# Cold Core Model of Earth's Structure

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### #1 modest

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Posted 02 June 2009 - 03:26 PM

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

### #2 stereologist

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Posted 02 June 2009 - 05:52 PM

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.

### #3 Cold-co

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Posted 02 June 2009 - 08:41 PM

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.

### #4 modest

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Posted 03 June 2009 - 12:01 PM

...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....2/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

### #5 stereologist

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Posted 03 June 2009 - 01:03 PM

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.

### #6 Cold-co

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Posted 03 June 2009 - 04:24 PM

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.

### #7 CharlieO

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Posted 03 June 2009 - 11:44 PM

http://www.cox.net/n...nson3/part0.ppt

### #8 Pyrotex

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Posted 04 June 2009 - 08:35 AM

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

### #9 Cold-co

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Posted 04 June 2009 - 05:05 PM

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.

### #10 stereologist

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Posted 04 June 2009 - 05:22 PM

No answers can be made unless you post something. Try using the LaTex offered by this forum. It's not that hard to use.

I for one have no idea what you mean by horizontal and vertical gravity.

### #11 CharlieO

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Posted 04 June 2009 - 05:53 PM

Re: Cold-co power point http://members.cox.n...nson3/part0.ppt

### #12 modest

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Posted 04 June 2009 - 05:57 PM

My question for this forum is, “Have I done something wrong mathematically?”

Without seeing your math it seems rather impossible to answer that question. If you'd like members to analyze your model then you should go to the Alternative theories forum and start a thread on its topic. I'd suggest showing how you calculate this "horizontal acceleration".

~modest

### #13 Cold-co

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Posted 04 June 2009 - 07:32 PM

Hi All:
Here are the results of my calculations.
Radius gv cold gh cold gv hot gh hot gv aver gh aver
6371 9.8331 10.0604 9.9307 8.0385 9.8253 9.7883
6370 9.8346 10.0715 9.8327 8.0434 9.8263 9.7998
6365 9.8381 10.1121 9.8399 8.0735 9.8306 9.8574
6359 9.8475 10.1943 9.8525 8.1349 9.8343 9.9264
6291 9.8625 10.9073 9.9326 8.6772 9.8173 10.6533
6116 9.5896 12.3340 9.8759 9.7748 9.3648 11.9878
6011 9.5305 13.0232 9.9336 10.2814 9.3448 11.9978
5961 9.5833 13.3707 10.0280 10.5344 9.2707 12.7471
5721 8.9975 14.9015 9.9450 12.6066 8.7429 13.8600
5671 9.0731 15.1980 10.6211 12.9735 8.8248 14.0673
5371 8.1993 16.4861 9.9468 13.2172 8.2447 15.0086
4871 6.8711 18.2949 9.9203 15.2165 7.5047 16.4868
4371 5.2188 19.6469 9.9668 17.3052 6.7336 17.8049
3871 3.1122 20.1582 10.2350 19.7396 5.9698 18.9608
3485 1.0304 19.0349 10.6725 22.3436 5.3613 19.6641
2900 0.8926 16.8419 9.2710 25.9242 4.4675 20.3327
2300 0.7240 15.9334 7.6043 28.8435 3.5435 21.0659
1700 0.5570 15.4330 5.7847 30.8358 2.6197 21.5241
1217 0.4372 15.2446 4.2219 32.2407 1.8750 21.8904
700 0.2639 15.1213 2.4838 32.8782 1.0785 21.9731
0 0.0000 14.9443 0.0000 32.0258 0.0000 21.5092

Gravitational accelerations (m/secsec) for the three models of Earth’s interior

### #14 Boerseun

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Posted 04 June 2009 - 11:22 PM

I suppose what Cold-co is saying with vertical and horizontal gravity, is the following:

Imagine there is a shaft going all the way to the center of the Earth.

You hop into an elevator going down the shaft.

For every kilometer you descend, there's a kilometer's worth of mass overhead, and one less kilometer's worth of mass between you and the Earth.

In other words, you will experience gravity from mass distributed all around you - below you, overhead, etc. And the further you go down, the "less" you will weigh as you are now being attracted upwards as well as downwards. You will always feel a positive pull towards the center, because there will always be more mass there, but the resultant gravitational pull you'll feel will become less and less until you reach the very center of the Earth, where there is as much mass in all directions - in which case you will be perfectly weightless.

And the question he is asking, is what would this do to the classic "iron-core under immense pressure" model? Isn't the seismic interface between the inner and outer core the area of densest pressure, rather than the center, and isn't the pressure gradient there responsible for the seismic interface, rather than composition?

Have I got that right, Cold-co?

### #15 modest

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Posted 05 June 2009 - 09:44 AM

Yeah, I'd also guess Cold-co is saying something along those lines. Cold-co, I don't see how these numbers make sense:

Radius       gv cold         gh cold     gv hot       gh hot     gv aver       gh aver
6371	9.8331	[B]10.0604[/B]	  9.9307	  [B]8.0385[/B]	9.8253	  [B]9.7883[/B]

I'm at about r=6371 km right now and I don't feel 10 m/s^2 "horizontal acceleration". The only acceleration I feel is downward. I would probably guess that you don't mean acceleration in the usual sense.

