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| 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 per day too slow. | 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]'' |
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| The angular velocity of the earth is 2π/84164sec or 7.4754e-5 radians per second. The moment of inertia is 8.034e37 kg-m^2^, so the angular momentum is 7.4754e-5 rad/sec * 8.034e37 rad-kg-m^2^, or 6.0e33 kg m^2^/s . That is decreasing (on average) by 6e33*5.13e-18 or 3.08e16 rad kg m^2^/s . If we could add that much angular momentum back from external sources, or decrease the moment of inertia by 4.12e20 kg-m^2^/s , we could counteract the steady increase. | The angular velocity of the earth is 2π/86164sec or 7.292e-5 radians per second. The moment of inertia is 8.034e37 kg-m^2^, so the angular momentum is 7.292e-5 rad/sec * 8.034e37 rad-kg-m^2^, or 5.86e33 kg m^2^/s . That is decreasing (on average) by 5.86e33*5.13e-18 or 3.01e16 rad kg m^2^/s . If we could add that much angular momentum back from external sources, or decrease the moment of inertia by 4.12e20 kg-m^2^/s , we could counteract the steady increase. |
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| 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 m^2^/s. To compensate for the long term angular momentum change, we need 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 m^2^/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 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 m^2^/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 m^2^/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. |
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| Randall mentions 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 m^2^/s of added angular momentum per second. Divided by 3.189e11 kg m^2^/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. | 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 m^2^/s of added angular momentum per second. Divided by 3.189e11 kg m^2^/s per kilogram of rock, that yields 539 thousand kilograms of comet per second, still a factor of 5600 smaller than Randall's number. |
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| 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 [[ http://server-sky.com/OceanStoreCO2 | sequester it in the Romanche trench ]]. Or launch 600 tonnes into orbit for every individual on the planet with [[ http://launchloop.com | launch loops ]]. | 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 [[ http://server-sky.com/OceanStoreCO2 | sequester it in the Romanche trench ]]. Or launch 80 tonnes to the moon for every individual on the Earth with [[ http://launchloop.com | launch loops ]]. |
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| 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! | 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. ----- |
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| || seconds in a 365.24 day year || 31556736 || seconds || || sidereal day || 84164 || seconds || || earth angular velocity || 1.493081e-4 || radians / second || || earth moment of inertia || 8.034e37 || kg-m^2^ || || earth angular momentum || 12.0e33 || kg-m^2^ / sec || || earth equatorial radius || 6378 || km || || earth average radius || 6371 || km || || earth black body emission || 1/1920 || Kelvin / Terawatt || || density of rock || 3000 || kg/m^3^ || || cometary impact speed || 50 || km/sec || || angular momentum per kg of comet || 3.189e11 || kg-m^2^/sec || || kinetic energy per kg of comet || 1.25e9 || J/kg || || electrical generation efficiency || 0.40 || 40% || |
|| 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 / second || 2π/N03 || || N05 || earth moment of inertia || 8.034e37 || kg-m^2^ || || N06 || earth angular momentum || 5.858e33 || kg-m^2^ / sec || N04 * N05 || || N07 || earth equatorial radius || 6378 || km || || N08 || earth average radius || 6371 || km || || N09 || density of cometary rock || 3000 || kg/m^3^ || || N10A || cometary impact speed || 50 || km/sec || || N11A || angular momentum per kg of comet || 3.189e11 || kg-m^2^/sec || N10A * N07 || || N12A || kinetic energy per kg of comet || 1.25e-3 || TJ/kg || 1/2 N10A^2^|| || N10B || NEO asteroid impact speed || 20 || km/sec || || N11B || angular momentum per kg of NEO.A || 