Re: ILE - The Ideal Lane-Emden Equation

On 21 Jul, 21:37, af...@xxxxxxxxxxxxxxxxxxx (John Park) wrote:
Thomas Smid (thomas.s...@xxxxxxxxx) writes:
On 21 Jul, 11:12, af...@xxxxxxxxxxxxxxxxxxx (John Park) wrote:

I think it makes sense to describe all these planets as consisting
largely of an ideal gas, because even for Neptune, the internal
temperature (as given by the gravitational potential over the virial
theorem) is still about 15,000 K, which should make it impossible for

Could you elaborate on that? What I've read suggest that the outer planets
are mostly slushes (with rocky cores).

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).

It's not clear to me. That's quite a bit denser than Saturn, which has
several times the mass.
>>> Which "virial theorem" are you
referring to, the one relating kinetic and potential energies?

Yes.

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.

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 page http://www.plasmaphysics.org.uk/research/sun.htm
is the electron/proton mass ratio)

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.

Well, the point is that these 'coincidences' confirm a mass-radius
relationship R~M^1/3, but invalidate other exponents.

They don't confirm anything. They suggest a mass-radius relationship, which,
if true, would imply that all other variables, such as composition,
temperature and thermonuclear activity, have no effect. That is
counter-intuitive, to say the least.

Again, the mass-radius relationship R~M^1/3 for solar system objects
(including the sun) stands as an observational fact, independently of
any physical theory. If a theory implies a different relationship,
then it is inconsistent with these observations. Unless you have an
explanation for this inconsistency, it invalidates thus the theory.
And given the formula I mentioned above R=[M(R)*(3-k)/4pi/n(R)/
m]^1/3 , it is apparent that the radius *does* indeed also depend on
the composition (as it depends on the atomic mass m) ,and potentially
the structure (through the index k for the radial density
distribution), but the data clearly indicate that the density
structure is identical for all the objects (including the sun).

You haven't enough points to exclude other possibilities.

There must be lots of stellar mass/radius data from eclipsing binaries. Why
don't you use those results?

Because they are not reliable and accurate enough. The determination
of masses and radii for binary stars depends on a host of assumptions
with regard to distance, orbit, limb darkening etc. These assumptions
are likely to be biased such that they conform with standard theories
regarding stellar structure, so they are by no means direct and
independent measurements (like they are possible for solar system
objects).

I'm not an expert, but frankly I don't believe you.


Do you at least agree that the sun together with the hydrogen planets
in the solar system fits more or less exactly a R~M^1/3 mass-radius
relationship, and that this is based on accurate and unambiguous
figures?

And what about
interferometric or other direct determinations of stellar radii?

There are no 'direct' determinations of stellar radii in the sense of
the word. They are all quite complicated and convoluted procedures
that make a large number of assumptions both with regard to the stars
as well as the measuring instruments and the intervening medium.


>>>> Put it another way: if Uranus and Neptune had the same masses but were 75%
hydrogen instead of 15% and had orbits inside Venus', would you still expect
them to fit your curve?

The fact that they fit the curve suggests that they *do* consist
largely of hydrogen.

You haven't got nearly enough data to claim that.

You can't really blame me that there not more planets in the solar
system (which is to date the only place where we can measure masses
and radii of objects accurately), but the point is that all objects
present obey a R~M^1/3 relationship.

I can blame you for leaping to an extravagant conclusion based on inadequate
data.

I am not making any conclusions. I am just stating that the sun fits
with the hydrogen planets into a R~M^1/3 relationship.


By the way, the smaller planets also follow a R~M^1/3 law for
themselves, but have a consistently smaller radius by about a factor
0.6 (Mars 0.7) (this agrees with their higher average density compared
to the sun and giant planets).

So the 1/3 power law applies to rocky bodies as well as ideal gases? How
do you rationalise that?

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.

Thomas




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