Cold Core Model of Earth's Structure

[modest]

However the Cavendish balance works in a plane perpendicular to the force of vertical gravity, which shows that horizontal pulls exist.

 

It shows directly that the mass of earth did not affect the horizontal forces of the Cavendish experiment.

 

The schematic you keep showing from my powerpoint is within the earth.

 

That seems arbitrary to me. The fact that you show nothing above [math]M_1[/math] seems to show that the shell could be the earth's surface. More importantly, the data you have given indicates clearly that your methodology expects this horizontal force to persist at least to the surface of the earth. Your table of values expects a horizontal force on the surface at ~10 m/s². And, if that 10 m/s² has no effect on the surface (and we know it doesn't by experiment) then it would be unreasonable to think it has any effect further down in the earth.

 

As for being able to measure the horizontal pulls, there isn't an instrument made that can do that, if there were we would have noticed it many years ago. Hence, my analysis is a word experiment that cannot be mechanically tested. However, we know from Newton's definition of gravitational force that horizontal pulls exist, but they don't have a directional vector that one can measure. They are just there. Undisturbed the earth shows no effect from them. Nor can you feel them, because they surround you. They certainly will not tip you over.

 

Right. And if they have no measurable effect then we cannot expect them to have any effect. The density and acceleration inside the earth is as PREM models it because there is no *measurable* gravitational force exerted by a shell on its interior. As Newton said in Principia's proposition LXX theorem XXX:

If to every point of a spherical surface there tend equal centripetal forces decreasing as the square of the distances from those points, I say, that a corpuscle placed within that surface will not be attracted by those forces any way...

 

Therefore the forces of these particles exerted upon the body P are equal between themselves. For the forces are directly as the particles, and inversely as the square of the distances. And these two ratios compose the ratio of equality, 1:1. Their attractions therefore, being equal, but exerted in opposite directions, destroy each other. And by like reasoning all the attractions through the whole spherical surface are destroyed by contrary attractions. Therefore the body P will not be any way impelled by those attractions. Q.E.D.

 

A distinction needs to be made between the force of gravity and the gravitational potential energy a particle has for being in a gravitational field. A particle can be in a gravitational field surrounded by a lot of mass and have gravitational potential energy yet feel no force of gravity. A particle in a hollow shell, for example, has potential energy, but feels no force.

 

Consider your diagram once again:

Considering only the point [math]M_1[/math] and the mass of the annulus we have the following:

The center of mass of the ring is point O. The point under consideration is P. The ring does two things for P. It gives P potential gravitational energy and it subjects P to the force of a gravitational field.

 

The potential energy of P is:

[math]PE = -\frac{GM}{\sqrt{a^2+r^2}}[/math]

And the force on P is:

[math]F = \frac{GMr}{(a^2+r^2)^{3/2}}[/math]

These equations along with the conclusions I'm about to make can be sourced here and here.

 

What I'm getting at is this: when r = 0 the point P is at the center of the ring. If you plug r = 0 into the two equations above you will find that the force is zero (since r is a factor in the numerator F will equal zero when r = 0) yet PE has value when r = 0. This means the mass of the ring affects the potential energy of P when it is at the ring's center, but there is no field strength at that point. I think this is what you are looking for when you say the following (part of which I bolded):

 

It goes back to the problem of the six Eiffel towers. The forces between you and the towers are there but there is no way to measure them. They can however be calculated and shown to balance. What I have tried to do is give these horizontal forces values that can be mentally visualized. In the process I've shown that the horizontal force produced by a gram mass on the surface of the earth pulls all the other gram masses in the earth with force that is equal to the vertical force of gravity on earth's surface.

Which leads to the question, Is the equation used by the geodesists a valid representation of the forces that need to be overcome before earth will bulge from rotationally induced movement of the internal mass.

 

The force of gravity (or the field strength / acceleration) of a point in the earth's interior would not be a valid representation of how much energy would be required to remove that point from inside the earth. The energy needed to lift a mass out of a deep well cannot be obtained by knowing the acceleration on the mass at the bottom of the well. Likewise, a particle in a hollow spherical shell has zero force. But, this does not mean it would take zero energy to remove it from the shell. Since the shell gives the particle potential energy, that is the equation which tells you the energy needed to remove the particle from inside the shell. So, perhaps that is what you're thinking.

 

Potential energy and gravitational force have a very intimate relationship. The force that a particle feels is a measure of the slope (or the rate of change in) potential energy. So, no matter where a particle is inside a hollow shell it has the same amount of potential energy—the same amount of energy is required to remove it from the shell no matter where it is in the shell. This means that the change in PE as the particle moves around the shell's inside is zero—which then means that the gravitational field in the shell is zero. The particle feels zero force (just as the change in PE is zero) in the shell.

