Curious phase transition

by flicflex
Tags: curious, phase, transition
Mentor
P: 21,674
 Quote by Ophiolite Hmm. Questionable. The gasesous components of the Earth are almost entirely (perhaps entirely) derived from degassing of the interior once formed from solid planetesimals and from volatiles provided by cometary and asteroidal impact. There is no evidence I am aware of that would involve a meaningful contribution of nebular gases to the proto-Earth. I stand ready to be corrected on provision of relevant citations.
Sigh. Yes, it is a simplification. Yes, we did have nebular gasses, but it was mostly hydrogen and most blew away during Earth's formation and the start-up of the sun. But whether we call the "degassing" process that created our current atmosphere part of the formation of earth or not is yet another hair I don't care to split. It still follows the description I laid out: Heavier substances sink and lighter ones rise. That's why all of the gases are now in Earth's atmosphere instead of its core.

http://en.wikipedia.org/wiki/History...stem_formation
Mentor
P: 21,674
 Quote by flicflex My question could be reformulated like this : trough a mathematical point of view, why does the solid part of earth has a precise radius of 6400km? If the question is too difficult, can you give the answer for the similar question to any planet easier to study?
The reason the Earth's radius is what it is is because that's how much solid matter there was floating around our proto-solar system, in range of Earth's gravity when it formed. There really is nothing more complicated about it than that.
 P: 13 Let me rephrase... which parameters determine the radius (in our case 6371km) of a planet and how (which function) ?
 Mentor P: 21,674 Depends on how specific you want to get, but I'd start with mass and density: V=m/d
 Admin P: 21,722 I am going to throw $V=\frac {4} 3 \pi r^3$ in.
 P: 13 well let's admit the mass, but the density has to be explained ! does earth internal inergy alone can parametrize earth density? In such a case, how do we determine earth internal energy? earth initial internal energy + energy lost in the external system (cooling within the universe) + energy brought by the external system (sun and asteroid crashes) ?
P: 21,722
 Quote by flicflex does earth internal inergy alone can parametrize earth density?
Looks like a word salad to me.
 Mentor P: 21,674 I'll translate, and I was initially going to include this: Temperature (internal energy), heat generation, heat dissipation, thermal conductivity, heat capacity and.....coefficient of thermal expansion. http://en.wikipedia.org/wiki/Thermal...sion_in_solids In other words, as the earth cools, it shrinks. How much? Probably not enough to be more significant than the effect of its oblateness or plate tectonics.
 P: 13 Yes indeed the evolution of earth internal energy determines the evolution of its diameter. But my question is rather how does earth internal energy determines its diameter (statically talking)? borek (numerology salad blabla), i apologize for my lack of vocabulary but I beg you to try to understand the topic behing my babblings rather than despise them
P: 2,490
 Quote by Borek 1. It doesn't. 2. Looking for a logic behind the Earth radius is a pure numerology. IOW: there is no scientific meaning behind.
What you could say about the Earth's radius is that the radius at any point on the surface along a line extending from the center of gravity to outer space is the point of greatest gravitational potential along that line. The acceleration due to gravity on a test object falls as you move toward the center of mass from the surface and also falls as you move toward space from the solid or liquid surface.

However, as pointed out, the solid and liquid surfaces of the Earth consist of different substances than the atmosphere, so there is no phase transition per se. Where the surface is water or water ice, the atmosphere is still mostly nitrogen and oxygen with proportionately very little water vapor.

http://www.syvum.com/physics/gravita...vitation2.html
 P: 246 flicflex, phase transitions in geophysics refer to structural changes in minerals, generally associated with changes in pressure, or temperature. So, phase transitions do play a role in determining the Earth's radius, in as much as they have profound effects on the interior stucture and behaviour of the Earth's interior. The original composition of the planetesimals forming the Earth, the siderophile, lithophile or chalcophile tendencies of the elements, the resultant differentiation into crust, mantle and core, the subsequent phase changes within mantle and core, the consequent initiation and evolution of convection, all of these things in combination have determined the particular radius of the Earth. (Other items could be added.) That said, I'm still not sure I have properly understood what it is you are trying to ask.
P: 3,015
 Quote by flicflex trough a mathematical point of view, why does the solid part of earth has a precise radius of 6400km?
As Borek had said, and I will try to reformulate, there is no special significance behind the numerical value (here 6400) of any physical quantity (here the radius of the Earth) when expressed in particular units (here km). Coming back to our example, one may say that it is so, because the kilometer had been initially defined as 1/10000 part of the distance from the North Pole to the Equator along the meridian passing through Paris.

Assuming the Earth resembles as a sphere, it means that 1/4 of a great circle has a length of 10000 km. But, a quarter of a circlular arc with radius R has a length $R \pi/2$. Then, solving for the radius, we have:
$$\frac{R \pi}{2} = 1.0000 \times 10^4 \, \mathrm{km}$$
$$R = \frac{2 \times 1.0000 \times 10^4 \, \mathrm{km}}{3.14159} = 6.3662 \times 10^3 \, \mathrm{km}$$

Similar examples to your question would be to ask why is the radius of the Earth 3400 nautical miles (look up the definition of a nautical mile), or why is the triple point of water at an absolute temperature of 273.16 kelvin (look up a triple point, and the definition of a kelvin as a unit of thermodynamic temperature).

