Cold Core Model of Earth's Structure

[Cold-co]

Modest:

You are absolutely right. I went back and checked my spreadsheet and it insists on putting 4.8 in that slot. I tried the same calculations in a new section and got the figure you got. I even emptied that slot and replaced the math function and it came up with the wrong answer. I may have to replace my version of excel.

[modest]

I suspected it might be a spreadsheet problem.

 

You can try the one attached to this post. I saved it in an earlier version.

 

~modest

[modest]

From the PowerPoint slides:

 

Gravity should not be shown as

 

It should be shown as

 

I agree. The wikipedia page Newton's law of gravitation has a good image consistent with yours:

The force [math]F_1[/math] should always be equal to [math]F_2[/math] no matter how much larger [math]M_1[/math] is than [math]M_2[/math]. Each mass is pulled with an equal yet opposite force. I agree with that, though I personally would not use the term:

It is an elastic force.

Gravity is like an elastic force (Hook's law) in that both exert an equal and opposite force on two masses. But, they are different in many more ways (e.g. Hook's law is linear while gravity is quadratic. With Hook's law, force does not depend on mass while the force of gravity is proportional to mass.) Nonetheless, that is an issue of semantics...

When all of the annulus (or all of the shell) is considered then [math]M_1[/math] has no horizontal vector. It is left only with a vertical component. To consider both sides of the sphere, you can extend your consideration to 3 mass points rather than 2:

The red vectors represent the fact that each mass is attracted to each other mass. The black vectors are the sum result—the total direction and total force with which each mass is attracted. The black vectors are what the mass actually feels. [math]M_1[/math] is pulled straight down. It feels a force in only one direction: down.

 

This is confirmed in the real world where objects on earth's surface (like [math]M_1[/math]} are not pulled apart horizontally. From the diagram you give in the quote above it looks like you might expect things above earth's surface to feel a horizontal (what you call an elastic) force. Yet, I thought I heard you say previously that you agree a point outside a sphere is attracted only to the sphere's center as Newton's point mass 1/r² proof says it should...?

 

~modest

[freeztar]

So, Modest, according to the illustrations, the moon would be elliptical if it had "horizantal gravity", correct? For that matter, we would expect the Earth to be much more stressed at the equator than we currently see. Yes?

[modest]

So, Modest, according to the illustrations, the moon would be elliptical if it had "horizantal gravity", correct? For that matter, we would expect the Earth to be much more stressed at the equator than we currently see. Yes?

I'm not sure. If the earth contributed horizontal gravity to the moon then I guess it would have made it eliptical.

 

But, I think my diagram may have been overly confusing. In case you mean this one:

You can think of the dotted line as the surface of the planet (or a hollow shell). It would be circular. I didn't draw it all the way around because it was getting in the way of the arrows. [math]M_1[/math] is off the surface out in space while the other two masses being considered are on the surface of the shell.

 

All this is really is showing is that if we consider 3 masses (and only three masses) and they are all attracted to each other as gravity does and they form an equilateral triangle then each mass will be drawn to the center of the other two masses. (the black vectors). This means that the mass which is off the surface ([math]M_1[/math]) will be drawn down toward the center of the shell. Even though [math]M_1[/math] has a mass to the left of it and to the right of it, it does not feel any horizontal gravity.

 

[math]M_2[/math] and [math]M_3[/math] look as if they are acting weird (being drawn away from the shell), but this is only because we've neglected the rest of the mass of the shell—considering only 3 point masses—one of them being rather distant from the shell.

 

You can rotate [math]M_2[/math] and [math]M_3[/math] around the annulus (the solid gray lines) to any degree and [math]M_1[/math] will still be attracted toward the center of the shell.

 

So, [math]M_1[/math] should feel no horizontal gravity from the shell. But, i think Cold-co might agree with this. From post #4 he believes horizontal gravity is only an effect inside a mass (I hope I'm not misstating your position Cold-co). But, that being the case, I really don't understand this:

Both [math]M_1[/math] and [math]M_2[/math] could be considered outside a mass in that case. If it shows horizontal gravity then it should be something we could test for. And, I know testing for horizontal gravity gives no results. Holding a scale vertically on earth's surface shows 9.8 m/s² of acceleration and holding it horizontally shows zero.

