Cold Core Model of Earth's Structure

[Pyrotex]

I'm sorry but in my calculations thay do not nicely go away as you claim. They get stronger the deeper you go into the earth. Do the math.

All you have to do is PLOT the strength of the force against the distance from the Earth's center, and show us the PLOT.

 

Your unwillingness to do this is strangely troubling.

 

Or is there some reason you do not wish to do this?

[Moontanman]

Is this cold core idea based in a desire to confirm biblical veracity? How cold are we talking about? So far all i see is a lot of disagreement on gravity vectors and other details that seem to be simply irrelevant to he idea of an Earth with a cold core. Why is a cold earth core necessary to explain what we currently see? Does the cold core theory explain anything better than the current "hot" core models? Does the cold core idea propose anything that cannot be explained by current the model? I would like to know why the cold core model is better, how it could be possible for a planet to accrete with a cold core, and if the core is indeed cold why is it hotter the further you go into the Earth? One more thing, how do you account for the mass and gravitational pull of the earth. The elements you propose as making up the core of the earth do not account for the mass of the earth or it's average density. The earth does not contain enough mass to make metallic hydrogen via gravitational pressure. The is no way to achieve metallic hydrogen on a body the size of the earth via accretion.

[stereologist]

Don't overlook the issue that radioactive decay adds a tremendous amount of heat to the interior. Rocks are poor heat conductors and the heat generated by radioactive decay remains inside the earth.

 

For all this, however, Marone says, the vast majority of the heat in Earth's interior—up to 90 percent—is fueled by the decaying of radioactive isotopes like Potassium 40, Uranium 238, 235, and Thorium 232 contained within the mantle.

This from Probing Question: What heats the earth's core?

[Pyrotex]

Yes, Mootanman is correct.

At the end of the day, we are still left with these facts:

The Earth's core is at least 5 to 10 times as dense as liquid or solid hydrogen could be.

The Earth's core has insufficient pressure to create metallic hydrogen.

The Laws of Thermodynamics require a temperature of at least 5000 K at the Earth's core.

All the irrelevant asides, like chemical bonding, and horizontal gravity, make no difference to the above.

There are no such things as an "elastic force" -- a force is just a force.

The mechanisms for the concentration of iron in the early formation of the solar system are not "fantastic" at all. Given the basic laws of physics, they are inevitable.

[modest]

It is similar to the pull produced by a rubber strand. My horizontal gravitational pull occurs in a 360 degree plane. It is similar to the restorative pull in the skin of a rubber balloon, which in a schematic pulls in both directions in a lateral plane. That is why you cannot feel its effect. But as one progresses into the earth the horizontal pull quickly becomes the dominant force.

 

:confused: What? Your "cold core" data has it as the dominant force from the surface to the core:

 

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

 

Starting at 10.0604 on the surface and ending at 14.9443 at earth's center.

 

This really seems like a very simple thing to disprove. If you hang two kilogram weights next to each other (a meter apart), will they indeed feel 10.0604 newtons of force between them as your data above claims? No they don't. Each mass is attracted to the earth's center at 9.8 kg•m/s^2. By simple vector analysis this means they should attract one another at [math]1.6 \times 10^{-8}[/math] Newtons. Experiment agrees with this while it does not agree with your numbers above.

 

Please tell me, if you drop two one-kilogram masses at sea level separated by one meter what will the acceleration between them be? You must recognize that the answer can't be 10.0604 m/s^2, so what possible meaning can your numbers above have?

 

By what method of experiment can we obtain the answer 10.0604 m/s^2? I'm sorry to be so unreserved here, but if you can't answer the sentence immediately preceding this one with a direct and straightforward answer then I'll be rather convinced that no physical meaning can be attached to what you're saying. :(

 

~modest

[Pyrotex]

I have to agree.

CharlieO and cold-co,

I've talked and chatted and worked and posted with many, many people over the years

who had an understanding of their work and what they did.

They could answer almost any question about their assumptions, their reasoning, their results.

Even if it was a field I never trained in, they were able to clear up all my misunderstandings.

 

CharlieO and cold-co,

you don't talk like them. you don't write like them.

You show very few signs of even understanding what it is you're doing,

and no ability to explain your results, other than just saying,

"this agrees with a cold core".

