Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Mansfield's Earth Formation Hypothesis: Evidence.

I wrote a lot, but since edited to cut it down. Long post made short - most of your critisisms were answered in the citations I've provided previously. Maybe you should try reading them, instead of ignoring them and then pretending those materials were never provided. I have only one point of yours I wish to address at this point. If we can get past this, than we can go back to your other points:


1) Prove the calculations are incorrect. Simply stating so doesn't make it so. For that matter, putting your faulty (see below) math up as a counter example only weakens your case.

2) They are 100% relevant, in that they put an upper limit on the degree of gravitational shift which could occur, using nothing more than newtonian physics. And they show that the most extreme possible shift you could have - the core sitting immediately under one side of the earth - the shift you get is smaller than the one you propose.

If you think Newton was wrong, say it. Otherwise, show where the calcs are wrong. If you don't then you are going the same route as pre - rejecting evidence that opposes your beliefs without a valid and demonstratable reason to do so.

That is wrong is so many ways:

1) You're treating the core as though it were the entire mass of the earth, which is completely false. The cores combined are ~32% the total mass of the earth. So even if we take your calcs at face value, the shift is 56% of 32% the earths mass, not 56% of the total mass. So your shift isn't 56%, but rather 56% of 32% = 17.92%.

2) Your calculations ignore that the area formerly occupied by the core will be filled by mantle. The mantle weighs, per unit volume, 4.5g/cm^2 while the cores average 11.5g/cm^2. Ergo, the effect of the core shift will be reduced by 4.5/11.5 = 39% due to replacement of the core by mantle. So the 17.92% now drops by 7.01% to 10.91%.

3) Your math is fundamentally flawed. Gravity is dependent on radius, not on diameter. Ergo, you cannot use diameter as the "measuring stick" by which you position objects.

4) Of course, none of that matters since your formula is completely and utterly wrong. Lets derive it from the beginning:

The force of gravity at earths surface is determined by:
Fg = G*m1*m2/r^2

Lets use a consistent test-mass (m1) of 1kg. In the case of the earth (and core), mass (m2) is constant and therefore the G*m1*m2 part of the equation stays constant. Lets use 'a' as the symbol for G*m1*m2. If 'a' for earth = 1, than 'a' for the core = 0.32. But sticking to symbols, our formula now becomes:

Fg = a/r^2, which is the same as

Fg = a(r^-2)

Unshifted, the core is b units (6 in your example, but lets stick with symbols) from the earths surface, therefore under this case the gravity will be:

Fg(start) = a(b^-2)

Shifted, the core is c units from the earths surface (8 in your example from pangea, 4 from the antipode. Under these cases:

Fg(end) = a(c^-2)

We're interested in the change of gravity during the shift, we need the delta(Fg), which is:

Delta(Fg) = Fg(end)-Fg(start)
Delta(Fg) = a(c^-2) - a(b^-2)

This can be written as Fg = a/c^2 - a/b^2

Note how different that is from your b^2/r^2.

To use your numbers we need to convert your units into fraction of an earth radius:

Unshifted = 6 units, therefore 1 unit of radius = 1/6

So unshifted we have 6*1/6 = 1 earth radi

Shifted = 8 units, therefore 8*1/6 = 1.3333 (4/3rds) earth radi

Using your numbers, and a core of 0.32 earth masses (i.e. a = 0.32):
Delta(Fg) = a/c^2 - a/b^2
Delta(Fg) = [0.32/1.333^2] - [0.32/1^2]
Delta(Fg) = 0.18 - 0.32
Delta(Fg) = 0.14,

Which is 14% of a G, a tiny fraction of the 54% your incorrect formula provided. And this formula still does not take into account the gravity of the magma which will fill the former position of the core.

The magma which fills the core will have a G*m1*m2 of:
a' = 0.32*0.39 = 0.1248

Since this "filler" is at the former position of the core, r = 1, therefore Fg(filler) = 0.1248/1^2 = 0.1248.

Delta(Fg) now becomes:
Delta(Fg) = Fg(end)- Fg(start) + Fg(filler)
Delta(Fg) = 0.18 - 0.32 + 0.1248
Delta(Fg) = -0.0152

Which is 1.52% decrease in gravity at pangea.

Bryan