Re: ILE - The Ideal Lane-Emden Equation

On 23 Jul, 15:08, af...@xxxxxxxxxxxxxxxxxxx (John Park) wrote:
Thomas Smid (thomas.s...@xxxxxxxxx) writes:

Yes, I have read this as well, but it is evidently wrong (an average
density of 1.2g/cm^3 clearly shows that hydrogen is the main
constituent).

The actual values appear to be 1.3 and 1.6 g/cc--Heptune is more than
twice as dense as Saturn, and more than three times as dense as
liquid methane. You haven't addressed this point.

One should be careful with using notions like 'liquid' or 'solids' for
those temperatures (10,000K+). I doubt that any of the molecular bonds
making up the liquid will survive under those conditions. What you
should have then is just a plasma of nuclei and electrons, and usual
density data will simply not be applicable here.

Under the assumption that Uranus was made of hydrogen I used the VT to
estimate a mean temperature of 5000 K,

My calculation according to E=G*M*m/R/2 yields about 13000 K for
Uranus.

I may check that and get back to you later. (Incidentally I don't believe
that's quite the self-gravitational energy of a sphere. Integrating
spherical shells gives me a factor of 0.6 instead of 0.5.)

The factor 0.5 comes from the virial theorem which says that the
average kinetic energy is 1/2 of the average potential energy.
Also, you shouldn't forget the 1/r^2 density structure. The latter
could explain the discrepancy between your and my calculation.

In any case
applying the VT here is circular--it essentially assumes approximately
ideal gas behaviour.

The Virial theorem holds for any bound system in a state of
equilibrium (or something that is close to an equilibrium).


which may or may not be consistent
with your core temperature of 15,000 K. But clearly the surface temperature
is much lower than that, and molecular gases--especially hddrogen--are quite
transparent in the optical range. So why aren't these planets emitting
viisble light?

The surface is only there in the fist place because collisional
processes cool the gas if the density is below a certain threshold. It
is cooled down by about a factor 1/1800 (which according to the
suggestion on my pagehttp://www.plasmaphysics.org.uk/research/sun.htm
is the electron/proton mass ratio)

How do collisional processes cool anything?

Yes, sorry, I wanted to say 'collisional excitation processes'. These
obviously cool the gas as particles lose energy by exciting atomic
transitions (which then results in radiation).

Incidentally I also estimated a VT temperature for the Earth, assuming it
was made of iron: 40,000 K. I wouldn't trust either estimate.

I wouldn't insert the atomic weight of iron here. The mass density of
the earth is only 5 times higher than for the hydrogen planets. So as
an effective value I would insert 5 times the hydrogen mass here. This
gives about 17,000 K.

Are you saying you don't believe the core of the Earth is iron? (And see
comment at end.)
[...]
Incidentally I find it a bit easier to think about your model as implying
that density is constant.

As indicated above already, at these temperatures it isn't really
appropriate to think in terms of usual materials. At least as far as
the mass-radius relation is concerned, it appears to better (or at
least sufficient) to use some effective average atomic weight here (as
given by the average density).
But anyhow, I don't want to push the case too far here by including
the denser (and less massive) planets as well in my ideal gas model.
It would be difficult to justify here without getting into details of
the physics of the corresponding materials for these conditions. It is
still remarkable though that even here a R~M^1/3 relationship roughly
applies.


Even for the smaller planets, the temperature in the interior is still
a couple of thousand degrees. Apparently, this is still sufficient to
make the material compressible like an ideal gas (which will thus also
result in a 1/r^2 density increase towards the center).

So how compressible is iron at 2000 K? And what's your evidence for the
temperatures?

The melting point of iron is about 1800K and the boiling point 3100K,
so at least for the latter value one should expect that all lattice
and other molecular structures in the material become destroyed. At
this point the matter should simply become a plasma of ions and
electrons and should behave like an ideal gas.

Direct measurements of the internal temperature are obviously not
possible. The figures one can find are merely estimates from model
calculations.

OK. I'm having to re-educate myself a bit. But how do you deal with the fact
that seismic data indicate that only 15% of the Earth's volume (maybe
30% of the mass) is fluid--presumably liquid not gaseous--and the innermost
core is _solid_?

Again, I can only say what I said above already: one should be careful
with using notions like 'liquid' or 'solids' for those temperatures
and densities. The behaviour of matter under these conditions isn't
really known, so those data are difficult to interpret.
The point is that apparently still a R~M^1/3 relationship roughly
applies even for heavier elements.

Thomas


.

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