<snip>
“In the case of a shifting core, it is going to shift to correct an inequity in the earths mass”
No, it will shift to correct “rotational mass” or moment of inertia. For example, a movement of equal continental mass to each of the poles will have little or no effect on core movement but movement of that total mass to the equator would. It is the radius or distance to the rotational axis for the shifted mass that is significant.
Which is a round about way of saying exactly what I said. In the event you have an uneven distribution of mass perpendicular to the axis of rotation, there will be a force applied along the plane of rotation (i.e. perpendicular to the axis of rotation). In the event of a solid earth this force would result in precession of the earths axis of rotation.
“This will reduce the gravity felt on pangea and increase the gravity felt in the oceans.”
I agree.
“Assuming equilibrium is met (i.e. the earths center of mass is returned to its center of rotation), and the earth remains spherical, the gravity on the surface will be equalized - as in pangea will experience exactly 1.0G, and the ocean side will experience 1.0G.”
No, the shifting of the core(s), and therefore the shift in center of mass, has created the equilibrium. The center of mass will not coincide with the center of rotation. As long as Pangea remained basically intact, the differential surface gravity would remain the same.
No, that is incorrect. With the core centered, and pangea off to one side, you start off in disequlibrium - mass is not evenly distributed along the plane of rotation, which creates a precessionary "force" across the plane of rotation. In this case you have "extra" gravity on pangea, due to the increased amount of mass beneath pangea, relative to the earth on average. As per newtons 3rd law, in the disequlibrium state described above, you also have an equal, but opposite force that would push the core away from the pressesion of pangea. I'll attempt to draw, all images representing a line drawn across the plane of rotation, as seen from above:
P = pangea
) = oceanic crust
o = core
- = unit of distance
<> = forces
^ = location of axis of rotation (axis would extend out of your screen)
. = place holders, since this forum removes superfluous spaces
Starting position (not at equlibrium). More gravity will be found at pangea, due to the additional mass on that side of the rotational plane:
P-----o-----)
......^
Force on the above system due to rotation(top)and the opposite force on core (bottom, newtons 3rd law and all that):
<<P-----o-----)>
........^
........< >>
Distribution & force after core moves:
<P------o----)>
........^
.......< >
In the bottom case we have equilibrium - mass is distributed evenly across the plane of rotation, thus eliminating precessionary "force". Because the mass is now evenly distributed along the plane of rotation, surface gravity is also equal along the plane of rotation. Assuming a perfect sphere, this will be roughly 1G.
Again, the reason why the oversized dinosaurs and sea reptiles and pterosaurs were able to develop was this lowered surface gravity and the reason they were gone near the K-T boundary was the rapid increase in surface gravity resulting from the breakup and dispersal of Pangea.
Sorry, that doesn't work for a number of reasons:
1) In disequlibrium, there would be less than a half-percent change in gravity on pangea. At equilibrium, that change goes away. In the former case the difference is so small (0.4%, or 0.004G) as to be meaningless in a biological context.
2) Pangea broke up ~150MYA, with the major breakup complete around 100MYA. The dino's went extinct and the KT boundary formed 65MYA. So the timing doesn't fit.
3) Assuming a slow breakup of pangea, the ~100MY period of time it took would have been more than sufficient for the dino's to evolve along with changes in gravity. Instead, we see even the largest of dino's making upto the KT boundary intact, and then suddenly disappearing.
4) There is no known mechanism which could lead to a rapid breakup and leave the crust intact. There is also a boat-load of evidence for a slow breakup of pangea.
5) The mineralogy of the KT boundary is consistent with chondritic meteorites, and not with the earths mantle, providing further evidence for the meteoric extinction model, and further evidence against a geological mechanism.
6) There is a crater (Chicxulub Crater) which is both of the right age, and right size, as was predicted for the dino-killing impact. Once again, consistent with conventional science, and in direct opposition to your alternate answer.
7) There is no geological evidence consistent with large-scale or rapid changes in the makeup of the earth around 65MYA.
8) There is no mechanism by which earths irridium, which is largely locked up in the core, to move to the surface. To get it to the surface would require sufficient force to destroy the earths core, or lift a portion of it to the surface. No irridium = no KT boundary.
9) There is no biological need for lower gravity - biophysical analysis of even the largest dino's (sauropods; see the references in my older post) shows that their physiology was more than sufficient to support their mass given earth-normal gravity. Same is true for the largest of flying petrosaurs - aeronautic analysis has shown that the second largest would have had no trouble flying at earth-normal. The largest isn't much bigger, and is expected to have very similar physics.
Like I've been saying, the evidence is against you model. The math shows that any variation in gravitational force will be small - a half % of 1G max. There is no evidence suggesting a catastrophic breakup
Bryan
EDIT: as I mentioned to pre, I'm gone the next few days for some R&R. Please reply, and I'll try to get back to you early next week.
