A circular track means you'd have to accelerate the vehicle inward proportional to the square of the velocity. If target exit velocity is 11 km/s, and the acce

A circular track means you'd have to accelerate the vehicle inward proportional to the square of the velocity. If target exit velocity is 11 km/s, and the acceleration of the straight portion of the track is insignificant compared to the circular portion, in order for the combined linear and centripetal accelerations to stay under 3G, I'd make each component 20.8 m/s^2. So if 20.8 m/s^2 = (11000 m/s)^2/r, then r has to be 5800 km. Closing the ring at that radius makes the circumference larger than the distance needed for a linear track. Untrained humans could not ride it up to Earth-escape velocity. LEO only needs about 8 km/s. So that requires a radius of 20.8 m/s^2 = (8000 m/s)^2/r, 3000 km. Still too big. What if we bump the G threshold to 5G? Then centripetal portion of the acceleration should top out at 34.7 m/s^2, and required radius drops to 1850 km. Okay, cargo only then, just to LEO. 25G max acceleration, 8000 m/s exit velocity. Radius 370 km, circumference 2300 km. Still more track than needed by a linear launcher. It's always cheapest to put all the acceleration towards moving the vehicle in the direction you want it to go, instead of wasting some energy pushing it in different directions that ultimately cancel out. You could use a spiral track, but the final leg of the spiral is going to have an effective radius so large it might as well just be straight the whole way.