Monday, March 1, 2010

Tornado Gun


Tornadoes are a big problem in much of the United States. In addition to ruining homes, cleaning up after them can cost the government millions of dollars each year. Perhaps there’s a way to shoot them down before they do any damage. A tornado is basically a giant pocket of air rotating really fast. Any mass moving in a circle must be accelerating. An accelerating electric charge releases power in the form of electromagnetic radiation, i.e. light.1 In principle, if you dumped a bunch of electric charges into the tornado and let them spin around, the tornado should lose energy. How many charged pith balls would you need to fire at a tornado to sap all its energy out in one second?

To solve this problem, it helps to know some physics. In 1897, Joseph Larmor derived an equation2 for the power P emitted by a charge as it accelerates,

P = e2 a2 / 6 π ε0 c3.

Here, e is the electric charge, a is the acceleration, π = 3.1415926... , ε0 = 8.85×10-12 F/m is the permittivity of free space3, and c = 3.00×108 m/s is the speed of light. Pith balls were one of the first materials used in an electroscope. Much like a balloon that you rub on your head, they can pick up a static charge. I’m imagining that we drop a bunch of pith balls onto a van de Graff generator to charge them and then fire them into the tornado. Pith balls can hold anywhere from 10-9-10-6 coulombs (C) of charge, so I’ll assume each one holds e = 10-8 C. To calculate how much power the charges will drain out, we need to know what their acceleration will be once they’re inside the tornado.

For simplicity, I’m going to assume the pith balls are light enough that they’ll be caught in the air stream and get flung around like Dorothy’s house in The Wizard of Oz. If that’s the case, then we can estimate the acceleration of the charges from the wind speed of the tornado. According to Wikipedia, tornados have winds between 64 and 177 km/h (~40-110 mph) and are approximately 75 m (~250 feet) across. They maintain contact with both the ground and a cumulonimbus cloud, meaning they can be anywhere from 150 to 3,960 m tall. To make things easy, I’ll assume the tornado is shaped like a 2000 m tall cylinder with a 75 m diameter and an outer edge that’s moving at a speed off 177 km/h. Since most of the pith balls will land somewhere in the middle of the tornado, they’ll be moving slower than the edges. I’ll assume that on average they’re traveling 100 km/h at a distance 20 m away from the center. From these numbers, we can estimate the acceleration of particles inside the tornado using the formula,

a = v2 / r
= (100 km/h)2 / (20 m)
= 39 m/s2.

By plugging this result into the Larmor formula, we can compute the power radiated by a single pith ball each second,

P = (10-8 C) 2 · (39 m/s2)2 / [6 π (8.85×10-12 F/m) · (3.00×108 m/s)3 ]
= 3.4×10-12 W per pith ball.

To calculate how many pith balls we would need to kill the tornado in one second, we’d need to know how much kinetic energy is in the tornado. The rotational kinetic energy of a cylinder is given by,

KE = I ω2 /2,

where I is the moment of inertia for a solid cylinder and w is the angular velocity given by,

I = m r2 /2,

and,

ω = v / r,

respectively. The mass of the tornado can be calculated using the density of air (1.2 kg/m3) and the dimensions of the tornado to get a total mass of 1.1×107 kg. From this and the equations above, we can solve for the rotational kinetic energy of a tornado,

KE = m v2 / 4,
= (1.1×107 kg) · (177 km/h)2 / 4
= 6.6×109 J.

From this number and the total power emitted by a single charged pith ball, we can compute how many pith balls we would need to stop the tornado in one second,4

# of balls = (energy) / [ (power per ball) · (time)]
= (6.6×109 J) / [ (3.4×10-12 W per ball) · (1 s)]
= 1.9×1021 pith balls

This number is huge. If each pith ball were a centimeter in diameter, they would fill up a cube 100 km on a side.5


[1] Since a charged tornado would emit light, you might wonder if actually doing this would look cool. In truth, the frequency of the light emitted will be about the same as the rotational frequency of the tornado. This would produce invisible radio waves.
[2] Rigorously speaking, this formula only works in vacuum. The main result won’t change, but in principle you should use the permittivity of air.
[3] The “F” here is short for “Farad”, a unit of electrical capacitance.
[4] To solve this correctly, you need more advanced mathematics since the acceleration of the charges will not be constant, (i.e. the charges will slow down as the tornado loses energy). It should be possible to get a better estimate using calculus.
[5] There’s another fundamental problem we haven’t considered. Namely, once you have a few pith balls in the tornado, it will be harder to get more in there since the charges will repel.

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