Capacitance from Earth to Moon

I believe in the KISS principle (don't use calculus if algebra will do).

Here is my solution to determine some limits. I used the formula for parallel flat plates using the diameter of the moon for the plates. Don't know where I got it then, but will use this one now from the book Electronic Communication by Robert L. Shrader.

C = 0.225 * K * A / S

Where C = capacitance in picoFarad

K= dielectric constant

A = area of one plate in square inches

S = spacing between plates in inches

The dielectric constant of space is 1, so it can be simplified to C = 0.225A/S

First we need the area of the moon as a flat plate in square inches. The area of a circle is pi*r^2, where r is the radius. Dividing the diameter by 2 we get 1079.5 miles. One mile = 5280 feet so 1079.8 * 5280 = 5,699,760 feet. One foot = 12 inches so r = 5,699,760 * 12 = 6.84E+7 in. A = pi*r^2 = 3.14 * 4.68E+15 = 1.47E+16 in^2.

Next we need to find the distance (S) in inches. 238,857 * 5280 * 12 = 1.51E+10 in.

Now we plug these values into the formula. C = 0.225 * 1.47E+16 / 1.51E+10 = 2.19E+5 (pF). One picofarad is 1 million microfarad so C = 0.219 uF. This is the extreme lower limit.

Now lets use the area of the earth as a flat plate for the plates. Radius = 7926/2 = 3963 miles * 5280 * 12 = 2.51E+8 in.

A = 3.14* 2.52E+8^2 = 1.98E+17 in^2.

Now plug into the formula. C = 0.225 * 1.98E+17 / 1.51E+10 = 2.95E+6 pF = 2.95 uF.

My manager explained that with a sphere, some of the "lines of force" can reach around to the back side, so using the area of a flat plate is not sufficient. Lets use the entire area of the earth as a sphere. A = 4*pi*r^2 = 7.92E+17 in^2.

Now C = 0.225 * 7.92E+17 / 1.51E+10 = 1.18E+7 pF = 11.8 uF. This is the extreme upper limit.

I submit that the correct value must lie between 0.219 uF and 11.8 uF. The second calculation looks about right so my second guess is 3 uF.

I wonder if using a formula based on the international MKS system of units would present the same answer.

I wrote a BASIC algorithm many years ago to do these types of calculations for me. I probably didn't get it right then and I would probably screw it up again.

One problem I see is that you are using 2 different plate sizes when the formula was to be used for a parallel plate capacitor where the two plates are of equal area. I would probably sum the two areas then divide by 2 and use that as my plate area.

I have no idea what the relative permittivity of earth's atmosphere is; it might be possible that the value would not be negligible even given how thin it is relative to the great thickness of the rest of the dielectric.

In any case I would use a standard MKS formula and compare it to Shrader's to see if they come close.

Speaking of capacitance.

What would the mass and dimensions be of a capacitor that could store the equivalent energy of a 1000 KG mass moving 30 m/s given today's dielectric materials?

If you could make those numbers manageable then it might be possible to solve the acceptance problem in electrical regeneration of brake energy in automobiles and trucks.

Gav

I wonder if using a formula based on the international MKS system of units would present the same answer.

I'm sure it would, as long as the constant was correct for the units used.

One problem I see is that you are using 2 different plate sizes when the formula was to be used for a parallel plate capacitor where the two plates are of equal area. I would probably sum the two areas then divide by 2 and use that as my plate area.

But that would give an approximate value. Not that I need accuracy in this case, it's the principle of the thing.

