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Casey Handmer May 2026
What?Elon has recently (late 2025, early 2026) been talking about building many terawatts of orbital AI compute and launching some components from Moon factories with a mass driver. This is an old idea, enabled by the Moon’s relatively low gravity and lack of an atmosphere. See, for example, The Moon is a Harsh Mistress and The High Frontier.

The fundamental problem with The High Frontier is that the set of products that can be made in space and sold on Earth while making money is very limited, due to the sheer cost and difficulty of accessing space. In general, they are observations and communication, which in both cases distributes the product using radio waves, which are much cheaper than physical return of artifacts from space.
In 2019, I wrote that Starlink was likely to be incredibly lucrative and I’m happy to see this is the case, with over $10b in revenue last year. Space AI takes this business model and ramps it up to 11. Why? Starlink has already established that converting a space solar photon into an electron and using it to relay bits of information around the world is extremely profitable. Space AI is a great way to vastly increase both the total demand for space data bits as well as the value per bit, as the tokens encoded by these bits have already proven to have stupendous economic value and apparently unlimited demand.
Why?Starship promises a near future with launch costs to LEO of perhaps $100/kg. With electric propulsion, large quantities of cargo, including solar powered AI can be positioned anywhere in cis-Lunar space for an incremental cost beyond that. For AI hardware, most of the cost is in the GPU/TPU die, which contributes almost no mass, while most of the mass is in the solar panel and radiator, which cost (relatively speaking) almost nothing.
Here’s a spreadsheet I put together last year with some basic first-principles analysis. At even $500/kg, launch cost is only 5% of the total satellite deployment cost, so a lunar mass driver is unlikely to drastically improve the economics of space-based AI, by reducing launch costs.
It’s also unlikely to have low start up costs!
Instead, we must look to a future where Starship costs stop falling from experience and economies of scale and rise to unaffordable levels, perhaps comparable to the Shuttle’s $50,000/kg, because of a constraint on launch capacity.
In my spreadsheet, I estimate that one Starship can deliver about 15 MW of solar power to orbit. Last year, China produced over 1 TW of solar photovoltaic panels. Supposing we weren’t constrained by chip fabrication and Starship was fully operational, it could launch 1 TW to orbit per year with just 67,000 launches, or one every 8 minutes.
This might seem like a lot, but the world currently sustains about 100,000 commercial flights per day! In a world where SpaceX can turn around a launch site in an hour, only seven or eight pads would be necessary to keep up with this rate of launch, requiring a fleet of perhaps 10 boosters and a few hundred Starships.
Nor would this launch rate defeat our global oil production. One Starship launch consumes roughly 10,000 barrels of oil (equivalent), and the world currently consumes 100 million per day. So 67,000 launches per year is less than a week of the world’s current supply of oil. It may require a few gas pipelines in Texas to be upgraded, and of course by the time this happens solar synthetic fuel will be a recognized and mature technology.
There has been some speculation about damage to the upper atmosphere of the Earth caused by huge launch volumes.
In any case, we’re talking about a launch volume of hundreds of thousands of Starships per year, or more than 10 million tonnes of cargo per year, with a total launch revenue of about a trillion dollars – equivalent to about six weeks of the global oil and gas industry!
The lunar mass driver must transcend this scale.
What does the lunar mass driver drive?The Moon is made of rocks. Primarily volcanic rock, similar to Earth basalts. As ores go, they are not preferred sources of metals on Earth. Though they contain nearly every metal – the net present value of the metal in 1 tonne of basalt is about $1300 vs the $20 price as crushed gravel – they’re mixed together and generally considered energetically infeasible to extract. We’re working on this at Terraform but the energy demand is a fact of life.
In one model, the lunar mass driver fires raw rocks into Lunar orbit, to be processed in space using copiously available space solar power. In another model, moon rocks are pre-processed to increase their metal content, or even converted into finished products, before launch. Blue Origin has demonstrated a process to convert Lunar regolith (dirt) into a functioning solar panel, but it’s not clear what the energy return on energy invested for this process would be.
