Cold Core Model of Earth's Structure

[Cold-co]

Modest:

I like your toy model, but to make it work trigonometrically you have to slice each orb into rings (annuluses) across the vertical axis you have shown. Then rotate the mass of each ring around the vertical axis to a chosen circumference. Now you have the masses of the rings in a position where you can deal with them trigonometrically. I chose to use 18 separate shells of different densities. Hence, I needed to have 18 different spreadsheets. I know the concept is hard to grasp, but it is the same concept that Newton used to prove all of an orb's mass can be considered to be located at the orb's center.

In column G the matrices are set for a gram mass on earth's surface. When you want to change its location to a new radius just highlight column G and do a remove and replace operation for the radius at which you want to find the gravitational forces. The trigonometrics will take care of the rest. It is a tedious itterative operation.

[stereologist]

Trimming a volume is an excellent way to handle points. What sometimes happens is that people use spherical coordinates to lay out shells of points. Just wanted to be sure that your huge calculation was good to go.

 

PS I haven't used MUMPS since I did a job back in 1987.

[Cold-co]

Modest etal:

There is a powerpoint presentation that shows how Newton did his trigonometric proof. It is located at http://members.cox.net/nchristiansonc/part2.ppt If you can spare a few minutes to view it I think you will get an understanding of how I did my calaulations.

Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

[modest]

Here’re the accelerations, in gs, for depths of 0, 0.1, 0.2 ... 0.9 radii, and depth-mass data from various sources...

 

(Modest’s “5,4,3,2” toy model)

1.0000 0.9633 0.9513 0.9108 0.8399 0.8070 0.6518 0.5197 0.3880 0.2060

 

If a g is 9.8 m/s² then this can't be right. The total mass of the model is 6.87x10²³ kg. At a distance of 4,000,000 m that's F=(6.87x10²³)(6.67x10^-11[)/(4x10⁶)² =~ 2.86 m/s². Do you, perhaps, mean for "1.0000 g" to represent surface gravity at any numerical value? If that's the case then let's see, the percentage acceleration at depth .5 would be... (2.30646 m/s²)/(2.86561 m/s²) = 0.8049 and yours has 0.8070 at a depth of 0.5. Nevermind, that's what you did.

 

Please ignore this post :shrug:

 

~modest

 

PS Excellent work Craig. The program is so compact and clever as to border on ridiculous. You always set the bar and there's never enough rep to go around. :shrug:

[modest]

Modest:

I like your toy model, but to make it work trigonometrically you have to slice each orb...

 

There are no orbs in the model as it stands now. I cannot do your method from a vague 3 sentence description. If you are unable to take a step forward (perhaps do the next step yourself) then I have no hope in moving forward on this. Sorry.

 

~modest

[Cold-co]

Modest:

As I understood your model, each circle represented an orb of constant density. By what you are now posting I really have no idea what your model represents. I can only recommend that you at least take a look at the site I recommended 22 hours ago.

[Cold-co]

Modest:

The attached shows how the matrices were derived and manipulated.

[modest]

Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

 

I don't know about pillorying, but it does look like you've gone wrong somehow. I once again stipulate that I'm not endorsing your method for finding pressure with the following post, just critiquing the math...

 

Here are the differences between what I get and what you got for the crust layer of the cold core model:

 

               yours              mine
               ------             ------
R               6.371x10[sup]8[/sup]          6.371x10[sup]8[/sup]
r               6.366x10[sup]8[/sup]          6.366x10[sup]8[/sup]
Density         3.8                3.8
Mass            0.0968x10[sup]26[/sup]        0.165x10[sup]26[/sup]  
Area            0.48x10[sup]16[/sup]          0.2001x10[sup]16[/sup]
Distance (CG)   6.37x10[sup]8[/sup]           3.184x10[sup]8[/sup]
Force           0.0007x10[sup]28[/sup]        0.0003858x10[sup]28[/sup]
Pressure (KBar) 1.5                1.828

And here's a comparison of pressures in KBar:

yours   mine
------  ------
2.33    1.928
45.4    47.395
94.0    103.534
59.7    87.732
35.0    35.270
202     236.445
46.7    66.234
304     417.329
492     407.908
482     390.400
480     367.730
352     377.618
8.77    7.842
7.59    6.440
5.98    5.035
3.56    3.616
3.21    25.139
1.48    10.874

 

I can't really spot why our numbers differ. Where the only problem I had with the math in your previous rendition was a conversion of units where it was off by a power of 10, the problem now seems to be more chaotic. I'll attach the excel spreadsheet I used in case you got excel and here is an image of the data:

 

http://i227.photobucket.com/albums/dd216/modest4U/coldpressure.png

 

Notice in Excel "E" means "x 10 to the power of" so that "5.00E12" is shorthand for 5.00x10¹².

