Hi Jorrie,

When I first read your blog I thought 11 times faster seemed slow to produce such a dramatic result. Then I realized that it's because you're talking about angular velocity. Can you do that calculation but for angular momentum?

Also, I think maybe you aren't accounting for the change in gravity at the surface at the equator due to that change in density (due to the flattening of the earth) when you you use GM/r_e²

In other words I think the M in your equation is a dependent variable too, especially since we aren't assuming the Earth has uniform density.

I hope that makes sense. Interesting post, I have some follow up questions regarding frame dragging but I'll save those till later.

Roger

Hi Roger, you asked about angular momentum. Yes it can be done, but I'll have to come back on that.

"Also, I think maybe you aren't accounting for the change in gravity at the surface at the equator due to that change in density (due to the flattening of the earth) when you you use GM/r_e²"

I did account for the reduction in density due to the flattening (the ellipsoid has a larger volume than the sphere, for the same mass). The mass I took to be constant, which is a good approximation, I think, because the speeds and gravitational field is not relativistic in this case. In neutron stars, there will probably be some change in effective mass, because angular energy will add to the 'relativistic mass', but then there will be a lot of other relativistic side-effects.

I'll come to that in a future post - still trying to understand the geoid properly for non-relativistic cases.

-J

Hi Jorrie, My suggestion that the mass as a dependent variable wasn't because I felt relativistic effects were involved. I guess the best way to explain it would be to say, suppose the mantle is what deforms and the iron core does not when the earth is bulging at the equator. It's true the earth is wider at the equator (due to the bulging) but much of that additional radius is mantle (lower density). Whereas it's true that at the polar region that the radius has been reduced, but now a higher percentage of that radius consists of iron core (high density). I don't think you account for that when you calculate the gravity at the equator and the poles. Roger

Hi Roger, agreed, I did not account for that. The total mass inside the geoid does however remain the same.

I'm looking at how the WGS 84 Geoid accounts for the elliptical effect. Will come back to you on that (must run now for the usual Friday chores).

-J

Hi Roger, I redid the calculations according to the Geoid equations.^(a)

The extra effects that they bring in cause only a small difference to the equatorial situation. My 11 times Earth spin rate for a 'zero g equator' become 11.18 times Earth spin rate, at an equatorial radius of ~8,539 km. However, their polar gravity is about half of the 2.2g that I got, despite the fact that the polar radius is only about 1% less than mine.

When I look at the radial density distribution graph^(b), and taken with what you wrote, it makes some sense. Only about 1/3 of Earth's mass sits in the core, with rest in the mantle/crust. With the flattening, the poles are closer to the core, but farther from large parts of the more massive mantle. Hence the polar gravity of the oblate sphere is less than for a perfect sphere with the same mass (think of the 'polar gravity' of a long thin bar with the mass of the sphere). This is despite the fact that the core itself does not bulge much. I must say, I was quite surprised by this result.

The equatorial surface of the oblate sphere, on the other hand, feels a little more gravity than the perfect sphere with the same mass. One brings some mass closer to any specific point on the equator and takes an equal amount of mass the same distance farther away. My gut-feeling agreed with this result.

I am still checking results for consistency, but (so far) I could not find any obvious errors. My integration spreadsheet for this case is still a bit of a mess (for private use only), but maybe I can clean it up and then ask for detailed comments...

-J

(a) WGS 84 Geoid eqs. 3-23, 3-26, 3-53, 3-63 to 3-66.

(b) http://en.wikipedia.org/wiki/Structure_of_the_Earth

Awesome. Unfortunately I can't seem to view the geoid links (I get a broken link when I click on them). Can you provide another link? Also, I hope you will do the Angular Momentum (how many times more it is than spherical Earth). My guess is it will be somewhat higher than the 11x seen for angular velocity. This is because the moment of inertia has increased (I think) as well as the angular velocity.

You use the term Geoid when describing the shape the Earth becomes at the high velocity. In my mind I've been picturing a Oblate Spheroid. Is that the same thing, or is the shape the Earth becomes more complicated then I'm realizing (due to the variable density)?

Hi Roger,

Sorry about link. The best is to Google: WGS 84 ELLIPSOID 3.1 ... It's a .PDF and those links are sometimes difficult to copy effectively.

