[Physics] Why can’t we use railguns to propel rockets once in space

rocket-sciencespace

Once in space, I guess rail guns can already achieve 1.5 km/sec easily. And I guess it is possible to push it up to 3 km/sec with tech improvements. If you are able to reach 5km/sec, you effectively have a LOX/RP1 engine right there. And you wouldn't have the heat and all those issues associated with ion thrusters. It seems this can work for orbit transfers between planets better than currently proposed tech. Since no one is trying it, I guess there is something wrong with this, so what is it?

Best Answer

A major problem with your idea is the following: If you're looking to eliminate the need for a supply of propellant with the railgun, it won't happen. The impulse imparted to the rocket is, by Newton III, imparted o on the railgun and so each launch will de-orbit the latter. It will need a constant supply of fuel to keep its orbit stable. A given quantity of fuel imparts a given impulse, so whether impulse is imparted directly to the rocket through chemical engines or whether to the launcher, you won't save on the need for propellant.

The other engineering issues aside (like rail wear cited by Dmckee):

I know railguns are cool and all, but the rail wear problems are still insufficiently resolved for them to move out of the laboratory. The current needed to get high performance just chews through the material of the rails.

another major reason that if you planned to boost humans in this way, or delicate instruments, the rails would need to be very long. Not impossibly so, but clearly a project hugely beyond the ISS, for example. Suppose we are in LEO and wish to boost to Earth escape speed: we need a delta-V of about $3.5{\rm km\,s^{-1}}$ ($11.2{\rm km\,s^{-1}}$ from LEO orbital speed of $7.8{\rm km\,s^{-1}}$). If you did this at constant acceleration $a$, then (calculating in a frame comoving with the LEO orbiting railgun):

$$s=\frac{(\Delta V)^2}{2\,a}$$

and limiting $a$ to, say, $5\,g$, I get a rail length of about $123{\rm km}$.

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