ImagingGeek,
In reference to your calculations, You asked:
“Show me where the math is wrong.”
“Show where the calculations are inconsistent with basic physics.”
I will show you where you have made errors in your basic assumptions that are inconsistent with basic physics, making all of your calculation irrelevant. Let’s start with basic physics:
In response to my stating that:
“BTW, your statement us incorrect.....gravity depends on the inverse square of distance, not on radius”, you responded:
“The inverse square of what? I’ll give you a hint - it isn’t the inverse square of distance!”
“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(sic). Radius, but(sic) the way, is the ‘r’ in newton’s universal law of gravitation:
Fg = G*m1*m2/r^2
.............^ That guy right there”
I’m shocked because what you just wrote is totally wrong. This is something one learns in high school physics.
Sorry, “That guy right there” IS THE VARIABLE FOR DISTANCE in Newton’s Universal Law of Gravitation:
“EVERY PARTICLE OF MATTER IN THE UNIVERSE ATTRACTS EVERY OTHER PARTICLE WITH A FORCE WHICH IS DIRECTLY PROPORTIONAL TO THE PRODUCT OF THE MASSES OF THE PARTICLES AND INVERSELY PROPORTIONAL TO THE SQUARE OF THE DISTANCE BETWEEN THEM.”
This equation is usually written as:
F = G mm’/r^2
Where F is the resultant force
m is one mass
m’ is the other mass
r is the distance between m and m'
(NOT RADIUS)
G is the gravitational constant
The letter ‘r’ is used, I assume, because Newton was dealing with the Earth (and maybe the Moon) when he developed this law. Therefore, ‘r’ was used in writing this law because the radius of the Earth was the specific value of the variable distance that he was concerned with.
Let’s derive my ratio of the lowest G (the ‘F’ in Newton’s Law stated above) to today’s G:
To avoid confusion between the gravitational constant G and the force of gravity (which we have been using ‘G’ to represent) let’s use lower case ‘G’ (i.e., ‘g’) to represent the force of gravity. In other words ‘g’ represents the weight of, in this case, of an object on the surface of Pangea (near Pangea’s center of mass). Also, for clarity sake, let’s use ‘d’ as the distance in Newton’s Law.
The weight of an object (today), of mass m using m’ as the mass of the Earth, on the surface is:
g =Gmm’/d^2 =Gmm’/r^2 Where the value of r is the radius of the Earth.
If the center of mass of the Earth shifts away from the object on the surface so that the distance between the object and the new center of mass of the Earth is ‘x’ then the new weight of the object is:
g’=Gmm’/d^2 = Gmm’/x^2
The ratio of the new weight to the old weight of the object is:
g’/g =(Gmm’/x^2) / (Gmm’/r^2) =r^2/x^2
Since I was using ‘d’ instead of ‘x’, remembering that ‘d’ is not diameter, my previous references were stated as r^2/d^2. I hope this clarifies things. As you can see, there is no conflict with Newton’s laws and I was not using diameter as a variable.
Next, let’s examine why your assumptions are not valid by going over my assumptions, most of which have already been stated:
1. LAZE’S ASSUMPTION #1:
The process of consolidation of the masses forming Pangea caused a wobble in the Earth’s rotation. This wobble was countered by a force (Newton’s Third Law) whose action was to move the core(s) away from Pangea. For simplicity, I’ll say that this net force could be described as a single vector force between Pangea’s center of mass and the center of mass of the core(s) (at least initially).
2. LAZE’S ASSUMPTION #2:
When the core(s) moved from the Earth’s center (also the axis of rotation), the inner core no longer remained at the center of the outer core; it could move independently within the molten core under the force of the previously mentioned vector force and also subject to the, now unbalanced, centripetal forces.
3. LAZE’S ASSUMPTION #3:
When the core(s) moved from the Earth’s center, the outer core, like the inner core, was subject to the vector force previously mentioned as well as the centripetal forces. These two forces would have distorted the shape of the outer core, the effect of which would be to shift the center of mass of the Earth further than if the outer core were able to remain spherical. An analogy would be a balloon that is depressed at its center (by the vector force) and the opposing surface spreading out due to the centripetal forces.
4. LAZE’S ASSUMPTION #4:
As in any planetary body, density increases with depth. The densest part of the mantle surrounds the outer core. When the outer core shifts antipodally from Pangea, the spherical volume that is left vacant by the shift is filled with the densest material from the mantle. Hence, an additional movement of the center of mass (COM) of the Earth away from Pangea. This is a nonlinear movement of the COM; the rate of increase increases with shifting of the outer core.
5. LAZE’S ASSUMPTION #5
When the outer core shifts antipodally from Pangea, the densest part of the mantle at the leading edge of the outer core (i.e., furthest away from Pangea) is shifted away from Pangea, again moving the COM of the Earth further away from Pangea. This is a nonlinear movement of the COM; the rate of increase increases with shifting of the outer core.
I believe all of the above assumptions are reasonable and are based on “basic physics.”
ImagingGeek, Your calcs are based on assumptions that are in conflict with those that I have listed above and are not “real world” assumptions. As a matter of fact, you have not even listed your assumptions; I can only guess what they are. For example, using my assumptions number 4 and 5, you cannot just switch the “filler” from one side of the outer core to the other ignoring the movement of the densest part of the mantle and resultant shifting COM which I describe. Your statement that the “filler” on the trailing edge of the shifting outer core where “most of its flow will come from the side where the core moves to” is wrong. The densest part of the mantle will always move directly toward the new COM.
If I may quote you:
“This is a science board. Show us your science. If you want to make non-specific whines about how mean scientists don’t buy your incorrect math, I can direct you to a creationist board or two.”
There are other quotes that you have made that apply to you as well but I won’t waste the keystrokes needed to repeat them here.
I know I’m starting to sound like a broken record, but your calcs must be discarded. I have shown that a shift in the COM of the Earth not only involves the cores but also the densest part of the mantle and the outer core does not maintain its spherical shape which is part of the basis for your facile analysis.
We can exchange views on the other things you mention in your last post including
Galileo and the Catholic Church, the Beatles and John Lennon, Creationism and angular momentum in a future post, after we can agree on the magnitude of a change in ‘g’, per the request in your next-to-last post:
“If we can get past this, then we can go back to your other points.”
Laze
