Re: Jupiter (was PERIHELION etc.)

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

Regarding point (2), I have a couple of questions that, while doesn't
affect any order-of-magnitude estimates, still might affect the overall
numbers...

* Does Jupiter emit 2.5 times as much energy as it *absorbs* or as it
*intercepts*?

* The 120 W/m^2 is intercepted over the cross-section of Jupiter (pi
r^2).  But then you calculate the energy assuming it is emitted over the
entire surface (4 pi r^2).  Does this present a problem with your
estimate of 180 W/m^2?

____________________________________________________
Robert Cohen; 570-422-3428; www.esu.edu/~bbq
East Stroudsburg University; E. Stroudsburg, PA 18301


> -----Original Message-----
> From: Forum for Physics Educators 
> [mailto:PHYS-L@lists.nau.edu] On Behalf Of Ludwik Kowalski
> Sent: Tuesday, January 06, 2004 1:01 AM
>
> > 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.