1.276e11 || kg-m^2^/sec || N10B * N06 || || N12B || kinetic energy per kg of NEO.A || 2e-4 || TJ/kg || 1/2 N10B^2^|| || N13 || electrical generation efficiency || 0.40 || 40% || || N14 || earth black body emission || 1/1920 || Kelvin / Terawatt || see below || . . || || scenario || speedup (A)||maintenance(A)|| Randall(B) || || N21 || time period || 10 || 1 || 1 || year || || N22 || daily error corrected || 800 || 14 || 178,152 || μsec || || N23 || correction per year || 80 || 14 || 178,152 || μsec/year || || N24 || correction per day || 219 || 38.3 || 488,000 || nsec/day || N23 / N00 || || N25 || correction per second || 2.54 || 0.444 || 5645 || psec/sec || N23 / N01 || || N26 || fractional correction of solar day per sec || 2.93e-17 || 5.13e-18 || 6.53e-14 || /sec || N25 / N01 || || N27 || angular velocity correction per sec || 2.14e-21 || 3.74e-22 || 4.76e-18 || rad/sec^2^ || N26 * N04 || || N28 || angular momentum correction per second || 1.72e17 || 3.01e16 || 3.83e20 || kg-m^2^/sec^2^ || 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 || |
| Line 36: | Line 63: |
| || scenario || speedup || maintenance || || time period || 10 years || 1 year || || daily error corrected || 800μsec || 14μsec || || correction per year || 80μsec || 14μsec || || correction per day || 219 nsec || 38.3 nsec || || correction per second || 2.53 psec || 444 fsec || || fractional correction of solar day per sec || 2.93e-17 || 5.13e-18 || || angular velocity correction per sec || 4.38e-21 || 7.66e-22 || rad/sec^2^ || || angular momentum correction per second || 3.52e17 || 6.15e16 || kg-m^2^/sec^2^ || || comet mass per second || 1.10e6 || 1.93e5 || kg/s || || comet power || 1375 || 241 || Terawatts || || electrical power || 550 || 96.4 || Terawatts || || earth heating || 0.72 || 0.05 || Kelvin || |
(*) 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. |
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| '''Note: Raising the blackbody temperature of the earth''' The temperature that matters is the average of the infrared opaque regions of the atmosphere, 250K. The power level that matters is the light reaching the ground, averaging 120,000 terawatts. From the black body law, P = (constant) T^4^ . Taking the derivative, dP = 4 (constant) T^3^ 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, than heat generated near the poles. | '''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) T^4^ . Taking the derivative, dP = 4 (constant) T^3^ 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 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. |
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
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 / second |
2π/N03 |
N05 |
earth moment of inertia |
8.034e37 |
kg-m2 |
|
N06 |
earth angular momentum |
5.858e33 |
kg-m2 / sec |
N04 * N05 |
N07 |
earth equatorial radius |
6378 |
km |
|
N08 |
earth average radius |
6371 |
km |
|
N09 |
density of cometary rock |
3000 |
kg/m3 |
|
N10A |
cometary impact speed |
50 |
km/sec |
|
N11A |
angular momentum per kg of comet |
3.189e11 |
kg-m2/sec |
N10A * N07 |
N12A |
kinetic energy per kg of comet |
1.25e-3 |
TJ/kg |
1/2 N10A2 |
N10B |
NEO asteroid impact speed |
20 |
km/sec |
|
N11B |
angular momentum per kg of NEO.A |
1.276e11 |
kg-m2/sec |
N10B * N06 |
N12B |
kinetic energy per kg of NEO.A |
2e-4 |
TJ/kg |
1/2 N10B2 |
N13 |
electrical generation efficiency |
0.40 |
40% |
|
N14 |
earth black body emission |
1/1920 |
Kelvin / Terawatt |
see below |
|
scenario |
speedup (A) |
maintenance(A) |
Randall(B) |
||
N21 |
time period |
10 |
1 |
1 |
year |
|
N22 |
daily error corrected |
800 |
14 |
178,152 |
μsec |
|
N23 |
correction per year |
80 |
14 |
178,152 |
μsec/year |
|
N24 |
correction per day |
219 |
38.3 |
488,000 |
nsec/day |
N23 / N00 |
N25 |
correction per second |
2.54 |
0.444 |
5645 |
psec/sec |
N23 / N01 |
N26 |
fractional correction of solar day per sec |
2.93e-17 |
5.13e-18 |
6.53e-14 |
/sec |
N25 / N01 |
N27 |
angular velocity correction per sec |
2.14e-21 |
3.74e-22 |
4.76e-18 |
rad/sec2 |
N26 * N04 |
N28 |
angular momentum correction per second |
1.72e17 |
3.01e16 |
3.83e20 |
kg-m2/sec2 |
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? 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 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.