# How much electricity could an iron core moon passing over a copper deposit on a planet generate?

Assuming the moon is the same size as ours and its core is generating a magnetic field, and the copper deposits are quite large, consisting of a tectonic plate containing huge deposits of native copper, in a fairly high percentage form?

Would this electricity make something like reverse lightning, or would the effects be negligible?

• I’m confused. Are the moon or the copper deposits charged? I don’t know of a way to get electricity simply by revolving uncharged metal around each other. Commented Aug 21, 2018 at 17:15
• If you are asking about "how much would x", you have to name numbers which as a nice side effect would demonstrate that you understand the issue in principal. You should maybe ask "how much could x" if you can't name numbers. Commented Aug 21, 2018 at 18:04
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– L.Dutch
Commented Aug 21, 2018 at 18:46
• @RonJohn Its orbit takes it over the copper field. I know it isn't floating, but I assumed people would understand the context. Commented Aug 21, 2018 at 22:04
• We can't read your mind. If you say "floating" and really mean "floating" by I think, "nah, he can't be that ignorant (or is English as a second language); he must mean 'orbiting'", then I'm going to answer something different from what you asked. Commented Aug 21, 2018 at 22:31

Yes. The reverse of this is seen with Io and Jupiter.

A conductor moving through a magnetic field will develop eddy currents within itself, aligned so as to produce a magnetic field opposing the one that created it. This is how eddy current brakes work for trains.

The huge magnetic field of Jupiter induces a current within its moon Io which "feeds on Io's orbital energy" - it is the movement of conductive Io through the field that gives rise to this current and consequent magnetic field.

The Io Dynamo

Of the large moons--comparable to our own moon or bigger--the outer three are icy spheres, but the innermost one, Io, is heated by tides, and as a result has volcanoes and an ionosphere which is a fair conductor of electricity. Jupiter itself like Earth is a magnet, but one that is 20,000 times stronger; as a result it has a large magnetosphere and a very intense radiation belt.

A dynamo is created in a magnetic field by an electric circuit, part of which is moving relative to the rest (additional conditions must also be met). The circuit may consist entirely of fluids (as in sunspots), but solid conductors can also be involved.

The conditions for a dynamo are fulfilled in the case of Io and and >Jupiter. Both are conductors, and they move quite differently--Io orbits, Jupiter rotates. Furthermore, the plasma between them conducts electricity very well along its magnetic field lines, which act as if they were wires connecting Io and the planet (drawing). One expects a continuous current to flow in this circuit, feeding on Io's orbital energy.

If your moon had a magnetic field (and it does not matter if it is an electromagnet like Earth or some giant fixed magnet) that extended far enough to encompass conductors on its planet it could produce current within those conductors as the moon passed by. Our moon is a long way away to do something like this but the moons of Mars are comparably very close and could serve as your model - a fast-flying magnetic Phobos.

The question then - what would the induced current do? One would think it would go to ground. However lighting comes from the ground up all the time. It has to do with charge difference - atmospheric charge plays a role too. https://earthscience.stackexchange.com/questions/580/why-does-lightning-strike-from-the-ground-up If there is a charge in the sky (from wind, dust - the usual suspects) and an opposite charge induced in the ground a path would form between ground and sky.

Its negligible. Your copper wouldn't be able to build up enough charge or even store enough current for us to notice.

Firstly, your moon is simply going to be too far away to significantly impact the copper.

Secondly the moon also takes roughly a month to orbit the earth, which means that it takes a month for you to go through your current range. As mentioned by @Mathaddict your going to have a frequency of ~4 x 10^-7 Hz which is nothing compared to our standard 50Hz.

Finally, your piece of metal is in contact with the earth. Literally. Any excess current it builds up is going to dissipate into the earth. Its called grounding.

• If I increase the factors contributing could the copper generate an electromagnetic pulse, or is there no way for an orbiting body to create a noticeable effect on the earth with magnetism? Commented Aug 22, 2018 at 0:53
• @ClayDeitas Not with something like what you described. Your setup generates AC current, which is caused by the electrons moving backwards and forwards. You can't actually store AC current, so instead you need a way to push excessive charge into the copper, like when you rub your feet on carpet. I doubt you would ever be able to generate and store enough charge to reach lightning... we can't even generate that much power with current technology Commented Aug 22, 2018 at 1:18
• I was hoping to have a annual natural phenomenon with like moving ferrous sand or soumething, but I guess its just not possible. Thanks anyways. Commented Aug 22, 2018 at 1:22

Just being iron doesn't mean that it produces a magnetic field. Only a moving magnetic field would produce a current.

• So if the core was generating a magnetic field with its north-south alignment perpendicular to its axis, and the axis was spinning perpendicular to the ground, so that the polarity of the magnetic field was reversing above the copper field while facing it, that would work? Commented Aug 21, 2018 at 22:01
• Given that the moon's orbiting, and assuming the planet is rotating, the moon and copper deposit would indeed be moving with respect to one another - assuming I'm visualizing this correctly. Commented Aug 21, 2018 at 22:25
• Yes, it would then create alternating current in the copper. However, the frequency of this alternating current will be equal to the lunar period (~4 x 10^-7 Hz), which makes it pretty unusable for most things. Commented Aug 21, 2018 at 22:29
• How did you calculate that current? Commented Aug 21, 2018 at 23:32
• @ClayDeitas Mathaddict gave you the frequency which is different from the current. The frequency is roughly 30 days and he converted it into Hertz (Number of times per second) Commented Aug 22, 2018 at 0:23