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Suppose we built a enormous torus shaped space station that encompasses and leaves about 330 kilometers between it and the earth. So it's radius is about 6,371 kilometers . Obviously, this structure will have to spin very quickly to not collapse. So let's say it's going 28,000 kilometers per hour so people in the spaceship will feel weightless.

If we now were to build a rail inside this spaceship with a rocket car on it, that holds on to the rail much like a roller coaster holds on to its rail and have the rocket car go in the opposite reaction the space station is spinning. Would it start to exert downwards force on the space station since it moving more slowly relative to the earth? And reach almost 1g when the car is going 28,000 kilometers per hour?

The boarder question I guess is, how is the absolute rotation of something determined. What's the reference frame you should use? Is there an absolute reference frame for rotation or should you maybe use the local curvature of spacetime as a reference?

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  • $\begingroup$ Well, the total angular momentum is going to be central here. I'd say a vital question is "what are you doing with the exhaust of the rocket?"... $\endgroup$
    – Kyle Oman
    Commented Jul 24, 2014 at 17:20
  • $\begingroup$ Maybe the rocket car is on the outside? I also feel like the rocket car's mass would be negligibly small compared to the space station. Why do you think the exhaust would be such a big problem? $\endgroup$
    – overactor
    Commented Jul 24, 2014 at 17:24
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    $\begingroup$ @overactor Rocket-cars on rails, spaceships, roller-coasters...Toss in a few more things and you'll have a movie script on your hands :) $\endgroup$ Commented Jul 24, 2014 at 17:25
  • $\begingroup$ Well, think about how a rocket works. It chucks some mass out the back to propel itself forward. Momentum is conserved, so the total momentum carried away by all the exhaust is exactly equal in magnitude but opposite in direction to the momentum of the rocket. If the exhaust stays in the station, this momentum will be dissipated into the station (exhaust particles colliding with the walls), and will exactly compensate any momentum imparted to the station by the rocket -> nothing will happen. $\endgroup$
    – Kyle Oman
    Commented Jul 24, 2014 at 17:28
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    $\begingroup$ @Andres Yeah, I realized what you meant right after I posted my reply. $\endgroup$
    – overactor
    Commented Jul 24, 2014 at 19:08

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Depends on the type of engine. If its a space engine, watch out for the exhaust. The rocket speed can be computed solely by its mass lost, but if the torus absorbs that then uhoh.

Also, the rocket engine speeds up the torus whilst accelerating, so finding when the two are equal is kind of tough. It is a fair assumption that the angular veloctiy added by the car to the torus is negligible though.

The answer is yes, just imagine a frictionless car attached to the rail as you said, where the rocket is stationary compared to the earth or perhaps rather co-rotating with it. The rocket can't help being pulled down.

I am a little worried with what you mean by 1g, since at the equator objects are both farther away from the center of the earth (it is an ellipsoid) and spinning more quickly. Orient your torus parallel to the equator and perhaps mean corotating as in rotating at the same speed (but not frequency) as the equator. Realize that equal frequencies will result in greater tangential speed for larger circles. Wish I could bust out some equations but I'm pretty bad at more than conceptual methods of viewing things.

Finally, about absolute rotation, realize that that is a thing. There is a difference between a spinning bucket and a non-spinning one, as Newton showed (just look at the water bulging). Just because I'm spinning with the bucket does not mean it will be the same reality, in fact it will be different. Special relativity states that physics are the same in inertial reference frames.

The answer is yes though, if you don't worry so much about the inverse square law of gravity, which I don't think you were. Good question. Thinking about these frames of reference will help your understanding of special relativity. Remember-accelerating frames are not equal to inertial ones.

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