Differences between revisions 1 and 5 (spanning 4 versions)
 ⇤ ← Revision 1 as of 2015-12-14 08:19:48 → Size: 3917 Editor: KeithLofstrom Comment: ← Revision 5 as of 2015-12-14 18:54:59 → ⇥ Size: 8969 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 1: Line 1: #format jsmath Line 2: Line 3: [[ Rws | Really Wild Stuff ]] Line 5: Line 8: I have a 0.67 acre (0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. It is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. I have a 0.67 acre ( 2711 m², 0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. The property is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. Line 7: Line 10: So - assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center? How much material is that, and how much is it worth, purified to commercial grade metals? But let's assume this volume is "average" for the Earth's composition, and make the same kind of calculations that are made for materials in asteroids. Assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center and lift everything out? How much material is that, and (assuming market prices did not collapse) how much is it worth, purified to commercial grade metals? Line 9: Line 12: ||<-7> Continental Crust 0.47%, my share is 1.43e11 kg, est. 20 km thick, lift energy 1.4e16 J |||| SiO₂ || 60.6% || 8.67e10 kg || Glass || 8.67e10 kg || $1.00/kg ||$8.6e10 |||| Al₂O₃ || 15.9% || 2.27e10 kg || Aluminum || 1.20e10 kg || $1.80/kg ||$2.2e10 || ||<-7> '''Continental Crust 0.47%, my share is 1.43e11 kg''' ||||!SiO₂ || 60.6% || 8.67e10 kg || Glass || 8.67e10 kg || $1.00/kg ||$8.6e10 ||||!Al₂O₃ || 15.9% || 2.27e10 kg || Aluminum || 1.20e10 kg || $1.80/kg ||$2.2e10 || Line 17: Line 20: || TiO₂  || 0.7% || 1.00e09 kg || Titanium || 6.00e08 kg || $4.30/kg ||$2.6e09 |||| P₂O₅  || 0.1% || 1.43e08 kg || Phosphorus || 6.24e07 kg || $76/kg ||$4.7e09 || || !TiO₂ || 0.7% || 1.00e09 kg || Titanium || 6.00e08 kg || $4.30/kg ||$2.6e09 |||| !P₂O₅ || 0.1% || 1.43e08 kg || Phosphorus || 6.24e07 kg || $76/kg ||$4.7e09 || Line 23: Line 26: || U || 1.8 ppm || 2.57e05 kg || Uranium || 2.57e05 kg || $140/kg ||$3.5e07 || || U || 1.8 ppm || 2.57e05 kg || Uranium || 2.57e05 kg || $140/kg ||$3.5e07 || Line 26: Line 29: ||<-7> Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km, lift energy xxxxxx J |||| Mg || 22.2% || 4.48e12 kg || Magnesium || 4.48e12 kg || $4.00/kg ||$1.8e13 || ||<-7> '''Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km'''              |||| Mg || 22.2% || 4.48e12 kg || Magnesium || 4.48e12 kg || $4.00/kg ||$1.8e13 || Line 29: Line 32: || Fe || 6.3% || 1.27e12 kg || Cast Iron || 1.27e12 kg || $0.20/kg ||$2.5e11 || || Fe || 6.3% || 1.27e12 kg || Cast Iron || 1.27e12 kg || $0.20/kg ||$2.5e11 || Line 32: Line 35: || P || 260 ppm || 5.25e09 kg || Phosphorus || 5.25e09 kg || $76/kg ||$4.0e11 || || P || 260 ppm || 5.25e09 kg || Phosphorus || 5.25e09 kg || $76/kg ||$4.0e11 || Line 36: Line 39: ||<-7> Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km, lift energy xxxxxx J |||| Fe || 85% || 8.33e12 kg || Cast Iron || 8.33e12 kg || $0.20/kg ||$1.7e12 || ||<-7> '''Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km'''              |||| Fe || 85% || 8.33e12 kg || Cast Iron || 8.33e12 kg || $0.20/kg ||$1.7e12 || Line 39: Line 42: ||<-6> '''Estimated market value of my minerals, 34 trillion dollars'''    || $3.4e13 || ||<-6> '''Estimated market value of my minerals, 34 trillion dollars''' || '''$3.4e13 ''' || Line 41: Line 44: -------- Line 42: Line 46: === Lift Energy compared to Asteroid