 

Now take the case of the earth. A graph of the PE in a solid sphere of uniform density is as follows:

-

source

 

Where a is the radius of the planet and V is the potential energy at r. A graph of the field strength in a solid sphere of uniform density is:

 

-

source

 

where a is again the planet's radius and E is the field strength (or the force felt at r).

 

So, we've been knocking you over the head in this thread telling you that the second graph above is the correct force in a solid sphere. But, it's entirely possible that the concept you're really seeking is the first graph above. While the force at the center of a solid sphere is zero, the PE is not zero. The PE at the center of the sphere is -3GM/2a.

 

The equation for PE of a solid sphere is:

[math]-\frac{GM}{2a^3}(3a^2-r^2)[/math]

So, please, take some time to consider these graphs, formula, and this post and you can read these two links:

• Gravitational potential due to rigid body
  
• Gravitational field due to rigid bodies
  

and see if this is not perhaps the concept you are describing. It really seems like you're saying that the mass above a point in the earth gives the point some value of energy even though the forces "cancel" at that point. That is the case with gravitational potential energy so I think reading about PE might expand your outlook on this subject and perhaps help you to revisit your conclusions.

 

~modest

[Cold-co]

Modest:

I think we are not communicating. I am treating earth as a collection of gram masses and trying to establish the gravitational forces acting on each gram mass. Newton addressed that in the attachment you posted and concluded that the horizontal and for that matter the vertical forces cancel out in a static model. But the bulge at the earth's equator is a dynamic, which I believe Newton would recognize as being different than the static state. Hence the horizontal pull between gram masses must be overcome just as the vertical pull must be overcome for earth to bulge. However, geodesists ignore the horizontal pulls in their flattening equation.

Five days ago I posted response1.doc that gives the schematic of forces I used for making my calculations of vertical and horizontal forces. If you will refer to it you will see the outline of earth's surface is clearly indicated. I did not post the schematic that was lifted from my PowerPoint.

I appreciate your efforts to educate me in the gravitational potential but that is not germane to the problem I'm trying to solve. It just muddies the waters of my simple brain.

In another vane, I found the problem in my calculations for pressure within the earth. I had transfered some of the data into the matrix, it must have come with a hidden function that threw off my end results. Yes I verified your calculations.

[freeztar]

I'm starting to think that "horizantal gravity" is simply normal gravity coming from a 90 degree angle to your "vertical gravity" via trigonometry. If we place a triangle with its right angle vertice at the center of the earth and draw a 30-60-90 triangle, I could see why the 90-60 ray would be misconstrued as "horizantal gravity" (or the 90-30 ray for that matter).

 

That's the only thing that makes sense. Otherwise, you are arguing against Newton's Prinicpia which demands that "horizantal forces" cancel out inside a perfectly circular shell.

[TheBigDog]

I have been reading along with this thread for some time and it reminds me of an old one about the Coriolis Effect but from a different perspective.

 

The crux of the argument that I read from ColdCo has to do with how the formula for calculating how oblong the earth becomes due to spinning on its axis. There is a claim that the calculation treats the planet as a liquid, not accounting for the crusty outer layer of the planet and how it should impact the deformation of the planet.

 

I think the the answer is that the crust of the earth is inconsequential in the calculation. The crust undergoes very little local deformation as a result of the earth flattening with spin. And the crust itself is only a small fraction of the total mass and is not responsible for the earth being round. It is in fact floating on the inner liquid and can only assume the shape given to it by the inner molten globe of rock and iron.

 

I do think that ColdCo is correct asserting that there is some effect from the crust, but I would venture that it is so little that it is not worth bothering with. The same way that a gravitational slingshot calculation does not bother finding the change in the velocity of the planet or moon being used to provide the added energy. The crust is in essence along for the ride like a blanket on an elephant's back. Adding or removing the blanket will not change the capability of the elephant in any measurable way.

 

All of this horizontal gravity stuff is just confusing the question.

 

Bill

[Zythryn]

Cold-Co, is this all just a matter of semantics?

What happens if you see 'horizontal' gravity and turn your head 90 degrees, is it then 'vertical' gravity?

How is 'horizontal' gravity different than any other 'type' of gravity?

[Cold-co]

Zythryn, TheBigDog and freeztar.

Thank you for joining the discussion. Indeed horizontal gravity is no different than vertical gravity and it does balance out. However, the physical reality of earth's cross section tells us it is still present even though it balances out. Newton is right, in a static model horizontal forces appear to cancel out, but for earth to bulge from rotation requires a dynamic movement of the internal matter. That internal movement is not static, it is dynamic.

