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I am not a physicist, just a curious mind. I was reading a novel by Iain Banks where it was mentioned, that shifting from artificial rotational "gravity" (in space, on a rotating space craft) to real gravity caused some level of discomfort.

And this has me thinking; is there any truth to that? I mean I am aware that reading a science fiction novel does not science make; however it also strikes me as an unlikely story line to inject in there if it was not founded on at least some real theory or actual reality.

So I guess it boils down to this. From the perspective of the individual experiencing it, is there any notable difference from being rotated and thereby experiencing a sensation of gravity, to a person experiencing real gravity (from the attraction of mass)?

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    $\begingroup$ How was the shifting accomplished? It's not clear from what you're said whether there was some difference that caused discomfort, or whether it was the shifting itself that caused discomfort. Also, were the strengths equal? If you're on a spaceship designed to mimic Earth's gravity, and you go to a planet with more than Earth's gravity, that could cause discomfort. Or if you're on the upper levels of the spacecraft, then you would be experiencing less gravity, so going to full could cause discomfort. $\endgroup$
    – Acccumulation
    Commented Nov 13, 2018 at 18:34
  • $\begingroup$ Good question. 3 pedantic points: on Earth, you are experiencing artificial rotational "gravity" forces. Because Earth is spinning. But that's negligible compared to what we normally consider as "normal gravity". (Also, Earth is rotating about the sun (and rotating about the center of the Milky Way...!)) $\endgroup$
    – user1521620
    Commented Nov 13, 2018 at 20:36

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I think a rotating frame would have both a centrifugal force, mimicking gravity, and what is called a Coriolis force. So, for example, if you would throw a ball straight up in the air in the rotating space station, you would see it move sideways too, because the outside of a wheel always rotates faster than the inside.

It's possible that the people in the space station could feel this Coriolis force, hence the reason for the discomfort.

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    $\begingroup$ Nice answer. For an instructive homework exercise, try analysing various gravity-testing experiments, such as what happens when balls are dropped from a tall tower set up in a huge rotating space station whose rim rotates with acceleration 1$g$. $\endgroup$
    – Andrew Steane
    Commented Nov 11, 2018 at 16:13
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    $\begingroup$ In an episode of the show "The Expanse", a phrase similar to "in the core where the Coriolis is really bad" is used. $\endgroup$
    – Davis Yoshida
    Commented Nov 11, 2018 at 17:24
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    $\begingroup$ Wikipedia has dug out a rule of thumb (belief?) that at 2RPM or below the Coriolis force would be tolerable. 2RPM comes to about $0.2$ radians per second. Meaning that $1g$ or $10 m/s^2$ requires a station with a radius of $250$ meters. $\endgroup$
    – Jyrki Lahtonen
    Commented Nov 12, 2018 at 4:31
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    $\begingroup$ @Dithermaster "So, much like changing eyeglasses - the discomfort isn't physical" it is absolutely physical, for both cases. Changing glasses results in the lens muscles having to work in different ways, which tires them. Having uneven forces on your body is also physical. $\endgroup$
    – UKMonkey
    Commented Nov 12, 2018 at 11:58
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    $\begingroup$ @Åsmund 10m/s is a dead sprint for a top athlete; most people aren't "walking" 5m/s 6-minute miles $\endgroup$
    – Ryan Cavanaugh
    Commented Nov 13, 2018 at 15:41
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I'm speculating, but the speculation is based on actual physics :).

Your physical experience of gravity on a planet and artificial gravity at the outside of a rotating wheel might be different based on the following.

The force you feel from a planet is $G*m_{you}*M_{planet}/r^2$ (Gravitational constant times your mass times the mass of the planet, divided by the distance $r$ from you to the center of the planet, squared.

The force you feel from the rotating wheel is $m_{you}*\omega^2r$ (your mass times the angular velocity (squared) times $r$, the distance from you to the center of the wheel).

So, suppose you are on a planet (which would normally have a very large value of $r$--meaning, you are a long way from its center), and you are seated, then you stand up. Your head has moved from $r$ meters to $r+1$ meters (your head is now 1 meter farther from the center of the planet). So, on earth, you've moved from about 6.4 million meters away to about 6.4 million meters...plus one! That's going to make a change in the force on your head that's probably way too small for you to notice.

