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#format jsmath  
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[[ Rws  Really Wild Stuff ]] 

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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 e12 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 e12 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. 
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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? 
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<7> Continental Crust 0.47%, my share is 1.43e11 kg, est. 20 km thick, lift energy 1.4e16 J   <7> '''Continental Crust 0.47%, my share is 1.43e11 kg'''  
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!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  
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<7> Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km, lift energy xxxxxx J   <7> '''Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km'''  
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<7> Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km, lift energy xxxxxx J   <7> '''Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km'''  
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<6> '''Estimated market value of my minerals, 34 trillion dollars'''  $3.4e13   <6> '''Estimated market value of my minerals, 34 trillion dollars'''  '''$3.4e13 '''   === Lift Energy compared to Asteroid Reorbit 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 assumeduniform 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/kgK (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 nonbeneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly  besides access to a nickelrich 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 e12 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 Reorbit 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 assumeduniform 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/kgK (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 nonbeneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly  besides access to a nickelrich 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.