We often see films with spinning space station that create artificial gravity by having the astronauts pulled outwards by centrifugal force.

I'd like to know if this would really happen, and if so, why is the following scenario not true:

  1. Take an astronaut in open space. He doesn't move.

  2. Put a big open spinning cylinder around him - surely he still doesn't move.

  3. Close the cylinder. I still see no reason for him to be pulled outwards.

  • 3
    $\begingroup$ Just to comment on your terminology: in the rotating frame of reference, it's the centrifugal force that pulls the astronaut outwards, and the centripetal force is the normal force that prevents the astronaut from falling through the floor. i.e., centrifugal force pulls outwards, centripetal force pulls inwards, and they are balanced. In the non-rotating frame there is no centrifugal force, only the centripetal force pulling inwards, making the astronaut rotate with the station instead of continuing in a straight line. So if you want to talk about the outward-pulling force, it's "centrifugal". $\endgroup$
    – N. Virgo
    May 25, 2016 at 9:15
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    $\begingroup$ If the astronaut was not spinning with the station (so, they're moving quite fast in the rotating reference frame) and there's no air in the station, and no walls to hit them (a long corridor all the way around the station), then it would spin around them! Note that something similar happens with real gravity too - we call that orbit. $\endgroup$
    – user253751
    May 25, 2016 at 10:10
  • 4
    $\begingroup$ In those movies you will usually find a ladder that the astronauts use to climb "down" towards the floor. While climbing down, the astronaut speeds up "sideways" and therefore feels increasing force/gravity (often not depicted in those movies). This also means, that if the centrifuge is too small (then using higher rotational speed to simulate full gravity), your head would feel less gravity than the feet... $\endgroup$
    – linac
    May 25, 2016 at 13:10
  • 2
    $\begingroup$ Go watch Bablon 5 where Sheridan jumps from a tram and is basally floating, till he smacks into "the ground" that is. $\endgroup$
    – coteyr
    May 25, 2016 at 15:09
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    $\begingroup$ It's important to note, centrifugal force is not really an applied force, in the traditional sense. It's the apparent effect of angular momentum (perpendicular to the center of rotation) combined with centripetal force (towards the center of rotation), which seems to be a force pushing outward. en.wikipedia.org/wiki/Centrifugal_force ('fictitous force') $\endgroup$
    – rich remer
    May 25, 2016 at 17:59

6 Answers 6


Put a stationary astronaut in a small room inside a large spinning cylinder. After an instant walls of that room will hit him, and suddenly he will have the same velocity as the room. Due to angular motion, the room accelerates towards the axis of the cylinder. Subsequently, through the support force from the floor (the floor is at the surface of the cylinder) accelerates the astronaut too towards the center of the cylinder.

If the room accelerates $9.81~\rm{ms^{-2}}$ towards center, this will be feel like the regular gravity.

Note that one cannot feel gravity or acceleration as such (except for tidal forces). The 'weight' one feels is the support force from surfaces. In other words, gravity feels like so that you are constantly being pushed by the floor, which accelerates you at the rate of $9.81~\rm{ms^{-2}}$. If you stand, your organs will be pushed down etc.

  • 17
    $\begingroup$ Most people don't recognize that weight is not the downward force from gravity, it's the upward force of the ground/floor/table/etc. This is why people are weightless in orbit or the Vomit Comet, despite the fact that there's lots of gravity. This is also why I despise the term "zero gravity" when what's really meant is "weightlessness". $\endgroup$ May 25, 2016 at 15:57
  • 2
    $\begingroup$ It may not feel exactly like regular gravity due to the Coriolis force. $\endgroup$
    – ntoskrnl
    May 25, 2016 at 16:13
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    $\begingroup$ @TonyK, you'd better respect that, or you'll be sent to the Equivalence Principal's office. $\endgroup$
    – Joe
    May 25, 2016 at 18:31
  • 3
    $\begingroup$ @TonyK "upwards" shall be whichever direction the artificial gravity would like you to believe "upwards" is. By disputing the definition of "upwards", are you openly admitting your supposed equivalent to gravity isn't actually acting like gravity at all? $\endgroup$ May 26, 2016 at 3:12
  • 3
    $\begingroup$ How do you throw your balls? $\endgroup$
    – Epanoui
    May 26, 2016 at 18:41

