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Could you use a hot wire (120v 60 hz) from earth and a copper rod inserted into another extraterrestial body (mars) to complete a circuit? If impedance wasn't something to worry about (pretend Mars is only a mile away or something).
I've asked this before on other places and got some really conflicting answers.
It depends what the voltage between Mars and Earth is, which we don't know. It is unlikely that this voltage is 0, and could be enormous. Your circuit would receive a voltage of roughly the same magnitude, resulting in what would probably be a spectacular failure of your interplanetary toaster.
Clearing the air
The idea, "for current to flow, there must be a connection to earth," is a common one, and it is totally false. If the misconception were true, circuits on airplanes and satellites wouldn't work since there is no connection to earth. Its quite obvious that such circuits do work, and that a connection to earth is totally unnecessary for some circuits to function properly. The misconception arises from the concept of grounding, but remember that ground is simply a reference voltage. It doesn't necessarily need to be Earth. It could be Mars or any other electrical potential.
Second, for current to flow, a circuit does not need to make a physically closed loop. If point A is fixed at \$ 0V \$ and point B is fixed at \$ V_b \$, they don't need to be physically connected for this property to be true. When we connect a resistor between point A and point B, the current from B to A will be \$I_{BA} = V_b/R \$, according to Ohm's law. It is sometimes helpful to symbolically wire every component to the ground reference voltage. This way we can think of ground connected to point A, and connected to a voltage source which connects to point B. In this case, the two points still don't have to be physically connected, but may be considered to be symbolically connected. In this framework, think of Earth as point A, with our reference voltage of \$ 0V \$, and Mars as point B with some unknown voltage, \$ V_{mars} \$.
Hooking up with Mars
Let's say we can make nearly ideal electrical connections to anywhere (suppose we have a handy little portal). The physical laws governing electricity are the same everywhere in the universe. So what happens when you hook up your circuit with your ground in the Martian crust using this portal? Actually, it depends what what \$ V_{mars} \$ is:
Case 1, \$ V_{mars} \approx V_{earth} = 0V \$
In this case your circuit works perfectly normally. Your circuit has no idea that it is connected to Mars and not Earth since they have approximately the same electrical potential.
Case 2, \$ V_{mars} \gg V_{earth} = 0V \$ or \$ V_{mars} \ll V_{earth} = 0V \$
In the case that the voltage of Mars differs significantly from the voltage of Earth, current will certainly still flow, but your circuit might not behave how you expect. It might blow a fuse, arc weld everything in the vicinity or simply vaporize our brave little toaster, depending on just how huge \$ V_{mars} \$ is.
Voltage between Mars and Earth
We don't really know what \$ V_{mars} \$ would be, since we don't know the net charge of Earth or Mars. There is a paper, "Discussion on the Earth's net electric charge" which gives us some clues:
...Integrated over a sufficiently long time, the net current to or from earth must be zero. If it were not, the potential of the earth would build up to such a magnitude that no force could "shoot" more charges up the potential slope- and once this state is reached, the net current would indeed be zero. This is a dynamic equilibrium.
The problem of a net charge on the solid (and liquid) earth (i.e. the globe) can hardly be answered by starting from the fact that current to that body is zero (always or in the average over a long time); not even within the framework of the "classical picture of atmospheric electricity". There does not seem to be a practical method to measure it.
Basically, we can assume that Earth and Mars have each reached their respective equilibrium charges, but we have no way of knowing whether these charges are net positive, neutral, negative, or what their magnitudes are. Since we don't know the net charges of the respective planets, we can't estimate the voltage between them.
For all intents and purposes, no.
"Earth" connections work by virtue of the fact that all the earth-rods inserted in the ground of a single planet are effectively connected together through the planet's crust. The connection isn't necessarily perfect (e.g. 0Ω, but it's fairly low impedance).
Since the earth and mars are electrically isolated from each other by many millions of kilometers of extremely hard vacuum, there would be no path for the current to return through. As such, you would wind up with something that acted very much like a capacitor, as Wouter van Ooijen said.
I think the answer is actually yes, although not for the reasons that you think.
There are a couple of practical issues even if you reach into your SF box of tricks and pull out 400 million km of indestructible cable.
The first of those is the "solar wind": a stream of high speed electrically charged particles from the sun. This will consist of free electrons and protons. They are affected by the Earth's magnetic field; their entry into the atmosphere results in aurorae such as the Northern Lights. The effect of the solar wind may result in the net charge of Earth or Mars not being zero. This may result in a very large lightning-like discharge on connecting your wire. See e.g. this question.
The wire is also a conductor moving in the magnetic field of the sun, which will therefore induce a current even if it's earthed (marsed?) at both ends. Given its length this may be large.
It's also an antenna, so it will pick up solar RF interference.
Suppose we ignore all of those. There is another effect that comes into play. The wavelength of 50Hz mains is only 6000km, so your cable will function as a transmission line. At any given point current will flow back and forth.
You could model it as a very, very large monopole antenna with a nonconductive sphere on the end, like a car antenna with a pingpong ball on. Assuming that you have a power source large enough to overcome resistive losses in the wire, you can just splice your AC applicance in and it will work. You will probably need much more than 120V on the driving end to achieve 120V at the middle of the wire, even if your cable is superconducting.
