# 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/86164sec or 7.292e-5 radians per second. The moment of inertia is 8.034e37 kg-m2, so the angular momentum is 7.292e-5 rad/sec * 8.034e37 rad-kg-m2, or 5.86e33 kg m2/s . That is decreasing (on average) by 5.86e33*5.13e-18 or 3.01e16 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 3.01e16/3.189e11 = 9.43e4 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 94 metric tonnes of comet per second to produce the 3.01e16 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.

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 7.292e-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 5.86e33*2.93e-17 or 1.72e17 rad kg m2/s of added angular momentum per second. Divided by 3.189e11 kg m2/s per kilogram of rock, that yields 539 thousand kilograms of comet per second, still a factor of 5600 smaller than Randall's number.

And we only need that because we are in a hurry. The "hurry" mass stream is delivering 674 Terawatts of heat energy. At 40 percent efficiency, we could make 270 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 94.3 tonnes per second, 118 terawatts of heat, 47 Terawatts of electric generation (current US per capita energy usage for 4.7 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. There is clearly a math error. The numbers are below, in the third Randall(B) column.

### A summary of the numbers

 earth parameters N00 days per year 365.24 N01 solar day 86400 seconds N02 seconds in a year 31556736 seconds N00*N01 N03 sidereal day 86164 seconds N01*N00/(1+N00) N04 earth angular velocity 7.292116e-5 radians/s 2π/N03 N05 earth moment of inertia 8.034e37 kg-m2 N06 earth angular momentum 5.858e33 kg-m2/s N04 * N05 N07 earth equatorial radius 6.378e6 meters N08 earth average radius 6.371e6 meters N09 density of cometary rock 3000 kg/m3 A: comet impactors N10A impact speed 5e4 m/s N11A angular momentum per kg 3.189e11 m2/s N10A * N07 N12A kinetic energy per kg of comet 1.25e-3 TJoule/kg 1/2 N10A2 B: asteroid impactors N10B impact speed 2e4 m/s N11B angular momentum per kg 1.276e11 m2/s N10B * N07 N12B kinetic energy per kg 2e-4 TJoule/kg 1/2 N10B2 power and heat N13 electrical generation efficiency 0.40 40% N14 earth black body heating 1/1920 Kelvin/TWatt
 scenario speedup (A) maintenance(A) Randall(B) N21 time period 10 1 1 year N22 daily error corrected 800 14 178,152 μs N23 correction per year 80 14 178,152 μs/year N24 correction per day 219 38.3 488,000 ns/day N23 / N00 N25 correction per second 2.54 0.444 5645 ps/s N23 / N01 N26 fractional correction of solar day per sec 2.93e-17 5.13e-18 6.53e-14 /s N25 / N01 N27 angular velocity correction per sec 2.14e-21 3.74e-22 4.76e-18 rad/s2 N26 * N04 N28 angular momentum correction per sec 1.72e17 3.01e16 3.83e20 kg-m2/s2 N27 * N05 N29 comet mass per second 5.39e5 9.43e4 3e9 kg/s N28 / N11 N30 comet power 674 118 600,000 Terawatts N29 * N12 N31 electrical power 270 47.2 240,000 Terawatts N30 * N13 N32 earth heating 0.35 0.061 80* Kelvin N30 * N14

(*) 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? For stabilization, you need to generate about 3.01e16 newton-meters of torque against a moment arm we can assume to be the radius of the earth, 6.378e6 meters, about 4.72 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 perfect mirror lightsail, outbound from the earth, with an area of 519 million square kilometers, equivalent to a disk 25,700 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 or ocean and converted to heat, averaging 120,000 terawatts. From the black body law, P = (constant) T4 . Taking the derivative, dP = 4 (constant) T3 dT = 4 P / T dT .

Rearranging, dT ≈ ( T / 4 P ) dP . . . T/4P = 1/1920 TW .

So an increase of 1920 TW average dissipation results in approximately a 1K ( = 1C ) rise in average temperature. In Real LifeTM 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.

leapsecond (last edited 2013-01-11 02:54:07 by KeithLofstrom)