# Can a circuit be powered from Earth with a return ground on Mars?

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.

-
I tried to test it one time but I ran out of copper wire about halfway through. – Phillip Schmidt Aug 19 '13 at 12:59
@JYelton I think the purpose is that this experiment highlights the commonly held belief that current must travel through a closed circuit. – Pace Aug 19 '13 at 13:15
@PhillipSchmidt, did you try putting Mars a mile away from Earth? – trav1s Aug 19 '13 at 13:39
This question seems more like a physics question rather than electrical engineering. – Lie Ryan Aug 19 '13 at 15:36
Every answer here (except for the top one) assumes that AC current requires a complete circuit, which I don't believe is true. I've asked this question on the Physics.SE; hopefully we will get an authoritative answer there. – BlueRaja - Danny Pflughoeft Aug 19 '13 at 21:20

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.

-
First, you're quite right that the charge between planets will change. If the current is relatively small, this can be neglected. For the second bit, if we're going to nitpick, the dielectric constant of even a vacuum is non-zero. So yes, technically everything in the universe is connected electrically. For practical purposes, its useful to regard things which are sufficiently electrically isolated as open circuits. Thinking in practical terms above the fempto-level pedantry, we can confidently say that "for current to flow, there must be a connection to earth" is indeed a misconception. – trav1s Aug 19 '13 at 15:48
What makes you think the potentials of the two bodies are anywhere near equal? They've been electrically isolated from each other for 100's of millions of years, with a plasma flow past them, which affects them differently becuase they have different magnetic properties. Just a cloud a few hundred meters above the earth can charge to 100'000's of V in a few hours. Why wouldn't the Earth and Mars have millions, billions, trillions or more V of potential between them? – The Photon Aug 19 '13 at 16:11
If it equalises to your generated voltage then it will not flow. – JamesRyan Aug 19 '13 at 16:16
I may understand this incorrectly, could you give me a hint? If I have two batteries that are not connected with each other and I hook my device with the positive pole of one battery and the negative pole of the other battery, it will not work. How is this different from the case of the two planets? – David Aug 19 '13 at 16:26
Apologies for any confusion, but is there really a distinction? This answer made a lot of sense to me. All I'm looking to see is if you can take a regular mains AC hot wire from earth and a wire from the ground of a different planet to complete a circuit. 'Current return path' sounds a bit like DC. I may have used "return path" in a way that was unintended. In AC current just oscillates back and forth. – user2476549 Aug 19 '13 at 19:01

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.

-
Similar to my comment on Connor Wolf's answer, the current need not flow back to earth. The current can flow from earth to mars with little issue that it does not return to earth. – trav1s Aug 19 '13 at 10:19
I disagree, given that this is an AC signal of wavelength shorter than the wire. It's not a closed circuit but an antenna, and you can pump AC current into it until it melts. – pjc50 Aug 19 '13 at 14:01
Even if Mars is only one mile from earth, as the OP assumes? – Wouter van Ooijen Aug 19 '13 at 14:46
Hmm, I'd missed that :( – pjc50 Aug 19 '13 at 15:16
"a (very very very small) capacitor" Seems to me two planets would make a very very very large capacitor, but who am I to judge? :) – Michael Kjörling Aug 19 '13 at 17:17

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.

-
"There would be no path for the current to return through" the current doesn't need to return anywhere. Current flows from high voltage to low voltage. 0V could be the the crust of the earth, which conceivably could be the same electrical potential of the crust of mars or HIP 5158. Unless we are talking about astronomical amounts of current, the voltage of mars won't be significantly affected by the charge, so the capacitive effect is totally negligible. – trav1s Aug 19 '13 at 10:14
Current doesn't "return" anywhere in AC, it just moves back and forth. – BlueRaja - Danny Pflughoeft Aug 19 '13 at 15:55
@BlueRaja-DannyPflughoeft Even in AC, current does return. While the generator produces the positive half-wave, current travels one way through your circuit, via the load and the return. While the generator produces the negative half-wave, current travels the other way round. AC is really not that much different from a DC battery being reversed every so and so often. This is even true for high-frequency designs - these have just the added benefit that you can think about waves traveling along a (coax) wire or the air. – zebonaut Aug 23 '13 at 8:28
@zebonaut Please see Does AC current require a complete circuit? - "AC is really not that much different from a DC battery being reversed every so and so often" A DC circuit that wasn't closed would still have a short burst of current while the voltage equalized. So, even your example of a DC circuit that reverses would have a current. – BlueRaja - Danny Pflughoeft Aug 26 '13 at 9:16

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.