On a man-made rotating wheel, you're going to have a much smaller value of $r$ (assuming the wheel is way less than the size of a planet). So $r-1$ meters (keep in mind, when you stand up inside the rotating wheel, your head is closer to the hub of the wheel, so it's a change to $r-1$ instead of $r+1$ as it would be on the planet) might be different enough from $r$ meters to be something you feel, and, if you spent a lot of time there, or were born there, or whatever, you would get used to things (like your head) being "lighter" when you stand up. If that was your "normal", then it might feel really strange to you when that didn't happen in Earth's gravity.

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    $\begingroup$ Isn't this why such craft have to be pretty large? $\endgroup$
    – RonJohn
    Commented Nov 12, 2018 at 2:10
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    $\begingroup$ The term of art for the effects you're talking about is tidal forces. $\endgroup$
    – Michael Seifert
    Commented Nov 12, 2018 at 3:57
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    $\begingroup$ @RonJohn Yes, but there's an economy to consider. E.g. it would be nice if trips to space didn't require such high acceleration as in modern rockets, but it's more economical to train a few specialists to handle those accelerations than to fly rockets at lower accelerations. The same way, the rotating ships would be built as small as possible for a given tolerable level of discomfort for most of their users. Maybe at a radius of 200 meters, noöne would notice the rotation - but 200 meters is a pretty bulky ship (and it would only work on the outer edge anyway!). $\endgroup$
    – Luaan
    Commented Nov 12, 2018 at 13:02
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    $\begingroup$ You should also account for "running widdershins" making your heavier. The effect seems to scale down with the square root of the radius, so it might persist longer than the linear height change impact. $\endgroup$
    – Yakk
    Commented Nov 12, 2018 at 20:20
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    $\begingroup$ @Luaan, on the topic of economy: Just because it has a radius of 200m doesn't mean it has to have a circumference of 1256m... The habitat could just be a couple of evenly weighted capsules on a 400m tether. $\endgroup$
    – Ghedipunk
    Commented Nov 13, 2018 at 18:05
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For a non-technical answer, remember when you were a kid on the playground? (Yes, I know I'm making what's perhaps a parochial assumption.) If you sat on the merry-go-round (this: https://en.wikipedia.org/wiki/Roundabout_(play) ) and got the other kids to push it around really fast, you could feel the "gravity" pulling you outwards. But because you were also going around in a tight circle, the fluid in your ears sloshed around, and so you got dizzy.

Now scale this up to a moderately-sized space station. You might still have some effect on the ears from rotation (how much depends on the size), but because you've been there a long time, your body has adapted to this as being normal. When you shift to "real" gravity, the rotation effect goes away, but to your body this is now NOT normal.

(Whether this would actually happen I can't say: AFAIK no one has tried it, but it's certainly plausible enough for SF :-))

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  • $\begingroup$ The distance scale could be such that the rotation rate is very small, say once per 24 hours. Ear-related effects would then be too small to matter. $\endgroup$
    – Andrew Steane
    Commented Nov 11, 2018 at 18:32
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    $\begingroup$ @AndrewSteane It depends on two things: 1) how big your habitat is and 2) how fast its spinning. The smaller it is, the faster it has to spin in order to generate 1 G of gravity on the outer surface as well as causing a steeper gradient (i.e. if your habitat is 12 feet in diameter, then your head experiences 0 G and your feet 1 G; an extreme scenario). $\endgroup$
    – Draco18s no longer trusts SE
    Commented Nov 11, 2018 at 19:02
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    $\begingroup$ @AndrewSteane one rotation per 24 hours would require a radius of ~2 million km for 1G $\endgroup$
    – trapper
    Commented Nov 12, 2018 at 3:13
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    $\begingroup$ Merry go rounds create acceleration that goes sideways - something ears are not used to. Gravity and rotating spaceships create acceleration downwards, something your ears are designed for. So how does this matter? $\endgroup$
    – famargar
    Commented Nov 13, 2018 at 19:05
  • $\begingroup$ My 24 hour example was a bit extreme. Better example of timing would be 90 minute rotation for a planet-sized space station; 1 minute rotation for a km scale space station. Not much dizziness in these cases I think. $\endgroup$
    – Andrew Steane
    Commented Nov 14, 2018 at 23:07
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You would be unlikely to notice any difference unless the spacecraft is fairly small.

