(Hi, please be kind - first poster. Also, didn't know if it'd be allowed to put hard-science on something like this, so going with reality-check for now, but formulae will definitely be appreciated!)

I'm 52 years-old X AE A-XII Musk, and I'm trying to become humanity's first home world builder away from home. I've mostly been pleased with the space transport imperium I inherited from my father, the late Elon Musk, which is providing the basic infrastructure for all of us to finally make Mars into the planet that Earth can't be anymore.

Unfortunately, around the year 2040 many of the resources required for launching my rockets into space have begun to run dry on Earth. It was not an immediate problem as I was able to replace most of them with what I found on the Moon and various asteroids, no thanks to the Western Alliance government, who tried to prevent me for too long from using those resources as required. They have been dealt with now, but I've also been looking for a better way--burning chemicals does seem like an awfully dated thing to be doing.

Recently, after one too many times of "only 20 more years", the first experimental fusion reactor to have achieved a solid sustained net energy output of around 150MW, ITER-III, has been acquired by one of my companies, and I'm now planning on having those people build me a bigger one in order to power a mass driver, which I'll call the Space Slide. Texas seems as good as place as any for such a venture and I have signed a contract to develop the area around Guadalupe Peak and the Salt Basin Dunes east of it, including the outskirts of El Paso, for the purpose.

The first Space Slide is intended to be used to launch the quaint, but still reliable Dragon II capsule, which, at maximum payload capacity, has a mass of twice (20402kg) as much as the original Dragon.

Unfortunately, because I've spent most of my life managing companies, I have no clue of either mathematics and physics and so would like to ask the following questions:

  • there's around 140km of distance between El Paso and Guadalupe Peak, which rises to 2667m above sea level. Is this long enough² and tall enough for a mass driver capable of launching well-paying (but otherwise ordinary) humans into space, or is it still too short to keep the g-forces down¹?
  • (a) how good is Wikipedia's info of 40MJ/kg (~816GJ for the Dragon II) to get to LEO?
  • (b) how much more than that would you need to get to the Moon (given suitable alignment of Earth and Moon and assuming the capsule's thrusters are good enough to compensate sub-optimal alignment)?

... and finally ...

  • how big does the fusion reactor have to be³ for (a) and/or (b)?
  • assuming it's not feasible to feed directly from the reactor to the mass driver, how big would each capacitor in a chain of super capacitors⁴ along the length of the mass driver have to be and how many of them will be required?

¹: Let's say, 5g max? Does that make sense?

²: Assuming 11.2km/s escape velocity

³: How big, as in, what sustained and/or maximum power output in MW does it have to be capable of, not as in the required footprint (which is probably unknowable from a 2021 perspective, aside from an educated extrapolation of something like Wendelstein 7-X...)

⁴: And also: how close are we in 2021 to being able to build such super capacitors?

  • 1
    $\begingroup$ Not an answer, but generally mass drivers don't use direct power anyway. You'd have banks of capacitors storing energy, so the size of the reactor is kind of immaterial. $\endgroup$
    – jdunlop
    Commented Aug 5, 2021 at 17:19
  • $\begingroup$ Note that I'd already mentioned banks of capacitors. Changed the title to include that and changed the description to make it a more integral part of the question. $\endgroup$
    – Sixtyfive
    Commented Aug 5, 2021 at 17:24
  • $\begingroup$ Well, the steady state output of the reactor will have a lot to do with determining the launch cadence (there's a Musk term!) of your Space Slide. Does it take a week to charge the capacitor bank, a day, or a couple hours? Or can you launch as quickly as you can cool stuff off and load in the next capsule? $\endgroup$
    – Zeiss Ikon
    Commented Aug 5, 2021 at 17:25
  • $\begingroup$ For now it's mostly about making the very first launch work, and about not killing people from too many g's. I'll wait for some more comments or perhaps a first draft of an answer before working this into the question, but let's perhaps go for an 8 hour launch cadence for now, assuming the caps can take it as fast as the reactor can pump it out, and also that the bleed on the first caps while they're waiting for the last ones to complete charging can be ignored for all practical intents and purposes. $\endgroup$
    – Sixtyfive
    Commented Aug 5, 2021 at 19:27
  • $\begingroup$ Massdriver from earth not necessarly the best, in a first place because of 8km/s bare minimum - turn is too sharp(horizontal to up angle), athmosphere - so handwaving some submersive anti-g solutions is a necessity. The rest is okayish matter to be estimated - quite good q as my taste goes. $\endgroup$
    – MolbOrg
    Commented Aug 5, 2021 at 20:09

