Suppose you’re in a cylindrical space habitat spun for 1g of “gravity.”
Are the quotes necessary? Disregarding the internal curvature, any experiment you could perform would show that force to be gravity, wouldn’t it?
But, in modern physics, as I understand it, gravity results from mass bending space, and might be mediated through hypothetical but as yet undiscovered gravitons.
So, does the spinning habitat curve space or produce gravitons? Does the spinning produce “artificial mass”?
Same question applies to a spaceship with a constant-boost drive, constantly accelerating at 1g, thereby producing shipboard gravity.
There would be Coriolis effect, which would be noticeable for anything falling or pouring more than perhaps five or six feet. Very noticeable at 20 feet or more. That assumes, though, that there is an atmosphere or other medium to transfer the hull’s spin to objects within it.
If the interior were a vacuum, there would be almost zero effect on objects placed in a stationary position above the floor. It is only the motion of the object that produces the force outward from the center of spin. If you were wearing mag boots, you’d spin with the hull and feel 1g; an apple let go with no imparted motion at head height would float there until microgravity drew it to the nearest point of the hull. If the hull were massless, the apple would never move.
That was the flaw, IIRC, in the Babylon 5 episode where the captain jumped from the central tram. I don’t believe there would be any acceleration as he fell; he would have hit with the same force as if he had jumped into a wall. Having a surface structure spin into him like a sideways flyswatter would have been more of a hazard.
The whole basis of general relativity, whence comes the notion of mass bending spacetime, is that gravity is indistinguishable from acceleration. So you’d be perfectly justified in leaving off the scare quotes.
That said, though, in either of the situations you describe, the gravity would be distinguishable from the gravity due to a mass like the Earth, if your laboratory were large enough.
The question about producing gravitons is problematic: The gravitons in question (whether from a mass or from acceleration) would be virtual, and you can’t really pin down whether virtual particles exist or not. Heck, sometimes it’s tricky even for real particles: You can have real particles that exist in one reference frame but don’t in another.
Neither would result in curvature of space. The gravitational field is often described as being a curvature of spacetime, but this is mistaken: The curvature is actually a special kind of derivative of the gravitational field, and is more closely related to tidal effects, which you won’t have in either of these situations. Likewise, neither would be described in terms of “artificial mass”.
In all cases, the same answers apply to the rotating space station as to the rocket under thrust.
You’re neglecting that the station was full of air, which would put a drag on him. It’d take a while for it to bring him to the floor (it could be indefinitely long, if he were balanced well enough in the middle), but once it eventually did, he’d hit with approximately the same speed as terminal velocity on Earth.
And the scale needed to observe Coriolis effects depends on the size of the space station, which the OP never specified.
I didn’t connect my answer very well, but addressing Chronos’ comments: I think the difference is that “gravity” has effect over distance while a spinning hull does not. A massless shell could spin forever and never attract an object within it; real gravity would of course pull over distance.
I’ll concede I could be wrong, but… why would he accelerate towards the interior of the hull? What force is pulling him that direction? There IS no “pull” from the shell; there is only centripetal force if you are attached to it (by sheer drag, if nothing else). If anything, the air drag would slow his fall from his initial acceleration, and pull him into sync with the hull’s rotation so that he didn’t get slapped upside the butt by a support stanchion.
This would seem to apply to the “rescue” in *Rendezvous with Rama *as well. Why would Jimmy Pak gain any down/outward acceleration from jumping off an interior cliff, when there is no true gravity pulling him that direction?
I haven’t seen the movie mentioned, where was the tram? Dropping a ball on a bus doesn’t cause it to fly to the back of the bus. It’s not just air resistance, either- the ball is going the same speed as the bus.
Yeah, they made the exact same mistake in Red Planet, when Matrix-girl gets the ship spinning again everything that’s floating just instantly (and essentially magically) drops to the floor… (scroll down to the middle of the page).
I think a better way of putting is that from the point of view of a person inside the spinning tube, what the relativists call a momentary co-moving reference frame, space looks curved. Imagine, for example, a beam of light that travels across the tube. To the outside observer, of course the beam appears to travel straight across the tube. But not to the guy inside the tube, travelng around its circumference. Let’s make it easy by supposing the beam enters the tube right where the guy is standing, travers the diameter of the tube, and takes the same amount of time to traverse the tube as the tube does to make one-half a revolution. The guy in the tube sees a beam of light enter by his feet, propagate to the center of the tube, then bend back and fall, exiting the tube right at his feet again. So his conclusion is that space is mightily curved, since light does not travel in a straight line, and he concludes there must be a big mass under his feet. (He could also conclude he is in a rapidly rotating reference frame, of course, but if he can’t see outside the tube the presence of a mass might be a more appealing explanation.)
It’s less correct to say that mass bends spacetime than to say that we interpret any curvature of spacetime we observe in our own reference frames as mass (or really energy). Others, in different reference frames, may just conclude that we’re accelerating and that’s why our spacetime looks curved to us. There’s no compelling reason (within the realm of gravity and acceleration alone) to choose one reference frame’s point of view over the other, except on grounds of simplicity and taste.
Also be aware that there is nothing surprising at all about motion producing “mass.” Recall Einstein: there is no meaningful distinction between energy and mass. What is “mass” (energy at rest) in one reference frame is “energy” (of motion) in another.