I don't know how you found those numbers, but I'll show you how I would do it. Consider a spot at the inner / outer core boundary r = 1,220 km (1220000 m). The mass beneath this spot is roughly $1.01 \times 10^{23}$ kg and the mass above it (at larger values of r) is roughly 60 times greater. The mass above can safely be ignored by the two proofs I gave the other day and the mass below follows:

$F = \frac{GM}{r^2}$

$F = \frac{(6.67 \times 10^{-11} \ m^3 \ kg \ s^{-2})(1.01 \times 10^{23} \ kg)}{(1220000 \ m)^2} = 4.53 \ m/s/s$

The gravitational acceleration at r= 1220 km (the inner / outer core boundary) is roughly 4.53 m/s/s toward the center. To calculate the pressure at this point we would need to consider the weight of the material above it.

Would you object, Cold-co, if I split this discussion about your model into it's own thread perhaps titled "Cold Core Model of Earth's Structure"?

~modest

### #16 Pyrotex

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Posted 05 June 2009 - 10:14 AM

I suppose what Cold-co is saying with vertical and horizontal gravity, is the following:...For every kilometer you descend, there's a kilometer's worth of mass overhead, and one less kilometer's worth of mass between you and the Earth. ...you will experience gravity from mass distributed all around you - below you, overhead, etc. And the further you go down, the "less" you will weigh as you are now being attracted upwards as well as downwards. ...

In an earlier post, I showed how to approach this problem.

Take a "shell" of the Earth. This shell will be a hollow sphere of some constant thickness. You may assume the outer surface of the shell is the surface of the Earth. We can relate this shell to the problem above by saying that the thickness of the shell is the same as the depth you have descended to in your vertical tunnel.

This shell, therefore, will contain ALL of the mass of the Earth that is "above" your depth. If you have descended 500 km down from the surface, then this shell will be 500 km thick.

Now, take any arbitrary point, P, anywhere on the inside of the hollow shell. Take any arbitrary point, s, that is actually in the mass of the shell itself. The point s can be specified by its radius from the Earth's center, r; the longitudinal angle, phi, from some given zero longitude; and the latitudinal angle, theta, measured from the equator plane of the Earth, with positive theta being the Northern hemisphere and negative theta being the Southern.

Calculate the force of gravity on YOU (at point P), from a tiny element of mass at point s.

There's nothing really complicated so far, we're just using Newton's law of gravity. The tricky part is the distance between the tiny mass at point s, and YOU at point P, must be expressed in terms of r(s), phi(s), theta(s), and also r(P), phi(P) and theta(P). We will use standard Polar coordinate system.

Now we use integral calculus. We assume constant density in the shell. This is not a problem, as you can always divide your shell into hundreds of really thin, thin sub-shells and give them each their own unique (but still constant) density.

We integrate over theta(s) from -90 degrees to +90 degrees.
We integrate over phi(s) from 0 degrees to 360 degrees.
We integrate over r(s) from r = (inner radius of the shell) to r = (outer radius of the shell)

We have now accumulated (integrated) ALL the gravitational forces on YOU at point P, from ALL the tiny masses from ALL points s, within the ENTIRE shell.

The sum total gravitational force is zero. (0) Nothing. None Nada Zipity-doo-da.

It does not matter where you choose your original point P, as long as it is inside the hollow of the shell.

This makes the whole problem SOOOOO much easier.
You can always ignore ALL the mass of the Earth at ALL depths less than YOUR depth.
That is, you can always ignore ALL the mass of the Earth ABOVE you.
Because it ALL cancels out, ALL the time, at ANY depth you like.
And this includes ALL gravitational forces, including those that are horizontal, vertical or anywhere in between.

The ONLY gravitational force you can possibly feel, will be from the sphere of the Earth at or below your depth. As you descend, that remaining sphere will get smaller of course, and will vanish as you reach the center of the Earth.

{{{ In my opinion, this is the most fascinating and awesome use of Integral Calculus in all of advanced freshman year physics. }}}

### #17 Cold-co

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Posted 05 June 2009 - 02:19 PM

To All:
When I was taking integral calculus in 1960 the prof said you can always check your answer with trigonometry. That statement is what prompted me to run a check on the gravitational forces within the earth. A few years back I was recovering from a heart bypass and had plenty of time to think about the problem and devised a way to trigonometrically calculate the forces (accellerations) of gravity from Newton's original proof that all mass can be considered to be located at the center, which makes it a point mass. Now, I could have easily looked at my schematic of forces diagram and concluded that horizontal vectors cancel out, as they do when using calculus, because calculus employes directional vectors; but directional vectors is not the nature of gravity, it is a bidirectional force and a bidirectional force cannot be cancelled by another bidirectional force, only matched in strength.
Charlie O posted a link to my PowerPoint presentation, which explains the reasoning behind my concern.
Modest: I know you would like to be rid of Charlie O and me, but we are talking earth science here. so I think this is where we belong.