Your FAQ is inconsistent in logic.

I'm going to refrain from pulling academic achievement and instead point out you still have yet to answer a single question. If what I'm saying's wrong, instead of the lazy "You're just ignorant!" how about explaining where I'm wrong and why?

I didn't call you ignorant. I said that "it seems as if you where ignorant about that concept", equals "it seems as if you don't know/understand that concept". You don't need to worry, i do not insult people.

You are wrong in saying that when I mention that center of mass is in the irregularity, that doesn't mean that center of mass represents the totality of the mass. I am still working with an infinite object with an irregularity that is the center of mass.

And, you know what? For fun I checked the source of that. Sure, it was about the universe rather than an infinite flat Earth, but physics is physics. Try reading the very next sentences.
 
" For instance, consider a cloud in the sky. Certain kinds of clouds don't have a well-defined boundary, but instead just stretch out in all directions, getting thinner and thinner. Even though the cloud stretches out effectively to infinity, the high density region of the cloud exists in a limited volume, so you can find a center of mass through a limiting procedure. Observations currently indicate that the universe is infinite in size. Although planets and stars do represent non-uniformities in the spacetime structure, on the universal scale, such uniformities are randomly dispersed. On average, therefore, the universe is uniform."

What reason do you have to say that an infinite Earth would not be similarly uniform?

Two reasons:
1) The universe in the round earth model is different than the universe in the flat earth model.
The irregularities in glove earth model's universe do not apply to flat earth model's universe because they are totally different.

2) We are discussing the gravitational "force" in an infinite disk/surface with a very clear irregularity (inside the wall of ice), that is very differentiated from the rest of the surface, and thus, we cannot say that the infinite surface is uniform.

Also, please note that the universe in the Flat Earth model is drastically different that the globe earth's model, and thus, you cannot apply globe earth's physics to your flat earth universe. Solely this argument will make invalid all your points, but still, let's ignore this fact.

Anyway, I still keep my position, and still asking how infinite gravity works in your model.

You are wrong in saying that when I mention that center of mass is in the irregularity, that doesn't mean that center of mass represents the totality of the mass. I am still working with an infinite object with an irregularity that is the center of mass.

Which was my whole point. So how, exactly, does that result in infinite gravity?

Or are you now admitting in would cause finite gravity, and in which case what is your problem?

Quote from: rabinoz on July 28, 2018, 04:24:10 AM

On an infinite plane the center of mass is "everywhere".

For a uniform finite thickness infinite plane gravity does work out finite, the same everywhere and pointing vertically down.
But it does not show the observed variations with altitude and latitude.

The problem that the infinite earth model has is that the earth has infinite mass, and thus, infinite gravity since gravity is directly related to mass:



If 'm' or 'M' is infinite, 'F' is also infinite.
How can your "gravity" behave like finite if the mass of one of the objects is infinite? Here the thickness of the object is not involved in the equation at all.

Nope! Sure the total M_earth might be infinite but to include the infinite M_earth you need an infinite r.

It is only with a spherically symmetric object that you can apply the simple F = G x M x m/r².
For other shapes, you need to do an sum each small δF = G x δM x m/d² over the whole region, though Gauss's Theorem can often simplify it.

The simple form, F = G x M x m/r², is OK as an approximation for other shaped objects when much further from the object than its size.

Quote from: rabinoz on July 28, 2018, 04:24:10 AM

But that "Infinite Plane Model" uses Newtonian Gravitation and results in:
"a finite gravitational pull equal to 2π G ρ h where G is the gravitational constant, ρ is density, and h is the depth of the slab. As we can see horizontal forces "cancel" out and we are left with a converging sum for vertical components."

According to that formula (2π G ρ h), gravitational force is related to the density of the object. If that were so, then the Earth would have a stronger gravitational force than Jupiter.

Earth density = 5.51 g/cm³
Jupiter density = 1.33 g/cm³

Note the density x depth

In addition, if your formula relates gravitational force to density and depth of the slab, what formula rules the gravitational force of spherical objects such as the rest of the planets?

It is not my calculation but due, I think, to John Davis and jroa.
For spherically symmetric objects, even if not of uniform density, it is correct to place all the mass that is inside the radius at the centre.
The earth is not exactly spherical but is very close and there are various formulae to account for these small variations.

Quote from: RoundGuy on July 28, 2018, 05:26:13 PM

You are wrong in saying that when I mention that center of mass is in the irregularity, that doesn't mean that center of mass represents the totality of the mass. I am still working with an infinite object with an irregularity that is the center of mass.

Which was my whole point. So how, exactly, does that result in infinite gravity?

Sorry, but I don't want to repeat what I already said.
You can try applying simple physics to it.
I will give you a tip: infinite mass with a defined center of mass. I am pretty confident you are clever enough to reach a conclusion that shows how things work.

Or are you now admitting in would cause finite gravity, and in which case what is your problem?

I was not aiming to admit anything. Only asking how things work in your flat earth model universe, when you apply globe earth model physics. Which sounds, at best, contradictory.

Quote from: RoundGuy on July 28, 2018, 05:26:52 AM

Quote from: rabinoz on July 28, 2018, 04:24:10 AM

On an infinite plane the center of mass is "everywhere".

