infinite plane?

Quote from: Its a Sphere on October 14, 2009, 10:29:10 AM

Also, what if any value is the acceleration due to gravity.  Does this not exist either?
How can an infinite plane (of thickness absolute or differential) possibly have gravity?  I've never heard of such a thing

http://www.theflatearthsociety.org/forum/index.php?topic=32568.0

I was looking for a value for acceleration due to gravity (or apparent measured value) as I see the equivalence theory referenced here oh so often.
It seems that g should be easily calculated using the time it takes an object to fall to the surface of the earth. Traditionally it is valued at:  Earth's gravitational acceleration at sea level, 9.80665 meters per second per second.  Do you not agree?

And mtarlo11 just claimed that gravitation acts towards the centre of an object. I was wondering how an object with multiple centres can gravitate to them all at once.

I was referring to my comment.  I ignore what he posts, though he does have a valid question of how can an object gravitate to multiple centers at once?  If the earth is an infinite plane it has infinite centers, by definition. (with each point being equally far from the edge)

Quote from: Parsifal on October 14, 2009, 09:53:05 AM

Quote from: mtarlo11 on October 14, 2009, 09:50:48 AM

Is that a serious statement? I don't even know what trying to say? If we had equal infinite gravity all around us, we would be crushed into 1 dimension.

Imagine you and a friend are holding onto either side of a rope. You are pulling one way, and your friend is pulling the other way with equal force. What is the net force on the rope?

Gravity acts towards the centre.... Exact opposite of what you just said...

Do you really need me to explain this?

Take the plus as the centre of the object
<----+----> was your analogy

---->+<---- is my analogy

That is in 2 dimensions. obviously gravity acts in 3 but I couldnt be bothered drawing a picture.

I was looking for a value for acceleration due to gravity (or apparent measured value) as I see the equivalence theory referenced here oh so often.
It seems that g should be easily calculated using the time it takes an object to fall to the surface of the earth. Traditionally it is valued at:  Earth's gravitational acceleration at sea level, 9.80665 meters per second per second.  Do you not agree?

Yes. That observation does not depend on the shape of the Earth.

Do you really need me to explain this?

Take the plus as the centre of the object
<----+----> was your analogy

---->+<---- is my analogy

That is in 2 dimensions. obviously gravity acts in 3 but I couldnt be bothered drawing a picture.

I know what you mean by it. I am asking you to justify that this statement is valid:

Gravity acts towards the centre

Oh, and that diagram is in one dimension, not two.

Gravitation on a large scale, electromagnetism on a small scale, strong interaction on an even smaller scale.

No, in your infinite plane model there is no gravitation on a large scale remember?

Try again?

No, in your infinite plane model there is no gravitation on a large scale remember?

Source?

Quote from: Crustinator on October 14, 2009, 11:19:28 AM

No, in your infinite plane model there is no gravitation on a large scale remember?

Source?

Gravity on a large scale would curve the flat earth into a sphere....

Gravity on a large scale would curve the flat earth into a sphere....

Not if it is infinite.

Quote from: mtarlo11 on October 14, 2009, 11:22:00 AM

Gravity on a large scale would curve the flat earth into a sphere....

Not if it is infinite.

We just proved it cannot be infinite.

This is over. You are argueing pointlessly.

We just proved it cannot be infinite.

No, you made the claim that it cannot be infinite and are refusing to justify it.

Quote from: Crustinator on October 14, 2009, 11:19:28 AM

No, in your infinite plane model there is no gravitation on a large scale remember?

Source?

Memory problems?

Gravitation is directional. Infinite mass on one side, and infinite mass on the other side in the same distribution will result in zero net gravitation.

Quote from: Its a Sphere on October 14, 2009, 11:15:55 AM

I was looking for a value for acceleration due to gravity (or apparent measured value) as I see the equivalence theory referenced here oh so often.
It seems that g should be easily calculated using the time it takes an object to fall to the surface of the earth. Traditionally it is valued at:  Earth's gravitational acceleration at sea level, 9.80665 meters per second per second.  Do you not agree?

Yes. That observation does not depend on the shape of the Earth.

