Mass drivers for launching cargo or spacecraft can be improved in two ways. Either one increases acceleration (which the wetware can't handle so well) or one increases the length of the mass driver. If one takes the lengthening of the mass driver to its logical conclusion, one gets a ring around the entire celestial object.

The particular design I have in mind in a long maglev rail, which simply releases the cargo, once it has the desired velocity and vector.

However I was wondering, how high a realistic launch speed would be. I don't believe that spacecraft can (or should) be launched from these mass drivers at velocities close to the speed of light. But what would be the limiting factor?

Assume that your energy budget is truely massive (any civilisation building these is probably close to being a K2), you got room temperature superconductors, decent shielding against magnetic fields (assume that they got some kind of advanced metamaterial) as well as carbon allotrope based construction materials.

  • $\begingroup$ What is the purpose of the launch? Is the payload required to slow down at the other end? Implied that biological beings are on board, but will they always be? Very different answers for launching a ballistic shot at an incoming warship, an unmanned one-way probe to explore the universe and a crewed spaceship that needs to dock with a space station in Jupiter orbit. $\endgroup$ Commented Aug 11, 2020 at 9:30
  • 2
    $\begingroup$ What circumference of ring? This is vital because past a certain point the g-forces on the meat bags inside will still exceed survivability, assuming of course your materials can hold up to the strain of getting something to that speed... $\endgroup$
    – Joe Bloggs
    Commented Aug 11, 2020 at 10:33
  • 1
    $\begingroup$ The faster your vehicle is going, the more it "wants" to ignore your track and go flying straight off into space, and the more force your track will need to exert on it to keep that from happening. At some point this will be too much for track and/or vehicle to bear without failing. (A similar argument applies to keeping your cargo from smashing into the side of the vehicle.) $\endgroup$
    – Cadence
    Commented Aug 11, 2020 at 11:45

2 Answers 2


There are a lot of assumptions here, such as Room Temperatur Superconductors and unlimited energy. But ok, lets go with it.


Assuming a K2 Civilisation, the only real limitation is Physics itself. For everything that follows, ill assume Earth as the launch Platform.

Earth to the Stars

Earth has a circumference of 40.075km . So or Maximum Rail (Assuming no spiral pattern) is 40.075km long. The Human body is fit for a Horizontal Maximum acceleration of around 6g for 10min. But that is only for 10min. So lets assume the normal, everyday Human can Handel 2G for unlimited time.

This gives us a Maximum Acceleration of 2G (=19.62m/s²) for a Track lenght of 40.075km. We can use the equation v² = u²+2as. In which v is or final velocity, u² is or inital velocity, a is or acceleration and s is or track lenght. All in SI Units.

So v² = u²+2as

v² = (0m/s)²+2*(19.62m/s²*400750000m)

v² = 15725430000m/s

v = sqrt(15725430000m/s)

v = 125401.07655m/s /:1000

v = 125.40km/s

So we can expect a final velocity of around 125.4km/s with such a Track and only one revolution.

The time for this one revolution of course decresses the faster we go. We can find the time t with the equation v = u+a*t. In which v is the final velocity, u is the inital velocity, a is the acceleration and t is the time.

v = u+a*t /-u

v-u = a*t /:a

(v-u:a) = t

v/a = t

125401.07655m/s / 19.62m/s² = t

t = 6391.4921789s / 60

t = 106.524869648min / 60

t = 1.77541449414hr

So it would take around 1.8hr for the first revolution.

But, i hear you ask, why is any of that important ? Well, because as you can see, one revolution really dosnt give us much in terms of speed. But, i am about to prove that wrong.

Gravity Time

The earth has an acceleration of 9.81m/s². And there is this force, you probably know it, the centripetal force. If you think about it, this Mass Driver is nothing else then a giant centripetal force generator.

So how great would this force be with a Velocity of 125.4km/s ?

Firstly, i know what you are thinking, "but dosnt the centripetal force point inwards ?" And yes it does but also, it dosnt matter since the outwards pointing vector is the same. Magic.

