Cold Core Model of Earth's Structure

[Cold-co]

Moderation Note: The following 18 posts were moved from the thread "Belief in Earth's Iron Core still puzzling" in favor of having their own topic of discussion here—hopefully allowing for more-useful and more-specific responses.

A MATTER OF GRAVITY

Some two hundred years ago, scientists longed to break free from Church control. To break free, they needed to unseat the Church approved, cold-core cross section that had been taught for over 5000 years. The sleight of hand they devised was so well disguised even the Jesuits, who the Pope directed to derail their efforts, could find no fault in their logic. In time, the Jesuits came to teach the scientist’s view. Today, we perpetuate the scientist’s sleight of hand every time we teach gravitational forces at work within Earth.
Gravity is a bidirectional (elastic) force—Earth pulls you with the same amount of force that you pull the Earth. But since Earth is so much larger than any freely moving body, on or above her surface, we treat her gravity as a directional force. This makes the force of gravity relatively simple, so we teach gravity before we teach elasticity. But in reality, elastic cohesion (the drawing together of particles) is the force identified by Newton in his law of universal gravitation, “Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”
In addition to our less than accurate treatment of gravity, seismic wave data show Earth’s upper shells to be solids down to her core. Yet, we treat her as a large liquid drop to calculate her moment of inertia from her rotationally induced flattening (f). Since she is thought to at least act like a liquid, her flattening is believed to be held on check solely by the equatorial acceleration of gravity (ge); hence, our flattening equation becomes f = 1.5(C-A/Ma2) + 0.5 rate of rotation squared times Earth's radius (a)/ge). In so doing, we conclude she has a low moment of inertia. In turn, she must have a molten interior to allow heavier particles to sink deep into her core to achieve the low moment of inertia demanded by our flattening equation. However, our hot-core model is valid only if we ignore horizontal elastic cohesion.
Elastic cohesion in a solid imparts a constant pull between all its parts. But, no one ever bothered to calculate the strength of that pull in Earth’s shells, because in a schematic of forces diagram the improper use of directional forces (the scientist’s sleight of hand) makes them appear to cancel out; but a gravitational pull cannot cancel out another gravitational pull—only balance. Their pulls are still present.
Now, if we treat Earth’s outer shell as a structurally sound, hollow sphere, subject to horizontal elastic cohesion; then the mass movement, created by her rotation, must also overcome that shell’s elastic cohesion before she will flatten. Trigonometric calculations of Earth’s gravitational forces show horizontal cohesion in her outermost shell to be an acceleration of equal value to the acceleration of vertical gravity on her surface, thus our flattening equation needs another component, +0.5 rate of rotation squared times earth's radius/gh. Or, since this acceleration is of equal value to the acceleration of vertical gravity, that vertical equatorial acceleration can be doubled to obtain Earth’s moment of inertia. When doubled, our flattening equation yields a moment of inertia equal to the summation of moments of inertia mathematically derived for the individual shells of a condensed, cold-core model, whose shell densities are proportional to the speeds of seismic waves passed through them. Serendipity!!!
The squeeze afforded by horizontal elastic cohesion, gives us a unique way to look at Earth’s mechanics—one of a contracting pressure vessel driven by an ever increasing packing pressure provided by horizontal cohesion. A pressure vessel capable of producing a natural heat-pumping cycle—like the heat-pumping cycle employed in diamond anvil devices used to determine the physical characteristics of solid hydrogen. Just as the test sample in a diamond anvil gives up heat to move to a denser phase, so too does the hydrogen crystal in Earth’s pressure vessel give up heat—heat that shows up as geothermal energy in or on her surface.

[modest]

Originally Posted by Cold-co

Earth pulls you with the same amount of force that you pull the Earth. But since Earth is so much larger than any freely moving body, on or above her surface, we treat her gravity as a directional force. This makes the force of gravity relatively simple, so we teach gravity before we teach elasticity. But in reality, elastic cohesion (the drawing together of particles) is the force identified by Newton in his law of universal gravitation, “Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”

<...>

However, our hot-core model is valid only if we ignore horizontal elastic cohesion.
Elastic cohesion in a solid imparts a constant pull between all its parts. But, no one ever bothered to calculate the strength of that pull in Earth’s shells, because in a schematic of forces diagram the improper use of directional forces (the scientist’s sleight of hand) makes them appear to cancel out; but a gravitational pull cannot cancel out another gravitational pull—only balance. Their pulls are still present.
Now, if we treat Earth’s outer shell as a structurally sound, hollow sphere, subject to horizontal elastic cohesion

Hi Cold Co, welcome to Hypography.

Would you agree that the gravitational potential inside a spherical shell of matter is constant meaning there is no gravitational force inside a spherical shell?

