Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Path: utzoo!linus!security!genrad!decvax!harpo!seismo!hao!hplabs!sri-unix!MJackson.Wbst@PARC-MAXC.ARPA
From: MJackson.Wbst@PARC-MAXC.ARPA
Newsgroups: net.physics
Subject: Re: what is the voltage (EARTH based) of ground on the MOON?
Message-ID: <14426@sri-arpa.UUCP>
Date: Thu, 8-Dec-83 11:14:00 EST
Article-I.D.: sri-arpa.14426
Posted: Thu Dec  8 11:14:00 1983
Date-Received: Tue, 13-Dec-83 01:29:06 EST
Lines: 55

Of course I know that earth and moon are not "nonpolarizable
insulators;" that was merely an example (reductio ad absurdum, if you
will) of the separability of "force" and "potential difference."

Look, your argument was that since the electrostatic force on the moon
due to the earth is small (whatever that means) then the potential
difference must be small.  That's just flat out wrong, in principle.  As
I pointed out in the second paragraph of my previous message,

	F = Qm*Qe/Rem**2

which is zero, for ARBITRARILY LARGE Qe [or Qm] if the other charge
happens to be zero.  Of course if one of {Qe, Qm} can be arbitrarily
large then the potential difference between the earth and the moon can
be arbitrarily large.  I didn't say that I thought this was the case,
just that your conclusion (V is small) didn't follow from your evidence
(F(electrostatic) is small).

(Actually, I don't think it is necessary for one of the charges to be
zero.  I did a quick calculation, perhaps not correct, that says that if
the charges are equal and opposite the force is Q**2 times 6e(-8)
nt/coul**2, but the potential difference is Q times 6e(+3) V/coul.  So
if Q were a thousand coulombs the force would be less than a Newton but
the potential difference would be megavolts.)

Reintroducing the rest of the universe, I'm puzzled by your assertion
that in the case of two interacting (induced) dipoles the "resulting
wavering of the Moon's orbit would be pretty noticable."  I don't have
time to do this calculation, however each dipole is proportional to the
external field E, and the force is proportional to the product of the
dipoles divided by the FOURTH POWER of the separation (times orientation
terms of order unity or less, as you point out).  So whereas

	V = E*Rem,

we have that

	Fdipole ~ (E**2)/Rem**4

and it is clear that for some range of E, V is substantial and F is
trivial.  (Again, I am not claiming that the physical situation is as
described above, just that your argument is incorrect.)

Addressing, at last, the original question, Lew Mammel Jr. makes a
number of cogent observations.  If in fact the solar wind equilibrates
the potential between planetary bodies, then a comprehensive answer is
at hand.  The solar wind will "see" the moon's surface and the top of
the earth's atmosphere (which is an excellent conductor).  Thunderstorms
keep the potential difference between earth ground and the top of the
atmosphere at about 400 kV; hence this would be the difference between
"earth ground" and "moon ground."  (The 400 kV figure, and an excellent
discussion of atmospheric electrodynamics in general, can be found in
volume II of the *Feynman Lectures on Physics,* chapter 9.)

Mark