The equation to get the force is just F_z = m*v² / r. In which, oh god we have a problem. We need mass dont we ? And thats where you are wrong. We just look for the centripetal acceleration. For which the mass is not needed. Because:

F = m*a

F_z = F

ma = mv² / r -> m cancels out

a = v² / r

a = (125401.07655m/s)² / 6371000 m

a = 2468.28284412 m/s

a = 2.4km/s

Yeah thats right, when you shoot along a Rail at 125km/s you get pushed upwards with an velocity of 2.4km/s Which for all inteneds and pruposses, is equal to an acceleration force because the Train is still moving. Meaning those 2.4km/s are your effectiv Gravity.

But we can do one better. Lets assume a Train without mass and only a 80kg Human riding along. In this case F_z = m*v² / r is true.

Fz = m*v² / r

Fz = 80kg*(125401.07655m/s)² / 6371000 m

Fz = 197462.62753 N

And as F = m*a, we can solve for a by doing the old swiggity swoody

F = m*a -> /a

f/m = a

197462.62753 N / 80kg = a (we get acceleration because N = kg*m/s²)

a = 2468.28275 m/s²

a = 2.4km/s²

So yeah, this should come as no supprise. The Momentary Velocity is the same as the General Acceleration. And as we know, Mass dosnt play into effect when it comes to Acceleration (in simple terms).

The Problems

So i think you can start to see the problem. In theory, a Rail along the Planet could push stuff put to very high speeds. But, the centripetal fore / acceleration really dosnt see it that way. Most electronics would die with an Acceleration of 2.4km/s². Not to mention that no human can get out of there alive.

El Answer

But your question was who fast this could get. Well, going back to the question how many G´s a Human can handel, its around 2G. All we need to do is solve for lets say 20m/s² to make it simple.

We can use a = v² / r and just solve for v². As this is or final velocity.

a = v² / r

a*r = v²

sqrt(a*r) = v

v = sqrt(20m/s)*6371000 m))

v = 11288.0467752/s

v = 11.288km/s

So a final velocity for such a system that is intended for humans is around 11.3km/s. Which is really fast.

Assuming normal Cargo can sustain around 100G´s we have a Velocity of:

a = v² / r

a*r = v²

sqrt(a*r) = v

sprt(1009.81m/s²6371000 m) = v

v = 79056.6316004 m/s

v = 79km/s

Ti Solve

Now, this all seems like a bit of BS. You got this long Track and the best we can pull of are 80km/s ? But really, the only reason why we are limited is because of the shape. If it was a line track with no curve, you could go as fast as you want. There is no Upwards force.

Not to mention that i would like to see a Rail Track hanging on earth with 100times the Gravity pulling on it once your super Train goes over it.

All in all, its a very unsmart idea to build something like that.

Sorry for typos, i am German D:

  • $\begingroup$ When solving for the final survivable velocity you don’t seem to have accounted for gravity’s effect on the apparent acceleration. I think you can get up to higher velocities than you think because one g of the centrifugal acceleration (yes, I’m working in the rotating frame :-) is counteracted by Gravity. Solving that equation for exactly one g should yield the orbital velocity at sea level, shouldn’t it? $\endgroup$
    – Joe Bloggs
    Commented Aug 11, 2020 at 15:34
  • $\begingroup$ You are right ! I forgot to include the effects. All we have to do is to incress the Maximum acceleration to 3G´s. One G gets cancled out because Gravity points in the other direction. Well let´s see, again we go with 3G´s. Ill use 30m/s² to make it simple. In the end, for Human´s to survive the trip you could get up to 13.8 maybe 14km/s. Which is well above the 7.9km/s needed for Orbital velocity. Tbh, just dont make it a Circle. Its pointless as you waste almost all of your Track. $\endgroup$
    – Erik Hall
    Commented Aug 11, 2020 at 15:54

Short answer:

  1. Orbital speed is when gravity is mutually compensated with centrifugal force
  2. So one excess orbital speed is +1g of centrifugal acceleration

It means that if we can tolarate 2g - we can go up to 8+16=32 km/s. If can tolarate 3g - 8+32=40km/s

Straight space lift is not well suited as a mass accelearator either. With length of 150 000 km and accelaration of +1g (so total acceleration, including gravity, at sealevel is 2g) you can get up to 27,4 km/s and with +2g (3g at sealevel) - 38,7 km/s (and I do not even consider Coriolis force)


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