The Moon’s surface itself can be a tough place to do anything energetic, because it is subjected to 14 Earth days of shade during the long lunar night, followed by 14 days of unrelenting sun during the day. Any serious infrastructure will require serious power, either from extremely large nuclear reactors operated near the poles to create functional shaded radiators, or from energy beamed up from the Earth or from Lunar orbit, or both.
For the following I’ll assume the driver is shifting dumb rock. It doesn’t change much but the g-tolerance of raw materials is a big plus!
How big is my mass driver?Let’s run some numbers around mass flow rate. We’re not going to the trouble of building a mass driver for no reason, we’re doing it to alleviate the burden on Earth of launching 1000 Starships every day, to push total human compute into the 10s of TWs incremental increase per year.
So let’s assume one mass driver launches 10 million tonnes of rock per year. Taking into account rock refining losses and expansion this implies a Lunar fleet of a few dozen mass drivers, but you have to start somewhere. 10 million tonnes per year is 1 tonne every 3 seconds.
We can calculate the total energy expenditure too. Delta-V to LLO is 1.6 km/s, and we assume that Delta-V is cheaper to come by in orbit, thanks to solar powered tugs. We only need the lunar mass driver to get rocks into orbit, where they are collected and moved to wherever they need to go.
So total kinetic power is 0.5*mdot*v^2 = 450 MW, assuming 90% driver efficiency. The 10% waste covers ohmic losses, cargo-sled recycle weight, and active cooling.
Why not stack some additional assumptions on top of this result?
Let’s assume an equivalent price of $10/kg, reflecting that random rocks orbiting the moon are not quite as valuable to the customer as a finished satellite orbiting the Earth – where all the customers are. This implies that each lunar mass driver makes $100b/year. Assuming that 10% of this revenue pays for the power plant, this works out to $2.50/kWh, which is about 10x higher than a typical US rate payer in 2026.
Can a 450+ MW nuclear power plant be built and operated on the Moon for a 10x cost increment relative to Earth? I’m not sure but it’s not forbidden by the laws of physics.
For reference, a reactor of this scale would typically cost $2-4b on Earth and weighs perhaps 1000 T.
I previously estimated that a radiator-constrained space reactor mounted into a Starship could generate perhaps 3 MWe, implying several hundred launches for the power of one mass driver. A much larger, monolithic space reactor of 500 MW scale would need to be completely re-engineered, requiring delivery, an enormous radiator (many hundreds of acres, assuming permanent shadow), and other hassles. But it could be done.
Orbit, what orbit?Low lunar orbits are, in general, unstable, due to the presence of gravitational anomalies called mascons (or mass concentrations) in various places corresponding to ancient impacts. There are, however, four classes of frozen orbits (~27°, 50°, 76°, 86°) on the Moon that are relatively stable, so provided the mass driver has a latitude and launch azimuth high enough to access these, the launched rocks won’t necessarily immediately return to obliterate the launch site two hours after launching. On a long enough timescale any passive payload launched from the surface of the Moon into Lunar orbit will run into the surface, so the trick is to launch payloads into converging bunches and scoop them up in orbit, performing circularization and/or relay transportation using some kind of orbital tug. It’s also possible to launch rocks from the Moon into more distant libration orbits, but I don’t cover that case in this post.
Technical implementation
The launcher in this image looks a bit like a rail gun. But rail guns pass current from rail to rail through the sabot (projectile enclosure) and suffer rail erosion at far higher rates than we can tolerate.
The lunar mass driver will be a maglev in disguise.
In this section I draw on my ancient experience as a levitation engineer at Hyperloop back in about 2016.
The job of the track is to accelerate a passive cargo-carrying sled to high speed over a short distance. The sled must be controlled over six degrees of freedom. Unlike a passenger maglev, where propulsion forces are a fraction of gravitational forces, a mass driver is optimized for highly efficient acceleration.
Assuming a launch speed of 1.6 km/s, v^2 = 2 a s gives track length and acceleration as inversely proportional.

If rocks can survive 1000 gs of acceleration (they can) then the launch track need only be 128 m long (or 256 m including the sled catching portion), greatly reducing its mechanical and construction complexity. Can an electromagnetic launch system deliver 1000s of gs?
