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Where is Earth's gravity stongest?

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posted on Jan, 13 2012 @ 05:49 PM
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reply to post by -PLB-
 


If the Earth were a perfect sphere, the maximal gravitational potential would always be at ground-level, because the ground-level and the gravitational surface would be the same thing. It still doesn't matter what density the core is, as long as it's symmetrical. The density of different layers inside the planet doesn't matter, only the distribution of their mass.

Even if you increased the mass of the inner core, the maximal potential would still sit at the surface. This maximum would increase, but it would stay right where it is.




posted on Jan, 13 2012 @ 05:57 PM
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reply to post by CLPrime
 


Imagine a uranium core and a polystyrene outer layer, both spherical. The maximal gravity would not be on the surface of the polystyrene, but closer to the core. Its the same argument why its not in the atmosphere, The atmosphere is not dense enough.

Though, the atmosphere could counter the effect of the denser core. So, i'm not certain of my answer.

Edit: on second thought, the atmosphere does not counter the effect, the net field anywhere inside a sphere is 0.
edit on 13-1-2012 by -PLB- because: (no reason given)



posted on Jan, 13 2012 @ 06:03 PM
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reply to post by -PLB-
 


The OP and I have been through this scenario already. The polystyrene layer and the inner (once iron, now uranium) core are a single mass. The maximum is at the surface of the polystyrene.
The atmosphere and the Earth are not a single mass, and the atmosphere has no surface.
These two cases are not comparable.



Edit: on second thought, the atmosphere does not counter the effect, the net field anywhere inside a sphere is 0.


By that logic, the entire planet should have no gravity wherever you are. If you're going to include the atmosphere, then it must add to the mass of the Earth, which would increase its total potential and somewhat counteract what you're suggesting.
Although, what you're suggesting doesn't happen, and the atmosphere doesn't add mass to the Earth.
edit on 13-1-2012 by CLPrime because: (no reason given)



posted on Jan, 13 2012 @ 06:10 PM
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reply to post by CLPrime
 


I disagree, and I think it wont be hard to show using not too complex calculations. I may give it a try tomorrow.

ps, sorry, I mean empty sphere (looking at just the effect of the atmosphere).



edit on 13-1-2012 by -PLB- because: (no reason given)



posted on Jan, 13 2012 @ 06:12 PM
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reply to post by -PLB-
 


That will decide it. I await your results. And I may just do some calculations of my own.

ETA: I know what you meant, but consider this. The atmosphere doesn't add to the Earth's potential because it's a hollow sphere. Now, put yourself 1 mile underground. Everything above you is a hollow sphere, so that doesn't add to the Earth's potential, either. Now, put yourself 1 mile from the center. Again, everything above you is a hollow sphere, so it still doesn't add to the Earth's potential. That means that none of the planet, according to this logic, will add to the Earth's gravitational potential.
This is obviously false. A hollow sphere has no net gravitational potential inside it, but it does add to the total gravitational potential of the system, which definitely has an effect on the location/magnitude of the maximal potential.
edit on 13-1-2012 by CLPrime because: (no reason given)



posted on Jan, 14 2012 @ 01:56 AM
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reply to post by CLPrime
 


The atmosphere does add mass to earth potential, because at surface atmospheric pressure is approximately 1 bar = "14.5038 psi (pounds per square inch)". That is the total weight of atmosphere at surface per square inch.

Earth gravity pool is strongest were a body enters earth atmosphere. As the body falls towards ground the body will start to reduce speed because the body encounters more pressure/compressed atmosphere.

A solid will weigh more at 100 000ft than it would at surface. Because of the pressure differential at those two altitudes.



posted on Jan, 14 2012 @ 03:56 AM
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The calculations turned out to be pretty simple. I use a core with density 1 and outer layer of density 0.5. Everything is normalized to simplify it.

Gravity for the core alone: g=r for r1
Gravity for the outer layer g=r/2-1/(2r^2) for 1/r/R

where R is the outer radius of the outer layer. Combined:

g=r/2+1/2r^2 for 1/r/R

(note that / is smaller than, but the website does not allow that character)

This gives

for r=1, g=1
for r=R=1.25, g=0.945
for r=R=2, g=1.167

So as can be seen, there is a dip (local minimum) between r=1 and r=2. You can calculate the exact location of this minimum by taking the first derivative, but that is not of much relevance here. In any case, this proves that, depending on the thickness of the outer layer, maximal gravity will be either on the "surface" of core or on the outer surface of the outer layer.

Now looking at the earth, for the core I assume a homogeneous perfect sphere with a density of 5,515kg/m3 and an r of 6360km (so mantel and core are both modeled as "core" here). The crust has a density of 3,000 kg/m3 and a thickness of 35km. (values from Wikipedia). So the core is approximately twice as dense (similar as the example above), and the crust thickness 35/6360km=0.55% compared to the core.

I think that without calculating any further, it can be seen that gravity is highest on the "surface" of the core. However, to know the exactly place where gravity it greatest, you need to know the density distribution of what I call the core. If the outer mantel is again of low density, it may well be that the highest gravity is even deeper.

Edit, I made a beautiful graphic to show what is going on:





edit on 14-1-2012 by -PLB- because: (no reason given)



posted on Jan, 14 2012 @ 04:00 AM
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reply to post by spy66
 


Just to add why this is not true (as is shown in the calculations above), the decrease of gravity as result of the mass of the earth follows the inverse square law. The increase of gravity as result of the mass of the air molecules does not make up for this decrease. The net result is a decrease in gravity as you move away from the surface. (until you reach a certain distance, but the atmosphere is not that thick)
edit on 14-1-2012 by -PLB- because: (no reason given)



posted on Jan, 14 2012 @ 09:02 AM
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reply to post by BagBing
 
I doubt this is relevant to your question but the article may hold some science answers you find helpful ...peace
A Matter of Some Gravity wattsupwiththat.com...



posted on Jan, 14 2012 @ 11:02 AM
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Originally posted by -PLB-
I think that without calculating any further, it can be seen that gravity is highest on the "surface" of the core.
Nice graphic PLB, but the lower right graphic you made shows a slightly higher gravity on the outer surface than on the surface of the core, doesn't it?



posted on Jan, 14 2012 @ 11:11 AM
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reply to post by -PLB-
 


Okay. Let's say I repeat those calculations with any arbitrarily defined layer - say I do them now with the solid inner core, the liquid outer core, and the mantle being a single chunk of mass. Compared to the crust, these have much more mass, so they should certainly cause the maximal gravitational potential to lie under the crust. If you repeat your calculations for this case, you'll find a much higher maximum at the top of the mantle as compared to the surface of the Earth.
And you can choose any layer you want, you'll get the same results - a maximum at the surface of whatever layer you choose. But, they all can't be right, can they? If so, then that indicates that there exist as many maxima as you want to make, and they are all valid.

Gravity doesn't work like that. The entire planet is a single mass, and its maximum lies at its surface.
edit on 14-1-2012 by CLPrime because: (no reason given)



posted on Jan, 14 2012 @ 11:22 AM
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reply to post by Arbitrageur
 


The specific sketch shows gravity for R=2. When you want to know g for R2 (at least, that is what g will converge to). In case of the earth, the crust is about R=1.0005. You can see in the sketch that at r=1.0005 gravity is lower than on r=1. I have to note that the sketch is a free hand approximation of the calculations and not 100% accurate.
edit on 14-1-2012 by -PLB- because: Corrected R



posted on Jan, 14 2012 @ 11:26 AM
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reply to post by CLPrime
 


You can get a local maximum at the surface of each layer, but that does not mean it is also a global maximum. There is only one global maximum.

And I beg to differ, gravity does work like this. Well, I am not trained these specific physics, but as far as I know it behaves the same as an electric field.
edit on 14-1-2012 by -PLB- because: (no reason given)



posted on Jan, 14 2012 @ 11:28 AM
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reply to post by -PLB-
 


Indeed. And the global maximum is at the surface.



posted on Jan, 14 2012 @ 11:29 AM
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reply to post by CLPrime
 


Both my calculations and sketch show that this is not the case.



posted on Jan, 14 2012 @ 11:31 AM
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reply to post by -PLB-
 


Again, you chose a layer which only included the inner core. A layer large than that will give you a greater maximum further out, but still below the crust. Taking the entire planet's mass into account, you will that the maximum maximum is at the surface.
Your calculations are skewed to produce the result you want. Not intentionally...but it's a bit of a self-fulfilling prophecy.



posted on Jan, 14 2012 @ 11:33 AM
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reply to post by CLPrime
 


I did no such thing. I explained what I meant with "the core". With core I mean all the mass of the earth exept for the crust. All mass is accounted for. My model isn't missing anything, and isn't set up to get a specific result.
edit on 14-1-2012 by -PLB- because: (no reason given)



posted on Jan, 14 2012 @ 11:38 AM
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reply to post by -PLB-
 


So, why stop at the mantle? Why not include the crust as well? The OP was speaking of the greater density of the inner core as compared with the outer layers. So, what justification do you have for not including the crust in your definition of "core", given that the densities of the two outer layers (crust and mantle) are quite similar, especially as compared with the core proper?



posted on Jan, 14 2012 @ 11:43 AM
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reply to post by CLPrime
 


For the simple reason that I do not know the densities of the other layers, but do know the density of the crust. If you know the densities and radius of the other layers, you can model those too, and see if another global maximum shows up.



posted on Jan, 14 2012 @ 11:50 AM
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reply to post by -PLB-
 


That's what Wikipedia's for.
I can tell you exactly what you'd find. You'd find that each local maximum, at the surface of each layer, decreases the closer you get to the core. That's because less and less mass is contained by the inner section the further in you go. And the opposite happens, of course, as you move out - the local maxima increase with increasing distance, because more and more mass is being contained by the inner section.
The only true global maximum occurs when the inner section includes the entire mass of the planet. This maximum is always located at the planet's surface.




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