 

Very disappointing.

[Boerseun]

I don't want to sound like the chihuahua who grips the lion's neck because all the other dogs are in on the act, but just the following:

 

We've danced and parlayed around this topic, but the one question asked which remains to be answered, is simply how do you propose your model to explain the Earth's measurable and demonstrable density?

 

It doesn't. But I would like to see you attempt an answer, nonetheless.

[Pyrotex]

Since CharlieO doesn't have the time to Plot his data,

I have done it for him. The attachment shows what it looks like.

 

The X-axis is radius of the Earth, apparantly in km, with the center of the Earth at the left.

 

The five plots represent horizontal gravity (gv) and vertical gravity (gv) for the hot and cold core models; and gvaver which is the average of vertical gravity between the hot and cold models.

 

Since most (?) of the plots wind up around 10 at the Earth's surface (right edge), we may assume that the vertical scale of the plots is gravitational accelleration in meters*sec-2.

 

Let's take a look at the upper two plots, which are both horizontal gravity, one for cold, one for hot. Notice that they are not the same at the surface. Indeed, no matter what "horizontal gravity" is :hihi: you would expect them to be the same at the surface, yes? because the total mass of the Earth is "inside" the surface. I suspect this is an error.

 

Why should the hot core yield a horizontal gravity more than 3 times normal surface gravity, so near the core? Hello? I suspect another error. Can anyone give any sensible interpretation to these two curves???

 

But also notice that ghcold is 10 meters*sec-2 at the surface, numerically equal to the normal force of gravity. Hunh? This obviously makes no sense at all. Any horizontal force due to, say, the tidal force from interior shells of the Earth would be about 2 orders of magnitude less than vertical gravity. It is obvious that NO objects on the surface feel a force of this magnitude or we would all be whirling around, smashing into mountains.

 

Notice that the average of gvhot and ghhot is a straight line. What? We're talking the AVERAGE of the vertical and horizontal forces for the hot core model yield a straight line.

The actual pull of gravity (vertical) as you approach the core is NOT linear. This looks like some kind of artifact arising from a poorly designed model, not a real relationship. Any comments anybody?

 

[[ Can somebody please whip out a plot of the actual "real" force of gravity as you approach the core? One with constant density, and one with a linearly increasing density? ]]

 

CharlieO & co-cold, see what you can deduce from plotting your data?

[Cold-co]

Pyrotex:

Thank you for doing the plot. I only wish the site's software would allow me to look at it. Seems I haven't the required number of posts to either view your plots or make posts of plots on my own.

 

I think you are starting to understand the physics of the problem I have posed. If you trigonometrically calculate the vertical gravity vector as you progress deeper into an orb, you will end up with a horizontal gravitational pull. Since I view earth as a collection of gram masses interacting gravitationally with each other, then the existence of the horizontal pulls are a permanent, but latent force. They do not go away, only balance their pulls. On the surface of the orb you are being subject to these pulls, but they cannot be detected sensually, because they are similar to the air pressure on a calm day. You don't notice air pressure because it distributes itself equally around your body, but add air movement and your senses immediately detect the change. The presence of horizontal gravity is similar to the air around you, only when the earth bulges from rotationally induced movement of matter within earth does the strength of the horizontal pull of gravity need to be considered. That is why I believe the flattening equation needs another component. The lack of which has produced a moment of inertia that dictates the earth's core be composed of heavy materials; a flood theories followed. The inclusion of which allows the distribution of mass to agree with the speeds of seismic waves, just as Oldam predictated, density equals speed. The lack of this component forced Beno Gutenberg to conclude that seismic wave speeds in earth's core drastically drop their speed due to melting. I think one condensation-solidification model fits all planetary bodies.

[Pyrotex]

Pyrotex:

Thank you for doing the plot. I only wish the site's software would allow me to look at it. Seems I haven't the required number of posts to either view your plots or make posts of plots on my own....

You do not need software at this site to make Plots, or to view the plot I made.

 

You have more than enough posts at Hypo to do whatever any member can do.

 

The plot I made was done entirely in Excel. I just copied your data into Excel, split apart the numbers, and clicked on the Scatter Plot button. That's it.