You are correct in that if the astronaut is undergoing no translational or rotational motion relative to the centre of rotation of the space station the astronaut will feel weightless as in diagram $A$ and will not touch the space station.
This is equivalent to jumping onto a rotating turntable with no friction acting.
That feeling of weightlessness is due to the fact that the astronaut feels no contact forces acting as the astronaut is not touching the space station.
If the space station is rotating what the astronaut will see the space station going past so the astronaut will know that the space station is rotating but the astronaut will not feel any contact forces due to the space station.

enter image description here

If the astronaut comes in contact with the space station as in diagram $B$ then a frictional force between the feet of the astronaut and the space station or the astronaut holding on to the space station will cause the astronaut to rotate with the space station and so the space station will exert a centripetal force on the astronaut to accelerate the astronaut.
So the astronaut is now subjected to a contact force either at the astronaut's feet or arms which is the same as as if the astronaut was on the Earth if the space station's rotational speed is correctly set up.

For example a space station of radius $160$ m would have to rotate at approximately 2.4 revolutions per minute to simulate a gravitation field strength of $10$ ms$^{-2}$.

To go back to a permanent weightless condition (no contact forces) the astronaut would have to engineer it so that the astronaut is undergoing no translational or rotational motion relative to the centre of rotation of the space station.

  • 2
    $\begingroup$ OK. So what if the astronaut jumps? There's no force to pul them back down. I can see that they'd have a sideways motion, so presumable they'd move sideways and hit the wall again? And if they jump in the opposite direction to the rotation? $\endgroup$
    – tarmes
    May 25, 2016 at 12:17
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    $\begingroup$ @Farcher, assuming an astronaut is moving in a straight line, he will eventually hit one of the curved walls, or more accurately bounce off/skim them $\endgroup$
    – Troyseph
    May 25, 2016 at 12:45
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    $\begingroup$ @tarmes When standing, you are experiencing acceleration "downward". You can jump up, but you are still accelerating downward, so you drop back to the floor (though you will be "downspin" of the point you jumped from). If you can jump with enough force to accelerate upward at a greater rate than the downward acceleration, then you will not drop back to the floor and will likely impact a side wall as it sweeps through its rotation. $\endgroup$ May 25, 2016 at 13:25
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    $\begingroup$ @tarmes You're forgetting the tangential velocity - you're still moving "forward" (in the direction of the rotation), and this will quickly bring you back to floor, just like with gravity. You'd also need to eliminate this velocity - i.e. jump hard enough in retrograde direction to zero out the velocity. Then, assuming there's no interior walls, you will float above the "ground", with the station whooshing around you. Again, this is analogous to gravity - orbits are pretty much the same thing, and as Einstein explained, there's no way to distinguish gravity from an inertial force. $\endgroup$
    – Luaan
    May 25, 2016 at 14:18
  • 1
    $\begingroup$ Funny how you've been the only one to mention something as fundamental as friction. $\endgroup$
    – valerio
    May 26, 2016 at 13:48

If there is no atmosphere, and the station is a relatively smooth cylinder, you can indeed float there as the exterior walls spin around you (in the middle, or just above a wall, or anywhere).

Now, suppose you start drifting towards a wall (maybe you threw your shoe the other way). You move towards the wall, but do not accelerate due to the rotation of the station. However, the exterior wall of the station is moving very fast as you approach it. You hit the wall, and both bounce and gain some horizontal speed in the direction of the wall. You'll also start to spin (as it will mostly produce torque on you).

This new velocity vector results in you intersecting with the wall again somewhere else, with more velocity normal to the wall and less horizontal usually. Each time you do so, you'll gain more and more angular momentum -- you'll spin both around your own center of mass (lacking a way to somehow counter it), and in a sense around the center of the habitat.

Assuming you avoid getting mashed to a pulp in your repeated hitting of a high velocity wall, you'll start travelling along with the wall more and more. As you do so, the wall will start seeming "more down" and less "moving fast", as your impacts with the wall will be closer to normal with the wall, and less glancing blows with something moving very fast.

If there was air in the station, the air would be moving along with the wall (for similar reasons as you'll end up doing so), so it will drag you along, like a strong wind. This drag will result in you falling faster towards the wall as you match its rotational velocity, compared to the above scenario -- basically, it moves the repeated collisions up in time and space (up, as in earlier, and up, as in further from the wall).

If there are large features, like buildings, those buildings will smack you on your side and speed you up to the rotational speed of the exterior wall. You can consider this a more aggressive form of wind.

Once up to speed, you start experiencing the pseudo-gravity of the rotating station when you are supported by the floor. When you are not, you'll experience pseudo-free fall, where the longer you are in the air the faster the ground will move relative to you (up to a point).

But what does it feel like, other than the being beaten to pulp "getting up to speed"?