Re-orbit Energy ===The material is in a tall, skinny, inverted pyramid. We will optimistically assume uniform density to make the calculation easier, though this is hugely less optimistic than the absurd idea that we can actually do anything like this.Assume a perfect sphere with a surface gravity of $g$, radius $R_e$, and density $\rho$. Assume the surface area is $A$. How much energy $d E$ does it take to lift a slab from depth $R$ with thickness $d R$ to the surface?The area of a slab at depth $R$ is $A_r = A ~ ( R / R_e )^2$. The mass of this slab is $d m = \rho ~ A_r ~ dR = \rho ~ A ~ ( R / R_e )^2 ~ d R$Gravity inside a uniform sphere is linearly proportional to radius, so the gravity at radius $R_y$ is $g_y = g ~ R_y / R_e$ . The amount of energy to lift the slab form depth $R$ a distance of $d R_y$ is $d E_R ~ = ~ g_y ~ d m ~ d R = ( g ~ R_y / R_e ) ( \rho ~ A ~ ( R / R_e )^2 ) ~ d_R ~ d R_y ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ~ R_y d_R ~ d R_y$ Let's integrate that over $R_y$ from $R$ to the surface $R e$ :$E_R ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ( {R_e}^2 - R^2 / 2 ) = ( \rho ~ g ~ A / 2 ) ( ( R^2 / R_e ) - ( R^4 / {R_e}^3 ) )$The total energy for the entire column of material integrates over $R$ from 0 to $R_e$:$E ~ = ~ \rho ~ g ~ A ( R_e )^2 / 15$The volume of the inverted pyramid is $A R_e / 3$, so the assumed-uniform density $\rho$ is the total mass $M$ divided by volume:$\rho ~ = ~ 3 ~ M / ( A ~ R_e )$Producing a very simple result for the whole column:$E ~ = ~ g ~ M ~ R_e / 5$If $g$ = 9.8 m/s², $M$ = 3e13 kg, and $R_e$ = 6371 km, the total lift energy $E$ is 3.7e20 J, about 50 minutes of earth solar illumination, or 9 hours of 10% efficient global PV. The energy per kilogram is 12.5 MJ, or about 3.47 kilowatt hours, or a $\delta V$ equivalent of 5000 m/s. Assuming that the heat capacity of earth mantle material is 500 J/kg-K (POMA), a kilogram brought up from the 4000 K lower mantle boundary would add about 2 MJ/kg, a total of 6e19 J, about 8 minutes of total earth surface solar energy.=== Asteroid Delivery Energy === If the volume of material was shaped into a spherical asteroid, it would have a radius $R_A ~ = ~ ( A ~ R_e / 4 \pi )^{1/3} ~ = ~ 1.11 km. But since we need reaction mass, let's assume we are tossing the material off a much larger asteroid with a superhigh acceleration launcher.Moving a chunk of asteroid from the asteroid belt into low earth orbit is difficult - let's assume we will merely change its orbit from a circular 2 AU to an elliptical orbit between 2 AU and 1 AU, and we will use the atmosphere to bring the material down, and that the material somehow remains intact and impacts somewhere safe. We will assume it has the same composition as our hypothetical wedge of earth and is equally valuable. The 1 AU earth orbits at 30 km/s, and the 2 AU asteroid belt orbits$ \sqrt{2} $slower, 21.2 km/s. An elliptical orbit between 1 AU and 2 AU has an eccentricity$ e $= 1/3, a semimajor axis$ a $= 1.5 AU, a characteristic velocity$ v_0 $= 26 km/s, an aphelion velocity$ v_a $= 17.3 km/s, and a perihelion velocity$ v_p $= 34.6 km/s . So, assuming a relatively small escape velocity from the source asteroid, the asteroid launch velocity is 21.2 - 17.3 = 3.9 km/s, only 61% of the energy of extraction of the wedge from the earth. Note that if both operations are solar powered, the Earth's surface gets about 2.8x more solar illumination.The asteroidal material arrives at the earth's orbit with a relative velocity of 4.6 km/s, an energy of 10.6 MJ/kg, which is added to the earth escape velocity ( 10.7 km/s ) energy of 57.3 MJ/kg, totalling 68 MJ/kg. That energy would be divided between impact energy and atmospheric heating, but would all become atmospheric heat after the impact energy dissipates. Total energy for 3e13 kg is 2e21 joules, equivalent to 4.5 hours of earth surface solar energy.