La Bourse / Volume VII / Nº VII.09 / Ten concept briefs

A watercolour of a young woman with an open book, flowers and leaves around her. — Plate VII.09 · Ten concept briefsThe Weight of the Bottle.The bottle holds one kilogram. On this bench it is worth a tenth of a penny. Four hundred kilometres above her head, the same kilogram is worth three thousand and fifty-nine dollars — and nothing about the water has changed.

TEN CONCEPT BRIEFS · Chapter VII.09 — Space, Materials, and Real Abundance

One page each. A reader who reads only these ten pages has the chapter.

BRIEF 1 — The Gravity Well Is a Price

The idea. A planet is not a place. For an economist it is a toll, and the toll is the specific energy needed to leave: ½v², where v is the escape velocity.

  Earth   11.186 km/s   ->   62.6 MJ/kg
  Moon     2.38  km/s   ->    2.83 MJ/kg
  Bennu    0.20  m/s    ->    0.02 J/kg

Earth costs 22.1 times what the Moon costs, and three billion times what Bennu costs, per kilogram lifted off it.

Worked example. You want a tonne of steel in orbit. Lifting it off Earth costs 62.6 gigajoules of theoretical minimum work and, in practice, $3,059 per kilogram of delivered launch — $3.06 million. Lifting the same tonne off a small asteroid costs twenty joules, which is one person pushing it gently.

Why it matters. Every argument about space resources reduces to which side of that toll booth you are standing on. Material already up the well is cheap to move; material at the bottom is not. The asymmetry is enormous and real.

You already know this because you have paid more to have a heavy thing carried up three flights of stairs than the thing itself cost, and you did not conclude that stairs were magical. You concluded that lifting is the expensive part.

BRIEF 2 — The Break-Even Price, P*

The idea. For returned material there is one number that decides everything: the price per kilogram below which nothing is worth bringing home.

  P*  =  (LEO mass required per kg returned)  ×  (launch cost per kg to LEO)
      =  1.049 kg  ×  $3,059/kg
      =  $3,208 per kilogram

That is a floor, not an estimate. It includes propellant and nothing else — no vehicle, no development, no refining, no capital. Give the return vehicle a dry mass of twenty percent of its payload and it rises to $7,313; amortise $500 million of capital over a hundred tonnes returned and it is $12,313.

Worked example. Copper trades at $9.50 a kilogram. It misses the floor by 338 times. Nickel at $15.00 misses by 214. Iron ore misses by four orders of magnitude. Gold at $112,528 clears by 35.1 times, platinum at $41,796 by 13.0 times, iridium by 45.1.

Why it matters. It converts an argument into a comparison. You do not need a view on the future of humanity to evaluate an asteroid-return proposal. You need a price list.

You already know this because you have decided against shipping something heavy home from a holiday on exactly this logic, without writing the inequality down.

BRIEF 3 — The Rocket Equation, and Why Outbound Costs More

The idea. A rocket's mass ratio is exp(Δv / v_e), where v_e is the exhaust velocity. It is exponential, and the exponent is unforgiving.

At a flight-proven specific impulse of 300 seconds, v_e is 2.942 km/s.

  LEO -> Bennu rendezvous   Δv = 5.10 km/s   ->   mass ratio  5.661
  Bennu -> Earth capture    Δv = 0.50 km/s   ->   mass ratio  1.1852

Worked example. Bringing home one kilogram burns 0.1852 kg of propellant at the asteroid — a splendid 5.40 kilograms returned per kilogram of propellant. But that propellant had to be carried out there, at the outbound mass ratio, so the true cost is 0.1852 × 5.661 = 1.049 kg in low Earth orbit per kilogram returned. And putting that kilogram into orbit took 15.17 kilograms of propellant on the pad.

Why it matters. The exponent runs on the leg you were not counting. Almost every optimistic space-resource sum quotes the return leg, which is cheap, and quietly assumes the outbound leg is free.

You already know this because you have costed a delivery and forgotten the van had to drive out empty first.

BRIEF 4 — Delta-v Is a Map, Not a Distance

The idea. In space, distance is nearly irrelevant and delta-v — the velocity change a trajectory demands — is the only geography that costs money.

  Earth surface -> LEO           9.40 km/s
  LEO -> GEO                     3.90 km/s
  LEO -> lunar halo orbit        3.40 km/s
  LEO -> Bennu rendezvous        5.10 km/s
  Bennu -> Earth (departure)     0.28 km/s
  easiest NHATS bodies           3.80 km/s

Worked example. Bennu is hundreds of millions of kilometres away and the Moon is four hundred thousand. Bennu costs 5.10 km/s to reach from low orbit; the Moon costs 3.40. The asteroid is a thousand times further and only half again as expensive. Meanwhile coming home from Bennu costs 0.28 km/s, because the atmosphere does the braking for free — the outbound leg is 10.2 times the return.

Why it matters. It is why "how far is it" is the wrong question and "what does the trajectory demand" is the right one. It is also why nearby things can be expensive and distant things cheap.

You already know this because you have flown across an ocean for less than the taxi to the airport, and understood that the ticket prices a route, not a length.

BRIEF 5 — Position Pricing: The Same Kilogram, Twice

The idea. A commodity's price is a fact about where it is. This is Chapter II.01's positional theory of scarcity, and space is its clearest case.

  water at a municipal tap           $0.001 /kg
  water in low Earth orbit           $3,059 /kg
  water in a lunar halo orbit        $9,717 /kg
  water in geostationary orbit      $11,517 /kg

The ratio from tap to low orbit is 3,059,211 to one, and the molecule is identical.

