I did write “Your inner/outer core calculations are superfluous and incorrect.” Your calcs in the last posting are also incorrect and based on erroneous assumptions.
Stating so does not prove it.
Show where the math is wrong.
Show where a calculation was done wrong.
Show where the calcualtions are inconsistent with basic physics.
Saying "your math is wrong" is no different then a kid throwing a temper-tantrum.
This is a science board. Show us your science. If you want to make non-specific whines about how the mean scientists don't buy your incorrect math, I can direct you to a creationist board or two.
You stated that “You’re treating the core as though it were the entire mass of the earth....”
Not so, I have used a lot of care in all of my prior posts to describe relevant distances as, for example, “the distance from the CENTER OF MASS of Pangea to the new CENTER OF MASS of the Earth” after the core(s) shift, I have not stated “to the shifted cores” when calculating gravitational changes.
And another math fail. From your post (#35896):
The Earth has a diameter of 12 units
The Earth’s center of mass shifts from position 6 to position 8 (i.e. 8 units away from Pangea)
The ration of the new G to the old G is r^2/d^2= 6^2/8^2= 36/64= 56%Where in the above calculation did you account for the core only weighting 32% the total mass of the earth? Where did you account for the back flow of mantle that would fill that space?
The answer, of course, is you didn't. Ignoring that your "formula" runs counter to the physics of gravity, the
only way you could get a reduction in gravity of that size is by moving the
entire earth. The core is not enough.
In fact, we can figure out exactly how far you would have to move the entier earth, to get Fg down to 0.56G:
Fg = G*m1*,2/r^2
0.56G = 1/r^2
r = 1.33631 earth radi
You'll notice that the size of the shift - 1.33 (AKA 4/3), is
exactly equal to the size of the shift in your example (8/6, AKA 4/3).
The math says you are wrong. Either prove that I did something wrong with the math, or show us the derivation of your d^2/r^2 formula.
You wrote: “If you think Newton was wrong, say it.” Definitely not, this whole theory is based on Newton’s laws.
Really? Then why is it you are calculating changes in gravitational force without actually calculating a change? Changes (deltas), since you seem to have forgotten your math, are always calculated via additive or subtractives; what you have is a ratio, not a difference.
And, for that matter, it is impossible to derive your formula from newtons law of gravitation.
And if I'm wrong on that later point -
PROVE IT. Derive your formula, and post it here.
I've put my chips on the table; strange that you have not.
You wrote:
“Your math is fundamentally flawed. Gravity is dependent on radius, not on diameter. Ergo, you cannot used diameter as the ‘measuring stick’ by which you position objects.”
I’m not sure if you are confusing my use of the letter ‘d’ for distance with diameter because nowhere in my posting do I use diameter as a variable to describe changes in gravitation.
Sure you do. Your D of 8 can only be mapped using diameter. If using radius, as you should be using, that would be 2 (or -2, depending on how you label your axis).
BTW, your statement is incorrect.......gravity depends on the inverse square of distance, not on radius.
The inverse square of what? I'll give you a hint - it isn't the inverse square of the diameter!
I cannot believe you actually wrote the above - gravity is dependent on the inverse square of the
RADIUS. To say gravity is not dependent on the radius is like saying John Lennon has nothing to do with the Beetles. Radius, but the way, is the 'r' in newtons's universal law of gravitation:
Fg = G*m1*m2/
r2..................^ That guy right there
Since you replaced CENTER OF MASS with CORE in your calcs, they must be discarded for comparison to my calcs, which are not “....completely and utterly wrong.” They conform 100% with Newton’s laws.
Prove it. Derive your formula from either Newtons law of gravitation, or general relativity, and post the derivation here. If you cannot, then your formula does not conform.
I'd point out that the formula I derived IS based on newtons law of gravitation, and I posted my derivation here. Lets see your response...
Even though your calcs must be discarded
And there it is folks - no reason as to why they should be discarded, only that they should. Of course anyone reading this thread knows why they must be discarded - because they conflict with your
beliefs. You're like the Catholic church, demanding that everyone ignore Galleleo proof of a heliocentric solar system. And your reason is the same as theirs - you've got some random scribbles which you believe over properly derived, and evidenced, science.
, I will point out another error:
You wrote:
“Delta (Fg)=Fg(end) - Fg(start) +Fg(filler)”
This is incorrect because you have ignored the effect of the filler before the shift. The equation should be:
Delta (Fg)=Fg(end) - Fg(start) +Fg(filler-end) - Fg(filler-start)
The reason I did not add in that number is because it is extremely difficult to deal with. That filler, pre-shift, would be in a roughly hemispherical shell. Because its flow is not even (i.e. most of its flow will come from the side where the core moves to), it cannot be treated as a point mass. A crude approximation would be to treat it as if it all were a point mass, located mid-way between core and the surface, opposite pangea. The mass will be the same as the mass of the filler, so:
a' = 0.32*0.39 = 0.1248 (from my last post)
r' = 0.6769, from pangea = 1.6769
Fg(pre-filler) = 0.1245/1.6769^2
Fg(pre-filler) = 0.04427G
And from before:
Fg(start) = 0.32
Fg(end) = 0.18
Fg(filler) = 0.1248
Delta(Fg) = Fg(end) - Fg(start) +Fg(filler-end) - Fg(filler-start)
Delta(Fg) = 0.18 - 0.32 + 0.1248 - 0.04427
Delta(Fg) = -0.05947
Still a small fraction of the 54% you're claiming; in fact its pretty much 1/10th of what you claimed.
The most important erroneous assumption that you have made is one that I also made at one time:
You, and at one time I, assumed that when the outer core shifts off center, it maintains its spherical shape. This would not be the case, centripetal forces would distort its shape; the greater the core shift the greater the flattening distortion. It would be comparable to putting a blown-up balloon into a somewhat larger rigid spherical container and then pushing the balloon against the side of the chamber. In the case of the outer core, this would shift the CENTER OF MASS further from the Earth’s center, i.e., further away from Pangea, than if the outer core retained its spherical shape.
No, the center of mass would remain the same; angular momentum dictates that - unless you're planning on breaking not only the laws of gravity, but also the law of conservation of energy as well.
As I've challenged 3 or 4 times above -
SHOW US THE MATH. Starting from known physical principals, derive d^2/r^2.
If you cannot show us that, then you don't have a leg to stand on. I'm gone for a week on holiday - that'll give you plenty of time to write out the math and post it here. If you can hammer it out in the next 6 or so hours, I may be able to reply before I leave (but I doubt it).
Bryan