=== Conclusion ===Overall, the two processes are within an order of magnitude of each other, both delivering gravitationally sorted but otherwise non-beneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly - besides access to a nickel-rich core, there many disadvantages compared to a mine 20 km deep and 67 acres in area, or a mine 200 meters deep and 6700 acres in area, which would require far less energy to remove. The product is the same - uninteresting rock, unless this was done around a concentrated ore body. '''Asteroid mining to provide raw materials to Earth is ridiculous.''' Backyard Minerals Space materials delivered to earth as manufactured goods do not make much sense. The materials are not beneficiated ores, and manufacturing requires a huge and interconnected network of factories. Let's compare it to something equally silly. I have a 0.67 acre ( 2711 m², 0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. The property is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. But let's assume this volume is "average" for the Earth's composition, and make the same kind of calculations that are made for materials in asteroids. Assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center and lift everything out? How much material is that, and (assuming market prices did not collapse) how much is it worth, purified to commercial grade metals?  Continental Crust 0.47%, my share is 1.43e11 kg SiO₂ 60.6% 8.67e10 kg Glass 8.67e10 kg 1.00/kg || 8.6e10 Al₂O₃ 15.9% 2.27e10 kg Aluminum 1.20e10 kg 1.80/kg || 2.2e10 FeO 6.7% 9.58e09 kg Cast Iron 7.45e09 kg 0.20/kg || 1.5e09 CaO 6.4% 9.15e09 kg Calcium 6.54e09 kg 2.00/kg || 1.3e10 MgO 4.7% 6.72e09 kg Magnesium 5.39e09 kg 4.00/kg || 2.1e10 Na₂O 3.1% 4.43e09 kg Sodium 3.29e09 kg 3.30/kg || 1.1e10 K₂O 1.8% 2.57e09 kg Potassium 2.14e09 kg 22/kg || 4.7e10 TiO₂ 0.7% 1.00e09 kg Titanium 6.00e08 kg 4.30/kg || 2.6e09 P₂O₅ 0.1% 1.43e08 kg Phosphorus 6.24e07 kg 76/kg || 4.7e09 Ni 90 ppm 1.29e07 kg Nickel 1.29e07 kg 3.90/kg || 5.0e07 Cu 68 ppm 9.72e06 kg Copper 9.72e06 kg 2.00/kg || 1.9e07 Nd 33 ppm 4.72e06 kg Neodymium 4.72e06 kg 60/kg || 2.8e08 Ga 19 ppm 2.72e06 kg Gallium 2.72e06 kg 200/kg || 5.4e08 U 1.8 ppm 2.57e05 kg Uranium 2.57e05 kg 140/kg || 3.5e07 In 160 ppb 22900 kg Indium 22900 kg 700/kg || 1.6e07 Se 50 ppb 7150 kg Selenium 7150 kg 150/kg || 1.1e06 Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km Mg 22.2% 4.48e12 kg Magnesium 4.48e12 kg 4.00/kg || 1.8e13 Si 21.2% 4.28e12 kg Glass 9.17e12 kg 1.00/kg || 9.2e12 Fe 6.3% 1.27e12 kg Cast Iron 1.27e12 kg 0.20/kg || 2.5e11 Ca 2.6% 5.25e11 kg Calcium 5.25e11 kg 2.00/kg || 1.1e12 Al 2.4% 4.85e11 kg Aluminum 4.85e11 kg 1.80/kg || 8.7e11 P 260 ppm 5.25e09 kg Phosphorus 5.25e09 kg 76/kg || 4.0e11 Cu 20 ppm 4.04e08 kg Copper 4.04e08 kg 2.00/kg || 8.1e08 U 22 ppb 4.44e05 kg Uranium 4.44e05 kg 140/kg || 6.2e07 In 13 ppb 2.63e05 kg Indium 2.63e05 kg 700/kg || 1.8e08 Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km Fe 85% 8.33e12 kg Cast Iron 8.33e12 kg 0.20/kg || 1.7e12 Ni 5% 4.90e11 kg Nickel 4.90e11 kg 3.90/kg || 1.9e12 Estimated market value of my minerals, 34 trillion dollars$3.4e13

