Is there a capacitance between the Earth and the Moon, and if there was enough potential difference, could a discharge strike occur?

  • \$\begingroup\$ Two spheres of unequal radius is going to turn into a really nasty equation. At the end of the day, there will be a \$\dfrac{1}{d}\$ term that makes the result extremely small. \$\endgroup\$
    – Matt Young
    May 23, 2014 at 14:40
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    \$\begingroup\$ I really like this question because it made me imagine the Moon randomly shooting the Earth with huge lightning bolts. I would guess the capacitance does exist, but due to the large distance between the "plates" (if you create a model where the two bodies are just flat plates), it is very minuscule. \$\endgroup\$
    – dext0rb
    May 23, 2014 at 14:42
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    \$\begingroup\$ This might be a good question for what-if.xkcd.com \$\endgroup\$ May 23, 2014 at 15:04
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    \$\begingroup\$ @dext0rb You are a such a maroon! Why would use a model of flat plates capacitor when the moon and earth are obviously spherical? \$\endgroup\$
    – dext0rb
    May 23, 2014 at 18:47
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    \$\begingroup\$ @NickAlexeev There isn't an approved migration path for that \$\endgroup\$
    – W5VO
    May 23, 2014 at 18:48

4 Answers 4


The capacitance between two plates varies as:

$$C = \frac{eA}{d}$$

in which \$d\$ is the distance between the plates, \$A\$ is the area of the plates and \$e\$ is the Coulomb constant. $$e = 8.9 \times 10^{-12}$$ Distance from earth to moon: $$d = 4 \times 10^8\text{ meter}$$ Approximate equivalent earth surface: $$A = (1.28 \times 10^4)^2$$ Therefore, $$C = \frac{8.9 \times 10^{-12} \times 1.64 \times 10^8}{4 \times 10^8} = 2.39 \times 10^{-11} = 10\text{ pF}$$

The numbers were truncated to the nearest the third place.

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    \$\begingroup\$ Your answer is 7 orders of magnitude smaller than the other one \$\endgroup\$ Dec 2, 2021 at 17:15

I remember that - in one of his columnes in "Electronic Design" - the late Bob Pease has shown how to calculate this capacitance. Just now I have found an addendum to the original contribution: Here it comes

Quotation R.A.Pease:

I received a lot of answers after asking the question, "What is the actual capacitance from the earth to the moon?" There were a few odd ones at 0.8µF or 12µF. But about 10 guys said it was 143 or 144µF. They used the formula:

$$C = 4x(\frac{l}{r_1} + \frac{1}{r_2} − \frac{2}{D})−l$$

valid for \$r_l, r_2 << D\$.

NOW, my original estimate of 120µF was based on this approximation: The capacitance from the earth to an (imaginary) metal sphere surrounding it, 190,000 miles away, would be 731µF. (If that surrounding sphere were pushed out to 1,900,000 miles away, the capacitance would only change to 717µF — just a couple percent less. If the "sphere" moved to infinity, the C would only decrease to 716µF.) Similarly, the C from the moon to a surrounding sphere 48,000 miles away would be 182.8µF. If the two spheres shorted together, the capacitance would be 146.2µF. I guessed that if the spheres went away, the capacitance would drop by perhaps 20% to about 120µF, so I gave that as my estimate. But removing those conceptual "surrounding spheres" would probably only cause a 2% decrease of capacitance. That would put it in close agreement with those 10 guys that sent in the 143µF figure.

But THEN 6 readers wrote in LATER — from Europe — all with answers of 3µF. I checked their formulae, from similar books, in several different languages. They were all of the form:

$$C = \frac{4\pi \times \epsilon \times ( r1 \times r2 )}{D}$$

multiplied by a correction factor very close to 1.0. If you believe this formula, you'll believe that the capacitance would be cut by a factor of 10 if the distance D between the earth and moon increased by a factor of 10. Not so! Anybody who used a formula like that, to arrive at 3µF, should MARK that formula with a big X.

