Re: ILE - The Ideal Lane-Emden Equation

On 25 Jul, 19:14, af...@xxxxxxxxxxxxxxxxxxx (John Park) wrote:
Thomas Smid (thomas.s...@xxxxxxxxx) writes:
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.

The point is that if Saturn is a "hydrogen planet" it's easy to explain
Neptune's density as reflecting a different composition. If Neptune were a
"hydrogen planet" how could you explain Saturn's density?

I don't know what causes Saturn to make an exception here. With
hydrogen being already the lightest element, it shouldn't really be
less dense. The only potential causes I can think of at the moment is
that either Saturn has an unusually thick atmosphere (which results in
the apparent radius being significantly larger than R (as used in the
equation above)), or that an internal magnetic field causes a
deviation from the simple gravity/pressure equilibrium.

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).

So where's the radiation?

Well, in case of the Sun it is obvious, Jupiter is also known to emit
radiation on its own, and for the other planets it may be just not
detectable due to its weakness.


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.

Particularly if you're expecting ideal gas behaviour.

At sufficiently high temperatures all substances should behave like an
ideal gas because all molecular and atomic bonds will be destroyed,
i.e. there will just be a high density plasma of individual nucleons
and electrons elastically colliding with each other. 10,000K or so may
probably not be enough to fully achieve this for iron, but it might
already partially behave like an ideal gas.

I wrote:

This is true for a 1/r potential which implies esentially an ideal gas. My
calculation gives V = k*G*m^2/R for the gravitational self-energy of a
sphere. If the density is constant k = 0.6; if the density goes as r^-2,
k = 0.3.

This was a mistake. If density goes as 1/r^2 your result for the energy seems
correct.

But: In that case the mass is: integral 4*pi*rho*(r^2)*dr ~ integral 4*pi*dr
= 4*pi*r. So where is your m ~ r^3 relationship?

Your linear relationship describes the partial (relative) mass, not
the total mass. The latter is in general (for a density distribution
n(r)~r^-k) given by

M(R) =n(R)*m*4pi* Int[dr*r^2*(R/r)^k](0...R) =

=n(R)*m*4pi/(3-k)*R^3 ,

where n(R) is the number density at the edge R and m the atom mass.

Thomas



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