What would make sense to ask is why does the ratio of two physical quantities of the same kind (eg. lengths) that are related to eachother, have a particular numerical value (but no units or dimensions). I emphasize the phrase "related", because, for example, you may take the Bohr radius (colloquially known as the radius of the hydrogen atom), which has a value $a_0 = 5.29 \times 10^{-11} \, \mathrm{m}$, and calculate the ratio
$$\frac{R}{a_0} = 1.20 \times 10^{17}$$
You may take the cube of the above ratio to find the ratio of the volume of the Earth to the volume of a single hydrogen atom
$$x = \left( \frac{R}{a_0} \right)^3 = 1.74 \times 10^{51}$$
Is there any significance to this astronomical number? Not unless you show a reason behind your choice of the Bohr radius as a length of comparison.

I may argue that the above number represents, at least to an order of a magnitude estimate, the number of atoms of which the solid portion of the Earth is made out of. Indeed, if you assign 1 a.m.u. of mass to each of these "atoms" (look up the atomic mass unit), then their combined mass would be $2.89 \times 10^{24} \, \mathrm{kg}$. Compare this to the mass of the Earth, $5.97 \times 10^{24} \, \mathrm{kg}$, and you are in the right order of magnitude range. But, by no means should you ask why the first result is nearly half of the second! It just turned out that way (we know that the Earth is not made out of hydrogen, nor can we pack spheres to occupy the whole space). But, what it should show you is that the Earth is made up of atoms, and that it is not a white dwarf or a neutron star.

Well, anyway, that was my long winded digression that I hope someone will read through.
P: 13
 Quote by SW VandeCarr What you could say about the Earth's radius is that the radius at any point on the surface along a line extending from the center of gravity to outer space is the point of greatest gravitational potential along that line. The acceleration due to gravity on a test object falls as you move toward the center of mass from the surface and also falls as you move toward space from the solid or liquid surface.
This is an interesting element of answer. But why does the materia doesn't accumulate anymore (isn't solid anymore) beyond that greatest gravitational potential field (6400km)?

 Quote by Ophiolite That said, I'm still not sure I have properly understood what it is you are trying to ask.
Calculate earth's radius with the help of some chosen parameters (something like internal energy, temperature and quantity of materia, magnetic field, i don't know something like that)

 Quote by Dickfore As Borek had said, and I will try to reformulate, there is no special significance behind the numerical value (here 6400) of any physical quantity (here the radius of the Earth) when expressed in particular units (here km). Coming back to our example, one may say that it is so, because the kilometer had been initially defined as 1/10000 part of the distance from the North Pole to the Equator along the meridian passing through Paris. Assuming the Earth resembles as a sphere, it means that 1/4 of a great circle has a length of 10000 km. But, a quarter of a circlular arc with radius R has a length $R \pi/2$. Then, solving for the radius, we have: $$\frac{R \pi}{2} = 1.0000 \times 10^4 \, \mathrm{km}$$ $$R = \frac{2 \times 1.0000 \times 10^4 \, \mathrm{km}}{3.14159} = 6.3662 \times 10^3 \, \mathrm{km}$$ There is your magic number!
Dickfore, you just said that $$\frac{2 \times \frac{R\pi}{2}}{\pi}= R$$

 Quote by Dickfore I may argue that the above number represents, at least to an order of a magnitude estimate, the number of atoms of which the solid portion of the Earth is made out of. Indeed, if you assign 1 a.m.u. of mass to each of these "atoms" (look up the atomic mass unit), then their combined mass would be $2.89 \times 10^{24} \, \mathrm{kg}$. Compare this to the mass of the Earth, $5.97 \times 10^{24} \, \mathrm{kg}$, and you are in the right order of magnitude range. But, by no means should you ask why the first result is nearly half of the second! It just turned out that way (we know that the Earth is not made out of hydrogen, nor can we pack spheres to occupy the whole space). But, what it should show you is that the Earth is made up of atoms, and that it is not a white dwarf or a neutron star.
Your comparison is interesting, but I'm looking for a more precise calculus (something close at least of 20% of the real radius). We obviously can't add the size of all the different atoms of earth, that's why I'm rather looking for a thermodynamic related calculus.
P: 21,722
 Quote by flicflex i don't know something like that
I am afraid that summarizes whole thread.
P: 13
 Quote by Borek I am afraid that summarizes whole thread.
Exactly. Which parameters can determinate earth radius and how.
P: 246
 Quote by flicflex Exactly. Which parameters can determinate earth radius and how.
I have answered that in a general way in post #29. If you wish to get a more precise answer then an extensive study into geochemistry, geophysics and planetology over a five to ten year period should provide the answer you seek. The question is not, in my view, interesting enough to warrant anyone yet having considered it. You could become famous.
 P: 13 Your answer was interesting, but I'm indeed looking for a thermal answer rather than a (geo)chemical answer. Do you think earth can be considered as some kind of solid gaz, or at least as complex system, in order to study it trough a thermal/energetical point of view? Is it what you call an extensive study? Why would'nt it be interesting? Such a method would save obviously a lot of calculus and measures in order to describe few characterisics of any planet.