A helpful intuition pump for this is to consider how much force a permanent magnet can deliver. A commercial neodymium magnet can easily support 20x its own weight, but 2000x is highly non-trivial. Bear in mind that no matter how cleverly built, the track will have to be toleranced with non-zero gap between the sled and the maglev track.
Add to this the fact that the acceleration is diluted by the mass ratio of the magnet to the rest of the system. I can imagine a launch sled which is 70% magnets, 10% structure, and 20% rocks, in which case even an optimal launch system would have to take a 30% haircut on acceleration.
There are a few ways to skin this cat, but probably the easiest is with a synchronous linear motor along a pair of parallel tracks with the sled suspended in between. At any meaningful level of acceleration, gravitational forces from the Moon round to zero. It is possible to include, for example, some null flux electrodynamical suspension system for guidance, but if you’re using the sled magnets for anything other than acceleration then you’re throwing away performance and making the track longer.
On this note, achieving anything like the necessary levels of magnetic shear force requires very very large, very flat magnets. In my model below, they are 20 cm wide, 2.8 cm thick, and 9 m long, entirely enclosed by an electromagnetic stator, and connected via high strength steel shear panels to the payload bucket.
It’s not impossible to build long tracks but, having seen this close up at Hyperloop, it becomes extremely difficult to achieve sub millimeter alignment precision over long distances. To give a flavor of this, the Shanghai Pudong transrapid Maglev operated with a ride height of about 1 mm. It is built on an elevated steel reinforced concrete trackway whose foundations were dug to a depth of 80 m into the Pudong silt. Shanghai, like many cities, is built on a river delta. Literally thousands and thousands of tonnes of concrete and steel to form the track way. You would think that it would be stable enough that the track could be calibrated once during construction and then would be fine, but in fact the track had to be re-aligned, by a custom designed track maintenance vehicle, twice per day, to account for such perturbations as tidal deformation of the crust of the Earth.
The Moon does not have saturated silty deltas or wildly varying tidal forces (being tidally locked) but it does have a very extreme temperature cycle over its 28 Earth-day day, so probably the best way to achieve dimensional stability, as well as a measure of meteorite protection, is to bury the track under a few meters of dirt.

(I asked AI to make a better version of this diagram but it wasn’t right. Motion is into the page. 200 kg of moon rocks can fit in a container 40 cm on a side.)

(Here’s an OnShape model. Below, diagrams of how the whole thing goes together.)


The cart oscillates back and forth on the track, launching rocks for ¼ of its cycle, with the rocks separating from the sled bucket at the midpoint of the track. The slow down step can recover some momentum as power, provided there’s a place to store it! At 1000gs, launch takes just 0.16s and the complete cycle (accelerate, release rocks, slow down, fly back to rock loading site) is as short as 0.64 s, not including time to reload rocks. At the 10 million tonnes per year rate, each launch would carry about 200 kg, with a total sled weight of 1000 kg loaded and 800 kg empty. This launch and recycle process would load the structure with up to 1000 T, that is 10 MN, of force, at about 1.4 Hz. That’s a pretty wild vibrational environment for the ground anchoring system to endure.
Power = force*velocity, peaking at 16 GW at the middle point of the 256 m long mass driver, a lot more than the average demand. Power consumption averages 1.75 MW per meter, but much of this is recycled. Assuming 3% loss as ohmic heating, we’re still looking at 50 kW per meter, easily enough to justify an active coolant loop. For the purposes of this post, I assume any waste heat pumped and dumped in a permanently shadowed radiator used primarily by the power reactor. Beaming power up from the Earth would require a receiver rectenna several km across.
Structurally, the entire launch rail must endure tremendous forces, variable loads (due to unbalanced cargo), impacts from micrometeorites and rock spray during rocket landings, electrostatic and thermal nonsense caused by the Lunar day night cycle, and lots of other things. Fortunately, it has no moving parts and no wear surfaces!
So while the mass driver could be brought to the Moon over several launches in pieces, it needs a sophisticated anchoring system to react out loads and cope with extreme thermal swings during the Lunar day/night cycle.
