Ridiculous Geoengineering

Eliminating the Leap Second through Earth Spin Adjustment

Even if this was remotely practical, the Earth's rotation does not vary smoothly, see http://www.ucolick.org/~sla/leapsecs/dutc.html

But lets play along with http://what-if.xkcd.com/26/, and see if we can figure out what numbers Randall Munroe is using, and whether his results are in the ball park. [note - I reported some incorrectly large numbers on the xkcd forums]

Let's pick 1.4 milliseconds per day per century as the average rate that the day is increasing. That means the day length increases 14 microseconds per year, or 38.3 nsec per day, or 4.44e-13 seconds per second. The solar day is 86400 seconds, the sidereal day is 84164 seconds, so the average fractional increase in day length is 4.44e-13/86400 (and decrease in angular momentum) is 5.13e-18 per second.

The angular velocity of the earth is 2π/84164sec or 7.4654e-5 radians per second. The moment of inertia is 8.034e37 kg-m², so the angular momentum is 7.4654e-5 rad/sec * 8.034e37 rad-kg-m², or 6.00e33 kg m²/s . That is decreasing (on average) by 6e33*5.13e-18 or 3.08e16 rad kg m²/s . If we could add that much angular momentum back from external sources, or decrease the moment of inertia by 4.12e20 kg-m²/s , we could counteract the steady increase.

Randall offers cometary impact at the equator as a speedup mechanism. Comet impact speeds are around 50 km/sec, and at a radius of 6378000 meters, the angular momentum added per kilogram is 5e4*6378000 or 3.189e11 kg m²/s. To compensate for the long term angular momentum change, we need 3.08e16/3.189e11 = 9.65e4 kg/sec . Randall claims a billion liters per second ( perhaps 3 million metric tonnes per second ) of rock is needed. WRONG. We only need to deliver 97 metric tonnes of comet per second to produce the 3.08e16 rad kg m²/s needed for stabilization. Further, if we could capture all the kinetic energy of incoming rock at 50km/s, and launch mass back out with it at 12.5km/s, we could launch 16 times as much rock with 4 times the angular momentum, so we would only need to deliver 19 metric tons per second. Not sure where Randall's error came from, but he probably forgot to multiply or divide by the Earth's radius somewhere.

Randall mentions 10 years and 0.8 milliseconds in his writeup. Instead of stabilization, perhaps he is talking about correcting the current 800 microsecond per day discrepancy in 10 years, a rate of 7.4654e-5*8e-5/((10*365.24*86400)) or a fraction of 2.93e-17 per second instead of 5.13e-18 per second. That would require 6e33*2.93e-17 or 1.76e17 rad kg m²/s of added angular momentum per second. Divided by 3.189e11 kg m²/s per kilogram of rock, that yields 552 thousand kilograms of comet per second, still a factor of 1800 smaller than Randall's number.

And we only need that because we are in a hurry. The mass stream is delivering 690 Terawatts of heat energy. At 40 percent efficiency, we could make 276 terawatts of electricity with it. We could use that power to separate all the excess CO2 from the atmosphere and sequester it in the Romanche trench. Or launch 80 tonnes to the moon for every individual on the Earth with launch loops.

The extra power would heat the earth about 0.4C over those ten years. After we got things stabilized, we can drop back to 96.500 kilograms per second, 121 terawatts of heat, 48 Terawatts of electric generation (current US per capita energy usage for 5 billion people), and 0.06C temperature rise. Better than coal!

Perhaps Randall is thinking about more mass and slower impacts; asteroidal NEOs rather than comets. That could slow the impact speed to 20km/sec, requiring 2.5x more mass for the same angular momentum transfer. However, that also reduces the energy to momentum ratio by 2.5x, too. However, even at that lower speed, 3 million tonnes per second generates a heck of a lot of angular momentum.

A summary of the numbers

(*) small dT approximation invalid, black body increase if heat spread to whole earth, heated to 329K . Since this would actually make an incandescently bright band around the equator, coupled by raging windstorms to the rest of the earth, it is quite difficult to say what the actual average temperature of the rest of the earth would be. If the primary heating was confined to a belt 63 km wide, with 20% of the energy reaching the rest of the earth, the incandescent belt would be 830K, while the rest of the earth would be 30K hotter. Probably survivable near the poles.

Howzabout instead we decrease the angular momentum by moving mass north? The optimum place to do this is around 45 degrees north, but lets instead assume we can frictionlessly slide mass down from the Himalayas at 6000 meters altitude and 28 degrees north, to Xinjiang desert at 1000 meters altitude and 40 degrees north. Assume the Earth's sea-level radius is around 6377 km in the Himalayas and 6372 km in Xinjiang. The radius of the Himalayas from the earth's axis is (6377+6)*cos(28) km or 5636 km. The radius of Xinjiang is (6372+1)*cos(40) or 4881 km. The angular momentum of a kilogram of rock in the Himalayas is 5.636e6² * 7.4754e-5 or 2.37e9 kg m²/s, and 4.881e6²*7.454e-5 or 1.78e9 kg m²/s. The difference is 5.9e8 kg/m²/s . If we can slide down 7e11 kilograms per second, about one cubic kilometer every 4 seconds, we can lower the moment of inertia rapidly enough. That sounds a wee bit harder ...

Howzabout light sails? You need to generate about 3.08e16 newton-meters of torque against a moment arm we can assume to be the radius of the earth, 6.378e6 meters, about 5 billion newtons. Light sail pressure off a perfect mirror is twice the optical power divided by the speed of light, 2*1.366e9W/km² divided by 2.997e8m/s, about 9.1 newtons per square kilometer. So we would need a light sail, outbound from the earth, with an area of 530 million square kilometers, equivalent to a disk 26,000 kilometers in diameter, twice the size of the earth. It would be in the night sky, destroying most of the night-time biosphere with huge amounts of light pollution. Not sure how you would support the cables, or deal with the rotation of the earth and the fractional availability of torque over the day. That sounds a wee bit harder still ...

Note: Raising the blackbody temperature of the earth The temperature that matters is 250K, the average of the infrared opaque regions of the atmosphere, not the surface temperature. The power level that matters is the light reaching the ground, averaging 120,000 terawatts. From the black body law, P = (constant) T⁴ . Taking the derivative, dP = 4 (constant) T³ dT = 4 P / T dT . So dT = ( T / 4 P ) dP . T / 4 P = ( 1 / 1920 TW ). So an increase of 1920 TW average dissipation results in a 1K ( or 1C ) rise in average temperature. In Real Life^TM this can vary quite a bit, depending on how the heat makes its way through the atmosphere and where it dissipates; heat near the equator may end up heating already-hotter upper atmosphere air, and dissipate slightly more effectively, compared to heat generated near the poles.