Simulated gravity via rotation

Factual Questions
1

I was reading the wikipedia page on “2001: A Space Odyssey”.

I came across this:

Is this statement accurate?

From what I understand, what happens is that in any single frame, any single instantaneous moment, a person is being accelerated outwardly from the center of a sphere or disc in a straight line.

The floor (what their feet are on, the outer edge of the disc away from the center) keeps them from moving in that direction, the normal force is applied, and they feel pulled against the floor - simulating a similar effect to gravity.

As the floor is constantly rotating, it keeps changing the direction in which the people’s bodies are “trying” to accelerate outwards due to momentum, hence they are perpetually pushed in a new direction that corresponds with the location of the floor under them. Since they’re always trying to accelerate off in a straight line, and the floor is always under them, they stay pushed against the floor.

I may have explained that badly - it’s tricky to explain - but I think I have a correct understanding.

Now what I don’t understand - if this is the case, why would the apparent simulated gravity be different for different areas on the body?

To make an analogy… imagine someone travelling through space in a straight line, feet first. Now imagine putting a floor under that person (that for our purposes is stationary and unyielding). That person would feel the force of their momentum against the floor similar to the simulated gravity scenario above - right?

Would that person feel a different simulated gravity at the level of their head as compared to their feet? I think not - yet a rotating disc seems like the same scenario to me, except it’s a constant acceleration against the surface rather than one impact.

If a rotating disc had multiple decks, I understand why an inner deck would have less simulated gravity than an outer disc (rotational velocities). But it seems to me that the simulated gravity comes from the normal force offsetting their acceleration - and that force should be uniform throughout their body.

2

Aren’t you treating the body as a single point of mass, rather than a distributed object? Your head is going to be pushing down on your shoulders less (relative to mass) than your thighs will be pushing down on your knees. I have no idea whether this would be noticeable whilst walking around.

Here’s a question that could illuminate the issue: in such a rotating wheel, would a person weigh more or less when (a) standing, or (b) sitting on a stool of a height to keep his torso at the same height?

I believe the answer is that apparent weight would be the same (barring issues of where the guy’s legs are) because the issue is how far from the centre point the various parts of the distributed mass of his body are, rather than at what point (floor or stool) his movement away from the centre point is being arrested. But I could be wrong.

3

Because they are traveling at different rates of speed. Let’s assume their feet make a 100m circle (the circumference of a hypothetical station; I’m just choosing a number) every minute. Your feet are thus travelling at 100m/min. The radius of the station in the example is about 16m (100m/pi/2*). Let’s assume you’re two meters tall. That make the radius that our head will travel about 14m. 14pi2 gives the distance your head travels; about 88 meters. So your head is going 88m/minute and your feet 100m/minte.

Since the gravity is determined by the speed you are traveling, your head and your feet are thus “falling” at different rates.

using circumference = 2pi*radius

4

I may misunderstand something basic about this situation.

The reason that they’re held against the floor isn’t simply because the floor is spinning, but because they’re constantly being accelerated in a new vector outwards from the center, but the floor gets in the way, pushing against them, giving them the feel of gravity, right?

In other words, their whole body is trying to move outwards in a straight line, but constantly being “caught” by the floor.

If the disc were to suddenly dissapear, they would go flying off in a straight line, correct?

If that’s the case, then it seems to me that the simulated gravity isn’t from the rotation itself, but simply from being accelerated towards the floor. The reason for the spinning is that it’s constantly changing the person’s vector, keeping them in a state of outward acceleration towards the floor.

5

Basically the head is being accelerated less than the feet. How much less depends on the radius, but the acceleration is proportional to the square of the distance to the center or rotation. The result is what we call a tidal force, which falls off as the cube of the distance. Incidentally, if you were to fall into a black hole feet first, the tidal force of anything but a supermassive black hole would tear you apart long before you crossed the event horizon. The reason it would be different for a supermassive black hole is that its event horizon could be so far away that there might not be such a strong tidal force there. The tidal force falls off much faster than the gravitational force. That’s why the earth is much more attracted to the sun than the moon, but the latter exerts a stronger tidal force.

6

When standing in a rotating space station, the force that keeps your feet on the floor is not gravity.

It’s centrifugal force.

In a sufficiently large space station, the problem would not be any difference between the pull exerted on your feet and head… it would be the tendency of a “falling object” to veer slightly to antispinward, due to velocity (I think – I’m no physicist).

7

Force equals mass times acceleration. On a uniformly rotating circle, the radial acceleration is, I believe, V^2/R, where V is the velocity at that point on the circle and R is the radius. Since both V and R get smaller as you go from your feet to your head in such a circle, the force gets smaller, too. I’m not too sure what the rate of change is over the course of your body, though. I’m sure I could figure it out if I tried.

