Question Regarding Gravity, Centrifugal Force and Equivalence

Tenspace · 05-19-2006, 01:58 PM

Okay, we've all read the sci-fi books and we have seen the movies where artificial gravity is generated using centrifigual force by spinning a structure to generate 1 gee at the inside surface. Think Ringworld, or 2001, or Peter Hamilton's hollowed-out asteroid habitat.

I've done a bit of digging on the internet; most sites I've found discuss the rate of spin necessary based on the size, the coriolis affect on the internal atmosphere... stuff like that. But, other than a simulation of a rotating body at the NASA site, I can't find anything that describes the effects of rotation as a function of distance from the surface.

Is this truly equivalent to gravity? In other words, would the gravitational force felt be proportional to the distance to the center of the spinning cylinder? Or would you be in freefall if you left the surface by any distance?

Inquiring minds want to know.

Tenspace · 05-19-2006, 03:36 PM

::bump::

Choob? Anybody?

Mog · 05-19-2006, 04:18 PM

Hey Ten, you check the wiki article about artificial gravity? It seems to have your answers.

http://en.wikipedia.org/wiki/Artificial_gravity

Tenspace · 05-19-2006, 04:25 PM

Quote:

wiki wrote

Gravity gradients: Artificial gravity levels vary depending on distance from the center of rotation. With a small radius of rotation the amount of gravity felt at one's head would be significantly different from the amount felt at one's feet. This could make movement and changing body position awkward.

Wiki's vague in this regard. They don't give any specifics, other than the amount of gravity felt would be different. There's no "why", if you see waht I mean...

Why would the rotational forces affect something not on their surface? If this rotating cylinder is in empty space, and the inside is zero gee as well, what force is keeping objects planted to the surface?

(FYI, I've read wiki, the Standford info, as well as the stuff from NASA. Only NASA seems to offer an explanation. I'm interested in the thoughts on this subject from others who might be able to provide some enlightenment.)

If centrifigual force is proportional to the distance from the surface, like gravity, then what is the centrifugal force affecting? Certainly not spacetime, like gravity.

PanAtheist · 05-19-2006, 04:26 PM

It is confusing that you say "surface"!
It's *inside* the spaceship, and is better called a "floor"!

Of course it is not *equivalent* to gravity - it is artificial! And is produced quite differently.

If you jump hard enough in the right conditions you could leave the floor and head to the centre.
Your ascent will appear to slow as the floor pursues you, and you will pass through the centre at a steady speed, experiencing no attraction to the floor (at the centre, no part of the fall falls towards you, except that bit which you launched yourself too) - and we are assusming that you jumped hard enough and that your velocity will be carry you through the centre and you'll pick up speed again in the direction you were already going, arriving at the "floor" again.

Of course, with practise, you can give yourself exactly the right amount of launch speed, so that you can get right to the centre - and stop there. And then you'd be buggered!

Everyone would watch, amused, as you do desperate swimming motions. :)

Tenspace · 05-19-2006, 04:29 PM

Quote:

PanAtheist wrote

If you jump hard enough in the right conditions you could leave the floor and head to the centre.

That sounds like a description of escape velocity.

Quote:

Your ascent will appear to slow as the floor pursues you, and you will pass through the centre at a steady speed, experiencing no attraction to the floor (at the centre, no part of the fall falls towards you, except that bit which you launched yourself too) - and we are assusming that you jumped hard enough and that your velocity will be carry you through the centre and you'll pick up speed again in the direction you were already going, arriving at the "floor" again.

If there is no attraction, then I wouldn't have to jump hard at all, would I? All I would have to do is remove myself from the floor, and I'd be in freefall according to that.

If you are deposited inside this cylinder, and you are placed near the center of rotation with no relative movement, away from any "floor", then you would remain in one spot, and not move relative to the cylinder?

PanAtheist · 05-19-2006, 04:32 PM

There is a wonderful sequence of *dancing* in zero-gravity in the film "Mission to Mars" .
It's worth checking out for this sequence alone!
A chunk of this film is really good - the sequence of the trip to Mars itself. I highly recommend watching this part of the film, and not bothering with any of the rest of it. (It is kind of offensive to sane people like ourselves!)

05-19-2006, 04:33 PM

As long as you are traveling at the same velocity as the rotating surface, the centripetal acceleration would be the same as gravity.