 

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|

 

S means surface. From post #4:

 

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.

 

~modest

[Pyrotex]

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

 

This is rather bogus. It becomes obvious from the PowerPoint presentation that either cold-co or CharlieO gave us. On slide 6 of 7 we have:

 

|gv| + |gh| = 2 |gvs| ... which leads to replacing

 

W^2 a / Ge

 

with

 

W^2 a / 2 Ge

 

In other words, they just arbitrarily stuck a 2 in the denominator.

 

Now, we all know that Physics and Math is just not done that way.

 

Can anyone draw a vector diagram (using REAL vectors, not bogus "elastic" forces) demonstrating the relationship between the 3 vectors in the equation at the top of this post?

[Cold-co]

Modest:

You have shown why I chose to use a two dimensional schematic. Since gravity is an elastic vector it cannot be cancelled out. Balanced, but never cancelled. It is present and pulling all the time, but what is its strength. No textbook that I am aware of gives even the indication that there is such a force as horizontal gravitational attraction. 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.

[modest]

If you can't measure it then what possible effect can we say it has? If it has no effect on our measuring equipment then we can't assume it has any effect on the planet. Yes, there are horizontal gravitational forces which balance. We consider them canceled because they have no net effect. A person standing on the surface of the earth is pulled in only one direction: toward its center. We are neither squeezed together nor pulled apart horizontally. There is no such effect of gravity.

 

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.

 

Yes, it is valid. It is precisely valid. We have launched rockets into orbit knowing exactly how much angular momentum to give them to overcome earth's gravity. The equations work.

 

~modest

[stereologist]

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.

 

Spaceships have traveled to Uranus, over a billion km away, and done their flybys with remarkable accuracy.

 

It is possible to represent any vector as the sum of two or more vectors. That's how vectors work. The vector equation |gv|+|gh|=2|gvs| can be reduced to

 

|v|+|h|=2|vs| since g is a scalar

 

So the lengths of 2 sides of a triangle is equal to what? Please explain.

[Cold-co]

Modest:

It didn't occur to me until later that you modified my schematic of forces. You claim that the forces on the left side of the schematic balance the forces on the right side, but in my schematic all the masses are on the right side. Otherwise it is easy to visualize that they balance each other. Your debunking of my method must stick with the model I define not revert to a different model. If you can show me how my schematic errors I will happily go away.

As far as launching rockets, I did that for years and the flight equations only considered the vertical vector of gravity for a good reason. Outside of the orb, vertical gravity is the only force needed to be overcome. But, we are dealing with the forces inside of the orb, forces that have never been established. Even though the horizontal effects of horizontal gravity can be visualized, no one has bothered to try to quantify them.

I want to thank you for bothering with my musings, most have already written me off as a crackpot, which I may well be.

[Cold-co]

Sterologist:

I'm sorry, your post came in while I was sending a response to Modest, I would have included a response to you at the same time. I think you have already concluded that vs stands for the force of vertical gravity on earth's surface. The relationship of |gv| + |gh| = |2vgs|. As you progress into an earth model of constant density the absolute value of vertical gravity plus the absolute value of horizontal gravity equals twice the absolute value of vertical gravity on the orb's surface. It is not a vectored equation It is just a relationship. But, remember it is my relationship based on the trigonometrical analysis of forces in a model of constant density. Since that relationship was established trigonometrically, it can be applied to an orb of hydrogen condensing in a molecular cloud. It gives a reason for why solar systems form.

[Pyrotex]

Modest:

You have shown why I chose to use a two dimensional schematic. Since gravity is an elastic vector it cannot be cancelled out....

Cold-co,

 

there is no such thing as an "elastic vector".

Go to Wikipedia and do a search for it.

Google it.

Get any freshman physics textbook.

 

For that matter, wiki-search on "vector", and you'll see what a vector really is.

 

A vector is defined by two arithmetic values: one number describes the magnitude of the vector, and the other number describes its direction. That is why it is usually drawn as an arrow.

 

A vector acts upon a point. A Force is properly represented as a vector.