 

Then I copied the plot and pasted into an ordinary M$ Word document. That' is what I attached to my post. All you have to do is double click on the document-name-link at the bottom of my post.

 

You DO have Word on your computer, don't you?

 

Do you have Excel?

[modest]

[[ Can somebody please whip out a plot of the actual "real" force of gravity as you approach the core? One with constant density, and one with a linearly increasing density? ]]

 

Of constant density, it should be linear:

-

source

where a is the radius of the earth and r(m) is the distance from center and E is m/s^2. But, I agree with you, where density increases with depth it would NOT be linear.

 

Of linearly increasing density... I can find no source so I'll attempt to work out a plot and show my thinking along the way. We can say the acceleration is always:

[math]g=\frac{GM}{r^2}[/math]

noting

[math]M=V \times D[/math]

we can solve for volume at any r:

[math]V_{®} = \frac{4}{3}\pi r^3[/math]

and account for a changing density with r:

[math]D_{®}=A- \frac{rB}{a}[/math]

where

• r = point under consideration
  
• a = radius of planet
  
• A = average density at r << a
  
• A-B = average density at r = a
  

This will mean density changes in such a way that the average density (for the whole planet) at earth's surface can be set and the average density of the very inner core can be set and the average density at r will change linearly between the two. This is not exactly the same as having the local density at r itself change linearly (with r), but is close enough for a rough plot (me thinks).

 

So then, mass is:

[math]M = VD = \left[ \frac{4}{3}\pi r^3 \right] \left[ A - \frac{rB}{a}\right][/math]

Which rearranges to:

[math]M = \frac{4 A \pi r^3}{3}-\frac{B 4 \pi r^4}{3a}[/math]

Multiply that by G and divide by r^2 and we should have a function of r giving gravitational acceleration with linearly increasing average density:

[math]\frac{GM}{r^2} = \frac{G 4 A \pi r}{3}-\frac{G B 4 \pi r^2}{3a}[/math]

It's already clear that the equation is quadratic and not linear (as I think we should expect). Subbing,

• G = 6.67428E-11 m^3/kg/s^2
  
• a = 6371000 m
  
• A = 14500 kg/m^3
  
• B = 9500 kg/m^3 (this A and B give us a average density changing from 14.5 g/cm^2 at the very inner core to 5.5 g/cm^3 for the whole planet)
  
• Pi = 3.14159
  

and that allows simplification...

[math]g_{®} = (4.0538 \times10^{-6}) r - (4.1688 \times 10^{-13}) r^2[/math]

//edit: where r is distance from the center in meters and g® is in meters per second squared

It plots as:

which I plotted at

this website

.

The right side is the surface and the left is the center and the red line is acceleration due to gravity. While this is very approximate, it should at least demonstrate that the curve is not linear but quadratic.

 

I've also created that plot as a spreadsheet, Pyro. It's attached.

 

~modest

[modest]

Pyrotex:

Thank you for doing the plot. I only wish the site's software would allow me to look at it. Seems I haven't the required number of posts to either view your plots or make posts of plots on my own.

 

Pyro's plot:

 

 

Like Pyrotex says, you really should get yourself a copy of Excel. I'd recommend it above all other software for what you're doing.

 

Have you given any thought to this question:

 

By what method of experiment can we obtain the answer 10.0604 m/s^2?

 

~modest

[Cold-co]

pyrotex etal

For the life of me, I cannot think of an experiment to verify my trigonometric findings. It is frustrating, because the horizontal forces described act in a 360 degree plane, only when you try to disturb that plane does the horizontal force show it ugly head. I can describe what it is like, such as comparing it to restorative force in the skin of a rubber baloon, but as for an experiment I'm wide open to any suggestions. However, it may be possible to fit a plyable sphere with strain gauges, then spin the sphere until it starts to flatten, then read the strains produced. Not having access to a lab, I'm stumped.

I do not have excel and that is a problem. At my age learning a new program would take some doing.

[modest]

pyrotex etal

For the life of me, I cannot think of an experiment to verify my trigonometric findings. It is frustrating, because the horizontal forces described act in a 360 degree plane, only when you try to disturb that plane does the horizontal force show it ugly head. I can describe what it is like, such as comparing it to restorative force in the skin of a rubber baloon, but as for an experiment I'm wide open to any suggestions.