The two bodies would form a (very very very small) capacitor, so the circuit would indeed be closed, and a (very very very small) current could flow.
That is the theoretically correct answer. In practice the current would be so small for nearly all practical purposes we could call it zero.
:)
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It depends on the return ground or on the question whether you complete the circuit, using the words from the question.
The confusing part in your picture is the A.C. generator: It is shown without its two terminals and surrounded by the equipotential waters of the Atlantic Ocean. Therefore, all the current it creates will stay right in the Ocean. Another potentially (pun intended) misleading detail of your question is the earth/ground symbol on Mars, and I suspect this is the root of your troubles. Thus, ...
I have taken the freedom to rearrange your setup, just using another picture for the generator, with two terminals. One end of the generator is connected to Earth (ⴲ), the other one goes to a light bulb and from there to Mars (♂), complete with the copper rod you mention. Note that the light bulb remains unlit ("off").
If you want current to flow and the light bulb to be on, you need a return wire, closing the circuit like this:
For the circuit to work, you don't need anything to be grounded or earthed, what you need is a whole circuit where the current can flow in a loop. For the fun of it, let's use the earth/ground symbol arbitrarily and create one circuit that's Earthed and one that's Marsed. Note that this is just an additional symbol used at our discretion to obtain a common reference for the entire thing - it is not something we need for the circuit to work.
What we really needed for the circuit to work was the return wire from Mars to Earth. This return wire actually puts Mars and Earth at the same potential, so we can even consider Earth to be marsed and Mars to be earthed - this is still the same arrangement, no matter if we use no, one or two Earth/Ground symbols:
Now, there may be one last question left: What happens if we use a Ground symbol in the first picture, the one without the return wire? Like this:
Or this:
Well, the bulb stays off. Again: It is not the symbol, it is the return wire that's needed, and we don't have it in either of the last two pictures.
Hmmm... If the circuit doesn't really care about the Ground/Earth symbol, why is it so commonly used by so many? What's the point?
Check this out:
The light bulb is on, but the return link is missing. This is possible because, using ground symbol magic, all parts of a circuit diagram that are attached to a ground symbol are considered to be connected, even if no one took the time to actually draw the wires that connect all the grounded bits of a circuit.
These are all the same, just using a different notation:
The ground/earth symbol is so handy that we use it often, and when we are sloppy, we sometimes don't even pay attention to properly calling it ground or earth. Strictly speaking, an earthed circuit is connected to Earth or safety ground (literally: a big rod next to your house or garden, driven into earth ground), and a grounded circuit can be connected to the (-) end of a battery driven device, with no actual connection to the real ground. This is confusing, but it's jargon and everyone uses it.
Armed with all this knowledge, you can figure out why this is plenty funny (image source)...
Note (1): I have used the astronomic symbols (ⴲ, ♂) for a reason. In doing so, I was free to use the electrical Earth symbol on either planet.
Note (2): The A.C. generator could also be a D.C. battery. For the purpose of understanding this setup, you can think of an A.C. source just like you would of a battery being reversed 50 or 60 times a second (depending on national standards). To explain why D.C. would not work well on earth and in the ground, with no physical return wire, made of metal, consider electrolysis and moisture in the ground, to start with... This is a whole 'nother story...
It's better to abandon the word "ground". One node of your circuit is earthed, the other is marsed. These are separate grounds.
We don't have to go to space to create disconnected grounds. In the following scheme, V1 supplies 120 VRMS at 60 Hz, and that XFMR1 and XFMR2 are 1:1 isolation transformers. Each one creates its own electrical domain where a node can be identified as ground, and another as "hot". Current cannot flow from the "hot" in one domain and the ground in the other, for the lack of a complete circuit, and so the lamp cannot glow.
simulate this circuit – Schematic created using CircuitLab
Unfortunately I have never been to mars so there is no way to test this theory. The 120v 60hz from a generator is calibrated from earth ground. In other words, we are using earth's charge as a zero reference. The reference can be analogizes of speed. For instance, how fast are you moving? That depends on what you're comparing it to: you're starting point; the center of the earth; the sun; the galactic center; or, the person next to you in a race. I postulate that voltage to ground is the same thing. So the net voltage for a grounded circuit would be 120v +/- the difference of charge between Earth and Mars. If I were able, I would test this theory by measuring the voltage of a grounded circuit on mars with a known voltage source. Because it calibrated to earth, the difference would define Mar's charge with respect to Earth's.
The problem is that the drawing you present does not contain a closed loop, so no current should flow.
This is however possible in some cases:
you can put to Earth some charge and the current will flow until charges of two planets are equal, for example you can take some electrons from Mars and send them to Earth,
if you use AC, there is a capacitance between them (see Wouter van Ooijen's answer), which closes the loop,
if Earth's charge is different than zero (case 1), there is a magnetic field around it, as it rotates around the Sun, which (somehow) could be used to induce voltage between planets. I can't imagine how one could control it.
My answer is, that yes - it is possible, but at the moment only in theoretical way.