What do you think, Claude?In which I roast myself with the latest AI so you don’t have to. Also some more technical detail.
Missing or under-treated
- Sled recycle losses. You bury the second 128 m of deceleration in “recover some momentum as power.” Empty sled at 1.6 km/s = 1.02 GJ. At 1.56 Hz and 90% recovery, that’s 160 MW of pure dissipation, not in your 450 MW budget. At 80% it’s 320 MW. You either need to argue >95% round-trip electrical recovery (hard given switching losses on multi-GW pulses), or your reactor is closer to 700–800 MW. [CH: Correct, there’s a very steep penalty for higher parasitic power losses.]
- Pulsed-power infrastructure. 16 GW peaks at 1.5 Hz with bidirectional flow. The capacitor banks / flywheels for this dominate the system mass and probably the cost. ITER’s pulsed power is tens of GW for comparable durations and it’s a building. On the Moon, this is the actual hard engineering problem, more so than the rail itself. Worth a paragraph. [CH: There are several ways to skin this cat.]
- Catcher architecture. 56,000 launches/day into LLO requires tug coordination at obscene tempo. Velocity dispersion at exit sets apogee error: 0.1% velocity error → ±5 km position error after a half-orbit. Either you hit launch precision <10⁻⁴ (very hard with thermal drift on a 256 m structure) or your catcher needs km-scale capture and active pursuit. The catcher fleet plausibly outweighs and outcosts the driver. [CH: I have also waved my hands about what to do with rocks in lunar orbit – they still need to be converted into solar computronium somehow!]
- Δv beyond LLO. Customer is in Earth orbit, not lunar orbit. LLO→GEO ≈ 3.9 km/s, LLO→LEO higher with aerocapture. You save ~9 km/s vs Earth launch but you’re not at the customer. The relevant comparison is (launch + tug from LLO) vs (Earth launch), not (launch) vs (Earth launch). Worth one paragraph stating the tug budget. [CH: As stated, the assumption is that Moon launch delta V is the expensive part we’re trying to overcome.]
- Chemical-lander baseline. A dumb LOX/LH2 (or LOX/CH4) lander from the Moon to LLO needs ~1.8 km/s, mass ratio ~1.7. If you have any propellant production at all, chemical wins until volume is huge. Mass driver’s case rests on (a) no propellant feedstock available at scale, or (b) volume past the crossover. State this; the reader will ask. [CH: There’s no good reason to synthesize prop on the Moon and the gear ratio makes flying chemically propelled shuttles to and from LLO pretty silly.]
- Reactor radiator. 450–800 MW thermal, polar shadow, ~700 K reject → tens of thousands of m². Your 2025 Mars-reactor post implies a tonne/m² class radiator. This is hundreds of tonnes of radiator alone, which dwarfs a 1000-T terrestrial reactor mass. The 10× cost premium claim wants more support. [CH: True, as a good rule of thumb, a space reactor’s radiator will weigh about as much as everything else put together.]
- Magnet survivability at 1000g. NdFeB is brittle. Sintered blocks 9 m long under 1000g shear and oscillating tension fatigue is not obviously feasible. Either bonded magnets (lower B), or segmented with structural steel, which eats into your 70% magnet mass fraction. [CH: Actually building magnets and a sled structure which can make this work is serious engineering.]
The Moon’s lack of an atmosphere means you can also operate a lunar sling. A tower with extendable tethers whirls around to obtain high speeds without as high g loads. I see no reason why entrepreneurs couldn’t build a range of lunar launching solutions!
ConclusionLunar mass drivers do not violate the laws of physics. Their tech can be developed and tested thoroughly in labs on Earth at prices that are reasonable in comparison to the complexity and cost of Starship development. They are unlikely to be able to compete with Starship flying at any level of volume from Earth (where the chip fabs are) to space (where the infinite sunlight is), unless Earth launch is supply limited in some way. Order of magnitude, this is beyond 100 TW per year deployment. For reference, 10 TW of compute is roughly equal to the total economic output of the entire world’s supply of natural intelligences. We live in interesting times!











