I'm used to the term "Geoid" for the theoretical sea level, but they use "Ellipsoid". The Geoid is the equipotential surface of a spinning mass in space, where the potential includes rotational and gravitational energy.

Once I'm reasonably confident with the shape, I'll try to convert values to angular momentum L.

-J

PS: here are the Geoid equations that I used:

The top two are the gravitational accelerations, m is the spin parameter and the rest are ellipticity or oblateness parameters.

Hi Roger,

I've done some angular momentum calcs from the equations in WGS 84 ELLIPSOID 3.1.The results for my 11.18 times spin, oblate planet, r_equator = 8,539 km and r_polar = 4,196 km (of post #9) are:

Angular momentum ratio L/L_earth = 20.45, from the moment of inertia ratio: C/C_earth = 1.83.

The moments come from these WGS equations:

with f the flattening, a the equatorial radius, b the polar 'radius', C the moment of inertia and m the spin parameter as in my prior post. I have not figured out how they derive all the equations, but since it is such a widely used system, I took them at face value.

The angular momentum is then simply: L = ωC.

-J

Hi Jorrie,

I haven't had time to fully read and absorb the geod .pdf, but I have access to it now (using the google search you suggested). In the meanwhile, I've found the following that may be helpful. It's called MacCullagh's Formula. Basically it's a formula to calculate the gravitational potential of nearly spherical bodies. For an oblate spheroid it gives (axially symmetric example in link below):

http://geophysics.ou.edu/solid_earth/notes/maccullagh/maccullagh.html

here is a text book explanation:

MacCullagh's Formula

Hi Roger, thanks for the links.

I quickly scanned and can saw the correlation between the MacCullagh's Formula derivations and the WGS equations for inertial moments. Unfortunately, it seems that MacCullagh's system does not deal with the density of the interior and changes due to changes in rotation.

The results of their gravity for the real Earth (eqs. 257) are the same as the WGS system. Will look into it further.

-J

Here is another explanation of MacCullagh's Formula and it works in spin as well (helpful to understanding in my opinion):

http://www.phys.um.edu.mt/NOTES/PHY2130/

Select:

PHY2130_notes_gravit..> 13-Oct-2008 06:57 1.2M

Hi Roger, what I called 'the spin parameter' m (actually the dimensionless ratio between the equatorial centrifugal force and the static gravity there) is used in all the MacCullagh's Formula links that I've looked at. My problem is more that they do not bring density distribution into it. This is necessary knowledge for 'spinning earth up', because density changes with oblateness. I suspect it gets lost in the "neglecting higher order terms" in the integration that gives MacCullagh's Formula, as shown in your reference PHY2130_notes_gravit..>. It is OK for the Geoid, because we measure things and set certain values. Not so if we want to extrapolate...

I've made the simplifying assumption that the effective density of the more oblate 'earth' stays around half the average density of the spheroid (as it is for real Earth). I think this is a conservative approach; my guts tell me that the proportion of low density mantle stuff goes up and hence the effective density should drop to below half of the average density of the flattened ellipsoid. However, I have no clue how to determine the real situation for a 'fast-spin-earth'.

-J

I have noticed a curious correlation between my results, using slightly different assumptions for different runs, e.g. density not changing with spin-oblateness, slightly different averaging values for the effective density for spinning oblate spheroids and highly simplified (approximate) gravity. Although the critical spin rates for zero equatorial gravity vary, the equatorial radius always comes out close to double the flattened polar radius (if we can call that a radius).

Then I remembered my Blog entry http://cr4.globalspec.com/blogentry/1372/Rotating-Black-Holes-the-Naked-Truth (2007), showing the flattening experienced by the 'static limit' of a spinning black hole. Lo-and-behold! For an extreme Kerr black hole, spinning as fast as allowed by present theory, the static limit at the poles is exactly half the radius at the equator. For ease of reference, here's the image:

A black hole cannot 'bulge' in the way a spinning solid sphere bulges at the equator and flatten at the poles. Its equatorial radius stays at the Schwarzschild radius and its 'polar radius' shrinks as it rotates faster, until it reaches the 'Kerr-limit'. I copied some equations from the said Blog entry here:

where a (eq. 2) is the spin parameter (directly proportional to spin rate). The 'Kerr-limit' is at a = GM/c², with R_inner = 0 and R_outer = GM/c², exactly half the Schwarzschild radius. This means that at the Kerr limit, R_{static(θ = 0)} = GM/c², exactly half R_{static(θ = pi/2)} = 2GM/c², which is the Schwarzschild radius, of course.