The geodesists use the force of gravity at the equator (earth's crust) as the deciding force in dermining earth's moment of inertia. However, the crust is what is being deformed by earth's rotation. Now, if you skip forward in your Physics text you should come to a section on surface tension. Quote: "All surface phenomena indicate that the surface of a liquid can be considered to be in a state of stress such that if one considers any line lying in, or bounding, the surface; the material on either side of the line exerts a pull on the material on the other side." The surface tension in a liqiud, S, is defined as the ratio of the surface force to the length along which the force acts. S=F/2l, or for the earths equator S=F/2C, where C is earth's circumference. I don't know what the surface tension in granite or basalt would be, but suspect it would be greater than that of the element mercury.

[TheBigDog]

I don't see the problem. You know the dimensions of the planet, you know the velocity of rotation on the axis. You know the formula. What is the margin of error between what the formula you don't like provides as the answer and the observed measures? Then simply allow that your crust surface tension may account for the difference. :rolleyes:

 

Bill

[Cold-co]

TheBigDog:

I'm not sure you have read the first post on this thread. It lays out the problem.

[Cold-co]

Modest, Pyrotec, etal.

In this thread we have been bantering about the force of horizontal gravity. During our debate we have touched on the Cavendish balance and how it was used to determine the gravitational constant. If the Cavendish balance’s four weights were of equal mass there would be no movement of the weights to allow us to derive the gravitational constant. We know there is a pull between those weights, but there is no resultant movement. We can however calculate the pull between the weights using Newton’s equation and we would find they balance out; or, as Newton claims cancel. From this word experiment we can conclude that the horizontal force of gravity only works when there is an imbalance in the weights of the masses. Hence horizontal pull is present but balanced out. Now if you were given an assignment to calculate gravitational accelerations of horizontal gravity at descending depths within the earth, how would you go about doing it?

[modest]

If the Cavendish balance’s four weights were of equal mass there would be no movement of the weights to allow us to derive the gravitational constant.

 

Cavendish's experiment would work fine with 4 equal masses. The masses would turn the balance and the amount of displacement could be used to determine G just as if the two sets of masses were unequal. Can you explain why you think otherwise?

 

We know there is a pull between those weights, but there is no resultant movement.

 

:eek_big:

 

We can however calculate the pull between the weights using Newton’s equation and we would find they balance out; or, as Newton claims cancel.

 

I don't at all follow.

 

-source

 

If m and M were equal, you think the force between them would be zero or somehow canceled? Why?

 

~modest

[Cold-co]

Modest:

I see your point, but we were taught that if they were of equal value the balance would be in a stable state of equilibrium, hence the smaller weights were needed to get movement. The balance itself being in a state of instability. Once the weights move to an equlibrium point then movement stops. I think that is what the accelleration of horizontal gravity does in planetary bodies. It restores equilibrium.

 

Now, we have beaten this problem into the ground and it appears we will never come to any agreement, so I suggest the monitors erase this thread and terminate further discussion.

[Southtown]

Are we talking about istocacy of sorts?

[stereologist]

Once the weights move to an equlibrium point then movement stops. I think that is what the accelleration of horizontal gravity does in planetary bodies. It restores equilibrium.

 

Equilibrium appears to be used in two different ways here. In the Cavendish balance it means no movement. In planetary systems it certainly does not mean no movement.

 

The equilibrium in the Cavendish balance is reached when the gravitational force is balanced by the force of the twisted cable on which the weights hang. So when the torsion balances the gravitational forces movement stops. An instrument is used to measure the small angle rotated as accurately as possible.

 

The reason for using smaller weights is simple to understand. First hang as large a set of weights as possible on the cable that do not damage the cable. Then place huge weights next to the balance to cause the suspended weights to move as much as possible.

[Cold-co]

Stereologist:

Referring back to the figure of the Cavindish balance that was provided by Modest, if the large weights (M) were suspended on the quartz string and the smaller weights (m) were on the fixed beam, then the movement of the heavier weights would be slight but noticable. Now if the weights all were M in value, I contend their would be no noticable movement because all weights are in perfect balance. I contend the Cavendish balance works only when the weights are of different masses. I may be wrong, because if each weight were suspended on a quartz string then they would be attracted toward each other and that movement would be measureable.

Southtown:

"Isostasy" is exactly what horizontal gravity provides, however it goes unrecognized because it sums to zero. but it does have a value that can be calculated. That is what I have attempted to calculate trigonometrically.

[CraigD]

Are we talking about istocacy of sorts?