 

A Force acts upon a point. If the point is part of a Rigid Body, then we can freely speak of the Force acting upon the body.

 

Vectors (and Forces) add together with vector arithmetic.

 

Two vectors of equal magnitude and opposite direction, acting on the same point (or rigid body) CANCEL OUT. That is the way vector math works. And you cannot change that any more than you can define 2 + 2 = 5.

 

"Elastic" is a property of material objects. It is NOT a property of forces.

 

There is no such thing as an "elastic force". Please deal with this.

[modest]

Modest:

It didn't occur to me until later that you modified my schematic of forces.

No. I posted your diagram exactly as it is in your power point slide. I created another diagram of my own in addition to yours.

 

You claim that the forces on the left side of the schematic balance the forces on the right side, but in my schematic all the masses are on the right side. Otherwise it is easy to visualize that they balance each other.

 

I'm glad we agree. Since the mass of the shell is symmetric, you should fix the diagram and add a mass to the left side.

 

As far as launching rockets, I did that for years and the flight equations only considered the vertical vector of gravity for a good reason. Outside of the orb, vertical gravity is the only force needed to be overcome.

 

Yes, I thought I had seen you say this before. But, you're not being very consistent. Your diagram clearly intends to show a horizontal force present for either [math]M_1[/math] or [math]M_2[/math], both of which are easily within reach of experiment. [math]M_1[/math] is outside the surface and [math]M_2[/math] is on it. Furthermore, your data given in post 14 says that we should expect 10.0604 m/s² of horizontal acceleration at the planet's surface.

 

Moreover, this equation: |gv| + |gh| = 2 |gvs| tells you that gh at the planet's surface should be equal to gv.

 

But, all of that is plainly and demonstrably false. If I hang a mass on a spring and hold it vertically I will measure 9.8 kg•m/s². If I hold it horizontally, I measure zero. The same is true of a person in a submarine at the bottom of the ocean.

 

How could a Cavendish experiment miss this force you are describing?

 

Tell me, experimentally, how do we measure the 10.0604 m/s² horizontal acceleration which you have claimed is present at the planet's surface? If there is no way to measure it then it can have no measurable effects.

 

~modest

[Pyrotex]

I hope this helps.

 

Cold-co, I have here one drawing out of the PowerPoint presentation. From page 3 I think.

It shows horizontal forces pulling on an "annulus section".

However, it shows ONLY the force from the matter in ONE direction.

The truth is, at every point on the annulus section, that point will be pulled in ALL directions by the matter in ALL directions.

 

See the drawing I made--it's attached.

 

Every point on the annulus will be pulled by EVERY element of mass that surrounds it.

If you use PROPER vector arithmetic, the SUM of ALL the force vectors at any one point on the annulus section will be a single net force pulling towards the center of the Earth.

 

All other vectors cancel out through symmetry.

[Cold-co]

To all:

You keep redefining my schematic, and then ask me to justify it. By placing all the mass(es) on the right side of the perpendicular axis, I have set up a two dimensional problem that can be solved trigonometrically. Your models cannot be solved at all, because the horizontal forces balance out. When you are standing on the earth you are pulling on the earth with the same force as the earth is pulling on you. You are also pulling horizontally on all the gram masses within the earth. You have agreed that there are horizontal gravitational forces, but you cannot quantify them because in your models they cancel out.

If you can show me the illegitimacy of my model I will apologize for bothering you, fold up my tent and steal into the night.

[modest]

You have agreed that there are horizontal gravitational forces, but you cannot quantify them because in your models they cancel out.

 

They cancel out in reality. The mass is symmetric in reality. Can you answer the questions in my last post?

 

~modest

[Cold-co]

Modest:

If the Cavendish balance could have detected the horizontal forces, then we wouldn't have anything to discuss. However the Cavendish balance works in a plane perpendicular to the force of vertical gravity, which shows that horizontal pulls exist. The schematic you keep showing from my powerpoint is within the earth. I probably should have shown a dashed outline of the earth's surface, but I didn't because I thought my narrative made it perfectly clear that I was sketching a shell, like the 2nd Bonded shell, within the earth and the gram mass was above that shell within the earth. It could also be below that shell and the trigonometric analysis still works.

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.