On the Babylon 5 station, which spun to produce 1g at its main circumference, there was an axial tram that ran the length of the inhabited portion. A bomb forced Sheridan to jump out. I’d guess (not having any B5 references at hand) that it was a half-kilometer fall, or around 800-1000 feet.
What would increase his speed towards the hull, since there is no true gravity? Wouldn’t he hit at the speed imparted by his jump, less whatever reduction caused by air friction? Would the rotational friction of the air (creating increased centripetal force) accelerate him to anything like a true fall’s speed?
You’re starting in the middle of the story. Consider how the object came to be in midair: was it thrown or was it gently released from the hand? Either way, the inertia of the object and of the system will have effects that most posters are ignoring. I have no time for details now, but I’ll try later.
I think you have to work out the math, but I suspect the kind of mass distribution that would produce the same apparent acceleration as a spinning hull (which would be very peculiar indeed) allows for the equivalent of “geostationary orbits” in which objects do not fall, and that would take care of the problem. I don’t think you can beat relativity any more than the Second Law of Thermodynamics. It’s always just a question of working out how you’re wrong.
According to (possibly fan-supplied and therefore maybe suspect) info, the “floor” of the station was rotating at 60 mph, so even touching down on flat ground would be disconcerting to say the least, akin to stepping out of your car while driving on the freeway.
The phrasing of the answer to this question depends on what reference frame we’re working in.
In the co-rotating reference frame (the frame which would appear to be stationary, to someone in board the station), he starts off by moving in small circles around the axis of the station (the equivalent, in this system, of an orbit), but air resistance causes his rotational movement to slow, and as his rotational movement is slowed to a stop (that is, matching the station), the centrifugal force starts acting on him, and pulling him to the floor. Note: This is centrifugal, with an f. Centripetal force is a meaningless concept in this reference frame.
In the external inertial reference frame, he starts off hovering in place, a bit off the axis of the station. But there’s a rotating station, full of rotating air, surrounding him. The rotating air exerts a force on him that causes him to start moving laterally. But since he lacks a centripetal (with a p; it is meaningful in this reference frame) force from a floor below him, he doesn’t move around in a circle-- Instead, he starts moving on a tangent to the circle, a path which will eventually intersect the wall of the cylinder.
In the limit of a vacuum, he’d have whatever small velocity he had when he jumped, plus the velocity of the rotating station, which I’d expect to be large.
In the limit of thick air, he’d be pulled along with the air. Since he’s being pulled rotation-ward, he’ll be pulled into the wall/floor of the station somewhere rotationward of where he was initially headed. But since the air is thick in this limiting case, it wouldn’t be very fast, since he won’t move very fast through thick air.
For air at atmospheric pressure, it would be somewhere in between. It’s not clear to me whether it would be near either limiting case, or even whether the speed he hits would have to be within the speed of those two limiting cases. I think you’d have to run the numbers.
If the guy was riding in a vehicle in a circular path and jumped out, wouldn’t his velocity, now tangent to his former circular path, be what it was before, only in a straight line since the centripetal force imparted by the vehicle isn’t holding him in a curved path?
It would look like a straight line to someone who was outside, watching the vehicle turn. But to his friend who is still on the vehicle, it would start off appearing to be a straight line, but then it would curve outwards, and would eventually seem to be along a radius from the axis of rotation.
This is really not at all different from a bullet which starts out cheerfully sitting inside a gun, travelling along a circle just a few feet larger than the circumference of the planet. Then he shoots the gun, and now the bullet seems to be going in a straight line tangential to the circle of previous travel. But it’s not really a straight line, is it? It only looks like a straight line in the beginning, but eventually it becomes an obvious curve towards the center of the planet.
The two cases are no different at all, except that in one gravity is pushing out, and in the other gravity is pulling in.
Anyway, the guy would still fling out towards the outer hull of the station due to momentum, right? He’s going to decelerate a bit, but not much, due to air resistance. I figure the air in the ship would be about sea-level Earth density. I can’t see that slowing the guy down much, but he definitely wouldn’t accelerate towards the outer hull after leaving the tram. Imagine a pole, like a spoke in a wheel, connecting the point of disembarking to a spot directly outwards from the tram. Due to the outer hull going faster than the tram track, he’ll land somewhere behind the pole’s position on the outer hull. This effect is noticeable if a kid on a playground merry-go-round tosses a ball from near the center to the edge.
Here’s one that would be different - get on a motorbike and ride it in the opposite direction to the spin (so you are stationary with respect to the system as a whole), and you will be weightless.
Ride your bike the other way (in the same direction as the cylinder is spinning and you will experience twice as much ‘gravity’
Spacetime curvature describes how gravity affects the paths of different objects in relation to each other. For example it describes how the paths of two free-falling objects which are parallel to each other ‘at infinity’ can converge due to the effects of a large mass. Spacetime curvature therefore doesn’t enter the equation when locally describing the inertial or indeed gravitational force felt by observers.
In general relativity the gravitational/inertial force felt by an observer is really a measure of how much they deviate from a free-falling observer, which can be mathematically described in terms of Christoffel symbols which are not directly dependent on curvature. In both artificial gravity situations described by the OP the force felt by those aboard the spacestation/ship can be seen as a direct result of the observers on board having non-free-falling paths.
Gravitons are really quantitized gravitational waves, so as there’s no gravitational waves involved in artificial gravity there’s no reason to look to them as particularly helpful in describing the situation.