chongo@nsc.UUCP (Landon Noll) (11/24/83)
what is the potential difference between the EARTH (Ve) and the MOON (Vm)? how much does Ve-Vm change, if at all? assume that you had some conductor of some given cross section connected between the EARTH's axis pole and the MOON's axis pole with enough slack to handle the distance changes: (and assuming that Ve <> Vm) what would the current in the conductor be? would the resistance of the wire prevent a good current from forming? what effect would be produced by this wire being dragged through the EARTH's magnetic field? what would the charges carried by the solar wind do? what effect would the Van Allen belts have on the wire? etc.? this is just an idle thought that i have had for a while. any discussion on this topic would be appreciated. chongo /\../\
leichter@yale-com.UUCP (Jerry Leichter) (11/28/83)
The potential difference between the Moon and Earth must be extremely small. If it weren't, the repulsive or attractive force between the two would rapidly be large enough, compared to gravity, to significantly affect the orbits. (Note that up until 20 years or so ago, we had little independent verification of the Moon's mass - although measures of Earth mass go back 150 years. Now that we have orbited satelites around the moon, we can get the moon's mass pretty accurately.) I have no idea what limits these calculations would actually place, but they must be pretty tight. -- Jerry
leichter@yale-com.UUCP (Jerry Leichter) (12/02/83)
"In particular, two nonpolarizable insulators in an intense electric field..." True, but it sort of begs the original question. If you assume both Earth and moon are truely non-polarizable insulators, then the external electric field has no effect on "the charge on the Moon" relative to Earth, whatever that means. We have to stand back a bit and try to make sense of the question: If the question is: If I consider the Earth and Moon to be more-or-less conducting objects - implicit in the notion of "ground" in the electrical sense, then what I'm asking is: What will a voltmeter read, if one side is attached to the Moon, the other to Earth? In this case, an external xxx (not relevant). Two components contribute to the reading: An Earth- Moon difference in charge - which must be small, by my previous argument - and an outside potential, which would also have to be small, since a conductor placed in such a field WOULD become polarized. If I consider the Earth and Moon to be non-polarizable insulators - which is definitly false, by the way; the Earth's core, at least, is a conductor, and most rocks are quite polarizable crystals - I then have the interesting question: Where is this external field coming from? SOMETHING is alleged to be generating a field between Earth and Moon - but both are moving. So any field would have to be very variable. In fact, such a field probably does exist, centered at the Sun, due to the solar wind, which consists of a lot of charged particles. However, since the Earth and Moon ARE quite polarizable, the field must be pretty small. In this case, you probably don't get nearly as sharp a bound, since you have to figure in the polariza- bility - capacitive constant? Permitivity? - It's been a while - but this is balanced out by the fact that the field must be changing - you would see an attraction whenever the Earth-Moon line was pointing toward the Sun, no effect when it was normal to a solar radius (cosine curve variation). The resulting wavering of the Moon's orbit would be pretty noticable. -- Jerry decvax!yale-comix!leichter leichter@yale
MJackson.Wbst@PARC-MAXC.ARPA (12/08/83)
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.)
MarkBILLW%SRI-KL@sri-unix.UUCP (12/10/83)
A simple thought experiment: consider two capacitors, one with a charge Q and voltage V, the other uncharged. Connect them in parrallel, so they each have charge Q/2, but still voltage V. Now dissconnect them and connect them in series, so that the total circuit has charge Q again, but Voltage V*2. Since the end plates are now twice as far apart, they exert less force on each other. [This is logically equivilent to pulling apart the plates of a capactitor apart. Since you have to do work to pull them apart (cause they attract each other), The voltage (which after all is a measure of potential energy) increases. Voltage and charge are different things, and are not related!] I think. Bill W
dya@unc-c.UUCP (12/18/83)
References: sri-arpa.14471 You have got to be kidding. If you have two capacitors with voltage V/2; and connect them in series, the potential difference at t=0- is either 0 or V (until you put your meter on it, etc.) Try it with two real world capacitors.......Last time I checked, KVL still works. dya