2 Answers 2


There are several isssues:

  1. The Mass Driver: Mass drivers are a very cool concept, but they, like all things, must consider friction. The FAA definition for space at just over 80 kilometers is based on the fact that objects may pass over that limit and remain in orbit. Any thing beneath that is producing enough drag to deorbit itself. To wit: 2667 meters above sea level is not nearly high enough to directly shoot something into orbit.
  2. NASA usually recommends not passing 3g of acceleration, though a well-trained person can sustain up to 10, and survive in excess of 40. Considering this and good old calculus, we can calculate a rough estimate of the necessary length of the driver(x = (v^2)/2a): at 3g you'd need 1380 km, at 5g you'd need 828 km and at 10g you'd need just 414 km. To do this acceleration in 140 km you'd need to accelerate at 30g. It is likely that most humans can't take that for a whole five minutes.
  3. Using energy expenditure to explain how you get to orbit is misleading. To be in orbit of earth you must be at a velocity of about 7.8 km per second and outside of most of the atmosphere. In ideal trajectories, this translates to a delta-v of about 9.1 km per second. Reality, unfortunately, is hardly ideal. If the maneuver to get to orbit includes a larger horizontal element, there is greater loss of thrust, therefore energy, to gravity. Additionally, atmospheric drag shaves off energy as well, increasing exponentially as you go though the atmosphere at higher and higher speeds.
  4. As far as power goes, by and large, where there is a will there is a way. We currently have the ability to charge magnets in the LHC to 10 GJ, so you should be able to do it with today's tech. The size of the powerplant just depends on how often you want to fire the thing (1 watt = 1 joule per second. Do the math.)

While this concept is somewhat flawed, it is not fatally so. The Mass driver can be used for initial acceleration etc. So don't give up little X, inherit the stars.

  • $\begingroup$ SoA: problem of the A - it does not address the question asked, for the most part, problems mention legit, but they, probably, should be accompanied with the answer to what has been asked if it is possible. $\endgroup$
    – MolbOrg
    Commented Aug 6, 2021 at 12:05

Mass driver, on the earth ground, is not necessarily a great idea, but there are still some ongoing projects from the gun to space and they may make sense, so let's crunch some numbers.

Launch energy to spend is

40MJ/kg is a reasonable number, this number represents kinetic energy at 8944 m/s. Going through the air for small, under ton projectiles, if I remember correctly from that ramjet gun project - losses can be up to 50%. For bigger projectiles situations tend to improve, the bigger(decreasing proportion of surface area/mass) the projectile, the lesser the losses, longer in air shallower the angle more the losses. But, let's go with 50% losses with that aspect, maybe too much but it depends on the specific trajectory, angle of ascent.

Storage losses get efficiency 70-80 percent - batteries, capacitors, mechanical storages + plus some transmission losses - 70 percent storing efficiency.

The track itself isn't an ideal machine, whatever happening there - self-cleaning procedures, straightening devices, cooling coils or conditioning the track or whatever - let's go 50%.

So we may end with a 17.5 percent efficiency and that 40 MJ/kg bloats to about 230 MJ/kg and total 4600 GJ or 1.26 GWh. Let's round it up a little, as you may need some additional sled(like in rocket sled) that does not go in space but keep the thing on track and say it 1.5 GWh(5.4 TJ).

So, launch parameters end up be:

  • 12.6 km/s velocity (insane), 57g (a lot), 140km acceleration track (short, not human-rated, kinda)

total energy to produce and to charge for launch - 1.5 GWh price with 0.1 bucks per kWh - 150 grand.

  • 30% loss in charge-discharge, 50% loss in track this way in capacitor banks it needs to store about 2 times more than it needs in an ideal case and again 2x due we need to shoot faster than if there would be no air - total 4x that of an ideal case
  • and overall we need to supply from power plant to the system 5.7x than of an ideal case, a total of 1.26(1.5) GWh
  • Low Earth orbit (LEO) Trans-Lunar Injection (TLI) 3.20km/s delta-v, this way you need about 3km/s extra, and it is around +57% in terms of energy.

reactor requirements

150MW, ITER-III - yep, pretty much typical ITER - 840m3 plasma volume, 500 MW heat with a big building and all that around.

And from the total, we see it may be sufficient to make 1-2 launches a day - so as a power source - it is okay.

more frequent launches and you may need proportionally more of those or get decommissioned nuclear plant - just one block gets you about 10 times more, and there are few of those in a plant. (or just pay electricity bills, not that expensive)

  • even a diesel(methane) powered generator will do, it definitely a more efficient solution than using that fuel in a chem rocket.

