Kapazität zwischen Erde und Mond

19

Gibt es eine Kapazität zwischen der Erde und dem Mond, und wenn es genügend Potentialunterschiede gibt, kann es dann zu einem Entladungsschlag kommen?

capacitance

— Skyler
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Zwei Kugeln mit ungleichem Radius werden zu einer wirklich schlimmen Gleichung. Am Ende des Tages wird es eine

Begriff, der das Ergebnis extrem klein macht.

\frac{1}{d}

$\dfrac{1}{d}$

— Matt Young

1

Ich mag diese Frage wirklich, weil ich mir vorstellte, der Mond schieße zufällig mit riesigen Blitzschlägen auf die Erde. Ich würde vermuten , die Kapazität tut exist, aber aufgrund der großen Entfernung zwischen den „Platten“ (wenn Sie ein Modell erstellen , in dem die beiden Körper sind nur flache Platten), es ist sehr winzig.

— dext0rb

6

Dies könnte eine gute Frage für what-if.xkcd.com sein

— Nick Alexeev

1

@ dext0rb Du bist so ein Kastanienbrauner! Warum sollte ein Kondensatormodell mit flachen Platten verwendet werden, wenn Mond und Erde offensichtlich kugelförmig sind?

— dext0rb

3

@ NickAlexeev Es gibt keinen genehmigten Migrationspfad dafür

— W5VO

-1

Die Kapazität zwischen zwei Platten variiert wie folgt:

C = \frac{e A}{d}

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

$d$ $A$ $e$

e = 8.9 \times 10^{- 12}

$e = 8.9 \times 10^{-12}$

d = 4 \times 10^{8} meter

$d = 4 \times 10^8\text{ meter}$

A = (1.28 \times 10^{4})^{2}

$A = (1.28 \times 10^4)^2$

C = \frac{8.9 \times 10^{- 12} \times 1.64 \times 10^{8}}{4 \times 10^{8}} = 2.39 \times 10^{- 11} = 10 pF

$C = \frac{8.9 \times 10^{-12} \times 1.64 \times 10^8}{4 \times 10^8} = 2.39 \times 10^{-11} = 10\text{ pF}$

Die Zahlen wurden auf den dritten Platz gekürzt.

— user66283
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13

Ich erinnere mich, dass der verstorbene Bob Pease in einer seiner Kolumnen in "Electronic Design" gezeigt hat, wie man diese Kapazität berechnet. Gerade habe ich einen Nachtrag zum Originalbeitrag gefunden: Hier kommt es

Zitat RAPease :

Ich erhielt viele Antworten, nachdem ich die Frage gestellt hatte: "Was ist die tatsächliche Kapazität von der Erde zum Mond?" Es gab ein paar ungerade bei 0,8 uF oder 12 uF. Aber ungefähr 10 Leute sagten, es sei 143 oder 144µF. Sie verwendeten die Formel:

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

$C = \frac{4 π \times ϵ \times (r 1 \times r 2)}{D}$ $C = \frac{4\pi \times \epsilon \times ( r1 \times r2 )}{D}$
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.

— LvW
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2

Now, how can we measure it? ;)

— dext0rb

1

What's all this electricity stuff, anyhow?

— HL-SDK

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.

— dext0rb

8

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.

— pjc50
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9

I like imagining a moon with 7.7 petavolt potential.

— mskfisher

1

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.

— mouseas

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.

— WhatRoughBeast

2

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!

— Gary Frazier
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