Would you also agree that an object outside a spherical shell is drawn toward the center of the shell with a force equal to a situation where the mass of the shell were replaced with an equal mass at a point in the center of the shell?

I ask because it sounds like you are disagreeing with this kind of vector analysis—believing that an element in the earth is affected gravitationally in a net way by the mass above it and next to it. I would certainly disagree with this, but before showing why I'd like to know if you are indeed saying that.

~modest

[stereologist]

Cold-co most of what you have written sounds like "Cold-co sleight of hand".

Today, we perpetuate the scientist’s sleight of hand every time we teach gravitational forces at work within Earth.

Sorry to hear you were not taught basic physics properly. Your experiences are probably not the norm.

What you wrote does not make sense especially the last half. I'm looking forward to your responses to Modest's questions who has clearly expressed my thoughts.

[Cold-co]

Modest:
Would you agree that the gravitational potential inside a spherical shell of matter is constant meaning there is no gravitational force inside a spherical shell?

Would you also agree that an object outside a spherical shell is drawn toward the center of the shell with a force equal to a situation where the mass of the shell were replaced with an equal mass at a point in the center of the shell?

The first part, my answer is no. According to a trigonometric analysis, the force of vertical gravity diminishes as one progresses into the earth, but the force of horizontal gravity increases. My trigonometric analysis of a hot-core model agrees with the analysis done by Dziewonski (Harvard). I get the same results he gets. In an orb of constant density the relationship is: the absolute value of the force of vertical gravity plus the absolute value of force of horizontal gravity at any level within the orb is equal to twice the value of vertical gravity on the surface of the orb.

The second part, I have no problem with Newton's proof. The whole mass of an orb can be considered to be located at its center for mathematical purposes.

[modest]

Originally Posted by Cold-co

...The first part, my answer is no. According to a trigonometric analysis, the force of vertical gravity diminishes as one progresses into the earth, but the force of horizontal gravity increases. My trigonometric analysis of a hot-core model agrees with the analysis done by Dziewonski (Harvard). I get the same results he gets. In an orb of constant density the relationship is: the absolute value of the force of vertical gravity plus the absolute value of force of horizontal gravity at any level within the orb is equal to twice the value of vertical gravity on the surface of the orb.

The second part, I have no problem with Newton's proof. The whole mass of an orb can be considered to be located at its center for mathematical purposes.

I looked up the fella at Harvard you're referring to, and I see nothing of any weird horizontal or vertical gravity ideas.

I don't have much time today, so instead of trying to construct a proof of the spherical shell thing, I'll just quote a couple sources that have proofs. The idea, I will reiterate, is that an object in a hollow spherical shell of matter feels no gravitation force in any direction regardless of where the object is inside the shell.

This means you can pick a spot in the interior of the earth and imagine it splitting the earth into two pieces. The first piece being the solid sphere that exists below the point. If the point is 2,000 km from the center then that would be a sphere of 2,000 km. The second piece is the tick shell of matter above the point. Basically, everything that is further than the point under consideration from the center of the object.

In asking how much gravity this point feels we can dissect the earth like this and disregard the hollow shell part above it and treat the part below as a point mass, because (as the proofs below show) the part above has no net effect on the point gravitationally, and, as you agree, the part below is well-represented as a point mass. That's not to say there are no pressure effects from the matter above it—there certainly are, but no contribution as far as the force of gravity.

So... sources (I'm finding a bunch all over the internet)...

Gravitational field due to rigid bodies:

... Gravitational field due to thin spherical shell...

Case 2 : The point “P” lies inside the shell. The total gravitational field is obtained by integrating the integral from x = a-r to x = a+r,...

This is yet another important result, which has been used to determine gravitational acceleration below the surface of Earth. The mass residing outside the sphere drawn to include the point below Earth’s surface, does not contribute to gravitational force at that point.

The mass outside the sphere is considered to be composed of infinite numbers of thin shells. The point within the Earth lies inside these larger shells. As gravitational intensity is zero within a shell, the outer shells do not contribute to the gravitational force on the particle at that point.

A plot, showing the gravitational field strength, is shown here for regions both inside and outside spherical shell :

Gravitational field due to rigid bodies

I cut out all the pertinent math above, but the conclusions are what matters I suppose. Here's another:

Field Inside a Spherical Shell:



This turns out to be surprisingly simple! We imagine the shell to be very thin, with a mass density p kg per square meter of surface. Begin by drawing a two-way cone radiating out from the point P, so that it includes two small areas of the shell on opposite sides: these two areas will exert gravitational attraction on a mass at P in opposite directions. It turns out that they exactly cancel.