Here is an attempt using metric units:

Using your original numbers,

Diameter of the Moon: 2159 miles (3.475e6 m, 3.475e8 cm)

Diameter of the Earth: 7926 miles (rounded off)(1.276e7 m, 1.276e9 cm)

Average distance between the two: 238,857 miles(3.844e8 m, 3.844e10 cm)

From the Handbook of Chemistry and Physics (66th edition, 1985):

The capacitance of a parallel plate capacitor is C=KA/4πd (1 Farad = 9e11 electrostatic units; capacitance will be in electrostatic units if distances are in centimeters)

We agree that the dielectric constant of space is 1, so this simplifies to:

C=A/4πd electrostatic units [π is pi - it does not look good in this font)

Using the cross-section of the moon for the area,

C=π(D/2)^2/4πd = (D/2)^2/4d

substituting the values, C=(3.475e8/2)^2/4*3.844e10 = 3.019e16/15.38e10 =0.196e6 electrostatic units.
0.196e6 eu/9e11 eu/F = 0.0218e-5 Farads or 0.218 microFarads


Now from the Standard Handbook for Electrical Engineers (5th edition, 1922):

The capacitance of an air capacitor [parallel plate] is C=A/4πt*9e5, where Capacitance is in µF, A in cm^2, and t in cm.

This is exactly the same formula as the one from HofC&P, except for the exponent 5vs11. Just a little factor of one million!

OK, I tried... Anyone else? Someone please check my math!

dk;

In the MKS system the unit for length is the meter.

Also; why do you show a square root function as an exponent? Where did that come from?

I would probably work the equation as C=(8.85E-12* (pi*r^2))/Distance.

C would be in Farads

r would be the radius of the moon in meters.

D would be the distance in meters.

It looks pretty straight forward.

I think its going to be about C farads.

Gav

Thanks for taking the time to look at my effort!

I would have preferred to use MKS, but both of my references (old and really old) specified cgs, so that is what I used. Its been a long time since I last did similar calculations. I can't believe I used to be able to read those reference books without glasses!

Where is there a square root? If I wanted to express a square root, I would either use √ (X) or (X)^0.5.

I do recognize the permittivity of free space in your equation, but don't remember why it is there... Please, what is the source of your equation?

I'm clearly open to learning more...

Using surface area seem reasonable to me.

If you had two circular plates and wanted to find spheres that have the same capicatance, you would end up comparing circle to sphere of rotation. So, in the calculation you get;

Rc = √(R_s³)

You said as much, but I was a bit concerned about the 'can reach around the back side' expression.

As to the two bodies being different sizes, I'd settle for average of the equivalent circular plates. In other words. Area to be used is the average of Earth and Moon's surface area.

Does that make any sense to anyone ? It does in my little world, so I'm content.

Last minute thought. I might settle for just the moon's surface area because, well, I don't exactly know how to put it. Oh well, I was content for a moment.

"I'd settle for average of the equivalent circular plates. In other words. Area to be used is the average of Earth and Moon's surface area. Does that make any sense to anyone?"

NO! According to your numbers, the surface area Earth/Moon ratio is 49.5:1. If you were to construct a parallel-plate capacitor with plates of equal width having that ratio, seen edge on it would look like this:

Although there are fringe effects, the ability to store charge is mostly limited by the area of the small plate. As a capacitor, the two plates must store equal and opposite charges. I suspect that the average area would be just about one order of magnitude too big!

Hence the last paragraph (that's the one at end of post);

Last minute thought. I might settle for just the moon's surface area because, well, I don't exactly know how to put it. Oh well, I was content for a moment.

Thanks anyhow.

ps - I didn't provide any numbers.

I must confess that as soon as I saw "Does it make sense to anyone?, I stopped reading and started my reply. I didn't read the last sentence at that time.

And yes, I should have said: according to the OP's numbers...

Dick

I appreciate your effort. I am not interested in their ability to hold charge, but the mutual capacitance. According to my figures in post 1, the ratio of the two as circular plates is 13.5 to 1. Since the total area is 4 times that for each, the ratio stays the same (13.5 to 1). Still, the average would be wrong number.

Doesn't capacitance imply storing equal and opposite charges on the electrodes/plates?

With such a large separation, I visualize the electric field of each as essentially normal to the surfaces at all points. If that is the case, we need to use the spherical surface, not the disc surface. I'm open to correction on everything!

