radius doesn't matter there

Makes sense, after a little reflection.

For a given density, the mass of an object increases in proportion to radius cubed. Its surface gravity increases in proportion to the mass, and decreases as radius squared - so for a given density the net effect is that surface gravity increases in proportion to radius.
For a given angular velocity, the centrifugal pseudoforce also increases in proportion to radius.
So once density and angular velocity are specified, the ratio of surface gravity to centrifugal "force" is automatically determined, because the two forces vary in proportion to each other. For the specific situation in which the two are in balance, radius becomes irrelevant.

Grant