While falling (with no air in the way) it feels like free fall. Same when jumping in the air. When "stationary" against the edge of the station, it will feel mostly like standing on the Earth.

You cannot, in general, directly feel gravity. Free fall, which you can experience to a lesser extent on a roller coaster, is the feeling you get when you are not being supported by some surface.

"Normally', you are supported by a substance, which pushes against you (the ground under your feet, the water in a pool, or the air as you reach terminal velocity). The parts supported by this substance then push against your organs and the rest of your body.

In free fall, no such support exists: you are still experiencing gravity, but not the "support".

The exception is tides, which if strong enough can produce forces directly on your body. No human has experienced tides (due to gravity) that strong.

A rotating station creates a pseudo-tide effect, because things closer to the axis have "less pseudo-gravity". If you are tall relative to the station radius, your head will feel less gravity than your feet. This could be felt directly while in pseudo-free-fall, but would more likely be felt as a tendency to rotate in free fall, or when standing on the ground.

  • 3
    $\begingroup$ Buildings move around at 40 m/s relative to the observer are a very aggressive form of wind indeed. $\endgroup$
    – gerrit
    May 26, 2016 at 11:21
  • $\begingroup$ When I lied down along the radius of a roundabout while my friends pushed it around as fast as they could, I did feel ultimately unpleasant forces on my body. Are those substantially different from (pseudo-)tides? $\endgroup$
    – gerrit
    May 26, 2016 at 11:25
  • $\begingroup$ @gerrit Probably. I'm uncertain how well it approximates a "real" tidal effect: I would have to do math. But the feeling where your head is being pulled up, while your feet are being pulled down, and your middle is weightless, is what tides should feel like. However, at that scale, you'd also experience enough rotation that your inner ear would be swirling, and your eyes (if open) would also be informing your brain. I don't know how big of an difference that would make in your experience. Plus, I don't know if the rate of change in pseudo-tides matches any real tide... $\endgroup$
    – Yakk
    May 26, 2016 at 13:37

All that you have written is correct.

Let's go further to the next point 4:
Once he comes to the cylinder's wall and stands on it, he gets the same angular speed as the cylinder, and he will also get a centrifugal force and rotational gravity as in the films.

  • $\begingroup$ The cylinder wall will have a relative velocity of 40 m/s relative to the astronaut at rest. He'd better very slowly climb some stairs down from the cylinder axis. $\endgroup$
    – gerrit
    May 26, 2016 at 11:26
  • $\begingroup$ @gerrit where did you get that figure from? $v=\sqrt{ar}$. For $r=10~$m and $a=10~$m/s$^2$, $v=10~$m/s... And $r=10~$m looks already fairly big to me! $\endgroup$
    – DarioP
    May 26, 2016 at 16:07
  • $\begingroup$ @gerrit, I agree. Method you suggest is anywhere safer. $\endgroup$
    – dmafa
    May 26, 2016 at 16:32
  • $\begingroup$ Actually, I researched this for a sci-fi book. Standing on the surface is easy to understand. "Freefalling" through an "atmosphere" in an O'neill cylinder is not so obvious. You fall like in a gravity field due to air pressure from the rotating air. The winds actually come to an equilibrium where there is just a good breeze at the "surface". It circulates from the center of the cylinder outward. Arthur C. Clarke had it right (as usual) in Rama. $\endgroup$ May 26, 2016 at 23:42
  • $\begingroup$ @DarioP Your own answer uses $r=160$ m at 2.4 revolutions per minute. That means the tangential velocity is $2 \cdot \pi \cdot 160 \textrm{m} / ((1/2.4) \textrm{min}) = 40.2 \textrm{m/s}$ $\endgroup$
    – gerrit
    May 27, 2016 at 8:58

This is a really nuanced issue, but it is not the spinning space station that "causes" the centrifugal force, but the spinning frame of reference. We begin to say things like "he feels a centrifugal force on him" at a point where the *best reference frame to describe his motion is a rotating frame.

You can model a system like your astronaut and a cylinder in any reference frame you please. That's one of the fundamental rules of how we model physics. You can even model your astronaut in a coordinate system consisting of North, East, and Down from the point of view of an observer on the ground watching the two fly by, and physics will still predict their motion accurately.

If you model it in an "inertial" frame, the motion will be described by the traditional Newtonian equations of motion (barring anything exotic like relativity, which will have a minimal effect at the speeds you are talking about). Objects will move in straight lines unless affected by a force from another object, like gravity or from contact with the surface of the cylinder. This is even true on rotating space stations! You can describe the motion of astronauts in a rotating space station without centrifugal forces if you describe their position in an inertial frame (such as ECI).