Lift Energy compared to Asteroid Re-orbit Energy

The material is in a tall, skinny, inverted pyramid. We will optimistically assume uniform density to make the calculation easier, though this is hugely less optimistic than the absurd idea that we can actually do anything like this.

Assume a perfect sphere with a surface gravity of g, radius R_e , and density \rho. Assume the surface area is A . How much energy d E does it take to lift a slab from depth R with thickness d R to the surface?

The area of a slab at depth R is A_r = A ~ ( R / R_e )^2 . The mass of this slab is d m = \rho ~ A_r ~ dR = \rho ~ A ~ ( R / R_e )^2 ~ d R

Gravity inside a uniform sphere is linearly proportional to radius, so the gravity at radius R_y is g_y = g ~ R_y / R_e . The amount of energy to lift the slab form depth R a distance of d R_y is

d E_R ~ = ~ g_y ~ d m ~ d R = ( g ~ R_y / R_e ) ( \rho ~ A ~ ( R / R_e )^2 ) ~ d_R ~ d R_y ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ~ R_y d_R ~ d R_y

Let's integrate that over R_y from R to the surface R e :

E_R ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ( {R_e}^2 - R^2 / 2 ) = ( \rho ~ g ~ A / 2 ) ( ( R^2 / R_e ) - ( R^4 / {R_e}^3 ) )

The total energy for the entire column of material integrates over R from 0 to R_e :

E ~ = ~ \rho ~ g ~ A ( R_e )^2 / 15

The volume of the inverted pyramid is A R_e / 3 , so the assumed-uniform density \rho is the total mass M divided by volume:

\rho ~ = ~ 3 ~ M / ( A ~ R_e )

Producing a very simple result for the whole column:

E ~ = ~ g ~ M ~ R_e / 5

If g = 9.8 m/s², M = 3e13 kg, and R_e = 6371 km, the total lift energy E is 3.7e20 J, about 50 minutes of earth solar illumination, or 9 hours of 10% efficient global PV. The energy per kilogram is 12.5 MJ, or about 3.47 kilowatt hours, or a \delta V equivalent of 5000 m/s. Assuming that the heat capacity of earth mantle material is 500 J/kg-K (POMA), a kilogram brought up from the 4000 K lower mantle boundary would add about 2 MJ/kg, a total of 6e19 J, about 8 minutes of total earth surface solar energy.

Asteroid Delivery Energy

If the volume of material was shaped into a spherical asteroid, it would have a radius \$ R_A ~ = ~ ( A ~ R_e / 4 \pi )^{1/3} ~ = ~ 1.11 km. But since we need reaction mass, let's assume we are tossing the material off a much larger asteroid with a superhigh acceleration launcher.

Moving a chunk of asteroid from the asteroid belt into low earth orbit is difficult - let's assume we will merely change its orbit from a circular 2 AU to an elliptical orbit between 2 AU and 1 AU, and we will use the atmosphere to bring the material down, and that the material somehow remains intact and impacts somewhere safe. We will assume it has the same composition as our hypothetical wedge of earth and is equally valuable.

The 1 AU earth orbits at 30 km/s, and the 2 AU asteroid belt orbits \sqrt{2} slower, 21.2 km/s. An elliptical orbit between 1 AU and 2 AU has an eccentricity e = 1/3, a semimajor axis a = 1.5 AU, a characteristic velocity v_0 = 26 km/s, an aphelion velocity v_a = 17.3 km/s, and a perihelion velocity v_p = 34.6 km/s . So, assuming a relatively small escape velocity from the source asteroid, the asteroid launch velocity is 21.2 - 17.3 = 3.9 km/s, only 61% of the energy of extraction of the wedge from the earth. Note that if both operations are solar powered, the Earth's surface gets about 2.8x more solar illumination.

The asteroidal material arrives at the earth's orbit with a relative velocity of 4.6 km/s, an energy of 10.6 MJ/kg, which is added to the earth escape velocity ( 10.7 km/s ) energy of 57.3 MJ/kg, totalling 68 MJ/kg. That energy would be divided between impact energy and atmospheric heating, but would all become atmospheric heat after the impact energy dissipates. Total energy for 3e13 kg is 2e21 joules, equivalent to 4.5 hours of earth surface solar energy.

Conclusion

Overall, the two processes are within an order of magnitude of each other, both delivering gravitationally sorted but otherwise non-beneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly - besides access to a nickel-rich core, there many disadvantages compared to a mine 20 km deep and 67 acres in area, or a mine 200 meters deep and 6700 acres in area, which would require far less energy to remove. The product is the same - uninteresting rock, unless this was done around a concentrated ore body.

Asteroid mining to provide raw materials to Earth is ridiculous.

BackyardMinerals (last edited 2021-04-18 00:58:37 by KeithLofstrom)