For a uniform finite thickness infinite plane gravity does work out finite, the same everywhere and pointing vertically down.
But it does not show the observed variations with altitude and latitude.

The problem that the infinite earth model has is that the earth has infinite mass, and thus, infinite gravity since gravity is directly related to mass:



If 'm' or 'M' is infinite, 'F' is also infinite.
How can your "gravity" behave like finite if the mass of one of the objects is infinite? Here the thickness of the object is not involved in the equation at all.

Nope! Sure the total M_earth might be infinite but to include the infinite M_earth you need an infinite r.

It is only with a spherically symmetric object that you can apply the simple F = G x M x m/r².
For other shapes, you need to do an sum each small δF = G x δM x m/d² over the whole region, though Gauss's Theorem can often simplify it.

The simple form, F = G x M x m/r², is OK as an approximation for other shaped objects when much further from the object than its size.

Of course, we are here using approximations and generalizations, otherwise we wouldn't be able to discuss much.

But I have to ask you something:

in that text that I marked in red, how far is "much farther". I can calculate the gravity over the surface of our globe earth, yet the distance between me and the center of mass (the core of the earth, lets say) is so small in comparison with any other galactic distance. Yet, I can still calculate it.

How about 1 light year from the earth? I can still calculate. How about 10000 light years?
Mind explaining? Because it seems that your whole point is somehow based on that statement.
Also, how close is "too close"? How about the gravity over Phobos (Mars moon), that is way smaller than the earth? or how about a tiny meteor, where an object on its surface is very close to its center of mass? It seems I can calculate gravity in all these cases.

The shape of the object doesn't mind as far as there is a defined center of mass, where in your FE model does exist, even when it is an infinite plane/surface/object/body/whatever (please read above where I explain it).

So, my overall point here is that both of you seem to be confusing terms or not applying them correctly, or even modifying parts of physics science to your own benefit.

The FE model proposes a universe where an infinite body (earth) has a defined center of mass, and thus, it has infinite gravity.

Or, if you want me to admit that there is no center of mass, then we can conclude that you need an infinite force to accelerate (gravity) the infinite-mass earth.

Either of them please choose and explain how that works in the FE model.

I will give you a tip: infinite mass with a defined center of mass. I am pretty confident you are clever enough to reach a conclusion that shows how things work.

You do not have a defined centre of mass.
An infinite plane, unless there is only a single irregularity will not have a defined centre.
Even if it did, the better way to treat it is as a uniform plane and an irregularity. That irregularity doesn't magically mean infinite gravity.
An infinite plane does not generate an infinite gravitational pull on a finite object.
The formula you are using only applies for objects that are spherically symmetric or can be adequately approximated as such; it does not apply to an infinite plane.

in that text that I marked in red, how far is "much farther". I can calculate the gravity over the surface of our globe earth, yet the distance between me and the center of mass (the core of the earth, lets say) is so small in comparison with any other galactic distance.

That depends upon how spherically symmetric the object is and what level of accuracy you want.
The distance would be measured relative to the object, not the galaxy.
For a fairly spherically symmetric object like Earth, at a distance greater than the radius, it would be fine to use that approximation if you wanted an approximate (e.g. 2 sig fig, i.e. 9.8m/s^2) value. If you wanted it to be more precise, then you would need to use a better formula which accounts for the non spherically symmetric mass distribution, or get much further away.
The exact distance will depend upon how accurate you want to be, and eventually errors in the distance will become more significant than errors in the distribution.

So for an infinite flat Earth you would need to be further than its "radius" which is infinite. But it isn't even approximately spherically symmetric. So I would say you would need to be many times the distance. So once you get much further than infinitely far away you can try using that formula. But good luck getting that far.

how about a tiny meteor, where an object on its surface is very close to its center of mass? It seems I can calculate gravity in all these cases.

This is another case where the distribution is significantly asymmetrical and thus you need to be some distance from it.

The shape of the object doesn't mind as far as there is a defined center of mass

But it does matter as to if you can use that simple formula or not when close to it.
The formula is meant to apply to a point particle, but due to some fun benefits of symmetry also applies to any spherically symmetric object when you are outside it.

So, my overall point here is that both of you seem to be confusing terms or not applying them correctly

That is wonderful, coming from the person who is failing to apply the formula correctly.

Quote from: rabinoz on July 28, 2018, 06:26:18 PM

Quote from: RoundGuy on July 28, 2018, 05:26:52 AM

The problem that the infinite earth model has is that the earth has infinite mass, and thus, infinite gravity since gravity is directly related to mass:



If 'm' or 'M' is infinite, 'F' is also infinite.
How can your "gravity" behave like finite if the mass of one of the objects is infinite? Here the thickness of the object is not involved in the equation at all.

Nope! Sure the total M_earth might be infinite but to include the infinite M_earth you need an infinite r.

It is only with a spherically symmetric object that you can apply the simple F = G x M x m/r².
For other shapes, you need to do an sum each small δF = G x δM x m/d² over the whole region, though Gauss's Theorem can often simplify it.

