Backyard Minerals
Really Wild Stuff  A Silly Argument About SoCalled Resources on Earth and in Space (copied from serversky wiki)
Delivering asteroidal 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. Gravel or sand is a few dollars per tonne. The asteroids represented by nickeliron meteorites  mostly metal  are not uniform alloys, and will need purification and blending to make an industrially useful feedstock. Extracting ppm impurities will be expensive and difficult, and will require large amounts of energy and solvents. We will do this someday, feeding the materials to projects in space, but it is unlikely they will ever be competitive with earth sourced materials on the Earth itself.
Let's compare "abundant space materials" 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 5.32e12 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 perhaps ten kilometers beneath me. There are a few dormant volcanos nearby. The hill that I live on is a longextinct volcanic cinder cone, topped with glacial till, debris, and silt from the Missoula floods.
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 out the pyramidal wedge? How much material is that, and (assuming market prices did not collapse) how much is it worth, purified to commercial grade metals and materials?
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 
Th 
6 ppm 
8.58e05 kg 
Thorium 
8.58e05 kg 
$250/kg 
$2.1e08 
Hf 
3 ppm 
4.29e05 kg 
Hafnium 
4.29e05 kg 
$1200/kg 
$5.1e08 
U 
1.8 ppm 
2.57e05 kg 
Uranium 
2.57e05 kg 
$140/kg 
$3.5e07 
As 
1.5 ppm 
2.15e05 kg 
Arsenic 
2.15e05 kg 
$1.43/kg 
$3.1e05 
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 

Addendum: Estimated platinum group metals in Earth's core: ratioed to iron 

Ru 
714 ppb 
1.13e9 oz 
Ruthenium 
$3.1e11 

Rh 
134 ppb 
2.13e8 oz 
Rhodium 
$5.3e11 

Pd 
557 ppb 
8.84e8 oz 
Palladium 
$1.1e12 

Re 
37 ppb 
5.67e6 oz 
Rhenium 
$4.7e9 

Os 
669 ppb 
1.06e9 oz 
Osmium 
$4.3e11 

Ir 
473 ppb 
7.52e8 oz 
Iridium 
$1.1e12 

Pt 
953 ppb 
1.52e9 oz 
Platinum 
$1.2e12 

Estimated market value of my platinum group minerals, 4.7 trillion dollars 
$4.7e12 
See: https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements and https://en.wikipedia.org/wiki/Prices_of_elements_and_their_compounds and related pages. For prices, check the spot markets. For rare earth prices, see https://www.metalsdaily.com/liveprices/pgms/.
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 .
Note: Magnesium is a sizeable fraction of the total. Not because magnesium minerals are particularly valuable, but rather because pure magnesium is difficult to extract from tenacious oxides. Semiconductor grade monocrystalline silicon is vastly more valuable per kilogram than sand, and so are hyperpure crystals of most of the elements listed, but that is even more ridiculous than this calculation.
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 from 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 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_{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/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.
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 slow the material and lower it gently, 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.
Launching mass retrograde from an asteroid will eventually speed it up prograde, increasing the size of the orbit and increasing the delta V necessary for launch (needs analysis). Keeping the asteroid in the same orbit requires balancing mass launched prograde. If the amounts are equal (wasting half the asteroid, preferably the less useful fractions) the total launch energy is doubled.
How many integrated circuits?
MOS FETs will continue to scale and exploit new technologies, but lets assume we stop shrinking at 10 nanometer dimensions, and build finfets out of bars of silicon 10 nm x 10 nm x 30 nm, with metal gates and 4nm thick hafnium gate oxides. We will ignore the conductor metal used, for now, we can probably use abundant aluminum and iron and magnesium if necessary. Plenty of oxygen in air, water, and granite.
The density of silicon is 2.33, HfO₂ is 9.68, and the hafnium fraction of that is 8.40. So the mass of a transistor is:
element 
Transistor volume 
density 
mass 
my resources 
Silicon 
10 x 10 x 30 e27 m³ 
2330 kg/m³ 
7.0e21 kg 
8.67e10 kg 
Hafnium 
3 x 10 x 10 x 4 e27 m³ 
8400 kg/m³ 
1.0e20 kg 
4.29e05 kg 
Transistor manufacture is limited by the abundance of Hafnium, not far more abundant silicon, so my share of the crust can produce "only" 4.3e25 transistors, a mere 40,000 times more than all the transistors in the world in 2015. Priced at 1e10 transistors per dollar, that is $4e15, or 4 quadrillion dollars. Hafnium at current prices constitutes 0.13 ppm of the cost of those transistors. The semiconductor industry's use of this rare metal is too small to impact its price.
How much fission energy?
Assume a technology like the integral fast reactor, that reprocesses fuel and sends the radioactive fission products back through the reactor for neutron bombardment and deactivation. Assuming 200 MeV per fission, 50 MeV lost to deactivation processes, and 33% thermal efficiency, we can expect 5e7 eV per nucleon plant output, or 4.8e12 J/mole.
I don't have numbers for thorium in the mantle, but assume it follows the same 3.33 to 1 ratio to uranium as the crust.
Crust Uranium 
2.57e5 kg 
0.238 kg/mole 
1.1e6 moles 

Crust Thorium 
8.58e5 kg 
0.232 kg/mole 
3.7e6 moles 

Mantle Uranium 
4.44e5 kg 
0.238 kg/mole 
1.9e6 moles 

Mantle Thorium 
1.41e6 kg 
0.232 kg/mole 
3.3e7 moles 
WAG 
Total 
4.0e7 moles 
4e7 x 4.8e12 = 1.9e20 J, or 5.3e13 kWh. Half the energy we need for the lift. Of course, if we lift only the crust and mantle, the energy needed will be smaller. OTOH, burying the waste (and everyone elses) at the earth's core might be worth the cost of exposing it. At 4 cents per kilowatt hour, a mere 2 trillion dollars worth of 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.
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.