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[[ Rws | Really Wild Stuff ]] - A Silly Argument About So-Called Resources [[ Rws | Really Wild Stuff ]] - A Silly Argument About So-Called Resources on Earth and in Space
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Delivering space materials to earth as manufactured goods does 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. Delivering space materials to Earth as manufactured goods does 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.
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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. If $ g $ = 9.8 m/s², $ M $ = 3e13 kg, and $ R_e $ = 6371 km, the total lift energy (at 100% efficiency) $ 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.
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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. 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.

Backyard Minerals

Really Wild Stuff - A Silly Argument About So-Called Resources on Earth and in Space

Delivering space materials to Earth as manufactured goods does 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. 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 5.32e-12 of the surface of the earth, and the same fraction 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

That's 50 trillion dollars an acre. The entire earth is worth 6e24 or six septillion dollars. Or nothing, if we wipe ourselves out trying to do this :-).


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_{lift} ~ = ~ \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_{lift} ~ = ~ 0.2 ~ g ~ M ~ R_e

If g = 9.8 m/s², M = 3e13 kg, and R_e = 6371 km, the total lift energy (at 100% efficiency) 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.

BTW, the above result is for lifting a tiny fraction of the Earth's mass to the surface. Launching it to escape velocity requires E_{launch} ~ = ~ 1.2 ~ g ~ M ~ R_e . Lift energy is a third of the energy of disassembling the entire earth and spreading it out into space, which is E_{dissassemble} ~ = ~ 0.6 ~ g ~ M_e ~ R_e .

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.

Asteroid mining to feed raw materials to simple manufacturing processes to produce objects used in the asteroid belt may be less ridiculous - except there is no factory infrastructure there. That infrastructure may grow from nothing to full local capability over hundreds or thousands of years - but please keep in mind that this growth will require new kinds of processes to manufacture new kinds of objects, and that will require a vast accumulation of new knowledge, and a vast infusion of capital to speed it up appreciably (in order to pay for all the mistakes and rapid obsolescence incurred during rapid development). It took 8 trillion dollars to develop the rocket fleet we have - use that to estimate the cost of vastly more ambitious projects.

Terrestrial industrial civilization took thousands of years to develop, with the whole human race participating. Please do not underestimate the effort required to recapitulate the process in a far more challenging extraterrestrial environment. It will never happen without understanding the realities of terrestrial extraction and production; the only advantage we have over our ancestors is accumulated knowledge, if we do not ignore it.

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