I don't understand your "total area is 4 times that for each"...

Doesn't capacitance imply storing equal and opposite charges on the electrodes/plates?

I may have misunderstood your intent. I thought you were referring to formula 1 of the second link of post 8 which gives the value 153 uF. The Cmutual formula (2) gives 3.2 uF. That's what I meant by capacitance from Earth to moon.

I don't understand your "total area is 4 times that for each

I had used the area for a flat plate with the diameter of the moon and the earth. The area of a sphere is 4 times the flat plate area.

Having worked with electronics for over 60 years, it had never occurred to me that a single conducting object would have capacitance! That is now clear...

Since by default I think of a capacitor in the charge storage mode, I think of the capacitance of the earth-moon system as in transferring charge from one to the other. I still have no idea what might be the significance of "capacitance from Earth to moon".

I had not yet read link 2 post 8. I have now read all of it (skipping across many of the longer formulas), and understand much of it. I will never pretend to understand it all!

And OMG, I did use the volumes of the spheres, not the surface areas. Your 4 is clearly correct!

not the surface areas

rofl - I made exactly the same mistake ! Just goes to show how important a team meeting is. We all drop a clanger now and again. Luckily in our cyberspace world it doesn't matter, but such things demonstrate how important teamwork is.

As there is no scope for beneficial use of the relationship for any practical purpose, does it really matter?

That is sooo discouraging. I was hoping to do a cheapo power factor correction scheme using earth/moon capacitance. (Well, the capacitor would be free, if not the wiring.) This just take the wind out from under my wings.

--Alice Kramden

Hey - here's a thought - can an installation on the Moon be earthed, or would it be mooned instead?

Although I have been mooned, I think you have a different idea of it. Considering the capacitance, a ground on the moon would be an AC earth for frequencies above 1kHz, yes? There must be an application such as a giant antenna for listening to the aliens out there.

I would think that trying to ground out on the moon would be impossible as it would be too dry to conduct unless you took a bucket of water with you but I'm guessing you'd spill most of it before you got there.

Bazzer

You solve that problem if you use the USA terminology: we don't earth our electric devices, we ground them. I believe the surface of the moon could also be called ground.

Now on the other hand, I believe what we call grounding is pretty much dependent on making contact with water in the ground. That might take a pretty deep grounding bar on the moon...

How interesting;

When I was in school I built a rail gun that fired ball bearings reasonably well using spaced coils a rotary switch and a car battery. I had a conversation in which I suggested using capacitors to energize the coils a bit faster without the risk of 'letting the smoke out'

I was told my dreams of a farad would be "the size of the earth".

The "self capacitance" of earth is approximately 700μF.

http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

Using the same calculation, the self capacitance of the moon is approximately 200 μF.

The calculation of the capacitance of earth to moon is fairly complex, but 1F is just too far outside any reasonable calculation.

Without going into details here, I'd accept values ranging from 1.5 μF to 155 μF depending on how you specifically define the system parameters.

See this paper for reference

http://itstriangle.com/~fme/spheres.11-03-99.pdf

Hi mjb (the coffee man?),

At first I thought that you first link was irrelevant, and then I read your second link. I found it very informative. I learned something, and that's why I am here on CR4. I found the section "The easy way" was informative but not that easy! So formula 2 is the one I was looking for, but it is useless without the "where" list that tells what each letter means and the units they are in. That's one of my pet peeves. It gives the correct answer (for ac coupling) to be 3.2 uF.

-S

Sadly, no connection to the coffee corp.

Both links provide some some interesting information regarding why finding a simple single answer is not as easy as one might think. In at least one calculation, the self-capacitance (or stray capacitance) of each body is a couple orders of magnitude larger than the mutual-capacitance. This is an unusual situation which does not often occur in practical lumped circuit design.

For papers like the one linked, and especially for Journal publications, one should assume SI units unless defined otherwise. The author of this paper assumes prior knowledge and does not provide a convenient table defining variables, constants, and units.