8

Technically, you’re being accelerated towards the center of the circle, not away from it. If the floor were to suddenly disappear, you would not accelerate, but you would move at a constant velocity tangent to the circle you were moving in previously.

If you’d like to see some of the math dealing with this, we can calculate how large the space station would have to be to accurately simulate gravity on earth.

It’s important to note that even here on earth, there is a difference between the gravitational force on your head and that on your feet. Let g be the gravitational acceleration experienced by your feet, and g[sub]r[/sub] be the ratio of the gravitational acceleration of your head divided by that of your feet, so that the gravitational acceleration of your head can be expressed as g*g[sub]r[/sub]. We can calculate g[subl]r[/sub] by using Newton’s gravitational formula, GMm/r[sup]2[/sup]. This simplifies to g[sub]r[/sub] = r[sub]e[/sub][sup]2[/sup]/(r[sub]e[/sub] + h)[sup]2[/sup], where r[sub]e[/sub] is the radius of the earth, and h is the height of the person.

Now, the centripetal (towards the center of the circle) acceleration attained by a uniformly rotating circle is w[sup]2[/sup]r, where w is the angular velocity or the space station, and r is the radius. We want the acceleration to be the same at the feet and the head so:
Feet: w[sup]2[/sup]r = g
Head: w[sup]2/sup = g*g[sub]r[/sub]
We solve these, and get r = h/(1 - g[sub]r[/sub])

Now, if I didn’t make any mistakes, I have approximated r[sub]e[/sub] as 6000 km, have set h = 2 m, and rounded g[sub]r[/sub] to 0.999999 (probably quite a bit more accurate than we really need). With this, I get r = 2000 km. Pretty damn big. Though we can probably knock a couple significant figures off that without much noticing it while living on the space station.

9

What’s the difference in principle between the different decks (at different distances from the centre of rotation) and the different parts of the body (also at different distances from the centre of rotation)?

As someone else said; the human body is not a point mass; it is a collection of masses connected together by flexible joints; is gravity (or fake gravity in a rotating spaceship) only acted to keep your feet on the deck, you wouldn’t need muscles in your neck to keep your head upright.

One other thing that is often overlooked in these sorts of devices is Coriolis effect (lateral forces caused by conservation of momentum when a body moves toward or away from the centre of rotation). If you stood up suddenly in Discovery’s artificial gravity, you would find yourself being pushed over in the direction of spin; if you sat down suddenly, you would find yourself tipping over in the other direction.

10

This is, basically, the same principle as Master Wang-Ka discussed with “falling objects”, correct? I did a few calculations with this, and it definitely seems to get worse as the radius increases. If I did it right, the distance between you and an object you “dropped” is approximately a function of the square root of the radius - slower than we get a more realistic simulation of gravity.

11

It gets worse as the radius increase for any fixed RPM. But if you increase the radius, the rotational speed required to simulate earth gravity becomes sufficiently lower that the coriolis force would actually decrease. I think.

12

Potato, potahto. General Relativity views the two as pretty much equivalent.

13

By my calculations, the distance between the “dropped” object and the person who dropped it is independent of the angular velocity. I’ll reproduce what I did here. It involves a drawing, which I will try to reproduce with words as best I can. If anyone is interested in checking my work, it would probably be best to attempt to draw the pictures.

The floor of our space station is centered at the origin of the xy plane, having radius r and angular velocity w, rotating counter-clockwise. That is, the floor is described by the equation x[sup]2[/sup] + y[sup]2[/sup] = r[sup]2[/sup]. A person is holding an object a height h from the floor, and drops it at the moment he crosses the x-axis - at the instant it is dropped, the object occupies the point (r - h, 0). I choose this point purely to simplify the calculations.

After it is dropped, the object experiences no net force (I assume that the gravitational force from the space station is negligable), and moves in a straight line parallel to the y-axis for a distance d, until it hits the floor again. The distance d is equal to the y-component of the point of intersection. It satisfies the equation:
(r - h)[sup]2[/sup] + d[sup]2[/sup] = r[sup]2[/sup].
Which simplifies to:
d[sup]2[/sup] = 2hr - h[sup]2[/sup]

We can define an angle, theta, that is the angle between the line connecting the origin to the intersection point (r-h,d), and the positive x-axis. This angle is:
theta = arctan(d/(r-h))

We must also determine how much the floor rotated in the time it took for the object to “fall”. The time it took to fall is:
t = d/v
Where v = w(r-h).

