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There is the popular question “What is the gravity at the center of the Earth?”. And the answer is zero, because the forces cancel out. And then the gravity increases linearly as you move to the surface. Would the same be true for a rotating cylinder?

I mean, if, for example, I have a cylinder with a radius 270 of meters, rotating at 2 rpm, you can calculate, that the gravity will be about 1.21 g on the surface of the cylinder, right? And then, if I am inside and I go deeper (towards the center) in the cylinder while it is still spinning, would the gravitational force decrease to zero (for example, when I'm 80 meters in, so in about 190 meters distance from the center, would be in a place with 0.8 g)?

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  • $\begingroup$ @Hobbes says it but I'llm add it here again for anyone who might miss it - this is not real gravity, it is (roughly speaking) the floor accelerating up towards you. So there is never any actual gravitational force due to the rotation, and the artificial gravity "disappears" as soon as you stop spinning around the axis.This is an interesting question! $\endgroup$
    – uhoh
    Commented Jan 7, 2017 at 20:19
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    $\begingroup$ @BenCrowell That applies to linear acceleration, but I don't think that applies to the centrifugal force created by rotation used as a source of artificial gravity. For example, the biggest problem with rotation-derived artificial gravity is the Coriolis force which would make it feel very clearly artificial, uncomfortable and hard to move around. You would know instantly you were in a rotating frame, without question. You can read the several good answers here, and this concise answer as well. $\endgroup$
    – uhoh
    Commented Jan 8, 2017 at 6:29
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    $\begingroup$ @BenCrowell here is a nice illustration of somebody "discovering" their local surroundings are a rotating frame. $\endgroup$
    – uhoh
    Commented Jan 8, 2017 at 7:08
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    $\begingroup$ @uhoh: The basic machinery of general relativity doesn't make any distinction between linear and nonlinear acceleration. The technical statement of this fact would be that field equations are invariant under any smooth change of coordinates. Transforming to a rotating frame is just one type of change of coordinates. The equivalence of a gravitational field to the choice of a noninertial frame is in general only a local equivalence. In your examples, the observer needs to explore a nonvanishing volume. Note that we are not disagreeing. I'm saying that all gravity is fictitious. You're [...] $\endgroup$
    – user687
    Commented Jan 8, 2017 at 17:34
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    $\begingroup$ [...] saying that not all fictitious forces can be interpreted as a uniform gravitational field over a large region of space. These are not logically in conflict. $\endgroup$
    – user687
    Commented Jan 8, 2017 at 17:35

2 Answers 2

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Yes and no. As long as you're on a solid surface: yes. 'Gravity' (centripetal force, really) decreases proportionally with distance to the center.

But if the cylinder has a vacuum inside (so no air mass that rotates with the cylinder), you'd be weightless as soon as you left the surface. If you were to cancel out your forward speed (=the rotation speed of the surface you were standing on), you'd be able to float just above the surface.

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    $\begingroup$ If you were driving anti- rotation direction and accelerated to a little over 100 kph, you and your car would suddenly be nearly weightless? Oh, the headwind would slow you down a bit then you'd regain traction. Sounds fun! $\endgroup$
    – uhoh
    Commented Jan 7, 2017 at 20:10
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    $\begingroup$ You refer to both leaving the surface and canceling the speed. The issue isn't leaving the surface, it's canceling the speed. Staying on the surface just makes it harder to cancel the speed. Maybe you could clarify this. For example, if you were above the surface and reached down and lightly dragged a finger across the surface, you would still be weightless. However, when your speed is not canceled, the normal force is how you sense the existence of the fictitious gravitational force. $\endgroup$
    – user687
    Commented Jan 7, 2017 at 23:55
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    $\begingroup$ If you leave the surface, you move tangentially instead of circular before. But a short tangential movement lets contact you the surface again very soon because you are inside the cylinder. You are weightless only for the very short time with no contact to the surface. $\endgroup$
    – Uwe
    Commented Jan 9, 2017 at 16:57
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Consider a huge, capped drum floating in space, rotating on its axis of radial symmetry. If you are free-floating inside it, no matter where you are, you experience no forces/accelerations due to the drum's rotation, because you are not moving with it.

Now, lets add a ladder attached to the inside of one of its end caps leading radially out from the center. At the "foot" of the ladder, you are standing on the interior surface of the drum and feel the artificial gravity created by the drum's rotation because you are moving with it. As you climb the ladder, you will feel lighter and lighter. While you may stop at any point and be "stationary" on the ladder, you are still moving with the drum's rotation, traveling along a circular path. When you get to the "top" (the center of the drum), you will feel no net acceleration (assuming your center of mass has arrived at the drum's axis of rotation). For the formulas to calculate the acceleration/force felt at any point up the ladder, see: https://en.wikipedia.org/wiki/Acceleration#Circular_motion

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