So then you can accept that the value (or perceived value so as not to get tied up in semantics) is 9.80665 meters per second per second?  That's at least somewhat refreshing.

Quote from: Parsifal on October 14, 2009, 09:43:11 AM

Gravitation is directional. Infinite mass on one side, and infinite mass on the other side in the same distribution will result in zero net gravitation.

There is still a normal gravitational field, which prevents it from falling apart by creating an equilibrium between its tendency to fall inwards due to gravitation and to expand outwards due to electromagnetism.

So then you can accept that the value (or perceived value so as not to get tied up in semantics) is 9.80665 meters per second per second?  That's at least somewhat refreshing.

Give or take a couple of centimetres per second per second, yes.

Quote from: mtarlo11 on October 14, 2009, 11:24:12 AM

We just proved it cannot be infinite.

No, you made the claim that it cannot be infinite and are refusing to justify it.

infinite area  = infinate mass        IMPOSSIBLE
                 = infinite gravitation  IMPOSSIBLE
                 = infinite energy        IMPOSSIBLE

or are you going to tell me otherwise? and then divide by zero?

infinite area  = infinate mass        IMPOSSIBLE
                 = infinite gravitation  IMPOSSIBLE
                 = infinite energy        IMPOSSIBLE

Why don't you try justifying one of these statements?

There is still a normal gravitational field, which prevents it from falling apart by creating an equilibrium between its tendency to fall inwards due to gravitation and to expand outwards due to electromagnetism.

No sorry. None of that made any sense.

No sorry. None of that made any sense.

Gravitation pulls it together, electrostatic repulsion pushes it apart. When the two are in equilibrium, it is stable. What doesn't make sense about that?

Quote from: mtarlo11 on October 14, 2009, 11:31:51 AM

infinite area  = infinate mass        IMPOSSIBLE
                 = infinite gravitation  IMPOSSIBLE
                 = infinite energy        IMPOSSIBLE

Why don't you try justifying one of these statements?

OK

E=mc2
c2 = E/m
c2 = infinity/infinity

infinity/infinty cannot be determineed

Gravitation pulls it together, electrostatic repulsion pushes it apart.

No. You've already stated that there's zero net gravity. Gravity can't "pull" anywhere. Infinite mass on one side, infinite mass on the other. Remember?

So what keeps it together?

Quote from: Its a Sphere on October 14, 2009, 11:28:35 AM

So then you can accept that the value (or perceived value so as not to get tied up in semantics) is 9.80665 meters per second per second?  That's at least somewhat refreshing.

Give or take a couple of centimetres per second per second, yes.

Ahhhhhh, I'm beginning to see the light now.

So let's walk through an examination of the data presented.

Given:
1) Earth is an flat infinite plane
2) The earth has infinite mass

Quote from: Its a Sphere on Today at 08:45:03 AM
And agreeing with the above, if A (Area) would be infinite, then mass too would be infinite.  Are you suggesting that the earth has infinite mass?
Yes.

3) The earth has a density(rho) that cannot be determined since the earth has neither finite mass nor finite area

Quote from: Its a Sphere on Today at 10:29:10 AM
I'm puzzled by your statement.  Knowing an objects volume (in the infinite plane of a finite thickness argument) and it's mass, it should be elementary math to discover it's density.
Only if its volume and mass are finite.

4) The development of the term governing the acceleration of gravity is governed by:
    http://www.theflatearthsociety.org/forum/index.php?topic=32568.0 which states: g= 2(pi)*G*rho

5) g is accepted as 9.81(roughly) m/s^2

From 4) and 5)
9.81 m/s^2= 2(Pi)*(6.67428E^-11 m^3 kg^-1 s^-2) *(rho)
rho = 9.81*2*pi*(6.67428E-11) [m s^-2 m^-3 kg s^2]
rho= 23,407,945,903 kg/m^2 ( a finite area density of the plane)
This disproves 1) and 2) and 3) and also means that equation in 5) (using Gaussian formulation approximating earth as a plane does not apply meaning the earth is not flat.

Now taking Newton's law of gravity and the Gaussian formulation of Newtonian gravity, the exact same formula you held to be true, so you must accept it only taking the earth to be a shape (let's try a sphere) both reduce to:

g = Gm/r^2

Substituting values assumed for G m and r:
g= 3.986005 E14 m ^3 s^-2/(6.378140E6 m)^2
g= 9.8 m/s^2 matching the observed numbers.

