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Where is Earth's gravity stongest?

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posted on Jan, 14 2012 @ 11:53 AM
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reply to post by CLPrime
 


That is not necessarily the case (but it is likely). But it seems your mind is set. The calculations are there. I made a nice image. There is not much more I can do
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posted on Jan, 14 2012 @ 11:56 AM
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Originally posted by -PLB-
reply to post by CLPrime
 


That is not necessarily the case (but it is likely).


It is always the case. The more volume you include in the inner section, the more mass it has, so the greater its gravitational maximum. You could define a layer in the middle of the crust, with the "core" being the inner core, the outer core, the mantle, and the bottom half of the crust. You'll find that the maximum in this case is in the middle of the crust.
Keep going out and the maximum will keep increasing until you reach the planet's surface and can't go out any further.

I will say, though, that your illustration was very good. I considered making one, myself, but figured I couldn't live up to the standard you set.



posted on Jan, 14 2012 @ 12:00 PM
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reply to post by -PLB-
 

CL Prime, if you have better math, please share it. I won't have time until later today to review PLB's calculations, but based on the graphics, it's about what I would expect.

The main problem I see with the model is the over-simplified assumptions, though I'm not faulting PLB for this, he stated his assumptions and made the calculations accordingly.

PLB, the over-simplification doesn't take into account the density gradient within each layer of the Earth's structure as shown here:

en.wikipedia.org...


Earth's radial density distribution according to the preliminary reference earth model (PREM).


Given the sudden decrease in density at around 3400km, I would expect to see a local peak similar to what your graphics show for the core surface, but I'd still expect gravity to be higher at the outer surface, with at least part of the reason being the non-uniform density distribution within each layer as shown here.
edit on 14-1-2012 by Arbitrageur because: clarification



posted on Jan, 14 2012 @ 12:06 PM
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Originally posted by CLPrime
You could define a layer in the middle of the crust, with the "core" being the inner core, the outer core, the mantle, and the bottom half of the crust. You'll find that the maximum in this case is in the middle of the crust.


That is possible. But the model would be less accurate, and so would the maximum you find.


Keep going out and the maximum will keep increasing until you reach the planet's surface and can't go out any further.


In this case, you are no longer modeling any differences in mass density at all. The model is simplified to such a degree that it gives incorrect results, and thus will lead to incorrect conclusions. The more accurate mass density information is included in the model, the more accurate the result will be.


I will say, though, that your illustration was very good. I considered making one, myself, but figured I couldn't live up to the standard you set.


I say give it a go
.



posted on Jan, 14 2012 @ 12:10 PM
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reply to post by Arbitrageur
 


That is a very nice graph, and would make it possible to calculate the exact maximum. I may give it a go. I will point out that for my specific model, the exact mass densities of "the core" do not matter. The field outside a sphere behave exactly like a point charge at the same distance. In other words the distribution of mass does not matter as long as it is symmetrical.



posted on Jan, 14 2012 @ 12:16 PM
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Originally posted by -PLB-
The field outside a sphere behave exactly like a point charge at the same distance. In other words the distribution of mass does not matter as long as it is symmetrical.
Outside the sphere, that's true.

Inside the sphere, I think the distribution matters, even if it's symmetrical.



posted on Jan, 14 2012 @ 12:30 PM
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Some layers work as barriers (caps) that can hold pressure/force from bellow and from above. That means you cant use G = P / density x height. To calculate the exact G at earth core.

Cap layers will zero out the downwards force.



posted on Jan, 14 2012 @ 12:36 PM
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reply to post by spy66
 


Gravitational force is affected by mass distribution. Pressure can affect mass distribution, so it can have an indirect effect on gravity but no direct effect.



posted on Jan, 14 2012 @ 12:55 PM
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Originally posted by Arbitrageur

Inside the sphere, I think the distribution matters, even if it's symmetrical.


No, it doesn't, which is what I've been trying so hard to explain since yesterday. It's asymmetrical mass distribution that affects the gravitational potential because it affects the location of the center of gravity. Beyond that, it doesn't matter if the core is denser or if the crust is denser - the maximum of the potential is always at the surface. Always.



posted on Jan, 14 2012 @ 12:57 PM
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Originally posted by -PLB-

... the model would be less accurate, and so would the maximum you find.


