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I've recently been contemplating things like artificial gravity in a rotating space ship (for example, the O'Neill Cylinder) and learning about the Coralis effect and other interesting fictitious forces that appear in rotating reference frames.

It occurred to me that, if you were living on a space station spun for artificial gravity, it might be possible to become weightless in such an environment by travelling fast enough against the spin of the ship. My friend, who I was discussing this with, thinks otherwise, that you would actually be under more g's as you tried to gain the speed to lift off from the inner edge of the ship.

In terms of concrete numbers, say the ship is 4 kilometers in radius (as an O'Neill Cylinder would). At this size, the rate of rotation to generate 1g in centrifugal force (the artificial gravity) is under 0.5 rpm. If a vehicle travels on the inside of the cylinder against the spin fast enough, eventually the vehicle would cease moving when looked at from a static reference frame. I would think at this point, the vehicle could simply push off from the edge of the cylinder and float towards the center of the station. Is this correct?

Furthermore, would this method of "getting into air", as it were, be easier than it would be to counteract real gravity on Earth (with things like planes that generate lift with their wings)?

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  • $\begingroup$ weightlessness is felt when you and your spaceship are both falling at the same rate towards a massive object preferably not a black hole, centrifugal force is pushing matters including you outwards and if you go against the flow you won't negate its effect. $\endgroup$
    – user6760
    Apr 12, 2015 at 15:24
  • $\begingroup$ I don't get your last question, can you explain in more detail what you mean? (Tough the answer is probably no ) $\endgroup$
    – mikuszefski
    Apr 12, 2015 at 15:26
  • $\begingroup$ @mikuszefski Using an airplane to get into the air requires a certain amount of force applied as lift. This requires wings of a certain size, and moving at a certain amount of speed. Take a Cessna, for example. If you put a Cessna on this theoretical space station (and the station was filled with enough air to mimic Earth atmospheric at a certain altitude conditions), would you have to go just as fast to get off the "ground", or could you, theoretically, only go as fast as the rotation of the station (which, may be less than the speed required to fly on earth). $\endgroup$
    – Tustin2121
    Apr 12, 2015 at 16:25
  • $\begingroup$ I guess this is indirectly answered already, as the vehicle in my answer does not have wings, hence its "take of velocity" on earth will be "higher". $\endgroup$
    – mikuszefski
    Apr 13, 2015 at 10:01

2 Answers 2

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Assume the vehicle is already there before the space station is built. So it is floating. If it is floating slightly above space station ground, this does not change if the station starts to spin (neglecting its acceleration due to air friction). In the reference frame of the station it will move with the velocity of the outer cylinder. If you get in the same position later be counter-accelerating the spin, the result is the same (does not depend on history).

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  • $\begingroup$ "neglecting its acceleration due to air friction" - also ignoring frame dragging, the effect of which would be much, much smaller than air friction: einstein.stanford.edu/content/education/lithos/litho-fd.pdf $\endgroup$
    – Ben Hocking
    Apr 12, 2015 at 22:25
  • $\begingroup$ @BenHocking Yes, and also the pure gravitational forces between cylinder and vehicle. $\endgroup$
    – mikuszefski
    Apr 13, 2015 at 11:53
  • $\begingroup$ @Tustin2121-Other people in the space station (stationary to it) will see you move (after you altered your velocity to the opposite angular velocity of the ship) in a circle above the floor. $\endgroup$
    – Deschele Schilder
    Apr 11, 2017 at 12:43
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The technique would work, but whether you would use it would depend on many many details.

Rotating frames of motion come with all sorts of counter intuitive bits, so lets look at it from an inertial frame outside the station. We perceive the station as rotating. The velocity vector of any object on the station is a tangent, so we see the result of every object pushing against the rotating body of the station. It looks roughly like a weird form of gravity.

Now a vehicle beings traveling in the opposite direction of the spin, faster and faster. From our inertial point of view, outside the station, it actually looks like it is slowing down! From what we know of physics, the centripetal acceleration is related to the velocity of the object, so we can assume they feel less and less of the effect of this artificial "gravity." Eventually, they can just push off and float up.

Interestingly enough, this happens on earth, thanks to the rotation of our planet. However, the effect is very small compared to the aerodynamic forces present, so we usually do not pay attention to it when deciding how planes will fly.

Let's go back to the O'Neil cylinder. In reality, people try to make transport as efficient as possible, so they will use these frame effects in whatever way is most ideal. Obviously, the faster you go against the rotation, the more wind drag you'll feel. However, there will be an "ideal" speed where the decreased apparent weight balances with the extra drag. The exact speed that is ideal will depend not only on your cylinder, but also on the characteristics of the aerodynamics of the vehicles you choose.

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  • $\begingroup$ hmmm. just wondering. as earth is a real source of gravity the centrifugal forces due to earth's rotation actually decrease $g$ and, hence, moving against the rotation of earth should increase "weight". $\endgroup$
    – mikuszefski
    Apr 12, 2017 at 11:57
  • $\begingroup$ @mikuszefski That is correct. The effect is very minor, but if you were driving against the Earth's rotation, the normal forces from the ground that hold you up would have to be greater. This effect is more obvious if you are in orbit. If you are in orbit and slow down, you will fall. $\endgroup$
    – Cort Ammon
    Apr 12, 2017 at 15:16
  • $\begingroup$ I mentioned it due to your line "...speed where the decreased apparent weight balances with the extra drag..." and I am not sure where your decreased apparent weight comes from. $\endgroup$
    – mikuszefski
    Apr 13, 2017 at 5:34
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    $\begingroup$ As your angular velocity (w.r.t an inertial frame) decreases, the centripetal acceleration that appears in that frame decreases. The difference between this effect in the O'Neil cylinder and the effect on earth is simply that with the O'Neil cylinder, you're on the inside, so the acceleration was pushing you towards the "ground," while on Earth the centripetal acceleration is away from the ground because you're on top of it. Same effect, just different topology. $\endgroup$
    – Cort Ammon
    Apr 13, 2017 at 5:39
  • $\begingroup$ After reading the last two paragraphs $n$ times, the problem, I guess, is that it is not clear that in the second last you jump in an earth-example and jump back to space in the last paragraph. Is this correct? I misinterpreted the last paragraph to deal with a situation on earth, in which case this would be wrong. Maybe you can clear that up a little. $\endgroup$
    – mikuszefski
    Apr 13, 2017 at 6:33

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