Whether the correlation with my 'fast-spinning-earth' is a 'deep truth' about spinning, gravitating bodies in general, or just a coincidence for Earth's particular values, I don't know...

-J

PS: I have updated the opening post to reflect the WGS 84 ELLIPSOID 3.1 values and equations.

I wrote in prior post: "Whether the correlation with my 'fast-spinning-earth' is a 'deep truth' about spinning, gravitating bodies in general, or just a coincidence for Earth's particular values, I don't know..."

I have now determined that it is indeed Earth's density distribution, more specifically, the fact Earth's 'effective flattening density' is about half of the average density, that causes the apparent correlation (flattening of ~50%). Black holes have a "simple" density distribution: all of the mass at or very near the center (the singularity) and nothing in the rest of the volume. Hence, they all display this 50% flattening of the static limit at extreme rotation.

This was an interesting diversion, but not of any significance. :(

-J

Hi Jorrie,

There seems to be a sizable difference between the average density of the core and the mantle. You could better approximate the density of the Earth with a step function consisting of the average density of the mantle then stepped to the average density of the core. Although I don't think this step function could be used for flattening (since it would create a discontinuity (a tearing of the mantle from the core), it could be used to approximate the gravitational field of the oblate spheroid and thus in a way improve your calculation.

Does that make sense?

Roger

Hi Roger,

Thanks for the heads-up; I've corrected the OP.

You wrote: "Although I don't think this step function could be used for flattening (since it would create a discontinuity (a tearing of the mantle from the core), it could be used to approximate the gravitational field of the oblate spheroid and thus in a way improve your calculation."

I have been toying with same idea, actually for calculating the flattening by numerically integration (eq. 2 this time), stepping through all the layers of Earth. I think one needs flattening to find the gravity - if I look at the WGS equations for gravity (e.g. post #11 above), I see no density, just various eccentricity parameters plus rotation.

Will give it a go soon.

-J

You wrote: "if I look at the WGS equations for gravity (e.g. post #11 above), I see no density, just various eccentricity parameters plus rotation."

Yes, I think because that gravity equation assumes a uniform density or that the gravity is being measured from far enough away that it appears uniform.

For the core / mantle idea I think I would treat the mantle and the core as two oblate spheroids (they don't have to have the same flatness). The mantle oblate spheroid would have a center oblate spheroid subtracted out, corresponding to the core oblate spheroid. Thus instead of your current equation which is:

Gravity of Oblate Spheroid (half density of earth). It would instead be:

Gravity of Oblate Spheroid (mantle density) - Gravity of smaller Oblate Spheroid (mantle density) + Gravity of smaller Oblate Spheroid (core density) = Gravity of Earth.

In essence you'll be treating the problem like the gravitational pull at the surface of the Earth is the result of the superposition of two objects with different densities (Mantle and Core).

Did that make sense?

Hi Roger, you wrote: "Yes, I think because that gravity equation assumes a uniform density or that the gravity is being measured from far enough away that it appears uniform."

I don't think so, because the WGS equations for gravity is what is measured at sea level at the poles and at the equator, so it seems to me that the actual density profile is included (not just an average).

However, the gravity is not where I get unstuck - I think I can worked that out in terms of a density profile, as you indicated. It is rather the influence of the density profile on the flattening where I have problems, because the only equation that I could find is for a body of uniform density. Without knowing the flattening for a non-uniform density and for a different spin, I have to use the gross approximation, i.e. ~50% of average density for Earth. And it is different for different planets, I found by looking at their data. If only I knew more about fluid mechanics...

-J

Maybe a rough way to approximate the flattening with varying density would be to spin an earth size sphere with mantle's average density and an Earth's core size sphere with core average density and then take the weighted average (based on respective original radii) of the flatness as your total flatness. Using the respective radii to weight the average flatness allows you to not overestimate the core's (higher density) effect on the overall flatness.

Essentially we're saying that the mantle "sticks" to the core and that the core resists some of the mantles flattening (since higher density given the same spin results in lower flatness values) or you could see it as the core "sticks" to the mantle and it's flattening is thus increased (since lower density given the same spin results in higher flatness values).

