What would it take to cause lightning to jump between the Moon and the Earth?

Question

This question comes from @Floris' speculation at the end of his excellent answer about what it would take to kill everyone on the Earth with electricity.

Doing all this in 1/10th of a second requires an instantaneous power of $7 \cdot 10^{27} W$ which is a bit larger than the power output of the sun (which is $4\cdot 10^{26}W$ according to Wolfram alpha

This being the case, I think we're pretty safe. The only way Dr Evil could get away with this plan is to do it in reverse: first pump the charge off the earth to the moon (slowly), then let it all flow back in a cosmic lightning strike. I am not absolutely sure that the moon would stay in orbit while we charge it up... electrostatic attraction would get pretty strong. But that might be the topic for another post.

Here's that other post!

• Can you cause lightning to jump between the Earth and the Moon?

• What scale of energy and charge would it take?

• Before the lightning cancelled out the charge, how much would the electromagnetic attraction alter the orbit of the Moon?

Answer 1

Based on the calculation in my earlier answer, we were going to try to charge the earth with $10^{12}C$ and put that charge on the moon. Sending all the charge back in a giant lightning strike would then cause such a rapid change in electric field (not to mention that it dumps all the energy of twelve suns for a tenth of a second...) that it would electrocute every human being on the planet not in a well shielded cage (and those would probably just fry instead).

I then speculated about the forces and fields that would arise from that charge...

The force between two spheres each charged with $Q=10^{12}C$, distance $R = 4\cdot 10^8 m$ apart, is

$$F_e = \frac{Q^2}{4\pi \epsilon_0 R^2}\approx 5\cdot 10^{16}N$$

By comparison, the force of gravity is

$$F_g = \frac{GMm}{R^2} \approx 2\cdot 10^{20} N$$

So it won't make the moon crash - but it might speed up the orbit a little bit. More full moons. Werewolves, rejoice!

Now as for the electrical discharge. Earlier, I calculated that the field strength of earth was about 200 MV/m. The dielectric breakdown of air occurs at about 3 MV/m - see this source. More precisely, if we look at the Paschen curve for air, it is given by

$$V_b = \frac{apd}{\ln(pd)+b}$$

Where for air, $a=4\cdot 10^7 V/(atm\cdot m)$, $b=12.8$, and $p= 1 atm$. For $d = 4\cdot 10^8 m$, the breakdown voltage (using the ridiculous assumption that the air pressure is the same all the way) would be $5\cdot 10^{14} V$ - and that was a very generous estimate. More realistic would be that the potential difference reached is such that the field reaches 3 MV/m - 1/70th of the desired potential difference.

What will happen long before we reach the desired potential difference is this - the atmosphere will be ionized on the side of the moon (where the field strength is greatest) and ions will start to be attracted to the moon (assuming that the potential difference was set up with the earth net positive, and the moon net negative). These ions will arrive at the moon with tremendous energy - enough to vaporize bits of moon on impact and create a plasma which in turn will be ripped apart by the electric field and run towards the earth.

Previously we calculated the energy associated with the full potential difference as $10^{26} J$ - but that was when the full charge was reached. At 1/70th of the voltage we will have about 1/5000th of the energy, so $2\cdot 10^{22}J$. If half of that is used to burn a hole in the moon, you can melt a big hole. How big?

Heat capacity of lava roughly 1 kJ/(kg K); latent heat of fusion of rock 400 kJ/kg (source), and boiling point around 2500 K (2230 C for quartz). I could not find the latent heat of vaporization of rock, but based on other silicon based compounds, I will put it at 8000 kJ/kg (somewhere between the value for iron and silicon).

So taking one kg of moon and vaporizing it takes roughly

8,000 + 2,000 * 1 + 400 ~ 10,000 kJ

Update:

According to this reference the specific energy of granite is 26 kJ/cm³, and the density of granite is about 2.6 g/cm³. That makes my estimate of the power required to vaporize rock surprisingly accurate.

This means that this lightning strike will vaporize $10^{22-7}=10^{15} kg$ of moon. At a density of about $3.3 \cdot 10^3 kg/m^3$ that is a volume of 300 cubic kilometers of moon - a sphere of about 8 km diameter. And all that matter will be vaporized, ionized, and hurtling about in space. The most spectacular fireworks you will ever see - and not be able to tell the grandkids about.

A similar hole will be made on earth, of course. I think the fact that we'll be getting electrocuted is dropping lower down on the list of causes of death - of the planet.

Answer 2

I will estimate the potential difference, and later add in the other considerations. I am taking the entire space between the Earth and Moon to be vacuum - so I am totally ignoring the effects of the Earth's atmosphere. The Swinger Limit (http://en.wikipedia.org/wiki/Schwinger_limit) is the largest electric field that can exist before nonlinear effects start to dominate, so let's take that as the electrical breakdown of space (these are roughly the same concepts). This is $1.3\times 10^{18}$ V/m. The Earth and the Moon are separated by ~380 000 km, so the potential difference required is

$$V_{max}=4.94\times 10^{26} V$$

To find the total energy required to set up this potential, I will consider the Earth-Moon system as a pair of conducting spheres, find their capacitance, and calculate $U=\frac{1}{2}CV^2$. One can find their capacitance using the method of images - the answer is an infinite sum, the first few terms of which is

$$C=4\pi\epsilon R_1 \left(1+\frac{R_1R_2}{R^2-R_2^2}+\frac{R_1^2R_2^2}{R^2(R^2-R_2^2-R_1^2)-R_2^2(R^2-R_2^2)}+...\right)$$

Where the radius of the Earth is $R_1$, the radius of the Moon is $R_2$, and their centers are separated by $R$ (if you want to see how to do this with two spheres of the same radius, check out this page. My generalization comes pretty easily from that). Fortunately, even just taking the first term here is good enough, because the second is $\sim 7.70\times 10^{-5}$. So the capacitance of this system is

$$C\approx7.12\times 10^{-4}\text{ F}$$

Actually not overly impressive, but with the potential above the energy required to charge the system is

$$U\approx 8.68\times 10^{49}\text{ J}$$

Assuming our supervillain had access to a power plant at least as powerful as the most powerful nuclear one on Earth (Kashiwazaki-Kariwa, 8000 MW), this would take $\sim 10^{32}$ yrs to do. Outside of current technological demands, to say the least.

So would there be other consequences of this? Well, using the method of images above, one can find the amount of charge on the Earth:

$$q_1=4\pi\epsilon R_1V_{max}\approx 3.52\times 10^{23}\text{ C},$$ and the Moon would be

$$q_2=-\frac{R_2}{R}q\approx -1.61\times10^{21} \text{ C}.$$

(The net charge here is not zero because I have made one of the conductors grounded - I'm not sure that makes too much sense in this context, but I doubt it would change the results significantly).

So the ratio of the gravitational force to the Coulomb force would be

$$\frac{F_G}{F_C}=\frac{\frac{GM_1M_2}{R^2}}{\frac{kq_1q_2}{R^2}}\approx 5.7\times 10^{-18}$$

So, as mentioned in the other answer, we would see what was happening on the Moon due to the buildup of charge far before the charge could go off. This should be expected of course - we are pushing the limits of the electromagnetic force in a perfectly classical gravitating system. I think this supervillain would crash the Moon into the Earth before even getting a chance to pull off his grand plan!

Answer 3

The maximum Voltage before a current flow will occur from moon to earth is not that Schwinger limit! It is simply the value when field strength at the moons surface (smaller radius than earth) will be high enough to start field emission. Due to roughness of the moons surface this value is not easily calculated.