This is because the ratio of the areas A1 and A2 at distances r1 and r2 are given by : A1/A2=r1^2/r2^2 since the cones have the same angle, if one cone has twice the height of the other, its base will have twice the diameter, and therefore four times the area. Since the masses of the bits of the shell are proportional to the areas, the ratio of the masses of the cone bases is also r1^2/r2^2. But the gravitational attraction at P from these masses goes as Gm/r^2, and that r^2 term cancels the one in the areas, so the two opposite areas have equal and opposite gravitational forces at P.

In fact, the gravitational pull from every small part of the shell is balanced by a part on the opposite side—you just have to construct a lot of cones going through P to see this. (There is one slightly tricky point—the line from P to the sphere’s surface will in general cut the surface at an angle. However, it will cut the opposite bit of sphere at the same angle, because any line passing through a sphere hits the two surfaces at the same angle, so the effects balance, and the base areas of the two opposite small cones are still in the ratio of the squares of the distances r1, r2.)

http://galileo.phys.virginia.edu/cla.../GravField.htm

You can also use General Relativity to prove that the spacetime in a hollow sphere is flat, thus no field strength.

If you want to discuss this in more detail then I suggest we make a thread dedicated to your idea. Otherwise, unless you can find some source agreeing with your conclusions, I don't know what you mean to accomplish in this thread.

~modest

[stereologist]

I also took a look at the website at Harvard and what I saw was the standard earth model. It was interesting to see that the location and structures of plates was being inferred from seismological information.

[Cold-co]

Modest:
I'm sorry I misled you. Dzeiwonski is the professor who calculated the strength of vertical gravity at descending depths within the hot-core model. His work is found in most texts on the subject. He did not work with horizontal gravity because he thought, like everyone else thought, horizontal components cancel out. My hot-core model's trigonometric results for vertical gravity produce equal results to those found by Dzeiwonski. The only reason I came up with the horizontal component is because of the trigonometric calculations I was performing on the standard hot-core model. They left me with a horizontal component whose force is like that of the restoritive force in the skin of a rubber balloon.
The references you have supplied work strictly with directional vectors of vertical gravity, they too ignore the horizontal component.
If you would like to take a look at the logic I used in working with Newton's gravity, you will find it at, a link that I am not as yet allowed by forum rules to post maybe Charlie O can provide it, just follow the thread through to see what a cold-core model looks like. Please excuse the commercials.

[CharlieO]

Cold-co asked me to submit the following for your evaluation.

http://www.cox.net/nchristianson3/part0.ppt

[Pyrotex]

My browser gives me a "page does not exist" error for that link, Charlie,
and an "HTTP 400 Bad Request".

[Cold-co]

To All:
Thank you for your responses, but I feel you have missed the point I made in my initial post. If there are other models of earth that can be considered then they must meet the moment of inertia found by the flattening equation. The low moment of inertia derived from that equation dictates most of earth's mass be located in her inner core. That is why I started looking into the components of the equation itself.
To analysis the packing accelerations at work within the earth, I used three different models—cold-core, hot-core, and average density. Each model uses the same eighteen divisions of seismically known shells: crust, lithosphere, asthenosphere, 1st bonded shell, 1st transition (phase change), 2nd bonded shell, 2nd transition, five divisions of the 3rd bonded shell, four divisions of the outer core and two divisions of the inner core. Except for the average model, which has the same density for each of its shells, density is proportional to seismic wave speeds in the cold-core model and, as required for the standard hot-core model, density is concentrated in the core. All models have a radius of 6371 km and all have the same total mass. See physical characteristics shown below.
To fill in a mental picture of what goes on with respect to gravitational accelerations, I used an adaptation of Newton’s model of Thin Spherical Shells. The model he used to prove gravitational forces acting on a small body (gram-mass) external to earth’s surface can be considered to be located at earth’s center. That model effectively rotates the total mass of an annulus (ring) around to a single point where the gravitational accelerations merge into a single acceleration. This merged acceleration can then be broken into two accelerations; a vertical acceleration (v) and a horizontal acceleration (h). Then, using trigonometric functions the values for h and v can be determined.
Just as Newton did, I set up my model’s eighteen separate divisions as individual spherical shells of zero thickness. Ninety annulus-masses for a selected shell-radius rotate around to concentrate at odd (1, 3, 5 ... 177, 179) degree points. After creating spreadsheets for each shell, I used a series of trigonometric functions to solve for horizontal, as well as vertical gravitational accelerations. By moving the radius at which the gram-mass is located and employing an iterative process, I solved for the vertical and horizontal gravitational accelerations produced by each individual division. Resultant gravitational accelerations for the radius selected for the location of the gram mass are shown below. Values for vertical accelerations in my hot-core model match well with values obtained by Dziewonski. This increased my confidence that my trigonometric approach is equivalent to his way of calculating vertical gravity for various levels within the earth.
My question for this forum is, “Have I done something wrong mathematically?”
I'm sorry but I cannot figure out how to insert the results of my calculations.