For the gravity approximation, I think we may see a not insignificant effect if we treat the Earth's gravitation as the superposition of the Mantle's gravity and core's gravity.

When you do that extreme spinning you're essentially moving mantle from the poles and moving it to the equator (the bulge due to flattening). So when you're standing near the poles there is a lot of high density stuff nearby, but when you're standing at the equator the high density stuff is farther away. I would imagine this inhomogeneity in the density makes a significant difference in the gravity from the current assumption of equal density everywhere.

Hi Roger,

I tried this approach, but I still get unstuck with the flattening part and also somewhat with the gravity.

There are some indications in the literature that fast spinning inhomogeneous bodies have not been solved in general, e.g. this Springerlink abstract:

On a method of computing the gravitational fields of inhomogeneous bodies: "These methods were also successfully applied by Tsirulskii and Golizdra [1, 2] in treating the homogeneous and inhomogeneous, two-dimensional direct problem by means of Cauchy's integrals. However, as regards three-dimensional fields a number of fundamental problems has not been solved in this respect".

Maybe you can get hold of the pdf for us through your library. I'm not even sure that they include spin for the 2-D case, but maybe they do.

-J

I'm puzzling over your comment that the atmosphere of the rapidly spinning planet would have escaped into space. I'm not saying you are wrong or that I disagree, I'm just trying to puzzle-out the physics of it.

Solar system objects that have lost their atmospheres (or mostly lost), such as the moon and Mars, lost them due to the low mass of the 'planet'. The mean free path for Jean's Escape is dependent on the mass of the planet and not (or not necessarily?) the rotation rate. The temperature also matters, of course, as does the composition. Light gases like H₂ and He are lost easily; O₂, N₂ and CO₂ are heavier and the average velocity of the gases is lower than H₂ or He for the same temperature, so fewer molecules at the top of the atmosphere will have the velocity needed for escape (assuming a Maxwellian distribution of their velocities).

Also, as you stated, the surface gravity near the poles is much higher than at the equator. So it seems to me that the oceans would actually 'pool' near the poles. This is almost counter-intuitive; one has to remember that your planet is not a rigid sphere spinning at 11X the Earth's rate; it is extremely oblate, and for any point not directly on the equator there is actually a component in the direction of the poles, where the surface gravity is 2.25X the Earth's surface gravity and many times larger than the surface gravity at the equator of your planet where it is essentially zero.

Well, that's where I am so far; as I said, I'm still puzzling through this. I suppose during the formation of such a planet oceans and an atmosphere would not have formed in the first place. Needless to say, I'd like to see your comments.

Thanks for the posting. Interesting food for thought.

What I wrote about atmospheres didn't come out very clear. O₂, N₂ and CO₂ are lost from the moon and Mars easily, along with H and He, due to the low mass of the moon and Mars. The Jean's Escape velocity is low. O₂, N₂ and CO₂ are much more common in Earth's atmosphere due to their higher masses, where the Jean's Escape velocity is higher and at the top of the atmosphere there are few O₂, N₂ and CO₂ molecules with high enough velocities to escape.

I'm still puzzling over the shape of the Earth and the ocean. The ocean would (should?) take on the shape of an equipotential surface. A plumb line should be perpendicular to the ocean's surface -- I think. At the poles the plumb line would point toward the center of the Earth; at the Equator the direction would be undefined. Hmmm....

Hi Usbport,

I'm also still puzzling over the shape of the "11 times spin geoid" (equi-potential surface). My problem is it depends on where one defines the surface to be. I think in the usual definition, the geoid is taken as a theoretical ocean surface, excluding the atmosphere. But then, the 1984 definition includes the mass of the atmosphere in the calculations, which is required for satellite calculations.

If at 'critical spin rate' for the equator, the surface air pressure there must be zero. Will parts of the atmosphere go into orbit there? Don't know.

I do not think the oceans will 'pool' at the poles; after all, the 'polar ocean' (without ice?) is at the same gravitational potential (V) as the equatorial sea, not so? I think it is only the gravitational acceleration (slope dV/dr) that differs. More study to be done, but that will have to wait for the weekend.

-J

Hi Jorrie,

You wrote:

we get from (1): f =0.0033489 and from (2): f =0.0034493, a difference of 3%

I think you meant (3), not (2).

Well nearly exonerated. ^{_{Thanks Roger}}