Last km's, peak power consumption is around 2800 GW (57g, top speed 12.6km/s, x2(due loss 50% in the track), 20000kg payload).

This is x40 of peak power of Germany(?) electricity production or about 242GW (average), for about 22.1 seconds, yeah a bit too much, for the most ways to generate energy - but, for a fusion reactor, it may be actually not that bad.

Not necessarily for Tokamak ITER - but in general, for fusion as it is now - they do work for a short time, as for now and the launch needs just a 22 seconds - so maybe exactly the necessity for such burst power is what makes it a good match to that time fusion technologies - they do not need to work too long, they can work in pulse mode as it exactly what the track may need. in some circumstances may be a match made in heaven, with some MHD generator, which in the mature stage may allow launch payloads with seconds intervals - fusion-powered rocket launches, how cool is that definitely something to look forward to.

  • this fusion direction requires some further investigation. it does not have to sustain the plasma, but work more like a prototype of a fusion rocket engine, be more open architecture, not tokamak, fewer stability issues(maybe) as they are blown out as exhaust.

  • it requires about 18'000 ITER's like in the q to power the thing directly if they can't give proper peak power, and if they can follow power profile in those 22 seconds then it may be 1600 of those blocks

  • with charging, once an hour launch requires about a 1.5 GW power line, which isn't that much. Do not get it wrong, it's a good meaty line, but power plants can be 1000's km away.

energy storage

fun fact first - per each meter of the track it needs an equal amount of energy. So energy storage may be a line of the same boxes along the track and in that sense, its volume does not matter that much, mass does as it correlates more directly to the price of a solution.

Thus we get our 1.5 GWh (5.4 TJ) and accounting our assumed 70% losses in charging(a bit wrong, but...), spec capacity per each meter of the track has to be 27 MJ (btw, a few kg's of propane, 49.6MJ/kg - can be an option, similar to EMP charge).

From wiki situation looks similar for Lithium batteries and Flywheels (good stuff btw) - about 57 kg of the stuff per meter of track. (output discharge is too low, however, so it is just a number for energy-storing)


trimmed table from wiki

specific energy is how much energy they can store, mass-wise. specific power - how fast or how powerful is the discharge

So energy-wise it needs about 1500 kg of super caps(18000J/kg, 5Wh/kg), but discharge wise it is a bit tricky.

with 10kW/kg specific power and total mass 210'000 tons of those super caps, total discharge power is 2100 GW, which is less than peak power of 2800 GW it needs more of those capacitors 280'000 tons or 2000 kg/m, realistically probably even more, but we got quite good safety margin previously with all efficiencies and losses, time to use it here.

  • conveniently the table contains numbers for more regular capacitors, about the same for them, 10 times better discharge power and 10 times worse in terms of energy-storing, but if they are 10 times cheaper it can be an alternative.

As a volume, let's take LSG/manganese dioxide with 42 Wh/L(150kJ/L) with 10kW/L discharge(which was used) and 10'000 cycles (suspicious dog.jpg, not in this setup, lol)

And we get 240L per meter of track or 12 of 5-gallon buckets(i guess).

In total, we get 33600 cubic meters, a lot, but a station each km will be 240 cubic meters per station, with spacing like few meters high 10x20m rectangle.

problems and non problems

@thewildnobody in his answer addressed them well enough.

However, there are still things to mention, maybe even positive ones.

NACA, an agency that predated NASA did some testings back in the days and submersive half-backed setup was used for testing and it did allow some testers to endure 30g(or something like that) for about 30 seconds - limits were due to centrifuge specs and holding breath limits of a tester. And max for short time was like 83g(not with the setup) - so in that sense, 57g aren't impossible, even if problematic they are.

And due to the atmosphere you have to be prepared for high g's, even if you have a longer track with moderate 5g as you wish it, then still there will be few seconds of declaration after the exit from massdriver, and it does not make it easier than it will be an opposite force, and this acceleration may be very high, in a few seconds, it may drop few km/s sooo, losses due to interaction with the air may be quite high, so the deacceleration. For some types of non-manned vehicles, it is still fine - but for humans at any rate you may need a good jelly box. All the viability depends on how good that jelly box is.

excess of jelly or water probably can be used for some ablative solution for those critical few seconds before the craft leaves the dense atmosphere, but it is not certain.

p.s. for some reason I stuck with numbers and made plenty of mistakes, now I believe it is more or less okay, but still, it is more or less an example of calculations.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .