@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 '18 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 '18 at 7:46
  • $\begingroup$ Oops, yes, missed a zero. $\endgroup$ – Ingolifs Nov 8 '18 at 7:59
  • $\begingroup$ 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 '18 at 8:34

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.


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