@Ingolifs' answer got me procrastinating thinking further.

What you're asking about is a mass driver space station in low earth orbit that a payload or ship launched from earth can dock with, and then be propelled into deep space.

The delta-V needed to go from LEO to an escape trajectory is about 3 km/s, and to go from there to a Hohmann transfer to the outer reaches of the solar system requires a further 5 or so km/s. I'm sure a mass driver could be made to provide such delta-Vs, though it would have to be rather long in order to not smush the payload with its massive acceleration.

One big problem I see is Newton's third law. Due to recoil, the orbital mass driver will have the same momentum backwards as the craft will have going forwards. This means its orbit will be altered...

The railgun's orbit will be of course altered, but if the rail gun is long enough to minimize acceleration for payloads like living humans, it will be subjected to some complex torques as the projectile accelerates along it and pushes it backwards. These will set this long structure rotating, and may even bend it under these transverse loads and tidal effects, unless it is made strong and heavy (and expensive) enough to withstand this.

Tumbling of something very long, heavy, and solid in LEO also means drag, reentry, and danger to folks on the ground, so you'd like to keep it re-oriented fairly nose-on to minimize drag as quickly as possible.

Is the best shape a straight line, or should it be somewhat curved in order to track what the spacecraft's trajectory would look like as it accelerated from circle to elliptical to hyperbolic tangents?

I would suppose that the rail-gun-as-reaction-mass would have to be much heavier than the projectile, but it isn't realistic to treat it as infinitely heavy.

With some suitable assumptions of length and mass and target (Moon, or Mars) what would be the ideal shape for this object in LEO?

note: I've added the and tags to indicate that I'm looking for a quantitative, reasoned answer, not just something like "it happens so fast that it wouldn't matter" type of hand-waving. Thanks!

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    $\begingroup$ My physics may be a bit rusty so i'm not sure if this is correct, if you assume you're trying to reach 8 km/s and the railgun accelerates at a reasonable 10g, it will take 81 seconds of acceleration to reach that speed. The distance travelled in that time is just over 32 km. That's how long the railgun would need to be. If you allow a much higher acceleration (maybe your probe is made of solid tungsten or something), you can have a much shorter length, if you're transporting humans, it would have to be much longer. $\endgroup$
    – Ingolifs
    Nov 8, 2018 at 7:41
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    $\begingroup$ @Ingolifs maybe 320 km? $x=\frac{1}{2} a t^2=\frac{1}{2} (10 \times 9.8) 81^2$ $\endgroup$
    – uhoh
    Nov 8, 2018 at 7:46
  • $\begingroup$ Oops, yes, missed a zero. $\endgroup$
    – Ingolifs
    Nov 8, 2018 at 7:59
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    $\begingroup$ A 32-km long gun only 3 times as heavy as the payload you're accelerating? That seems unlikely. And if the gun really is that light, 3 km/s imparted to the payload means the rail gun accelerates by 1 km/s in the opposite direction. $\endgroup$
    – Hobbes
    Nov 8, 2018 at 8:34

2 Answers 2


I've answered some aspects of this, but considering a 1km railgun in my answer to the "parent" question. A much longer railgun doesn't make a lot of difference to the calculations there, except for the acceleration and power. The issues with reaction de-orbiting the railgun are the same. Concerning the shape of a longer railgun. Let us consider a 10g 4 km/s delta-V launcher (basically a longer version of the one in my other answer) which would need to be about 80km long. It's shape will be some sort of blend between the original circular orbit and the hyperbolic orbit of the probe when it is launched. I can't work out the actual curve, but the 80km long launcher will deviate from straight by a few kilometers.

What is sadly true is that that shape will be quite unstable due to tidal forces. It is long and thin pointing broadly but not exactly along the orbit. It will experience significant tidal forces pulling it to a radial orientation. It's also big enough that lunar tides may be a consideration.


Having asked and thought about a similar question, I think the answer is simple. Don't build accelerators for humans. Build them for cargo.

The total mass of cargo that a human needs to be comfortable in space is hundreds or thousands of times her/his own mass. Efficient systems should deal with cargo.

Most cargo can be somewhat g-hardened without much hassle to survive 100g's. The Sprint missile carrying a nuclear warhead accelerated at 100g's with 1950s technology. In fact humans can sometimes survive brief 100g impacts. See https://en.wikipedia.org/wiki/G-force#Typical_examples

If an object can support 100x its own weight, then it can survive 100g. Most cargo items like clothes, shoes, fuel, water, structural materials pass this test easily, and in fact you can push them far further and you are limited by the structural integrity of a ship to put them in. More complex items like a laptop, smartphone, etc could easily be g-hardened to survive 100g. Some items that are delicate like solar panels can be disassembled and packed flat.

When your accelerator can use 100g accelerations, the structural and orbital problems simplify a lot. E.g. for an 8 km/s delta-v, we have:

$s = v^2 /(2a) = 8000^2 / (2 × 1000) = 32km$

Building a 32km long accelerator in orbit is hard, but at that size it is possible to make it rigid and tidal forces from the moon will be very small. Not easy, but possible. Of course it depends on how much mass you have, but a 30km tube can be rigidified with guy wires connecting to a central ring. Consider the KVLY-TV mast - 600 meters tall in earth's 1g, and exposed to substantial sideways wind forces that are much greater than sideways forces in space. This kind of design could be scaled up, and it would work provided the mass was increased proportionally and the g-forces it was exposed to were smaller (e.g. 1/30g axially). Structures typically get heavier like $x^3$ but only stronger like $x^2$, so if a 1000m mast works on earth, a 10km mast of the same design in space will be 1000x heavier and will be able to withstand $0.1g$. In fact it might be possible to cut the space version down a little in mass, because wind loading on earth is a lot more than 10x stronger than any sideways loads in space on such a structure.

If the track is only 30km long, you can make it straight and the deviation from the circular orbit will only be about 15m - negligible. For a 30km track firing a projectile with a 100g acceleration, the track will need to be at least 300 times more massive than the payload, possibly 1000x. As you make it lighter it becomes weaker, and it experiences a stronger g-force. The exact point where it becomes impossible to go further is unclear to me, but the breaking length of aluminium in tension is 20km in 1g, so a 30km long tower definitely cannot withstand 1g of compression regardless of whether or not it buckles.

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    $\begingroup$ very nice answer, thank you! Yes this approach seems to reduce the effect substantially. $\endgroup$
    – uhoh
    Aug 2, 2019 at 21:52

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