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Re: Jupiter (was PERIHELION etc.)



If I'm not missing something, the K-40 necessary is (would be) 0.3 % of
Jupiter's core -- not unreasonable. I am missing something -- I haven't
compared the heat quantities. However, I'll continue by noting the
earth's is v.~ 0.6 TW due to, again, v.~ 95 ppm (K-40) in the core.
Another thing I'm missing is Jupiter has a very much greater surface
area. O, well.

http://physicsweb.org/article/news/7/5/4

I think this is it:

"The rate of loosing energy must thus be equal
to 180*6*10^16=1.1*10^19 J/s"


0.6 TW is ~ 2 E7 low; with 10 X mass ~ two million times insufficient
if the same %age content of K-40

JM's is off by only 10 X?

bc

p.s.

"That is about 3% of the mass of our planet.
What would the mass of Jupiter be if one
Ignored hydrogen and helium (whose escape
velocities on Jupiter are much larger than on
Earth)?"


My reference gave the mass of jupiter's rock and ice core as ~ ten earth
masses.

Jupiter doesn't have a surface as we (I) think of one. Its
atmosphere's density smoothly increases.

However, here's some data from CRC: mass 318; radius (mean) 10.9;
escape speed 60.2 / 11.2

Enuff; it's midnight.



Ludwik Kowalski wrote:

I asked: how to explain the net emission
of energy by Jupiter?



Bernard speculated: Th, U, etc.? Would K-40
do it? It's earth's claimed energy source.



Let me try to answer Bernard's question. I will
assume that K-40 is the only source of energy.
Similar (Fermi-like) calculations can be done
Th and U.

1) Our solar constant is 1400 W/m^2. Jupiter
is 3.4 times further from the Sun than we are.
Therefore its solar constant must be 3.4^2
times smaller. It must be ~120 W/m^2

2) It is known that Jupiter emits 2.5 times
as much energy as it receives from the Sun.
This means the net emission is 1.5*120=
180 W/m^2. Can it be due to K-40?

3) The radius of Jupiter is 7*10^7 m. This
means that its area is 6*10^16 m^2. The
rate of loosing energy must thus be equal
to 180*6*10^16=1.1*10^19 J/s

4) The energy released by an atom of K-40
is 1.3 MeV. Assuming that all of this energy
becomes heat (ignoring that something like
one half of it is taken away by neutrinos) I will
estimate Jupiter's total radioactivity A.
Expressing 1.3 MeV as 2.1*10^-13 J one has:

A=1.1*10^19/2.1*10^-13 = 5.2*10^31 decays
per second.

5) Knowing the half-life of K-40 (1.3*10^9 years
= 4.18*10^16 s) one has lambda=ln2/4.18*10^16
=1.66*10-17 s^-1.Therefore the number of atoms
of K-40 in Jupiter must be A/lambda=3*10^48.

6) The mass of each K-40 atom is 40 amu or
40*1.66*10^-27 = 6.64*10^-26 kg. The total
mass of K-40( in kg) must thus be equal to
3*10^48*6.64*10^-26=2*10^23 kg.

That is about 3% of the mass of our planet.
What would the mass of Jupiter be if one
Ignored hydrogen and helium (whose escape
velocities on Jupiter are much larger than on
Earth)?

Is it reasonable to accept the hypothesis that
2*10^23 kg of Jupiter's mass is due to K-40?
I do not know. Is it reasonable to assume that
compositions of all planets were approximately
the same 4.6 billion years ago (when they were
formed from cosmic dust etc.)? I do not know.
Ludwik Kowalski