 

If you draw two dots on the surface of an inflated balloon and use some method to try to force the spots together or farther apart you'll find a large force is necessary. This is not the case with a gravitational field. If you set two weights next to each other on the surface of the earth and try to push them toward each other or to increase the distance between them, you will find it does not require the force which you are claiming exists.

 

There actually is a force between the two of them because they are both being drawn to the center of the planet (if they both could freefall to earth's center, they would collide. You can solve for this force using trig. The angle (theta) between the line connecting the two weights and the line of earth's radius is arccos(.5/6378000) where the weights are 1 meter apart. The angel is 1.57079625 radians (89.9999956 degrees). Where each weight is forced in a direction of the radius at 9.8 m/s^2 each has a force toward the other weight of cos(theta)/9.8 which is 7.836 times 10^-9 m/s^2.

 

The force between two one-kilogram wights at sea level is therefore 1.567 times 10^-8 Newtons (twice the force of each weight). This is the proper way to solve what you are describing and it is nowhere near the values you came up with. Your value at earth's surface is 10.0604 where I believe it is 0.00000001567—and you can use experiment to confirm this. Two weights dropped from the earth's surface which are 1 meter apart in a vacuum don't accelerate toward each other in freefall at 10.0604 m/s/s, but rather 0.00000001567 m/s/s.

 

Do you see what I mean? Acceleration of gravity (g) is acceleration. It has physical meaning (change in velocity). If you drop something from sea level it will change velocity downward at 9.8 meters per second squared. To stop it from falling like that you must push it with an upward force of 9.8 Newtons (assuming it is a 1 kilogram mass). There is no such force of anywhere near that magnitude needed to keep masses at sea level from attracting toward one another. The force you are claiming exists can be shown with experiment not to exist.

 

~modest

[Moontanman]

Coldco, so far I have read all sort of esoteric reasons why you think the Earth has a hydrogen core. vertical gravity, horizontal gravity, among others but so far you have ignored the $64,000,000,0000 question. Where does the observed mass of the earth come from? All of your arguments are nonsense unless they explain the mass and gravitational pull of the earth at the surface. What happens as you go toward the center is meaningless unless it explains the mass of the earth first. I am not the only person to ask this and so far you have ignored the question. WHERE DOES THE OBSERVED MASS OF THE EARTH COME FROM? A metallic hydrogen core is totally impossible under the current observed facts of the mass of the Earth. Can you explain this under your cold core model?

[HydrogenBond]

Abstract. Improvements in our knowledge of the absolute value of the Newtonian gravitational constant, G, have come very slowly over the years. Most other constants of nature are known (and some even predictable) to parts per billion, or parts per million at worst. However, G stands mysteriously alone, its history being that of a quantity which is extremely difficult to measure and which remains virtually isolated from the theoretical structure of the rest of physics. Several attempts aimed at changing this situation are now underway, but the most recent experimental results have once again produced conflicting values of G and, in spite of some progress and much interest, there remains to date no universally accepted way of predicting its absolute value.

 

The Newtonian gravitational constant: recent measurements and related studies

 

Because there is no solid theoretical framework to get consistent numbers for G, the calculated mass of the earth is only as good as the assumptions we wish to make for G. Changing G would lead to the need to change what we assume for all situations. If you change the earth's composition, the sun and planets also would have to change. I am not saying this is not possible, but only that much work will be avoided even if one has good arguments for the earth. But G is sort of soft and not yet cast into stone at the theoretical level.

 

We still use the basic number for G generated by Cavendish in 1798. One might think this was only a good starting point. But based on that ancient science, we fit the earth to make it work out. That in turn, fits the suns, stars and galaxies, etc. and underlies all the assumptions for formation and internal compositions.

[Pyrotex]

...Because there is no solid theoretical framework to get consistent numbers for G, the calculated mass of the earth is only as good as the assumptions we wish to make for G. Changing G would lead to the need to change what we assume for all situations. If you change the earth's composition, the sun and planets also would have to change.....

Gosh, HB,

usually I can find something, even if only one sentence, in your posts that I can agree with.

But not this time. :naughty: :steering: :doh: :smart: