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Is the moon leaving, or are we shrinking by 38 mm/year
So again, how much kinetic energy, and/or in terms of thrust (ISP
or kg/s in terms of fuel/energy consumption), would it take for a satellite the size and mass of our moon, to be accelerated so as to escape Earth by a rate of 38 mm/year? Thanks so much for all the feedback. I'll try a little more input into my three remaining brain cells and see if the amount of recession energy can be realized. The amount of energy needed to raise the moon's orbit by that 3.8cm/year is easy.. Just plug into basic orbital mechanics, out pops an answer of some 1.25 * 10^20 joules per year, easier to visualise as a constant power input of some 4 terawatt (4 * 10^12 watt) The problem is that this is indeed the *resultant* of all forces applicable. Just how big are the other forces working on the moon,like frictional drag against the not-quite-empty vacuum of local space.(keep in mind the moon orbits earth at about 1km/sec, and the earth whizzes around the sun at about 30km/sec)? |
#2
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Is the moon leaving, or are we shrinking by 38 mm/year
Marvin wrote in message ...
So again, how much kinetic energy, and/or in terms of thrust (ISP or kg/s in terms of fuel/energy consumption), would it take for a satellite the size and mass of our moon, to be accelerated so as to escape Earth by a rate of 38 mm/year? Thanks so much for all the feedback. I'll try a little more input into my three remaining brain cells and see if the amount of recession energy can be realized. The amount of energy needed to raise the moon's orbit by that 3.8cm/year is easy.. Just plug into basic orbital mechanics, out pops an answer of some 1.25 * 10^20 joules per year, easier to visualise as a constant power input of some 4 terawatt (4 * 10^12 watt) The problem is that this is indeed the *resultant* of all forces applicable. Just how big are the other forces working on the moon,like frictional drag against the not-quite-empty vacuum of local space.(keep in mind the moon orbits earth at about 1km/sec, and the earth whizzes around the sun at about 30km/sec)? Terrific feed back, at least 4 terawatts plus something for drag, thanks much. BTW; here's something I picked up off the net: http://www.europhysicsnews.com/full/.../article3.html Hydrodynamics of planetary nebulae: 10e10 atoms per m3. and of much lower density, typically 10e7 atoms per m3. Perhaps the friction aspect can be roughly derived, as there's got to be some worth in that much surface being driven through at the 1.025 km/s. The part about keeping up with Earth I'd think is a given, as pushing ahead (into the solar wind) should be fully counteracted by the exact opposit of traveling down-wind while being taken along by the momentum of Earth's traveling about the sun. |
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Is the moon leaving, or are we shrinking by 38 mm/year
Obviously there's tidal gravitational forces continuously at work, as
otherwise something as massive as the Earth/Moon union would become one. Since I've learned from Marvin ) that lunar recession is presumably taking roughly 4 terawatts/year, while I'm thoroughly confused about calculating something other upon lunar drag into similar terms of continuous energy requirement, I've decided to do a little shock and awe reverse engineering, based upon yet another of my village idiot guestimates of lunar drag consuming another 1e12 W, a terawatt worth of tidal forces that's having only to offset for drag in order to sustain the lunar orbit speed at zero recession. If we were to allot said 1e12 (terawatt) to represent the necessary energy to overcome a lunar drag coefficient. 1e12/38e12 = 26.3 mw/m2 If we were to further suggest that the space weather environment in which the moon travels about Earth is representing but 1/1000th that viscosity and kinetic energy of the ISS orbit. If we were use an ISS surface reference area of 1000 m2, as well as the 1000 times greater atmosphere as multiplier factors: 26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp) Thus it's taking 26.3 kw worth of continuous energy applied in order to sustain the likes of ISS in a fixed (non-recession) orbit, in other words of overcoming friction as that based upon the consideration of the ISS environment being of 1e3 greater viscosity than lunar space, as well as for having to travel roughly 11 fold faster than the moon, which may actually compute as an overall 11e3:1. Obviously if lunar space were merely 100 times thinner, then the reverse engineering calculations for ISS compensation drops to a mere 2.6 kw. Perhaps 26 kw is simply too much continuous energy applied for ISS but, I'd have to bet that 2.6 kw is not sufficient, thereby the atmosphere and/or space weather difference may actually become 400~500 times thinner for that of the moon traveling through at 1.025 km/s. Of course absolutely none of this is sufficiently correct but, at least it gives myself something to go along with those 4 terawatts worth of recession energy, making the overall tally for continuous gravitational tidal forces per year applied onto the moon as representing perhaps 5 terawatts. 4e12 + 1e12 = 5e12 W If there's now 5e12 W to draw upon (disregarding solar PV conversions and/or induced solar plasma electrons and/or EMF factors taken from solar weather as well as from our Van Allen zone of death, all of which should actually be worth quite a great deal, so much so that it seems exactly the sort of tether dipole extraction potential that I'm thinking can be safely applied into those massive counter-rotating flywheels situated at the ME-L1 (gravity-well null). Just for a little further shock and awe argument sake, lets say that the additional energy input (besides the overall recession energy) provides us with one additional terawatt resource, now all toll were looking at 6 terawatts, whereas tapping into 50% of that energy yields 3 terawatts, enough energy to run 20 of those 100 GW laser cannons plus another terawatt to spare. Obviously only 2 of those terawatts are those being extracted from the recession energy. 5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted We certainly can't take away everything, as that could reverse things by pulling the moon into Earth, a seriously bad sort of thing to be doing. Although, if the bulk of energy taken is what's converted into laser cannon energy focused upon relatively small portions of Earth (not that Earth isn't already getting a little too hot under the collar), say quadrant targets zones of as little as 1 km diameter along with a 10 km safety buffer zone, at least some of this extracted energy should return itself as a slight repulsion factor. As we manage to run ourselves out of natural petroleum as well as other natural resources and remain too dumbfounded to safely utilize nuclear energy (like those smart ass French have been doing for decades), and way too energy inefficient to rely upon wind and ocean tidal resources (too polluted and otherwise too greenhouse for solar PV), thereby having insufficient energy for exterminating the remaining populations we don't happen to like; By properly using laser cannons of either near UV or near IR, or perhaps both spectrums so that at least humans aren't blinded while the least amount of thermal energy is contributed upon Earth (this is actually where I'd favor the near IR [750~800 nm] that shouldn't blind the nocturnal species), this tactic being where those multiple 100 GW cannons can sort of light your fire from afar, just might do the trick. If those 20 or so 100 GW laser cannon beams are being efficiently transmitted and converted, I'd tend to think the overall input/output conversion that's obtained on Earth might eventually reach 25%, thus 2 terawatts of input becomes 500 GW, which isn't all that bad for the hundreds of billions it may take to pull that one off. Regards, Brad Guth http://guthvenus.tripod.com |
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Is the moon leaving, or are we shrinking by 38 mm/year
Dear Brad Guth:
"Brad Guth" wrote in message om... Obviously there's tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one. A real sidelight, but I don't think they would "become one". I think that in the existing system, there is enough angular momentum that there is no way a single "partially liquid" body wouldn't spin out one or more lobes. You might jam the two together, but I don't think they'd stay together. David A. Smith |
#5
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Is the moon leaving, or are we shrinking by 38 mm/year
\(formerly\)" dlzc1.cox@net wrote in message news:ovzub.12009$vJ6.10338@fed1read05...
