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A spaceship which uses centripetal rotation to create artificial gravity.

  1. Must it always be accelerating? (The rate of rotation for it to work?)

  2. If no, and even if yes, if one were to place themselves right at the center of this spinning structure, would one feel no gravity?

  3. If one were to be moving from a part of the ship with no centripetal rotation and thus 0 gravity to the part with it what would the experience be like transitioning? Would one technically be able to just fly over the entire rotating thing? If you never touched it in the first place?

Image: enter image description here (Sorry for the poor drawing :(, I should mention that the opening is connected just to the edge of the spinning living section so that this window is almost perfectly centered around this spinning living space) If you think of this living space as a circle he is in the center (the door).

Description: Person coming from the 0-gravity part of the ship about to enter the centripetal rotation large living space. But doing so from a high point... Would they be Superman?

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  • $\begingroup$ If you're a visual learner then this video may help - the example give is why water in a bucket doesn't fall out when spinning vertically but if you replace the water with a person and remove gravity, the direction of intertia is all you're considering (and the resistance of the floor/bucket). $\endgroup$ – Lio Elbammalf May 12 at 9:57
  • $\begingroup$ That image is the best! Had to write this comment. Will delete it soon. Probably. $\endgroup$ – Marko36 May 12 at 12:28
  • $\begingroup$ What do you mean by "centripetal rotation"? "Centripetal" means toward the center, so it's not clear what centripetal rotation would be. Maybe you mean rotation producing centrifugal (away-from-the-center) force as a substitute for gravity? $\endgroup$ – LarsH May 13 at 7:55
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    $\begingroup$ Note that artificial gravity doesn't exist. SIMULATED gravity exists, it's not gravity but some other force that imparts a feeling of weight on an object. $\endgroup$ – jwenting May 13 at 8:25
  • $\begingroup$ I wonder if OP has just watched "2001:A space odyssey" which is possibly the canonical fictional example of simulated gravity through rotation of a space station. $\endgroup$ – pbhj May 13 at 10:53
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If you were in a zero-gravity environment (e.g., in earth orbit or in a coasting trajectory en route to Mars), alone in your space suit, you would feel no gravity.

If a big pipe were placed around you in your zero-g environment, you would still feel no gravity. The pipe would have no effect on you at all.

If the pipe were spun with you inside it but not in contact with it, you would still feel no gravity, because you and the pipe would not be interacting. It makes no difference whether the pipe's spin rate is steady or accelerating.

However, if you moved to the spinning pipe's inside surface and grabbed hold of it, you would feel yourself first yanked by whatever you were hanging onto (because after all it is moving relative to you), and then you would feel yourself being pressed against the inside surface of the pipe. This is for the same reason that a rock swinging around in a circle on the end of a string makes the string taut: centripetal force exerted on the rock by the string tension causes the rock to move in a circle instead of fly off at a tangent.

From the above you can see the answer to the third part of your question: If you are floating in zero-g and enter a spinning chamber, you are not attracted to the walls of the chamber. To someone pinned to the wall by centripetal force, watching you enter the chamber, you will a) seem to be rotating, and b) float like Superman.

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    $\begingroup$ Appreciate that you tied it back to Superman lol!. One question: Wouldn't the air molecules be spinning as well given that they would be pushed by friction at the bottom and thus over time cause them all to spin so you would be forced by their friction to spin, and thus not be Superman? $\endgroup$ – Outsider May 12 at 2:52
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    $\begingroup$ You are right: if the chamber were filled with air, ithe air would end up spinning with the chamber; and as he entered, Superman would feel the air rotating if he spread his arms wide. If he were exactly on the axis of rotation, he would just stay on the axis and gradually begin spinning at the same rate as the chamber. But if he were even a little bit off-axis, he would drift to the wall. A balloon filled with gas whose mass density is less than air, though, would drift to the axis. $\endgroup$ – S. McGrew May 12 at 3:59
  • $\begingroup$ Ok, Thanks for clearing this all up for me and staying with the theme! $\endgroup$ – Outsider May 12 at 4:28
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    $\begingroup$ "To someone pinned to the wall by centripetal force, watching you enter the chamber, you will a) seem to be rotating, and b) float like Superman." And if you are rotating, at just the right speed, you'd appear not to be rotating. :-) (Like our moon.) $\endgroup$ – T.J. Crowder May 12 at 13:07
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    $\begingroup$ I'd add to this answer that if you're in the air in your room in Earth, you won't feel any gravity either. For a short moment, at least. The difference to floating inside an artificial gravity chamber is that on Earth, you'll eventually hit the ground, but while you're on your way there, there's no difference on how you feel. $\endgroup$ – JiK May 12 at 18:38
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It works exactly like a merry-go-round. In a merry-go-round you feel an outward force. If you lie down with your head toward the center, you feel an downward force toward your feet. That is exactly what artificial gravity would be like.

1) A spinning object is accelerated. Each point on the merry-go-round or spinning spaceship is accelerated toward the center, making the point travel in a circle. With no acceleration, each point would travel in a straight line at a constant speed.

2) At the center, you would feel no acceleration, no centrifugal force.

3) The farther from the center, the stronger the centrifugal force. It grows smoothly as you move outward.

You could fly over the ship or merry-go-round without feeling the centrifugal force if you are not spinning or flying in circles. You can stand next to a merry-go-round or run by it without feeling a centrifugal force. If you rode a bicycle around the merry-go-round, you would have to lean in because you would be continually turning. You would feel a centrifugal force.

