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/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

see below

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)