I don’t think so, but applaud you, Southtown, for bringing us what appears to be a truly obscure – if its google search (results 1 - 3 of 3 for istocacy) is used as a measure – term, istocacy – though I’m uncertain if its truly that obscure a term, or if somehow most search engines (I tried all the non-google-driven ones I could think of off the top of my head, with similar results) are somehow troubled by it (possibly because it’s a substring of a misspelling of aristocracy??).

 

Whatever term’s used, the theory referred to by the term istocacy is the widely accepted one that plate tectonics cause irregularities in the thickness and density of the earth’s crust, and that this can effect local gravity, sometimes in peculiar ways, such as causing a long plumb line to be measurably deflected from true geological vertical, as one would measure with a star sighting and a precise chronometer. Though I’ve not in this thread’s long history fully grasped what Cold-co is thinking, I’m pretty sure it’s not about this.

 

Speaking of Cold-co’s thinking, I’d like to join those that have tried clearing up an apparent misconception about torsion balances in general, and the Cavendish experiment’s use of it in particular.

I see your point, but we were taught that if they were of equal value the balance would be in a stable state of equilibrium, hence the smaller weights were needed to get movement. The balance itself being in a state of instability. Once the weights move to an equlibrium point then movement stops.

Referring back to the figure of the Cavindish balance that was provided by Modest, if the large weights (M) were suspended on the quartz string and the smaller weights (m) were on the fixed beam, then the movement of the heavier weights would be slight but noticable. Now if the weights all were M in value, I contend their would be no noticable movement because all weights are in perfect balance. I contend the Cavendish balance works only when the weights are of different masses.

As Modest and others have explained, the Cavendish experiment for measuring the force of gravity doesn’t require that the moving lead spheres be smaller than the fixed ones – though in Cavendish and Michell’s ca. 1897 experiment apparatus, they were, much – 2 inches and 1.6 pounds vs. 12 inches and 348 pounds. The reason for this is purely practical engineering, to keep the weight of the moving beam, including the 2 inch spheres, low enough that they can be supported by the thinnest practical wire, to maximize the rotation produced by a give force.

 

The force of gravity between 2 bodies is, for practical purposes, directly proportional to the product of their masses, and inversely proportional to the square of the distance between their two centers of mass – that is, [math]F = G \frac{m_1 m_2}{r^2}[/math]. So, had the original Cavendish experiment been done with spheres of equal size producing the same force, they would have been about 5.4 inches and 23.7 pounds.

 

Torsion balances apply a force equal and opposite to their loads proportional to the angle through which they twist, divided by the length of their beam – that is [math]F = K \phi l[/math]. Cavendish’s, and most similar experiments, however, didn’t allow the beam to be motionless, but rather measured the period of its oscillation. The experiment could, in principle, have allowed the beam to be motionless, but would have been more complicated and less accurate.

… I think that is what the accelleration of horizontal gravity does in planetary bodies. It restores equilibrium.

Throughout this thread, I’ve been struck by the strangeness of Cold-co’s considering the force of gravity on a body in neither of the two common ways: as a single net force vector (which, close to above or below the Earth’s surface, is nearly toward its center), or as a large number of vectors between the body and a collection of point masses representing the Earth’s body (in “A brute force calculation”, I used about 1,000,000. Ideally, one should use a point for every massive fundamental particle, though using merely one for each atom – about 10⁵⁰ should be good enough ;))

Now, we have beaten this problem into the ground and it appears we will never come to any agreement, so I suggest the monitors erase this thread and terminate further discussion.

If there are no objections within 24 hours, I’ll close this thread.

 

Except in cases of extremely offensive or illegal activity-promoting posts, we don’t as a policy erase hypography threads.

[modest]

Stereologist:

Referring back to the figure of the Cavindish balance that was provided by Modest, if the large weights (M) were suspended on the quartz string and the smaller weights (m) were on the fixed beam, then the movement of the heavier weights would be slight but noticable. Now if the weights all were M in value, I contend their would be no noticable movement because all weights are in perfect balance. I contend the Cavendish balance works only when the weights are of different masses. I may be wrong, because if each weight were suspended on a quartz string then they would be attracted toward each other and that movement would be measureable.

 

You may not be picturing the experiment quite right. I can't imagine how you would conclude that this has a force between the masses wanting to rotate the freely rotating set:

 

Overhead View

 

while this does not:

 

 

I think you might be thinking the experiment is arranged like this:

 

 

In that case, you'd be correct that none of the masses would move. However, in this case even if the two sets of masses were different sized there would still be no propensity for any of the masses to move:

 

 

In the image above there is no net force rotating the arm, but this is not how the experiment is arranged.

 

~modest

[stereologist]

Now if the weights all were M in value, I contend their would be no noticable movement because all weights are in perfect balance.

This is wrong. The force between the masses due to gravity is proportion to the product of the masses.