If you left the surface, your 'acceleration' back towards it would be a parabolic arc defined by your velocity and the rotation of the surface you just left. But since you are unaware of your velocity matched status, your view is that of acceleration towards the surface.

You are, in fact, in 'freefall' during this period. Just as if you jump up into the air here on earth or leap from an airplane.

If you left the surface AND reduced your velocity such that the surface was spinning but you were stationary above it, you would float there. weightless.

There mey be some quantum frame thing that is different between the two scenarios... (On earth or on the spinning surface) but I don't know anything about that.

PanAtheist · 05-19-2006, 04:37 PM

Quote:

Tenspace wrote

If you are deposited inside this cylinder, and you are placed near the center of rotation with no relative movement, away from any "floor", then you would remain in one spot, and not move relative to the cylinder?

Yay. It all might be spinning! Makes no difference!

Quote:

Tenspace wrote

If there is no attraction, then I wouldn't have to jump hard at all, would I? All I would have to do is remove myself from the floor, and I'd be in freefall according to that.

There is NO "attraction" in any of this.
The floor is coming at you and colossal speed!
And it gets faster, as you move away from it!
Boy, you're going to have to jump hard if its spinning so fast it's producing 1G!

05-19-2006, 04:41 PM

LOL Pan, a problem with that scenario...

For that to happen (be able to leap to the center of the room) the gravity would have to be VERY low or the room very small.

With a small room, the disorietation caused by tidal factors (Differences in gravitation) would be insane. You'd be sick all the time and probably have trouble standing.

With a big room going super slow, the gravity wouldn't be high enough to be useful.

It's assumed that the surface is spinning fast enough to provide decent, usable gravity and you would have no way to achieve escape velocity without external means.

05-19-2006, 04:43 PM

Ten, you're in freefall if you jump here on earth, too. You still have WEIGHT, which is defined by your mass and the net acceleration between you and the referance plane (ground here on earth, the spinning floor in the ship)

Tenspace · 05-19-2006, 04:44 PM

Let's say we have a cylinder that is 600 meters in diameter, and it is rotating about its axis at 1rpm, generating roughly 1 gee at the inside surface (a/k/a floor).

The angular momentum I exhibit is along a line from my position toward the direction of rotation. If I am hoisted an inch above the floor at the the moment (arm, not time) that matches my angular momentum, you're saying I would remain in freefall an inch above the floor while the cylinder rotated around me?

05-19-2006, 04:49 PM

No no! You would be in freefall until you hit the surface again. 'Freefall' is NOT the same as weighless.

Since you, once you got an inch from the surface, are no longer being pushed into a circular path by the floor, you would continue along a straight line until the floor curved up under you again.

PanAtheist · 05-19-2006, 04:54 PM

Freefall is a misleading term
Best use both Freerise and Freefall.
Both meaning nothing's doing any pushing.

And in space remember, their both only virtual.
In space you really just Free!

Or perhaps its best to think of freefall as a acceleration down, rather than a velocity down.
I have no idea about the official usage.
Actually I recommend that you don't use them at all in this situation.
They are not applicable, as this is not about gravity.
You need to think in terms of rotations, and rotational inertia.

Honestly, it's not about falling!
That floor really is coming to get ya!

Choobus · 05-19-2006, 04:54 PM

pan atheist is dead wrong acceleration is equivalent to gravity. That is a very important concept in the general theory of relativity and is known as the Einstein Equivalence Principle.

To answer your question Ten, acceeleration is defined as the rate of change of velocity. If you spin something round at a constant rate the magnitude of the velocity vector is constant, but the direction is constantly changing, which results in a centripetal acceleration the magnitude of which is given by v^2/r (v is the magnitude of the velocity vector, r is the radius. This is only true for uniform circular motion. Also, F=ma. Acceleration is the response of mass to a force, so if somethjing is acceleratinf there must be a force acting on it, and since force and acceleration are vectors, while mass is a scalar, the foce and acceleration must act in the same direction. In the case of uniform circular motion they both act towards the center of the circle. Therefore, in a spinning spaceship you will experience a force that is completely equivalent to the force of gravity towards the center of the rotation, and that force will be inversely proportional to the distance from that center, and be proportional to the square of the rotational speed. I wonder what space anal is like.