Let me go over the beginning of my original
note. The 1400 W/m^2 for Earth and and
120 W/m^2 for Jupiter are intercepted energies.
The statement "Jupiter emits out 2.5 more energy
than it receives from the Sun" allows us to
calculate the total energy emitted, provided the
radius of Jupiter is reasonably defined. For the
radius of 7*10^7 m the intercepting (projected)
area is Pi*R^2=1.5*10^16 m^2 (it is not 6*10^16
as I posted in the first message). Consequently,
the intercepted energy rate is 120*1.5*10^16
= 1.8*10^18 J/s.
The unexplained energy emission rate is 1.5
times larger, or 2.7*10^18 J/S. This is four times
less than in my first message. Using 4*Pi*R^2
instead of Pi*R^2 was a mistake. The final
conclusion should thus be modified. The amount
of potassium needed to account for the
unexplained excess energy should be four times
less than 2*10^25 kg; it is nearly equal to the total
mass of our planet.
There one more calculation I would like to make.
But I do not have the information needed. What
is the fraction of the mass of our planet due to
potassium? Knowing this number, and assuming
the percentages are the same on both planets
(if elements with Z<4 are ignored) I would
calculate the mass of Jupiter due to elements
above lithium. Comparing the result with the
known mass of Jupiter (all elements, including
H and He which are dominant) would help to
argue for or against the hypothesis that K-40
is responsible for the unexplained energy
emitted by Jupiter. But I must stop and go to
the airport now. I will be away for six days.
Ludwik Kowalski