The simple form, F = G x M x m/r², is OK as an approximation for other shaped objects when much further from the object than its size.

Of course, we are here using approximations and generalizations, otherwise we wouldn't be able to discuss much.

Let me make one point clear.
I believe that the earth is very close to being spherical and think that the idea of an "infinite earth" is ridiculous.
All I'm trying to explain is that on that "infinite earth model" the surface gravitation above it is finite and the same everywhere.

But I have to ask you something:

in that text that I marked in red, how far is "much farther". I can calculate the gravity over the surface of our globe earth, yet the distance between me and the center of mass (the core of the earth, lets say) is so small in comparison with any other galactic distance. Yet, I can still calculate it.

The earth is very near a sphere and so gravitation outside it can calculated with acceptable accuracy by assuming all of the mass is at the centre - the centre of the earth.

The inaccuracy arises because the "center of mass" calculation and the effective centre of "gravitational attraction" coincide for objects with spherical symmetry but not for other shapes.
This is because the center of mass calculations involve mass x distance, and gravitation calculations involves mass/(distance squared)[/b]. so the centres do not necessarily coincide (Any more and I might confuse myself).

This might help: Hyperphysics, Gravity Force of a Spherical Shell.
That site has "all sorts of interesting things", have a look: About HyperPhysics, Rationale for Development and
here is the bit on gravity Hyperphysics, Gravity

How about 1 light year from the earth? I can still calculate. How about 10000 light years?
Mind explaining? Because it seems that your whole point is somehow based on that statement.

A great distance away is no problem, though the gravitation very soon might become negligible - it is close where the error comes in.
Of course for this hypothetical flat earth no distance is "far enough away".

Also, how close is "too close"? How about the gravity over Phobos (Mars moon), that is way smaller than the earth? or how about a tiny meteor, where an object on its surface is very close to its center of mass? It seems I can calculate gravity in all these cases.

What matters is the distance from tne object compared to its size. How far depends on the accuracy needed.
For example you are 1000 km from the mid-point of two 1000 tonne masses, one 900 km away and one 1100 km away.
The gravitation from that pair would be the same as from one 2000 tonne mass 996 km away, now 1000 km as the centre of mass might indicate.

On a much larger scale, the solar system is very large but to calculate its gravitation a lightyear away it would be quite in order to lump all of its mass at the centre.

The shape of the object doesn't mind as far as there is a defined center of mass, where in your FE model does exist, even when it is an infinite plane/surface/object/body/whatever (please read above where I explain it).

I don't quite follow, but the shape of an object certainly does matter for gravitational calculations.

So, my overall point here is that both of you seem to be confusing terms or not applying them correctly, or even modifying parts of physics science to your own benefit.

No, I'm not confusing anything or "modifying parts of physics".

The FE model proposes a universe where an infinite body (earth) has a defined center of mass, and thus, it has infinite gravity.

No, an infinite flat body of finite thickness has finite gravitation at its surface - not that any serious scientist thinks the earth is such a shape.

Or, if you want me to admit that there is no center of mass, then we can conclude that you need an infinite force to accelerate (gravity) the infinite-mass earth.

I don't mind what you admit but there is no centre of mass for the calculation of the gravitation does due to a non-spherical body.

Either of them please choose and explain how that works in the FE model.

The infinite FE model "sort of works" but only partly explains what we observe. There are many things, including satellites that simply do not fit.

Let so kind flat earther explain more. It'd not my problem.

Quote from: Jane on July 28, 2018, 05:32:50 PM

Quote from: RoundGuy on July 28, 2018, 05:26:13 PM

You are wrong in saying that when I mention that center of mass is in the irregularity, that doesn't mean that center of mass represents the totality of the mass. I am still working with an infinite object with an irregularity that is the center of mass.

Which was my whole point. So how, exactly, does that result in infinite gravity?

Sorry, but I don't want to repeat what I already said.
You can try applying simple physics to it.
I will give you a tip: infinite mass with a defined center of mass. I am pretty confident you are clever enough to reach a conclusion that shows how things work.

*sigh*
Except you literally just said the centre of mass did not represent the 'totality of mass.'

Look, you have multiple round earthers who think that the idea of a flat earth is ludicrous coming in and telling you that no, you're wrong, the maths for gravity on an infinite Earth check out. Stop being so bloody arrogant.

Break down. You might not want to repeat yourself, but apparently I have to. Tell me which of these statements you disagree with; all I want is one number, so you can't just evade that with 'I don't want to repeat myself.'

1. A uniform infinite flat Earth is not going to have any defined centre of mass.
2. On a non-infinite, huge disc Earth, all matter is going to be drawn to the centre of mass. Therefore if you stand at somewhere near the rim, that distance from the centre of mass means that you will be under a much smaller gravitational force than if you stood over the centre.
3. Without a centre of mass, a uniform infinite flat Earth is therefore not inherently going to exert infinite gravity.
4. Creating an irregularity in said uniform infinite flat Earth, altering density by say two atoms in one location, will not make the gravity exerted jump to infinity immediately.
5. An irregularity in density in a finite location is therefore not going to represent an infinite Earth's mass.
6. Thus, an infinite Earth can exert finite gravity.