However, there's a catch here. The equations of motion may not be simple in such an inertial frame. Yes, you get rid of the centrifugal forces; you'll end up with some normal forces if they are standing on the inner surface of the cylinder. However, rotational motion like that means you're going to have to bring in all sorts of sine and cosine terms to describe the motion. The effect of the astronaut on the space station will be small but it may be remarkably hard to prove it is small, leading you to have to consider things like the space station wobbling about its axis.

In such a rotating situation, it can be more convenient to use a rotating frame of reference -- in this case, one attached to the space station or cylinder. When we do this, we get to avoid all of those sine and cosine terms because they end up getting bundled up into the motion of our frame. The math becomes tractable.

There's a price to pay for such rotating frames: centripetal forces and Coriolis effects. The laws of motion defined by Newton are for "inertial" frames, which are non-rotating. If you attempt to apply them in rotating frames, you get the wrong result. For a real life example, go to a playground with a ball and find a roundabout.
Get two people, put them on opposite sides of the roundabout while it isn't spinning, and have them toss the ball back and forth. This will work as expected. Now start spinning the roundabout and toss the ball back and forth. The people on the roundabout will observe the ball curving away from them at a right angle to its motion. Those on the ground will simply see the ball moving straight, and the people rotating out of its way. I can't link a gif here, but Wikipedia has a nice little gif showing this effect.

These odd effects come because the reference frame is rotating. If you dig into them with calculus, you discover that the equations of motion have some extra terms in them due to the continuously changing coordinate system. The first of these is the centripetal term, which has the appearance of a center-fleeing force. The second is the Coriolis term, which affects moving bodies at right angles to their motion, and is responsible for a great deal of the weather patterns we see on Earth. (note: I switched from centrifugal force to centripetal. The actual effect, when written in math form, is better described by a centripetal acceleration term, which goes towards the center. The centrifugal force is the effect is actually the "equal and opposite reaction" to the centripetal accelerations.)

These forces are often called "ficticious" forces, because they're not being caused by anything in the system. They're being "caused" by the mathematics of the rotating frame you chose. Simply put, in a rotating frame, the "correct" equations of motion include terms that don't appear in an inertial frame.

So in your example, we can view the astronaut and cylinder two ways. We can treat it with an inertial frame of reference or a rotating frame of reference. In the inertial frame, the astronaut and the cylinder are simply flying along in straight lines, decoupled. The cylinder has a few sines and cosines in its equations because it is rotating, but the astronaut does not. If we choose the rotating frame instead, we find the cylinder's motion is simple (it isn't moving with respect to our frame), but the astronaut will appear to be rotating at great speed around the central axis of the cylinder. Why? Because the astronaut isn't rotating with the cylinder, so as the frame of reference rotates around him/her, it will appear as though the astronaut is rotating. The exact equations of motion describing this astronaut in this rotating system will include a centripetal acceleration term which will ensure the astronaut is continuously accelerating towards the central axis just right to keep him/her on a circular orbit. In this case, we will find that both approaches describe the motion of the astronaut correctly, but the inertial case is much simpler looking.

However, let's change the situation a bit. Let's let the astronaut interact with the cylinder more. Let's either let them hold onto the surface, or fill the cylinder full of air (which will start rotating, as friction with the outer surface takes hold). Now the equations of motion are not so simple in the inertial frame. You have all of these extra forces to account for, each with ugly sine and cosine terms because they're all rotating. The equations get extremely messy in the inertial frame, but you do get the right answer.

Change to a rotating frame, and those complications go away. The air is (roughly) motionless, and the cylinder is motionless. The forces on the astronaut are much easier to write down now. Most of them even go away! However, we pay a price. The price of using a rotating frame is that the equations of motion for a rotating frame must include centripetal and Coriolis terms. You end up with the exact same result as you would have with the inertial frame, just with less mathematical pain and agony.

In many cases, handling these "fictitious forces" is easier than handling the forces that one has to include for an inertial frame of reference. Choose the right reference frame, and you cancel out as many of the ugly details as possible, leaving only an easy to solve equation!


In your scenario, your 3 statements are correct, and if nothing changes, your astronaut will not move from its spot as the wall of the cylinder moves past him. However, if somehow the astronaut "attaches momentarily" to the cylinder wall (the floor), then he will acquire the tangential velocity of the spot he attaches to, and this tangential velocity is what keeps him "attached" to the cylinder wall (the floor).


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