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. According to Randall, we are currently 0.8 msec or 800 microseconds too slow.
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 7.4754e-5 radians per second. The moment of inertia is 8.034e37 kg-m2, so the angular momentum is 7.4754e-5 rad/sec * 8.034e37 rad-kg-m2, or 6.0e33 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 3.187e11 kg m2/s 3.08e16/3.187e11 = 9.7e4 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 rock 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 makes reference to 10 years. Instead of stabilization, perhaps he is talking about correcting the current 800 microsecond per day discrepancy in 10 years, a rate of 2*π*8e-4/((10*365.24*86400)*(84164)) or 1.89e-16 per second instead of 5.13e-18 per second. That would require 6e33*1.8e-16 or 1.136e18 rad kg m2/s of added angular momentum per second. Divided by 3.187e11 kg m2/s per kilogram of rock, that yields 3.56 million kilograms of comet material per second, still a factor of 800 smaller than Randall's number.
And we only need that because we are in a hurry. The mass stream is delivering 4.45e15 watts of heat energy. At 40 percent efficiency, we could make 1800 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 600 tonnes into orbit for everyone on the planet with launch loops.
The earth collects 120,000 terawatts from the sun, and has an effective black body temperature of 250K, so the extra would heat the earth about 2.3C over those ten years. After we got things stabilized, we can drop back to 97,000 kilograms per second, 2.5e14 watts of heat, 100 Terawatts of electric generation (US energy standard of living for 10 billion people), and 0.13C temperature rise. Better than coal!
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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 of 1.78e9 kg m2/s, and 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 teensy weensy bit harder ...