Conclusions
1) Earth is not a flat plane
2) Earth is spherical.

OK

E=mc2
c2 = E/m
c2 = infinity/infinity

infinity/infinty cannot be determineed

Infinity is a limit, not a number, and this must be taken into account. Using your working as an example:

E = mc²
c² = E/m

Since E = mc² gives us the definition of rest energy, we have:

c² = lim_m→∞ (mc²) / lim_m→∞ m
c² = lim_m→∞ (mc² / m)
c² = lim_m→∞ c²
c² = c²

No. You've already stated that there's zero net gravity. Gravity can't "pull" anywhere. Infinite mass on one side, infinite mass on the other. Remember?

I stated that there is zero net gravitation where there is equal mass on all sides. This is not generally true in the infinite Earth model.

From 4) and 5)
9.81 m/s^2= 2(Pi)*(6.67428E^-11 m^3 kg^-1 s^-2) *(rho)
rho = 9.81*2*pi*(6.67428E-11) [m s^-2 m^-3 kg s^2]
rho= 23,407,945,903 kg/m^2 ( a finite area density of the plane)
This disproves 1) and 2) and 3) and also means that equation in 5) (using Gaussian formulation approximating earth as a plane does not apply meaning the earth is not flat.

This works only if the area density of the Earth is uniform, which it clearly is not. What you have calculated is, at best, an approximation as to the mean area density near where your value of g was measured.

I stated that there is zero net gravitation where there is equal mass on all sides. This is not generally true in the infinite Earth model.

One minute you're arguing for a uniform infinite earth. The next you're arguing that it's not uniform.

Shifting the posts is weak Robo.


Of course, I'm forgetting that you don't actually see any compelling evidence that the earth is flat to warrant believing it. Right?

One minute you're arguing for a uniform infinite earth. The next you're arguing that it's not uniform.

Shifting the posts is weak Robo.

It's very nearly uniform in two dimensions. Being a three-dimensional object, however, makes it non-uniform in general.

Of course, I'm forgetting that you don't actually see any compelling evidence that the earth is flat to warrant believing it. Right?

Irrelevant.

It's very nearly uniform in two dimensions. Being a three-dimensional object, however, makes it non-uniform in general.

Sorry you've drifted into nonsense again.

Quote from: Crustinator on October 14, 2009, 12:15:34 PM

Of course, I'm forgetting that you don't actually see any compelling evidence that the earth is flat to warrant believing it. Right?

Irrelevant.

Quite relevant. I've never seen so much conviction in someone with so much self declared lack of conviction. It's uncanny.

Quote from: Its a Sphere on October 14, 2009, 12:01:12 PM

From 4) and 5)
9.81 m/s^2= 2(Pi)*(6.67428E^-11 m^3 kg^-1 s^-2) *(rho)
rho = 9.81*2*pi*(6.67428E-11) [m s^-2 m^-3 kg s^2]
rho= 23,407,945,903 kg/m^2 ( a finite area density of the plane)
This disproves 1) and 2) and 3) and also means that equation in 5) (using Gaussian formulation approximating earth as a plane does not apply meaning the earth is not flat.

This works only if the area density of the Earth is uniform, which it clearly is not. What you have calculated is, at best, an approximation as to the mean area density near where your value of g was measured.

Correct, mean density impling it is finite.

How is it clearly not constant?

http://www.theflatearthsociety.org/forum/index.php?topic=32568.0

It can exist.  The pull would be roughly normal to the surface.

http://www.theflatearthsociety.org/forum/index.php?topic=32568.0

It can exist.  The pull would be roughly normal to the surface.

We already covered this.

Quote from: John Davis on October 14, 2009, 01:14:11 PM

http://www.theflatearthsociety.org/forum/index.php?topic=32568.0

It can exist.  The pull would be roughly normal to the surface.

We already covered this.

Then I don't see what the issue is.  Its solid.  Its correct.  You can't have an argument against it, because if you do you are wrong.  It clearly shows that it would not collapse to a point, or to many points.