Capacitance between Earth and Moon
I remember that - in one of his columnes in &quot;Electronic Design&quot; - 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, &quot;What is the actual capacitance from the earth to the moon?&quot; 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 &lt;&lt; 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 &quot;sphere&quot; 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 &quot;surrounding spheres&quot; 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.