The angle rotated by the floor is:
theta[sub]floor[/sub] = wt = d/(r-h)

Finally, the distance (along the circumference of the circle) between the person and the object is given by the radius times the difference in the angles:
distance = r(theta[sub]floor[/sub] - theta)

We can do some simplification by assuming that r is much larger than h. Thus:
r-h ~ r
2hr - h[sup]2[/sup] ~ 2hr
d/(r-h) ~ sqrt(2hr)/r ~ 0

With this, our distance formula simplifies to:
distance ~ sqrt(2hr)

I’m not sure if this is correct, but it’s what I’ve come up with.

In somewhat simpler terms, the rotational speed doesn’t matter because both the speed of the dropped object and that of the person who dropped it depend on it. They cancel out.

14

Oops. That was bad of me, that last approximation is wrong.

I believe we can simplify it down to approximately:
distance ~ sqrt(2hr) - r*arctan(sqrt(2h/r))
which does appear to slowly decrease as radius increases.

Gorsnak, I believe you are correct.

15

Nope. A rotating frame of reference non-inertial (under acceleration), so you can definitely differentiate between gravity and the “centrifugal” force, via its Coriolis component. (As for GR and rotating reference frames, check out Frank J. Tipler’s “Rotating Cylinders and the Possibility of Global Causality Violation.” Physical Review D 9 (1974): 2203-2206.)

I once worked out this problem, including the effects upon the labyrinth, and while I don’t have the notes at hand, I recall that the effect manifested itself at a radius of about 8m for a simulated 1G of outward force. Beyond this, the effect was sufficiently small as to be insignificant for normal movement, though you still might have to be careful which way you’re pouring your coffee.

Discovery’s rotation hull looked to be ~10m in diameter; definitely small enough to pose a problem. I’m not sure if there’s a scene that allows a measurement of its rotation rate, but by reducing the spin you square the reduction of the Coriolic component.

The variation in force between head and feet is real and significant for small radii, and a possible long-term health problem (with blooding tending to pool in the feet), but not a serious issue for short-term habitation.

Stranger

16

Just go to the Cage of Death and watch the guy on the motorcycle for a while.

Or go to the spinning drum where the floor drops away.

Now take away outside references.

Now enjoy your chess game.

17

[hijack]

If gravitons are involved in the force of gravity, are they also involved in the centrifugal/centripetal force?

[/hijack]

18

No. First of all, centrifugal force is a so-called “ficticious” force; that is to say, while it is a real enough force to someone within the non-inertial frame of the rotating system, it doesn’t exist to someone outside the system (the objective observer in the “privilged” reference frame). It’s merely the result of inertia of the body in question resisting the change in velocity by being forced to rotate. (This isn’t to say, as some will mistakenly do, that there is no such thing as centrifugal force, only that it isn’t definable statically; like the D’Alembert force, itj occurs only within a non-inertial, i.e. accelerating, reference frame.)

Second, the notion of gravitons is an attempt to interpret gravity into an integrated (and hopefullly renormalizable) quantum field theory via string theory. Even in that context, it’s wrong to think of gravitons as being spit out by one body and absorbed by another. Akin to weak interaction, the gravitons are exchanged between masses, creating an attractive field. Hence, gravity is a conservative force; the gravitons are never “lost”. The reality (like, I suspect, most of quantum theory) is merely a crude approximation–and likely conceptually completely wrong–of what is actually occuring. In any case, centrifugal force can be adequately described by Newtonian mechanics (as long as you pick an inertial reference frame) without getting bogged down by discussions of GR or gravitation.

Stranger

19

Care to help out an ignoramus? I don’t understand how, given the scenario under discussion, anything is being accelerated towards the centre. Not saying you’re wrong. Just saying I don’t understand this and need help understanding why it’s true or how it works.

This takes me back to a major headache of mine in high school. Some sources would offer a demonstration of so-called centrifugal force (e.g. turning a glass of water in a broad circular arc overhead and none of the water falls out; loose items on a turntable that scatter towards the rim when it’s turned on). Then someone else would come along and say actually, no, there’s no such thing as centrifugal force; the force is actually centripetal (spelling?) force, and it acts towards the centre, but in accordance with one of Newton’s laws there must be an equal and opposite reaction and that’s what you’re labelling as this so-called centrifugal force. I tried a zillion times to ask the right questions and clarify all this, but never got anywhere. Help!

20

I’m having trouble intuitively understanding how a rotation simulated gravity system works in concept still. I think I clearly misunderstand a fundamental component.

To clear it up:

It’s basically the same thing as the carnival rides that spin you in a circle so fast that you stick to the wall, right?

It seems to me that the reason that happens is because the rotation of the object is pushing your body in a new vector every instant, a vector that would, if there was no wall, send you flying off in a straight line away from the center of the rotation object, but since the wall is there, you essentially perpetually get thrown into it.

Filling a bucket with water, and revolving it in a large circle to keep the water “pinned” inside of it - is that the same principle?