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  • $\begingroup$ "It works exactly like a merry-go-round" yes but a merry go round comes with the downward force of gravity pushing you towards the ground and thus your feet are pulled because of touching the ground. Does the air molecules in the spaceship act like the equivalent friction? $\endgroup$ – Outsider May 12 at 2:48
  • $\begingroup$ @Outsider To ensure I understand your question, let me pose a scenario. Say I have a desk located on the wheel. I'm sitting at the desk, and an apple is resting on the desk. I pick up the apple, hold it above the floor, and let go. I now ask: why does the apple hit the floor? Would the answer to this question answer your question? $\endgroup$ – Ian May 12 at 14:15
  • $\begingroup$ I know the answer and explanation for that. but I got the answer anyway, thanks for the response though $\endgroup$ – Outsider May 13 at 22:51
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Lots of good answers for the title but not the enumerated questions.

  1. Must it always be accelerating? (The rate of rotation for it to work?)

The angular rotation can be constant, but even at a constant speed it is still accelerating.

  1. If no, and even if yes, if one were to place themselves right at the center of this spinning structure, would one feel no gravity?

They would feel no gravity at the center.

  1. If one were to be moving from a part of the ship with no centripetal rotation and thus 0 gravity to the part with it what would the experience be like transitioning? Would one technically be able to just fly over the entire rotating thing? If you never touched it in the first place?

It would be similar to being lowered down into one of those fast spinning carnival rides (example pictured). Grabbing onto it would be a similarly unpleasant experience.

enter image description here

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Ever tried to rotate a bucket by tying it with string and filling water in it?

If you do it you will find that if you swung the bucket with enough speed the water won't spill on you, but as you stop rotation the water spills out (depending on position of bucket).

Somewhat similar process is going on here too.

You will have artificial gravity till the wheel is rotating, as soon as it stops the artificially induced gravity also drops to zero.

The centripetal force is usually denoted by $F=m\omega^2r$.

You can see as the radius $r$ increases, the force on you will also increase. Hence when you are at the centre of wheel you face no force.

But as you start moving outwards the force increases in terms of $r$ (radius) assuming the angular velocity $\omega$ of the wheel is constant.

So in case you had a ladder to move on the wheel, you would feel that the artificial gravity you are facing (which is nothing but centrifugal force) is also increasing as you are descending on ladder.

Hence in this case your astronaut ( I suppose this is outer space) will not feel anything when he goes from zero gravity to spinning region.

Condition? He enters from the centre.

In case he enters from some other point on wheel he won't feel anything but as soon as he touches the rotating wheel (which is going to hit him hard) he will feel the artificial gravity.

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Simulated gravity occurs because an acceleration is being applied. If you are in contact with the rotating portion of a rotating-drum style artificial gravity spacecraft, you are being carried along a circular path, which means you are constantly accelerating towards its axis of rotation, so you "feel gravity". When you are not in contact with the rotating drum, without any other interactions, you would be in non-accelerating uniform motion - you would be weightless. What happens next depends upon the relationship between your velocity vector and that of the ship you are in. If the velocity vectors are identical, you would appear to be floating/flying, depending on the observer's frame of reference; to someone moving with the drum, you would appear to be either flying in a circle or hovering at its axis. If the velocity vectors are not identical, you'll at some point run into something in the ship; your path will be linear, but it won't appear so to an observer rotating with the drum because his true path is curved due to the drum's rotation.

If you were on the inside surface of such a rotating drum and started running really fast in the direction opposite the rotation, you would reduce your effective rotation rate and consequently the radial acceleration and the simulated gravity acting on you. You would feel lighter. If you ran as fast as the drum's tangential velocity, you would effectively bring your rotation rate to zero and become weightless. You could take off ad "fly" like Superman. This assumes you can run as fast as the tangential velocity of the drum. See here for a calculator. For an artificial gravity spacecraft to be practical and comfortable, you'd want close to 1g and a low rotation rate (less than 2 rpm). That would require about a 225m radius yielding a tangential velocity of about 47m/s - no human could run that fast.

If you were standing on the drum's interior surface and jumped "up" (toward the axis), you would be momentarily weightless, just as if you jumped up from Earth's surface. In the spacecraft, the instant you break contact with the drum, you will have a vector of uniform linear motion which will remain constant (air inside might affect your motion a bit, but let's ignore that for now) until you make contact with something. You have a small radial component of motion (from jumping) and the tangential velocity you had at the point you broke contact with the drum. That net vector will intersect with the drum surface at some point. From the perspective of someone standing next to you at the point you jumped, it will look almost the same as if you jumped on the Earth's surface, except that you won't land at exactly the same spot on the drum.

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To complement the other answers, I would like to answer number 2 and 3 in a scenario where there is atmosphere (or really, any gas) inside the sealed cylinder.

There are 2 important parts in this artificial gravity: the outer wall rotation, and the inertia of the "falling" bodies/masses. If there is no tangential inertia of the inside bodies in the direction of the rotation, there is no feeling of gravity. Thus, if you consider there is air in your space ship:

  1. and you are standing floating on the center, depending on the radius, you will feel the air rotating around you, and eventually the turbulence will throw you out of equilibrium towards a random direction closer to the wall. Once you are out of the center, the "wind" will drag you, add inertia and slam you onto the outer wall. If the radius is big enough, the air in the center will be rarefied, pretty much like at high altitudes on earth, and you would be able to hold your stable floating position for longer.

  2. and you enter the rotating section, the wind will quickly transfer the rotation motion to you, matching your velocity to the tangential ground velocity, throwing you to the ground. The procedure is likely to be violent, turbulent and messy, so I would suggest you to quickly attach yourself to the wall mounted ladder as soon as you enter the rotating section in order to avoid injuries.

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