A ground observer (Ivanova) sees the explosion and says:
*He’s more or less weightless, but the ground is rotating at 60 mph. If we can’t catch him, he’ll be killed by the impact.
*
She doesn’t say what the impact is, but it’s certainly plausible to infer that she means objects rotating into him from the side.
Shoot a rocket eastward at a sufficient speed, and it will be in orbit, weightless. Shoot it westward and you can do the same thing, but it will have to go an awful lot faster. Exactly the same principle.
True, but not quite exactly the same principle, as Earth’s gravity is not a side effect of its rotation.
But if you can’t tell the difference, then that’s exactly what the OP was asking.
Here’s another point: Many people understand that as you get closer to the axis of the spinning space station, the less gravity you feel. But many don’t realize that as you get closer to the center of the planet, there’s less gravity there too. Again, the reasons are totally different, but (as Spock once said) a difference which makes no difference is no difference.
But you can tell the difference.
You can perform experiments that determine Earth exerts gravitational attraction independent of your movement relative to it.
And you can perform experiments that determine the centrifugal gravity of a rotating space station is entirely an effect of its rotation.
Rotating reference frames are not relative.
On earth, gravity pulls you toward the center; you stand on the outer surface. The pull gets weaker the closer you get to the center (for all points beneath the surface). But it also gets weaker the farther you get from the center (for all points above the surface).
On a rotating space station, “gravity” pulls you away from the center; you stand on the inner surface. The “pull” gets weaker the closer you get to the center. But the “pull” gets stronger the farther you get from the center.
Wait, are you saying that even if the apple was previously spinning with the hull it would still float? That doesn’t make sense. It would still have all it’s previous angular velocity, so it’ll get flung outwards when it’s released, no? Whether or not there was air inside doesn’t change that when it’s let go it would quickly hit the hull.
Yes, but. It’s complicated.
The differences between the pseudo-forces* experienced on a spherical mass, a rotating spherical mass and a rotating cylinder can only be discerned if you have a large enough “laboratory” and sufficient time. For example, the Coriolis pseudo-force on Earth can only be discerned over scales on distance-sizes like hurricanes or speed-magnitudes like ballistic projectiles. Absent those, you’d need to be able to run very sensitive experiments over long time intervals to be able to learn that you’re on a rotating vs non-rotating mass.
If you had a cylinder with the radius of the Earth rotating at a speed to cause a 1G acceleration, you’d need similarly large and/or sensitive experiments to discern the source of the pseudo-forces. So, yes, one can always in principle learn the source of local “gravity”, but in practice it is not always easy or simple.
*Pseudo-forces are “forces” caused by using a non-inertial frame of reference (like the room you’re sitting in). Remember also that the “forces” experienced by ballistic objects on Earth do depend on their motion. This can be very counter-intuitive, and becomes very obviously so when trying to understand more extreme ballistic motion like orbits.
No, relative to the space station, he’s going to keep on accelerating (that is, going faster and faster) until he hits the ground.
How is the guy any different from a rock flung from a sling?
I haven’t seen the episode of Babylon 5 in question, but I always wondered about that in that scene in Rendezvous with Rama. It’s taken for granted in the book that close to the center point of the cylinder is a zero-G environment, while the ground level is … 0.8g or whatever it was. Which is why his moon-bike was able to pedal across as long as he didn’t drop too far from the centerline.
It seems like only things directly in contact with the ground would have anything close to full 0.8g. So if you were standing on it, of course you’d feel normal, and the air close to the ground similarly would be accelerated with it and feel normal (since it’s not just a smooth cylinder, it had tons of buildings and structures to push the air around.) But above, let’s be generous, a few hundred feet that force on the air would be diminished to nothing and it’d just be a zero-G environment with very weak air currents… right?
Of course the cheat there is that they later reveal Rama does have some form of gravity-altering technology, but that still doesn’t explain why the astronauts who landed on it don’t point out the obvious problem. And the book doesn’t make it clear that Rama uses the gravity-altering technology to keep things rooted to the “ground”, but they only discuss using it to change the ship’s course.
If you swing a weight around on a rope and let go - it goes flying off on a tangent. this is somehing anyone can relate to.
Similarly, if you are standing on a rotating space platform and drop an object, it will go flying off on a tangent. Of course, you, standing on the platform, will see it appear to rop down away from you in a relatively straight line - which from your point of view, will turn into a curve the longer you watch.
If you stand in the rotating cylinder and throw one way, along the axis direction, you will appear to throw a curve ball. the ball will appear to curve counter-rotation as it flies away from you. Similarly, throw into the rotation, it will fall faster than you’d expect, throw counter rotation, it will drop slower. (Throw just the right speed and absent air resistance, it will float forever…)
This is coriolis “force”. Like centrifugal force, it is not a real force but rather a construct to explain what we see happening to bodies motion if we assume ourselves to be in a standing still frame of reference.