-
I'm sure that it is safe to assume at this point that the cable used is very well-shielded. Of course, you'd only ground the shield on one end; otherwise, this whole discussion becomes moot. – Ben Miller Aug 19 '13 at 16:30

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...

-
This answer is cute, but unfortunately is also completely wrong. AC current does not require a return-path. – BlueRaja - Danny Pflughoeft Aug 19 '13 at 23:14
@BlueRaja-DannyPflughoeft, If you cut the neutral tab off the plug of your lamp, will it still light? – The Photon Aug 20 '13 at 0:38
@ThePhoton I hope it doesn't - otherwise it may have a potentially deadly insulation fault. And if it does, the return path is formed via the enclosure, possibly a human body (shock hazard!), and the floor. No return path, no light - loosely quoting Bob Marley... – zebonaut Aug 20 '13 at 5:39
@BlueRaja-DannyPflughoeft A "sufficiently large charge carrier" - you mean a giant electron (i.e. a sufficiently large majority charge carrier) or a giant hole (i.e. a sufficiently large minority charge carrier)? ;-) Seriously, a circuit needs to be closed, even an antenna acts like load resistor as seen from the transmitting amplifier, closing its output circuit... – zebonaut Aug 23 '13 at 8:22
This is the best answer. Vote it up highly. Several of the other answers are also correct, But @zebonaut has done a great job illustrating this and explaining it in simple terms. – SWB Dec 30 '13 at 10:01

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

-
This would be right if earth_gnd and mars_gnd had no capacitance, i.e. the voltages were completely floating. But this is not the case, these planets have huge amounts of capacitance and will strongly resist change in voltages. If mars_gnd is coincidentally the same voltage as earth_gnd, even if they are not physically connected, that lamp will light up as current flows from the source to mars_gnd. It will charge mars_gnd, but the planet is so huge that the voltage wont change significantly. – trav1s Aug 20 '13 at 5:37
@trav1s A capacitor is a two-terminal device; you cannot charge it using one wire. What you can do here is to regard the planets themselves to be the plates of a capacitor. But the distance between them is so great, that you have very little capacitance there. This capacitance is in series with the planet capacitances. Capacitances in series combine such that the smallest one dominates: $1/C_{\text TOT} = 1/C_1 + 1/C_2 + ... + 1/C_n$. – Kaz Aug 20 '13 at 14:06
quite right. When reading my comment, it is necessary to note that any object that can be electrically charged exhibits capacitance, regardless of the number of terminals. – trav1s Aug 20 '13 at 14:35
Sorry no. Capacitance is a phenomenon between two nodes in a circuit, which relates the voltage between them and current flowing between them over time. A capacitor that is connected with only one leg is simply an open circuit; there is no capacitance there. For example, the charge on an acrylic sweater is related to the phenomenon of capacitance, but it doesn't make the sweater a capacitor. Note that a capacitor does not store a net charge. It stores a charge separation: for every negative charge on one side of the capacitor, there is an opposite one on the other side. – Kaz Aug 20 '13 at 18:12
What you are describing is mutual capacitance, and what I am describing is self-capacitance. They are both a type of capacitance. – trav1s Aug 21 '13 at 0:42

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:

1. 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,

2. if you use AC, there is a capacitance between them (see Wouter van Ooijen's answer), which closes the loop,

3. 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.

-
There is no new information in this answer that hasn't already been posted here, and the information restated here is mostly false. – trav1s Aug 21 '13 at 3:04