Worked example. The value of water in orbit is not a guess. It is the launch it avoids: if delivering a kilogram there costs $3,059, then a kilogram already there is worth $3,059, minus whatever discount the buyer needs. That is a benchmark price, published, falling at 5.05 percent a year, and it is the only honest way to price anything in orbit.

Why it matters. It tells you what the space-resource business actually sells. Not material. Position.

You already know this because you have paid four times the shop price for the same bottle of water at a festival, and never once believed the water was different.

BRIEF 6 — Closure, and the Multiplier `1/(1 − p)`

The idea. If p is the fraction of a consumable that comes round again, each kilogram you bring does the work of 1/(1 − p) kilograms.

  p = 50%  ->   2.00x
  p = 85%  ->   6.67x
  p = 93%  ->  14.3x
  p = 98%  ->  50.0x

It is convex. The last few points are worth more than all the earlier ones together.

Worked example. The International Space Station recovers 98 percent of its water. Seven people at three kilograms a day need 7,670 kg a year open-loop; at 98 percent closure they need 153 kg of make-up. At $3,059 a kilogram that is $23.0 million of launch avoided every year, by plumbing.

Why it matters. This is II.01's circulation multiplier arriving in orbit, and in orbit it dominates. Raising closure from 93 to 98 percent beats a factor of three in launch price. Fund closure before you fund extraction; that is where the derivative is.

You already know this because you know that a deposit-return scheme changes the economics of a bottle more than a new glass factory would.

BRIEF 7 — The Inelastic-Demand Trap

The idea. Marginal revenue is P × (1 + 1/ε), where ε is the price elasticity of demand. When demand is inelastic — ε between 0 and −1 — that expression is negative, and selling more earns less.

Short-run platinum demand runs about −0.40, because autocatalyst use is technically determined. So 1 + 1/(−0.40) = −1.50: every extra tonne reduces total industry revenue.

Worked example. To book one billion dollars of platinum at today's price you must land 23.9 tonnes — 13.3 percent of world mine supply. That moves the price by −33.2 percent. You realise $668 million, not a billion, and you destroy $2.50 billion of value in the existing industry: $3.74 destroyed for every dollar earned.

Why it matters. It is the hidden fault under every asteroid-mining pitch. The only materials that clear the transport floor are the ones whose price is their scarcity — so a successful mission abolishes the condition that made it worth flying.

You already know this because you have watched a bumper harvest ruin the farmers who grew it.

BRIEF 8 — Grade, and the Mass You Must Handle

The idea. An ore body is judged on grade — the mass fraction of the thing you want — because grade decides how much rock you must move per kilogram sold.

  platinum in carbonaceous chondrite       1 ppm
  platinum in iron meteorite              10 ppm
  total PGM, Merensky Reef (Bushveld)      6 g/t

The best asteroid is 1.67 times the grade of a mine that already exists on Earth. Not a hundred times.

Worked example. To land 23.9 tonnes of platinum at 10 ppm, at perfect recovery, you must process 2,392,577 tonnes of metal — 100,000 kilograms handled for every kilogram returned — in microgravity, where there is no weight to settle a slurry, no atmosphere for a flotation cell, and no gravity to run a mill.

Why it matters. The popular claim — one asteroid holds more platinum than has ever been mined — is true and is a statement about total mass, not about grade. A body a kilometre across contains a great deal of everything. So does a mountain.

You already know this because seawater contains about twenty million tonnes of dissolved gold and nobody mines the sea.

BRIEF 9 — Kessler's Condition, and Capacity as a Bracket

The idea. Collision rate scales with the square of the number of objects. A shell is at carrying capacity when collision-generated fragments equal the fragments atmospheric drag removes:

  0.5 · k · n*² · F  =  n* / τ        ->      n*  =  2 / (k · F · τ)
       k = σ · v̄ · T / V

Worked example. For the 200–2,000 km shell — volume 1.269 × 10¹² km³, mean relative velocity 10.0 km/s, F = 2,900 catalogued fragments per break-up (the mean of 3,500 from Fengyun-1C and 2,300 from Iridium–Cosmos, both measured), residence time fifty years:

  assuming 1.00 m objects   n* =   4,412
  assuming 0.30 m objects   n* = 196,111
  geometric mean            n* =  29,417
  catalogue today                40,500

The honest output is a bracket 44.4 times wide, and the catalogue sits inside it — 1.38 times over the central estimate.

Why it matters. The width is the finding. A single number invites a single counter-number; a bracket forces the argument onto the assumption underneath — σ, F or τ — which is where the disagreement actually lives.

You already know this because you have stood in a room that was not yet full and known perfectly well that nobody could say exactly when it would be.

BRIEF 10 — The Orbital-Use Fee and the Disposal Bond

The idea. A congestible commons with no price is depleted by people acting entirely reasonably. Two instruments fix it, and neither is exotic: a fee on occupancy and a bond against disposal.

Worked example. Rao, Burgess and Kaffine compute an optimal orbital-use fee of $14,900 per satellite-year in 2020, rising to $235,000 by 2040 — an implied 14.8 percent a year — and find it raises industry value from $600 billion to $3,000 billion, a factor of 5.0. At its 2020 level that fee is 0.21 percent of the annual cost of a $50 million satellite over a seven-year life.

The bond is sized at removal cost. The only removal ever contracted is ClearSpace-1, at €86 million for one object; a fleet-scale target of $5 million is 10 percent of that satellite. Posted as restricted cash at licence, released on verified disposal.

Why it matters. The fee does not shrink the industry — on those endpoints it multiplies it fivefold, because what is being priced is congestion, and congestion is what destroys the asset. Ostrom's point exactly: a commons is not saved by exhortation, it is saved by a rule with a number in it.

You already know this because a mine posts a rehabilitation bond, a tenant posts a deposit, and neither arrangement is controversial anywhere on Earth.