If we were in the Rama cylinder, and we jumped from the half-way ring down to the outside, what actually happens? We’re going around in circles due to our contact with the half-way-down ring. So we’re going aorund at half the speed, experience half the gravity.
Step off the edge, and we are essentially like the ball on a chain let go - we go sailing off in a straight line at the half-speed rate. Meanwhile, the cylinder rotates around you. The guy standing beside you but smart enough not to jump off sees this - you start to drop, the further away you get, the more you spiral counter-rotation… until you hit the floor or get smacked by the wll of a building. (This does not account for any air resistance issues)
the only exception to this is, yes, if you step off in the exact center, there is no momentum taking you away from the center. But the least bit off center, attached to a rotating part, means you have a “sling” effect.
I understand that the cylinder (surface of Rama) would continue to spin beneath you, due to air resistance, but what I don’t understand is what would pull you towards it. But the air resistance would also quickly slow your angular velocity, and you wouldn’t necessarily fall downward when that happens, would you?
What you’re saying makes sense in a vacuum, but it seems to be that you’d step off the central shaft, drift maybe 20-30 feet, then just be kind of stuck there. Right?
No, because being just kind of stuck there would give you a significant speed relative to the rotating air.
And very noticeable when simply tossing a ball.
As the circling air pulls him in the direction of rotation, he’ll begin to fall toward the floor. To someone standing on the floor, he would appear to travel in a spiral out from the center of the cylinder toward the floor.
Bingo. For an even simpler and more obvious example, toss the ball at right angles to the direction of rotation. Toss it to your buddy: to hit him, you’ll have to lead, in the direction of rotation. When he tosses it back, he’ll have to lead in the same direction. The ball will appear to curve in the opposite direction of the
This is why I would use scare quotes. While both gravity and acceleration have the same mass, there’s a huge difference between rotational acceleration and linear acceleration.
If you were in a ship that provided artificial gravity by constant linear acceleration, you couldn’t tell that from gravity by any simple obvious experiements. The best you could do is to measure the G force at the lowest and highest height. The difference would be zero. If you could estimate the probable error in your measurement, that would give you a lower bound on how far away the center of mass of the planet would be, if it’s a planet rather than a spaceship.
Let me try that another way. On the space ship, if it were arbitrarily long, you’d measure the same value of G everywhere. On a planet, the value of G diminishes with the square of the distance from the center of mass (providing the mass is entirely below you. If it’s not, the forumula is different, but it still diminishes as you go up.
oops, I was oversimplistic above. The leading you’d have to do wouldn’t be in the direction of rotation. That’s already imparted by inertia. I guess md2000’s example is better: you’d have to aim high in the forward direction and low in the reverse direction. I’m still trying to figure out what would happen in the orthogonal direction.
For throwing a ball parallel to the axis of rotation, you’d get almost no Coriolis effect. There’d still be some, but only higher-order effects that would be negligible compared to those you’d get from throwing spinward or antispinward.
Again, though, you could just make your space station big enough that you wouldn’t notice it.
The difference is that “can’t distinguish between acceleration and gravity” rule applies to steady acceleration. If you were in a box, you could not tell if you were experiencing gravity or the box was attached to one (linearly) accelerating rocket. Circular motion involves an acceleration that is constantly changing direction (i.e. always perpendicular to direction of motion). What you are detecting with coriolis effects is the change of direction, not the difference between gravity and acceleration. Of course, the change in direction is a giveaway that it is rotational acceleration.
If you throw a ball in an axial direction on Rama, at the rim, what would happen? Same as if you dropped it, but with a additional velocity in the axis direction. Drop a ball - it appears to go straight down. But, the coriolis effect -as it recedes from the center, it has the same velocity as when you let go, but the ground about 4 feet away (sorry, 1.5m) is actually travelling a bit faster. The ball will appear to curve counter-spin as it drops. Probably not a lot, but this is the fictional “coriolis force” at work.
Same idea as Joe jumping off the balcony halfway up the climb to the rim, only the spiral effect and apparent velocity differential, and hence the deflection from apparent vertical, is less.
So to toss the ball to Fred along the axis, you have to throw a bit up so it doesn’t hit the ground before it reaches him. if the ball arcs up, at 10 feet “up” it will be travelling faster than the spinning station and appear to curve spinward. As it appears to come down again (as the station spins to intercept its straight line) on the downward arc it curves counterspinward. Instead of a vertical arc, the trajectory will look to the rim-standing observer like a leaning arc. Anyone who’s played Frisbee will recognize this curve, although the Frisbee does it for a very different reason.
Or so I think…
On planet, gravity is proportional to the square of distance from the center. On a rotating spaceship, gravity is linearly proportional to the distance from the center. Three simple measurements can tell you which one you are on.
But why would the air, hundreds of feet above and not in contact with any moving surface, be rotating? I mean, I concede it might be-- but what force would cause that?
I mean I get it for a small capsule, like the Red Planet thing, but Babylon 5 and Rama are miles in diameter.