Why? How is it any different from what you're doing?



... you are no longer modeling any differences in mass density at all. The model is simplified to such a degree that it gives incorrect results, and thus will lead to incorrect conclusions. The more accurate mass density information is included in the model, the more accurate the result will be.


Mass density doesn't matter as long as it's symmetrical. And, what makes my example simplified? Again, how is it any different from what you're doing? Do you think that a large density gradient actually matters? Because it doesn't.



posted on Jan, 14 2012 @ 12:59 PM
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reply to post by CLPrime
 

The oversimplified model PLB presented seems to show otherwise and you've been invited to present better math if you have it. I haven't yet confirmed PLB's math but at first glance it looks about right.



posted on Jan, 14 2012 @ 01:01 PM
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reply to post by Arbitrageur
 


And yet I've explained why PLB's example is wrong. Over and over again.
I need a break.



posted on Jan, 14 2012 @ 01:24 PM
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reply to post by CLPrime
 

Sorry my friend, this is one of the few times I've seen you be wrong, you're usually right!

This seems to confirm PLB is right! Good job PLB. This is a plot of the earth's gravity right next to the other graph I found of the Earth's density, if I wasn't in such a hurry I wouldn't have missed it the first time.

en.wikipedia.org...

I got that from Wikipedia but they list the source as:

7.^ a b Dziewonski, A. M.; Anderson, D. L.. "Preliminary reference Earth model". Physics of the Earth and Planetary Interiors 25: 297–356.



posted on Jan, 14 2012 @ 01:26 PM
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Aha!




It would seem from the above graphic that maximum gravity is at the edge of the outer core!


EDIT - beaten to the post!
edit on 14-1-2012 by BagBing because: (no reason given)



posted on Jan, 14 2012 @ 01:29 PM
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reply to post by BagBing
 

Yes good intuition there BagBing, better than mine, though I never claimed my intuition was all that great, that's why I have to do the math LOL.



posted on Jan, 14 2012 @ 01:32 PM
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reply to post by Arbitrageur
 


I literally just sat down to write a program to model the earth's gravity, but I wasn't sure of the mass at different locations. I saw your previous image , with the mass distribution, and thought I had everything I needed!!!

No need to bother now, thankfully - I image a detailed enough model would take some serious compution!
edit on 14-1-2012 by BagBing because: (no reason given)



posted on Jan, 14 2012 @ 01:35 PM
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reply to post by CLPrime
 


All I can say is show the math. I do think density distribution makes a difference and I showed my math. I think my math is correct. I do have to say that I use my knowledge of electric fields on gravity. As far as I know Q vs E and m vs g are analogous. I did briefly read this page. It says


The form of Gauss's law for gravity is mathematically similar to Gauss's law for electrostatics.



posted on Jan, 14 2012 @ 01:36 PM
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The answer I gave originally, was it would be strongest at the surface at each pole. Like CLPrime, I was certain that would be the case.

But a little while later I thought some more and concluded it must be underground, because of the density variation. However, to be honest I didn't think it would be all the way down to the outer core.



posted on Jan, 14 2012 @ 01:44 PM
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reply to post by Arbitrageur
 


Ok, I had not read this post yet, I guess its settled then
. I would have taken me quite some time to produce a similar graph (although a fun challenge). The result is different from what I had initially expected, which is that gravity would be highest just below the surface. Good find again (I guess Google makes almost all math and physics obsolete, its making us stupid).



posted on Jan, 14 2012 @ 02:02 PM
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reply to post by -PLB-
 

It's different than what I expected too. This was a fun thread where I learned something as did others. I understand your point about the internet making us stupid, since we look up the answers instead of doing our own calculations. What if this is wrong? And it does say preliminary model...so it's probably just a best guess at this point, but I think it's better than my best guess was. But you could also say the internet is making us smarter if we can find an answer to a question like this, which may have gone unresolved in a pub discussion over a few beers.

Probably 20 years from now (give or take) someone will post a more refined version of this graph using more accurate data, and by then it hopefully won't say "preliminary".
But this preliminaty model is probably good enough for the purposes of this thread.



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