 

I'll also point out one other thing which didn't seem to cause problems, but might in the future. The formula you gave for the center of mass of a half shell is correct except that you wrote it in a way that is missing a necessary set of parentheses. You gave:

[math]3/8(R^4-r^4)/R^3-r^3[/math]

which really would give the wrong answer if followed as shown since division comes before subtraction. It should be:

[math]3/8(R^4-r^4)/(R^3-r^3)[/math]

or, you could write it:

[math]\frac{3(R^4-r^4)}{8(R^3-r^3)}[/math]

 

If you want to show your calculations on one of the layers, I could show mine and we can see where they differ. How, for example, did you find the mass and the surface area for the crust layer? If you write out the calculation complete with numbers we can compare notes.

 

~modest

[Pyrotex]

Modest etal:

There is a powerpoint presentation that shows how Newton did his trigonometric proof. It is located at http://members.cox.net/nchristiansonc/part2.ppt If you can spare a few minutes to view it I think you will get an understanding of how I did my calaulations.

Also I'm attaching an update of the pressures in the earth. I'm prepared to be pilloried again.

I read the document you linked to.

 

First, that was NOT the way Newton did his calculations.

 

Newton used real forces, which were simple pushes or pulls on a point of mass.

 

Replacing the arrow of force in a diagram with two arrows, heads together, makes no sense and the author gives no justification for this at all.

 

There is no such thing as an "elastic force". If you really know what the words "force" and "elastic" are, then the concept of "elastic force" makes no sense at all.

 

Elastic applies to material bodies. It means that if a body is deformed by any combination of forces, then when those forces are removed, the deformed body will resume its original shape. That is what "elastic" means.

 

How can a force be elastic? Can you deform a force with other forces? No. It makes no sense.

 

Furthermore, even if you accept (or overlook) that idea, it becomes obvious from his diagrams that the author is doing vectors completely wrong. He has forces pulling on an annulus ring outward from all sides. He ignores the forces that would be pulling inward. He essentially is selecting from ALL the possible forces (real ones) only those forces he wants and ignores the others. He confuses force with strain. He confuses force with pressure.

 

Sorry, cold-co, but it looks bogus all the way down.

[stereologist]

I did not bother to look at the PPT until Pyrotex commented.

 

This is a bogus collection of whatever.

 

The title page is the warning. The force formula states that the gravitational force between two objects is a function of the masses of the objects and the distance between them. To state that "A STRANGE PHENOMENA: WE PULL EARTH WITH THE SAME AMOUNT OF FORCE THAT EARTH PULLS US." is a misleading statement to begin a presentation.

 

Wouldn't it be odd if the forces were not the same? Plug this force into F=ma and we see that a=F/m causes us to appear to fall and the earth to be motionless.

 

To call this formula strange suggests that the author does not understand the material they are about to present. This is material that has been reviewed and used for hundreds of years and even put spaceships within a few miles of their target positions billions of kilometers away.

 

Forget how ridiculous the term "elastic force" is. Also, forget how odd the vector diagrams are. Realize that the force is an attractive force along the line between the masses involved. We're dead in the water if we stop at each gross mistake.

 

On the next page is a diagram showing a force labeled Fg. I assume that this was meant to be a subscript. Call the force F. For unknown reasons, there are no explanatory notes, the force F is resolved into a force h and v, I suppose for horizontal and vertical.

 

In the lower right corner there is no explanation of the "horizontal forces acting on an annulus section". There is no similar diagram for the vertical forces acting on an annulus ring. It is the combination of the forces that are the force F on the annulus ring.

 

On slide 4 is a totally wrong formula. First, I do not see s defined. Second, this is not an equality, rather an inequality. Since slide 4 has a completely bogus formula in it, then its use on slide 6 results in useless malarky.

 

Slide 5 is baloney as well.

[Cold-co]

Modest:

In going over your spreadsheet, I notice you only use the distance from the hemisphere's cg to its base. For gravitational purposes you need to double that figure. I tried to download your work but for some reason the mathematical functions would not transfer to my system. Maybe the two systems are of two different versions.

 

Pyrotex:

I guess I'll just have to write to John E. Prussing and Bruce A Conway, the authors of, "Orbital Mechanics," 1993 Oxford University Press and tell them they don't know how Newton did his calculations, because Pyrotex and Stereologist say they don't. Even though neither provided an mathematics to back up their claim. I'm beginning to get the feeling here, that if it isn't in the textbook your are currently familiar with, it must be untrue.

[modest]

I notice you only use the distance from the hemisphere's cg to its base. For gravitational purposes you need to double that figure.

 

It's doubled in the formula for calculating force G(m/2)²/(2D)².

 

I'll start with a single example since you don't seem willing to show any of your calculations.

 

The area of a ring with inside radius r = 6.366 x 10⁸ cm and outside radius R = 6.371 x 10⁸ cm is:

[math]\pi(R^2-r^2) = 3.14159 \left[ (6.371 \times 10^8)^2 - (6.366 \times 10^8)^2 \right] = 2.001 \times 10^{15} \ cm^2[/math]

That is the "Area" I calculate for the crust layer. It should be 2.001 x 10¹⁵ You show 4.8 x 10¹⁵. All the numbers throughout your spreadsheet seem similarly problematic.