Dear Brad Guth: "Brad Guth" wrote in message om... Obviously there's tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one. A real sidelight, but I don't think they would "become one". I think that in the existing system, there is enough angular momentum that there is no way a single "partially liquid" body wouldn't spin out one or more lobes. You might jam the two together, but I don't think they'd stay together. David A. Smith That's certainly a good thing to know about. Any better notion upon my village idiot efforts at obtaining the kinetic energy requirement for just sustaining the lunar orbit? My guestimate of adding in another terawatt of continuous energy, as a 26 mw/m2 seems somewhat reasonable, though what do I know? Perhaps this following context is just the tip of this tether dipole iceberg. Obviously there's tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one. Since I've learned from Marvin ) that lunar recession is presumably taking roughly 4 terawatts/year, while I'm thoroughly confused about calculating something other upon lunar drag into similar terms of continuous energy requirement, I've decided to do a little shock and awe reverse engineering, based upon yet another of my village idiot guestimates of lunar drag consuming another 1e12 W, a terawatt worth of tidal forces that's having only to offset for drag in order to sustain the lunar orbit speed at zero recession. If we were to allot said 1e12 (terawatt) to represent the necessary energy to overcome a lunar drag coefficient. 1e12/38e12 = 26.3 mw/m2 If we were to further suggest that the space weather environment in which the moon travels about Earth is representing but 1/1000th that viscosity and kinetic energy of the ISS orbit. If we were use an ISS surface reference area of 1000 m2, as well as the 1000 times greater atmosphere as multiplier factors: 26.3e-3 * 1e3 * 1e3 = 26.3 kw (in other words 35.8 hp) Thus it's taking 26.3 kw worth of continuous energy applied in order to sustain the likes of ISS in a fixed (non-recession) orbit, in other words of overcoming friction as that based upon the consideration of the ISS environment being of 1e3 greater viscosity than lunar space, as well as for having to travel roughly 11 fold faster than the moon, which may actually compute as an overall 11e3:1. Obviously if lunar space were merely 100 times thinner, then the reverse engineering calculations for ISS compensation drops to a mere 2.6 kw. Perhaps 26 kw is simply too much continuous energy applied for ISS but, I'd have to bet that 2.6 kw is not sufficient, thereby the atmosphere and/or space weather difference may actually become 400~500 times thinner for that of the moon traveling through at 1.025 km/s. Of course absolutely none of this is sufficiently correct but, at least it gives myself something to go along with those 4 terawatts worth of recession energy, making the overall tally for continuous gravitational tidal forces per year applied onto the moon as representing perhaps 5 terawatts. 4e12 + 1e12 = 5e12 W If there's now 5e12 W to draw upon (disregarding solar PV conversions and/or induced solar plasma electrons and/or EMF factors taken from solar weather as well as from our Van Allen zone of death, all of which should actually be worth quite a great deal, so much so that it seems exactly the sort of tether dipole extraction potential that I'm thinking can be safely applied into those massive counter-rotating flywheels situated at the ME-L1 (gravity-well null). Just for a little further shock and awe argument sake, lets say that the additional energy input (besides the overall recession energy) provides us with one additional terawatt resource, now all toll were looking at 6 terawatts, whereas tapping into 50% of that energy yields 3 terawatts, enough energy to run 20 of those 100 GW laser cannons plus another terawatt to spare. Obviously only 2 of those terawatts are those being extracted from the recession energy. 5e12 + 1e12 = 6e12 * 0.5 = 3e12 W extracted We certainly can't take away everything, as that could reverse things by pulling the moon into Earth, a seriously bad sort of thing to be doing. Although, if the bulk of energy taken is what's converted into laser cannon energy focused upon relatively small portions of Earth (not that Earth isn't already getting a little too hot under the collar), say quadrant targets zones of as little as 1 km diameter along with a 10 km safety buffer zone, at least some of this extracted energy should return itself as a slight repulsion factor. As we manage to run ourselves out of natural petroleum as well as most other natural resources and remain too dumbfounded to safely utilize nuclear energy (like those smart ass French have been doing for decades), and way too energy inefficient to rely upon wind and ocean tidal resources (too polluted and otherwise too greenhouse for solar PV), thereby having insufficient energy for exterminating the remaining populations we don't happen to like; By properly using laser cannons of either near UV or near IR, or perhaps both spectrums so that at least humans aren't blinded while the least amount of thermal energy is contributed upon Earth (this is actually where I'd favor the near IR [750~800 nm] that shouldn't blind the nocturnal species), this tactic being where those multiple 100 GW cannons can sort of light your fire from afar, just might do the trick. If those 20 or so 100 GW laser cannon beams are being efficiently transmitted and converted, I'd tend to think the overall input/output conversion that's obtained on Earth might eventually reach 25%, thus 2 terawatts of input becomes 500 GW, which isn't all that bad for the hundreds of billions it may take to pull that one off. For the continuing entertainment sake, plus on behalf of those coming into this topic without benefit of nearly three years worth of my efforts, I've added another infomercial page (GV-LM-1) and edited upon a couple of others. If I were to be suggesting upon the sorts of wild and crazy things, as if this is what makes your life worth living, especially if those were to be of the sorts of horrifically spendy and somewhat lethal agendas like the Mars or bust and ESE fiasco, in that case I've got lots other to say about utilizing the moon as well as Venus. http://guthvenus.tripod.com/gv-lm-1.htm http://guthvenus.tripod.com/gv-cm-ccm-01.htm http://guthvenus.tripod.com/space-radiation-103.htm http://guthvenus.tripod.com/earth-moon-energy.htm http://guthvenus.tripod.com/gv-lse-energy.htm http://guthvenus.tripod.com/vl2-iss-joke.htm http://guthvenus.tripod.com/gv-illumination.htm http://guthvenus.tripod.com/moon-sar.htm http://guthvenus.tripod.com/gv-town.htm plus a few dozen other pages, with more on their way. |
#6
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Is the moon leaving, or are we shrinking by 38 mm/year
Dear Brad Guth:
"Brad Guth" wrote in message om... \(formerly\)" dlzc1.cox@net wrote in message news:ovzub.12009$vJ6.10338@fed1read05... Dear Brad Guth: "Brad Guth" wrote in message om... Obviously there's tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one. A real sidelight, but I don't think they would "become one". I think that in the existing system, there is enough angular momentum that there is no way a single "partially liquid" body wouldn't spin out one or more lobes. You might jam the two together, but I don't think they'd stay together. That's certainly a good thing to know about. Any better notion upon my village idiot efforts at obtaining the kinetic energy requirement for just sustaining the lunar orbit? Zero additional energy should do OK. The solar wind boosts one side, and retards the other. The earth would have to freeze solid for all eternity, of course. All the way down to the core. Might as well tidally lock it as well. Otherwise, it'll just continue to gain angular momentum from the Earth. Should pretty much stop the recession... David A. Smith |
#7
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Is the moon leaving, or are we shrinking by 38 mm/year
\(formerly\)" dlzc1.cox@net wrote in message news:rHTub.19859$vJ6.15193@fed1read05...
Dear Brad Guth: "Brad Guth" wrote in message om... \(formerly\)" dlzc1.cox@net wrote in message news:ovzub.12009$vJ6.10338@fed1read05... Dear Brad Guth: "Brad Guth" wrote in message om... Obviously there's tidal gravitational forces continuously at work, as otherwise something as massive as the Earth/Moon union would become one. A real sidelight, but I don't think they would "become one". I think that in the existing system, there is enough angular momentum that there is no way a single "partially liquid" body wouldn't spin out one or more lobes. You might jam the two together, but I don't think they'd stay together. That's certainly a good thing to know about. Any better notion upon my village idiot efforts at obtaining the kinetic energy requirement for just sustaining the lunar orbit? Zero additional energy should do OK. The solar wind boosts one side, and retards the other. The earth would have to freeze solid for all eternity, of course. All the way down to the core. Might as well tidally lock it as well. Otherwise, it'll just continue to gain angular momentum from the Earth. Should pretty much stop the recession... David A. Smith I agree about the solar wind being a given "zero". From: MLuttgens ) http://groups.google.com/groups?hl=e... site%3Dgroups "To calculate the effects of the deceleration on the orbit of planets (or satellites)" "According to LLR data, the Moon is receding from Earth at a rate of about 3.8 centimeters per year. Such increase of orbital distance, attributed to tidal effects on Earth, could mask the present small decrease of 1.87 cm/year." Unless I've misunderstood as usual, this above page seems to be indicating that the coefficient of lunar orbital friction is actually quite a fair percentage upon the overall scheme of things, as 1.87 cm/year is nearly half of the 3.8 cm of reported recession, thereby the overall energy necessary in order to impose the 3.8 cm/year recession may in fact become a factor of 3.8 + 1.87 = 5.67 cm/year, which in turn might further suggest 6 terawatts worth instead of my initial guestimate of 5 terawatts that was based upon adding one terawatt to what Marvin ) specified as to his calculations being 4 terawatts worth of recession energy (excluding matters of friction) that was capable of inducing the 3.8 cm/year. A satellite of the rough area of the moon, cutting through at the very least viscosity of 6e6 atoms/m3 at 1.025 km/s (more likely density 6e9/m3) will require some portion of said tidal energy just for sustaining the status quo. If it were not for the tidal gravity doing its thing, seems like an orbit degrade of 1.87 cm/year is sufficiently reasonable (nothing is forever). Regards, Brad Guth / IEIS |
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