For example with 50m radius there is only a 2% difference between 50m and 49m. The station in this case would be spinning at 4.25 rpm to generate 1G.

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    $\begingroup$ 2% per metre is quite a lot. A 2m tall, 80kg person upon standing up, would be thrown forwards with a 3kg force and vertical as sensed by their inner ear would vary by up to 18 degrees as you did so, depending upon which way you were facing relative to the direction of travel. That should be enough to stumble or fall if you expected it to go one way and it went the other. $\endgroup$
    – Robert Frost
    Commented Nov 13, 2018 at 4:53
  • $\begingroup$ I can't even imagine what kind of math lead you to those conclusions. $\endgroup$
    – trapper
    Commented Nov 13, 2018 at 5:21
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    $\begingroup$ Sine X approximates X for small X so simply multiply mass by percentage difference for an instant approximation. Simples. $\endgroup$
    – Robert Frost
    Commented Nov 13, 2018 at 5:37
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    $\begingroup$ You can’t just multiply numbers randomly though. 3kg is not a ‘force’, and your centre of mass while standing is at hip level, not 2m off the ground. $\endgroup$
    – trapper
    Commented Nov 13, 2018 at 5:47
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    $\begingroup$ You're obviously right re kg not being a force but if rotation is generating 1g at the circumference then mass at the circumference is isometric with weight on earth, so I was talking in terms of the weight of 3kg on Earth. $\endgroup$
    – Robert Frost
    Commented Nov 13, 2018 at 8:25
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Experiencing rotational forces and fixed direction gravity at the same time would be weird.

A person under the influence of gravity experiences a constant acceleration. A person in a rotating reference frame experiences a constant magnitude acceleration, but the direction is changing constantly.

This means that if you are experiencing both at once, and the axis of rotation is not parallel to the direction of gravity, the total acceleration that you feel will be constantly fluctuating. It's more or less equivalent to the fact that if you swing a bucket on a rope in a vertical circle, the tension in the rope is higher when the bucket is near the ground than when it is at the top of the swing.

Depending on how fast the rotation of your station is, this could make the transition period feel like a rollercoaster.

Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second. That would avoid the roller coaster effect. But if they skipped that process then I could easily see people emptying their stomachs during the process.

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    $\begingroup$ Sorry, but this is incorrect. Imagine swinging a bucket on a rope in a horizontal circle. $\endgroup$
    – Beta
    Commented Nov 12, 2018 at 2:15
  • $\begingroup$ @Beta, well, it depends on which way the station is rotating. You could organize the transition in a logical, non-rollercoaster manner. But you don't have to. $\endgroup$
    – Arcanist Lupus
    Commented Nov 12, 2018 at 2:55
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    $\begingroup$ I hope you aren't referring to orientation relative to a planet -- the only linear acceleration on the station would be due to its translational rocket engine. $\endgroup$
    – amI
    Commented Nov 12, 2018 at 8:11
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    $\begingroup$ The tension on a rope on a bucket increases and decreases because you are standing on a planet experiencing its gravitational field. That does not apply in this situation. $\endgroup$
    – msouth
    Commented Nov 13, 2018 at 13:13
  • $\begingroup$ "Of course, the logical way to transition reference frames would be to leave one, enter zero-g, then enter the second." Why would the transition matter after you are in the centrifuge? $\endgroup$
    – JiK
    Commented Nov 13, 2018 at 21:23
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Fist of all, let me apologize for the post, indeed i was just browsing around and this sparked my interest.

In my opinion there is mechanical difference in which the rotation affects you in those two cases (you rotate on planet while not on poles). On planets surface the mass pulls you inward and the planetary rotation lessens the force applied to you. On the station the rotation works the other way, basically creating gravity from nothing.

Have a nice day.

So to explain myself further: I was thinking, what difference would I feel on such station? The vertical movement is one thing. As previous answers stated delta g on one meter differs for the station when compared to the planet. Movement on the floor of the station, I presume, would feel different when walking against the rotation. In such case my angular velocity is lower than otherwise. Would I feel lighter if walking in one direction? Could this be the disorienting factor? And so on.

As for the first post. I was trying to be brief and oversimplified. Also please forgive me for slaughtering English language, I am not a native.

Best regards.

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