 

~modest

[Pyrotex]

...I guess I'll just have to write to John E. Prussing and Bruce A Conway, the authors of, "Orbital Mechanics," 1993 Oxford University Press and tell them they don't know how Newton did his calculations,....

Newton did his calculations with geometry.

All of them.

Pure geometry.

No trigonometry.

Then he used goemetry to create The Calculus.

Then he used calculus to verify his calculations.

 

You go ahead and write to Prussing and Conway.

I dare you.

Because the BS you are posting did NOT come from anything they wrote or that Oxford University Press printed.

 

Go ahead. Let's see copies of entire pages out of their book with those silly bogus diagrams. I'm waiting.

[Pyrotex]

...I'll start with a single example since you don't seem willing to show any of your calculations....

Did it occur to you that cold-co doesn't show us his calculations, because he doesn't have any. His mastery of math may not even be at the algebra level. I would be seriously surprised if it were.

[modest]

The Newtonian theorem under discussion is Proposition LXX in Principia. The translated text is:

 

PROPOSITION LXX. THEOREM XXX.

 

If to every point of a spherical surface there tend equal centripetal forces

decreasing in, the duplicate ratio of the distances from those points ;

I say, that a corpuscle placed within that superficies will not be attract

ed by those forces any way.

 

Let HIKL, be that sphaerical superficies, and P a

corpuscle placed within. Through P let there be

drawn to this superficies to two lines HK, IL, inter-

cepting very small arcs HI, KL ; and because (by

Cor. 3, Lem. VII) the triangles HPI,LPK are alike,

those arcs will be proportional to the distances HP

LP ; and any particles at HI and KL of the spheri

cal superficies, terminated by right lines passing through P, will be in the

duplicate ratio of those distances. Therefore the forces of these particles

exerted upon the body P are equal between themselves. For the forces are

as the particles directly, and the squares of the distances inversely. And

these two ratios compose the ratio of equality. The attractions therefore,

being made equally towards contrary parts, destroy each other. And by a

like reasoning all the attractions through the whole spherical superficies

are destroyed by contrary attractions. Therefore the body P will not be

any way impelled by those attractions. Q.E.D.

 

Full text of "Newton's Principia : the mathematical principles of natural philosophy"

 

It's explained here:

 

Newton's Principia for the common reader - Google Books

 

~modest

[stereologist]

You have a source you are comfortable with. Please explain the meaning of s in this formula and the justification for the | | operators making this an equality.

 

|gv|+|gh|=2|gvs|

[modest]

Newton did his calculations with geometry.

 

I agree. The relevant proof was done with geometry, not calculus. The link I gave earlier (Newton's Principia for the common reader - Google Books) comments on this on page 273:

Littlewood was, of course, repeating a legend that has often been told, namely, that Newton first constructed proofs of most (if not all) of his propositions by calculus and then transliterated them into ‘his’ geometrical language. I do not believe this legend... his physical and geometrical insights were so penetrating that the proofs emerged whole in his mind: ‘he was happy in his thoughts’ (qualifying de Morgan)... For my part, I am not surprised that ‘to the Newton of 1685’ the geometrical construction ‘that must have left its readers in helpless wonder’ came quite naturally.

 

Despite rumors that Newton first did his proofs with calculus then published them in a geometric form, there can be no doubt but that the proof as published in Principia is entirely a geometric construction. I'll quote one more source for good measure dealing specifically with proof LXXI [i should appologize, having now looked at the proofs in detail I've realized I was wrong in saying the relevant proof is LXX in my last post. It's the next one: LXXI, which proves that a point outside a sphere is gravitationally equivalent to a point mass at the sphere's center.]

 

It's hard not to admire the ingenuity that Newton displayed here and throughout the Principia, producing time after time a clever synthetic demonstration of a fact that would (today) be treated by calculus. On the other hand, it's interesting that although Newton was in possession of his fluents and fluxions since the mid 1660's, he never took the occasion to settle his question about the force of attraction of a sphere with an inverse-square law - at least not until 20 years later, and then he resolved the question by means of a synthetic demonstration, rather than by explicit use of fluents and fluxions. It has sometimes been suggested that Newton wrote two versions of the Principia, first using calculus, and then re-writing it in the synthetic style. However, in the case of Proposition 71, it seems to me that Newton's published method is not very suggestive of the way this problem would be approached using calculus. The whole strategy of taking the ratios of the forces on two distinct points, and of splitting up the demonstration into two parts for internal and external points, seems (to me) to imply that this synthetic demonstration was the one that Newton referred to in his comment to Halley, i.e., this is the thought process by which he (after 20 years) arrived at this conclusion. Despite being one of the inventors of modern analysis, the classical synthetic mode of thought seems to have been more to his liking.

Newton's Proposition LXXI

 

Modest:

The attached shows how the matrices were derived and manipulated.

 

Thank you. I've looked this over now as well as your PowerPoint slides. I'll write a response tomorrow.

 

~modest :shrug: