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Density Of Earths Inner Core = 232.3 G/cm3


granpa

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You are using an average density of the mantle that is based on what, basalt or granite?

 

Do you realize that only applies to the upper mantle, to a depth of about 700 km? Below that, and extending to a depth of over 2,000 km the mantle is denser peridotite-type rock.

 

The average density of the entire mantle is close to 5,000 kg/m^3. That results a mass of about 4.1 E24 Kg for the mantle, or about 69% of the earth’s mass.

 

Now, in considering the outer core, do you think it is reasonable to assume the entire outer core has the density of 7200 kg/m^3, or would it be more reasonable to think the density is graduated. After all, you claim the density of the inner core averages 232,000 kg/m^3 so are you claiming there is an abrupt change from that large number to 7,200 kg/m^3? Sorry, that makes no sense. The average density of the outer core is more like 11,000 kg/m^3, resulting in a mass of about 1.75 E24 Kg or about 29% of the Earth’s mass.

 

That leaves only 2% of the Earth’s mass for the inner core, or only 1.2 E23 Kg. With a volume of 7.6 E18 m^3, that works out to a density of about 15,000 kg/m^3, which is a little higher than the 13,000 kg/m^3 I gave earlier, due mainly to rounding approximations, but it agrees well with the generally accepted model.

 

Anyway, it is becoming clear you didn’t post this to have a reasonable discussion. You are convinced you are right and will do whatever it takes to defend your idea, including shoehorning the numbers to make it fit and I am wasting my time.

(as usual) :laugh: 

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The density increases abruptly because the outermost electron shell of the iron becomes degenerate. That was the whole point of the thread. That's why I pointed out that each electron shell is three times bigger than the previous shell

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Density of nickle =8.90 g/cm3 (fcc)

Density of iron = 8.6 g/cm3 (fcc/close packing) or 7.87 g/cm3 (bcc)

Density of inner core = 234.65 g/cm3 = 27 * 8.69 g/cm3

27 times denser means that the iron atoms in the inner core are exactly three times smaller.

This means that the iron atoms are no longer being supported by the third electron shell.

The third electron shell has become degenerate and the iron atoms are now being supported by the second electron shell.

Iron has 4 electron shells but the 4th is already degenerate (which is why its a metal and a conductor)

Radius of iron atom in core = ((232/(1.2*0.125))*(4/56))^0.3333 = 1/4.79757 helium radii.

The radius of an iron atom supported by the 1st electron shell would be expected to be 1/13 helium nuclii.

It would appear from this that each electron shell is three times larger than the previous shell.

 

Edited by granpa
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Again it seems like something you could have easily done it yourself

It seems the density figures for the Earth are too vague to back up your interesting idea of a super dense inner core of the Earth because using the average density from the upper and lower figures of a leading Google site: http://hyperphysics.phy-astr.gsu.edu/hbase/Geophys/earthstruct.html  only gives a figure of 12096kg/m3 (see below). Manipulating the uncertain density figures available for the various layers below the Earth’s surface can allow virtually any desired result for the density of the relatively small inner core. That said I did find this a thought provoking post, especially considering that the gravity field strength falls off to zero when travelling down beneath Earth’s surface and yet the time dilation effect keeps on increasing to reach a maximum at the center of the Earth.  

Crust                     40km thick, density 2700kg/m3, volume 2.03x1019m3, mass 5.481x1022kg

Upper Mantle   600km thick, density 3500kg/m3, volume 2.744x1020m3, mass 9.60x1023kg

Lower Mantle    2300km thick, density 5000kg/m3, volume 6.193x1020m3, mass 3.0965x1024kg

 Outer Core        2211km thick, density 11000kg/m3, volume 1.6157x1020m3, mass 1.777x1024kg    

Inner Core          1220km thick, density 12096kg/m3, volume 7.606x1018m3, mass 9.2x1022kg

 

Earth Totals        6371km thick, av. Den. 5515kg/m3, volume 1.0832x1021m3, mass 5.98x1024kg

Note:  Density = mass/volume. Density units can be written as 2700kg/m3 or 2.7g/cm3.

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It seems the density figures for the Earth are too vague to back up your interesting idea of a super dense inner core of the Earth because using the average density from the upper and lower figures of a leading Google site: http://hyperphysics.phy-astr.gsu.edu/hbase/Geophys/earthstruct.html  only gives a figure of 12096kg/m3 (see below). Manipulating the uncertain density figures available for the various layers below the Earth’s surface can allow virtually any desired result for the density of the relatively small inner core. That said I did find this a thought provoking post, especially considering that the gravity field strength falls off to zero when travelling down beneath Earth’s surface and yet the time dilation effect keeps on increasing to reach a maximum at the center of the Earth.  

Crust                     40km thick, density 2700kg/m3, volume 2.03x1019m3, mass 5.481x1022kg

Upper Mantle   600km thick, density 3500kg/m3, volume 2.744x1020m3, mass 9.60x1023kg

Lower Mantle    2300km thick, density 5000kg/m3, volume 6.193x1020m3, mass 3.0965x1024kg

 Outer Core        2211km thick, density 11000kg/m3, volume 1.6157x1020m3, mass 1.777x1024kg    

Inner Core          1220km thick, density 12096kg/m3, volume 7.606x1018m3, mass 9.2x1022kg

 

Earth Totals        6371km thick, av. Den. 5515kg/m3, volume 1.0832x1021m3, mass 5.98x1024kg

Note:  Density = mass/volume. Density units can be written as 2700kg/m3 or 2.7g/cm3.

The only way to achieve densities like that would be to make the atoms smaller. Atoms don't become smaller under pressure.  Atoms become degenerate.

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The density increases abruptly because the outermost electron shell of the iron becomes degenerate. That was the whole point of the thread. That's why I pointed out that each electron shell is three times bigger than the previous shell

"Degenerate" in quantum chemistry means that 2 or more states have the same energy. That does not seem to be what you mean. So what do you mean?

 

The electronic structure of atomic Fe is [Ar] 3d6 4s2. The 4s electrons are those principally involved in metallic bonding in metallic iron, as the 3d orbitals are quite strongly pulled in by the nuclear charge, though they do contribute to some extent. This is discussed, along with the phase diagram of iron under high pressures, in this Wiki article on iron: https://en.wikipedia.org/wiki/Iron 

 

The article points out that the high pressure phase diagram has been extensively studied, precisely because of its importance to the structure of planets such as the Earth. There is no mention of any such collapse of iron atoms leading to a new phase. 

 

I can think of no grounds for thinking that anything would happen to the orbitals of the iron atom under pressure, other than that the electrostatic and exclusion principle repulsion between electrons would go up, raising the energy levels of both the atomic orbitals and the molecular orbital "bands" that are formed from them in the metallic structure.

 

If you assert that some of these orbitals change in a fundamental way, you will have to explain what would give rise to such a thing and how it would take place.  

Edited by exchemist
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 In metals it is useful to treat the conduction electrons alone as a degenerate, free electron gas while the majority of the electrons are regarded as occupying bound quantum states. This contrasts with degenerate matter that forms the body of a white dwarf, where all the electrons would be treated as occupying free particle momentum states

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 In metals it is useful to treat the conduction electrons alone as a degenerate, free electron gas while the majority of the electrons are regarded as occupying bound quantum states. This contrasts with degenerate matter that forms the body of a white dwarf, where all the electrons would be treated as occupying free particle momentum states

It may be useful in white dwarves, but it is actually wrong in most commonly encountered instances. The observed properties of metals are generally better accounted for by treating the bands as arising from a continuum of molecular orbitals, formed by merging of the outer atomic orbitals. When you do this you do not use the notion of an "electron gas", as there is no ionisation. 

 

I see you are quoting without attribution a passage from the following Wiki article: https://en.wikipedia.org/wiki/Degenerate_matter, so I am reporting you for plagiarism.  

 

The centre of the Earth is nowhere near the conditions for a white dwarf. It is modelled using perfectly conventional phases of metallic iron, as the linked article in my previous post makes clear. And there is no evidence in the high pressure phase studies I referred to of any partial collapse of the atoms at such pressures.

 

But at least I now understand what you mean by "degenerate" in this context. You are referring to the idea of degenerate matter, evidently.  

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said nothing about 6,500,000 bar

..because no such pressure is present at the Earth's core.

 

Addendum: It occurs to me that what you have done is start from a hypothesis for which there is no evidence, namely this notion of iron atoms collapsing under certain pressures, yielding a very high density phase of iron and then you have assumed this collapse in order to create the gravitational pressures necessary (according to your model) to bring it about. 

 

But the whole house of cards depends on this idea of the collapse of the iron atoms. You need to do a lot more on that to show it makes any kind of sense. 

 

Here is the Wiki article on metallic bonding which indicates the degree to which the free electron model has been superseded: https://en.wikipedia.org/wiki/Metallic_bonding

 

though it does say it is still used a lot in education.

Edited by exchemist
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