Gravity at the center of the Earth?

More
21 years 9 months ago #3705 by dholeman
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
From my understanding of the Meta view of gravity, if there was a room sized hollow at the center of the earth, a person there would be floating in free-fall with no net effect from gravity.

Is that correct?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Correct for any model of gravity not just the Meta Model.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
What would the "standard" explanation be?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The standard model says that because you would be equally surrounded by matter in all directions the gravity emanating from that matter would exert an equal force <i>'pulling' on you</i> in all directions - hence a net zero directional force, or 'weightlessness'. The Meta Model says that the matter surrounding you shadows you from gravitons <i>pushing on you</i> equally in all directions to the same net effect.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I've always heard that the pressure at the center is extreme... but where does that come from, since there is no weight at the center?
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Just because the net vector of gravity - that is, the 'direction' of gravity - equals zero at the center does not mean the total force of gravity is zero. The total force is quite large because it is the the <i>sum</i> of the gravitational forces acting in all directions.

A simple thought experiment illustrates the point. Have someone split the earth in half and hold the two pieces apart a couple of feet from each other. Insert yourself in the middle centered between the pieces. Then have your partner let go of the pieces and wait a few moments. The two pieces would be attracted to each other and squeeze you with a total force equal to the sum of the gravitational force that they exert on each other.

All the Best,
Don



Please Log in or Create an account to join the conversation.

More
21 years 9 months ago #3786 by Jim
Replied by Jim on topic Reply from
I think the example that shows that the pressure at the center of the Earth (or any sphere of great mass)is not related to the zero gravity is incorrect and the true pressure at the center of a sphere is zero. The reason is simply that pressure requires mass on an area and the mass of a sphere at the center is resting upward in all directions. So, I agree with glittle and not dholeman.

Please Log in or Create an account to join the conversation.

More
21 years 9 months ago #4629 by dholeman
Replied by dholeman on topic Reply from Don Holeman
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I think the example that shows that the pressure at the center of the Earth (or any sphere of great mass)is not related to the zero gravity is incorrect and the true pressure at the center of a sphere is zero. The reason is simply that pressure requires mass on an area and the mass of a sphere at the center is resting upward in all directions. So, I agree with glittle and not dholeman.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

I'm not surprised. <img src=icon_smile_8ball.gif border=0 align=middle>

In the first place Glen asked a question, so there's nothing to agree with him about.

I pointed out the distinction between gravitational force and the pressure due to gravity because they are two different things. You are certainly correct in saying that the mass at the very center of the earth is 'resting upward in all directions', which basically repeats what I said. Hence, it is 'weightless'. That does not mean it does not experience pressure though. It carries all the weight of all the matter above it in all directions. In order to know what that adds up to you need to first determine the pressure (force per unit area) and then multiply it by your surface area.

I'm not smart enough to do the math but if I were I would express the pressure as the integral of the gravitational acceleration acting on the mass present in a column of constant cross section with the radius of the earth expressed as the differential of density with depth but we can simplify the problem by assuming a constant for earth's density. We can also take advantage of somebody else's work in putting all the funny squiggly Greek characters on the web for us:

[url] burro.astr.cwru.edu/Academics/Astr221/SolarSys/hydrostat.html [/url]

It comes to something like 350 gigapascals, which is the same order of magnitude force that can be achieved by a diamond anvil.

Please Log in or Create an account to join the conversation.

More
21 years 9 months ago #4764 by tvanflandern
For those unfamiliar with physics, there is a major difference between "force" and "pressure". As we stand here on Earth, we experience a downward force of 1 g. But we also experience a pressure in all directions of 14 pounds per square inch. (Because the pressure is equalized, we don't notice it much except when we try to go to high altitude, or when we try to remove the air from a container.)

Likewise, in the ocean, the downward force of gravity is the same, but the bouyancy of fish easily cancels that. Nonetheless, the deeper one goes, the greater is the water pressure in all directions. There comes a theoretical depth when it could crush anything. -|Tom|-

Please Log in or Create an account to join the conversation.

More
21 years 9 months ago #4633 by Jim
Replied by Jim on topic Reply from
The pressure at the center of a sphere is no greater than the pressure at the surface. This is a very important fact that is now not known in the field. Assuming as you do that the accepted math is correct is not going to show this, however. First let me say why this issue is a vital detail. Being a low pressure zone a mass center will not support current models of fusion.

Please Log in or Create an account to join the conversation.

More
21 years 9 months ago #4634 by MarkVitrone
Replied by MarkVitrone on topic Reply from Mark Vitrone
If some fusion reactions were occuring at a planet's core, what would the gravity be at the core assuming that the fusion was under extreme high pressures? Would there be exponential shielding effects as the gradient of pressure increased? MV

Please Log in or Create an account to join the conversation.

Time to create page: 0.323 seconds
Powered by Kunena Forum