The source of Earth's magnetic field is a dynamo driven by convection current in the molten core. Using some basic physics principles (Maxwell's equations, fluid mechanics equations), properties of Earth (mass, radius, composition, temperature gradient, angular velocity), and properties of materials (conductivity and viscosity of molten iron) or other relevant facts, is it possible to estimate the strength of the field to order of magnitude (about one gauss)?

Descriptions I've seen of the geodynamo all refer to extensive numerical computation on a computer, but can we get a rough idea with simple estimation?

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    $\begingroup$ Interesting question. My instinct is no, there's no simple explanation that can be made without some high-level assumptions about the speed of iron currents within the Earth's core and the magnetization of the iron etc. But I can't be sure. $\endgroup$ – David Z Nov 3 '10 at 22:33
  • $\begingroup$ Well, I don't think the iron is magnetized because it's molten. Perhaps the speed of the currents could be estimated based on things like the temperature gradient, viscosity, and density, but I don't really know much fluid dynamics to try to answer that. $\endgroup$ – Mark Eichenlaub Nov 4 '10 at 4:56
  • $\begingroup$ @MarkE Could you revise the question title? See the comment below. $\endgroup$ – Mark C Nov 4 '10 at 22:52
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    $\begingroup$ @MarkE Perhaps it could read "How can I make an order-of-magnitude estimate of the earth's magnetic field?" As it is stated, I simply understood you to be asking for a "numerological"-type question which boils down to why the unit "Gauss" is defined as such. I almost came back here from this page to rail at how 0.30 is not "roughly one Gauss". $\endgroup$ – Mark C Nov 4 '10 at 22:59
  • $\begingroup$ I agree with Mark C's latest comment, retitling the question as suggested would make it much clearer what you're really asking about. $\endgroup$ – David Z Nov 7 '10 at 8:07

Try looking at the figures in a 2005 simulation by Takahashi et.al. in Science magazine, that at least show recurring reversals at http://www.sciencemag.org/cgi/content/full/309/5733/459?cookietest=yes.

Given that the viscosity, structure, and heat generation of the core are all to some degree unknown, and that the process may depend upon parametric amplification and even parametric resonance, this seems a pretty good beginning.

  • $\begingroup$ Thanks for the link. I was not able to understand much of it, though. It looked like that were reporting on the methods and results of a computer simulation only. $\endgroup$ – Mark Eichenlaub Nov 8 '10 at 18:07
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    $\begingroup$ I decided to mark this answer as accepted because it accurately points out what we know about this phenomenon. I asked Sterl Phinney, an astrophysicist who teaches an order-of-magnitude physics class at Caltech, the question in person. He said that although we can estimate several relevant factors (and went into details that I didn't fully understand), there's no known way at this time to make an order-of-magnitude estimate. The supercomputer simulations are pretty much the knowledge we have. $\endgroup$ – Mark Eichenlaub Nov 14 '10 at 2:28

Live on earth is protected from solar wind by the earth's magnetic field. Charged particles from the sun (mostly) penetrate the earth's atmosphere with great velocity. These particles can be trapped by a magnetic field to follow circular path's around the magnetic field lines, thereby losing their energy due to collisions or bremstrahlung.

From first principles we can try to make an estimate of the strength of the magnetic field required to trap charged particles arriving with great velocity.

Starting with a lorentz force and a circular movement we have: $Bqv = m\frac{v^2}{r}$, so $B= \frac{mv}{qr}$.
$v$ is the velocity of the particle, approx. light velocity, order of magnitude $10^8$ m/s. $q$ is the charge of the particle, elementary unit order of magnitude $10^{-19}$ C. $m$ is the mass of the particle, approx. proton mass, order of magnitude $10^{-27}$ kg. $r$ is the radius wherein the particle has to be trapped, at the most 10 km (height of the atmosphere) $10^4$ m. This gives $B \sim 10^{-4}$ T $= 1$ Gauss.

We have to appreciate the intelligent design ...

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    $\begingroup$ Ha! Very amusing Gerard. Order-of-magnitude estimate based on the anthropic principle! (If the field were weaker, we wouldn't be here to observe it). $\endgroup$ – Mark Eichenlaub Nov 8 '10 at 18:06
  • $\begingroup$ Right! I am glad you can appreciate. $\endgroup$ – Gerard Nov 8 '10 at 21:39

One answer might be Benford's law which state that the probability of lediding digit being 1 among natural constants is approx 30%

How to interpret Benford's law:
Any constant and measurment of "new" phenomen can be in any range. Ok max magnitude for known universe is approx 10^60, but still. Picking random number from this range is not evenly distributed but logarithmic distributed.

example: We pick random number from 1 to 10^10
There is equal probability to pick number from 1 to 10^5 and from 10^5 to 10^10, this way we can prove that there is most likely to pick number that begins whit 1 and least likely whit 9. We can check this rule just by looking on logarithmic scale or doing some math.

Measuring of earth field, gravity or something else obeys this law and this is the reason that is most likely that you will measure 1*10^n.

Other reason that field is 1 gauss is that 1 is just magnitude/order of field. (by the way wikipedia article about this topic says that it is between 0.3 and 0.6 gauss) http://en.wikipedia.org/wiki/Earth's_magnetic_field#Field_characteristics

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    $\begingroup$ Haha, interesting. I never knew about this "law". $\endgroup$ – Noldorin Nov 3 '10 at 22:17
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    $\begingroup$ It is interesting but it doesn't really answer the question - for instance, why 1 and not 11, 120, 0.000015, or something like that? Besides, it's a probabilistic statement, not a reason for why a particular number has the value it does. So I think I have to downvote this. (Sorry! It's nothing personal) $\endgroup$ – David Z Nov 3 '10 at 22:25
  • $\begingroup$ I think that this kind of question is like question why is earth gravity 10 ms^-2 and speed of light 3*10^8 ms^-1 and h/ is 10^-34 Js and why on earth is earth pressure 1 bar. $\endgroup$ – ralu Nov 3 '10 at 22:43
  • $\begingroup$ Yeah, and Benford's law doesn't address any of those numbers either. $\endgroup$ – David Z Nov 3 '10 at 22:58
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    $\begingroup$ @MarkE Then perhaps you should re-title the question. For example, "Why is the mass of the proton rougly 1E-27 kg?"" could be the same kind of question, but, as given, they both are asking why our units have he magnitude they do. $\endgroup$ – Mark C Nov 4 '10 at 22:51

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