The ones relating to equation (2) are:
eo = 8.854e-12 (F/m)
r = radius (m)
d = distance (m)
pi ~3.14159 (no units)
k = fudge-factor (uncertain value and no units)

quick units check...
C=k*4*pi*eo*r1*r2/d=(F/m)*(m)*(m)/(m)=(F) ... units are consistent

Using SI units, I can also calculate earth-moon capacitance values ranging from 0.22uF to 159uF. These are consistent with similar calculations in other unit systems. The large range of results is due to various assumptions and simplifications of the system parameters. For now, it seems that a practical answer lies somewhere in the range of 1.5uF to 159uF.

The 1F value was probably a verbal or memory error from your retired manager (not you).

I give you a GA for the "where" list. The constant e₀ turns out to be the permittivity of a vacuum. I have to assume the k to be the dialectic constant which is 1 in this case, so we can ignore it. Solving equation 2 here:

C = 4*pi (12.56636) * 8.854 * 1.7344E6 * 6.378E6 / 3.844E8= 3.202E6 = 3.2uF

Now if we double the constant, then we can use the area of each 'front side' hemisphere rather than the whole sphere area. If we multiply it by 4, then we should have the formula for flat plate capacitors of different sizes. This can be tested in the lab. This still leaves the question of how much do the 'back side' hemispheres contribute to the total C (mutual).

"The 1F value was probably a verbal or memory error from your retired manager (not you)"

I appreciate the confidence, but admit to some memory errors from time to time.

I agree with your boss that with two nearly spherical bodies this far apart that the effective surface area is the whole surface of the smaller sphere, but this only increases the area by a factor of 4 from the 0.22 uF you calculated using the moon's cross sectional area to almost 1 uF. The added fringe effect of having the Earth's surface area being much larger than the Moon may bring some added small capacitance to get 0.88 uF closer to 1 uF. I suspect that 1 uF is what your boss actually calculated. I suspect an understandably fuzzy memory dropped the micro (10^-6) scalar.

Now the area of the Earth maybe slightly larger when you consider the ionizing layer of the Van Allen Belts but this will still only change the fringe effect. I do not see this making six orders of magnitude difference.

"I agree with your boss that with two nearly spherical bodies this far apart that the effective surface area is the whole surface of the smaller sphere"

He didn't say that. I got the impression that the far side of each sphere would have a smaller effect than the close side, and that it was a fringe effect.

"The added fringe effect of having the Earth's surface area being much larger than the Moon may bring some added small capacitance to get 0.88 uF closer to 1 uF"

I find your use of the word fringe to be strange in this case. If the earth's area is 3 times the moon's then there would be capacitance from the moon to all 3 areas. They would add up, would they not? So the three 1uF capacitors in parallel would be 3 uF.

"I suspect that 1 uF is what your boss actually calculated. I suspect an understandably fuzzy memory dropped the micro (10^-6) scalar."

I may be old, but my memory has not been known to drop decimal places. I suspect that his instructor did not know the actual value, and graded him on his method which was surely exotic mathematically. Or he may have been mistaken when he told me.
Here is an example of my memory: In the 6th grade I read in the book All About Light that the speed of light is 186,262 miles per second. That value differs from todays value of 186,282 mps because it was copyright in 1926 as indicated by the Roman Numerals. (It referred to an earlier measurement of c, I found out later.) The book was near the middle of the north wall in the library at the elementary school. It was the one on the far left of a set of "All About" books. The 3rd one from the right was All About Radio. That's where I got my start in electronics. My teacher (Mrs. Leavitt) was distressed when I chose my next book from the east wall: Bartholomew and the Oobleck by Dr. Seuss.

OK, now that we've got the Capacitance sorted out... How's about the Working Voltage?

I want to thank everyone who participated in this thread. It was a success. Not only do I have the formula I was looking for, but I learned a lot about the earth-moon relationship. Some of you are thinking about things you never thought about before. Who knows where that will lead? God speed to you all.