Finally, one guy sent in an answer of 159µF. Why? Because he entered the correct radius for the moon, 1080 miles rather than 1000. That's the best, correct answer! / RAP

Originally published in Electronic Design, September 3, 1996.

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    \$\begingroup\$ Now, how can we measure it? ;) \$\endgroup\$
    – dext0rb
    May 23, 2014 at 17:18
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    \$\begingroup\$ What's all this electricity stuff, anyhow? \$\endgroup\$
    – HL-SDK
    May 23, 2014 at 18:09
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    \$\begingroup\$ I guess you could scale all the dimensions down and put two charged spheres in a vacuum? But maybe there are some weird effects in space. \$\endgroup\$
    – dext0rb
    May 23, 2014 at 18:27

I believe the answers are

1) Edit: see other answer about Bob Pease

2) There's no theoretical reason why not, but there are a number of practical reasons:

  • It requires a colossal amount of charge. Wikipedia claims the breakdown voltage of vaccum is 20 MV/meter. The moon is 384,400,000 meters from the earth. That puts the minimum voltage at 7,688,000,000,000,000 volts.

  • Where would this charge come from?

  • The "solar wind" contains a constant stream of charged particles moving at speed. On entering Earth's atmosphere this results in the Northern Lights. On encountering a planet with a very large non-neutral charge it will tend to attract opposite charges and repel like charges, gradually reducing the net charge to zero.

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    \$\begingroup\$ I like imagining a moon with 7.7 petavolt potential. \$\endgroup\$
    – mskfisher
    May 23, 2014 at 17:32
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    \$\begingroup\$ I am imagining a massive discharge between the first moon-lander and the moon, then again when it returned to Earth...definitely what-if.xkcd material. \$\endgroup\$
    – Mar
    May 23, 2014 at 20:57
  • \$\begingroup\$ Actually, the Electric Universe folk made exactly that claim, although with respect to the Deep Impact mission. There are web sites that claim imagery of the collision shows 2 flashes, which they claim consist of a "preflash" caused by electrical discharge followed by the energy released by the actual collision. Also, Velikovsky made the same claim about arcs between earth and Venus during the close approach by Venus after it was ejected from Jupiter (!). It is also entertaining to calculate the attractive forces resulting from the 7.7 PV potential. Interesting orbits result. \$\endgroup\$ May 23, 2014 at 21:10

It is straightforward to calculate the capacitance of any two conductors. Place equal and opposite amounts of charge on each conductor then calculate the voltage between them. By definition, C=Q/V.

In the case of the Earth and Moon the calculation is difficult because the charges are not distributed over perfect spheres but oblate spheroids. To a reasonable approximation though we can assume that they are spheres.

With this approximation, the electric potential difference is roughly (to about 0.3%) equal to the difference of potential of each body at its own surface. This is a bit strange, but because the Moon is so far away the electric potential of say the Earth at the Moon is very small when compared to the electric potential of the Moon itself.

The mutual capacitance is quite small compared with the self capacitance of the Earth and Moon separately. The self capacitance of the Earth is about 709 microFarads and that of the Moon is about 193 microfarads. The effective capacitance of the pair is 1/709+1/193=1/Ceq, so Ceq=152 microfarads. Again, it is odd that the capacitance between the Earth and Moon is not dependent upon the Moon's orbital radius, but that is the answer.

To do this problem exactly requires you to integrate the electric field between the Earth and Moon over any path between them then divide this voltage into the charge that you used to create the field. This will show a small dependence upon separation. As a last comment, this is a nice problem in that it shows that the conductors themselves hold charge and store energy in their respective electric fields. Capacitance must account for all of this energy.

Normally, the mutual capacitance dominates as in a parallel plate capacitor with a small gap between the plates. But the capacitance of a parallel plate capacitor, where the plates-size-to-gap ratio is very small, is just the sum of the capacitance of each plate in isolation!


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