Does artificial gravity based on centrifugal force stop working if you jump off the ground?
In an answer to another question of mine, concerning gravity, there was a link to a video about creation of artificial gravity, based on rotation.
The question I have might be silly (or with an obvious answer), but it puzzles me non the less. As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'. If an astronaut walks on the inside of the 'rim' (like here in the video), the contact with the 'rim' is maintained via legs, thus centrifugal force is in action.
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
newtonian-mechanics forces reference-frames free-body-diagram centrifugal-force
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
|
show 14 more comments
In an answer to another question of mine, concerning gravity, there was a link to a video about creation of artificial gravity, based on rotation.
The question I have might be silly (or with an obvious answer), but it puzzles me non the less. As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'. If an astronaut walks on the inside of the 'rim' (like here in the video), the contact with the 'rim' is maintained via legs, thus centrifugal force is in action.
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
newtonian-mechanics forces reference-frames free-body-diagram centrifugal-force
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
31
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
2
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
9
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
1
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
4
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13
|
show 14 more comments
In an answer to another question of mine, concerning gravity, there was a link to a video about creation of artificial gravity, based on rotation.
The question I have might be silly (or with an obvious answer), but it puzzles me non the less. As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'. If an astronaut walks on the inside of the 'rim' (like here in the video), the contact with the 'rim' is maintained via legs, thus centrifugal force is in action.
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
newtonian-mechanics forces reference-frames free-body-diagram centrifugal-force
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
In an answer to another question of mine, concerning gravity, there was a link to a video about creation of artificial gravity, based on rotation.
The question I have might be silly (or with an obvious answer), but it puzzles me non the less. As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'. If an astronaut walks on the inside of the 'rim' (like here in the video), the contact with the 'rim' is maintained via legs, thus centrifugal force is in action.
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
newtonian-mechanics forces reference-frames free-body-diagram centrifugal-force
newtonian-mechanics forces reference-frames free-body-diagram centrifugal-force
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
edited Dec 13 '18 at 11:57
knzhou
42.6k11117205
42.6k11117205
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
asked Dec 12 '18 at 18:18
Filipp W.Filipp W.
32827
32827
31
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
2
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
9
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
1
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
4
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13
|
show 14 more comments
31
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
2
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
9
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
1
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
4
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13
31
31
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
2
2
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
9
9
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
1
1
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
4
4
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13
|
show 14 more comments
9 Answers
9
active
oldest
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Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it hits the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!
The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.
Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge tangential velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
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If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
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show 2 more comments
As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.
Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.
But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)
When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.
From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.
The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.
For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.
Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.
All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
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This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.
The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.
You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
add a comment |
Here's a really simple experiment you can perform right now.
You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.
Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.
Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.
But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.
This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
add a comment |
You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.
So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.
This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
add a comment |
Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.
So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.
However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
add a comment |
Just a simplified example scenario depicting what the angular velocity answers say:
Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.
Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.
When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.
This also means, intuitively, that if you ran with the spin of the station you would get heavier.
Once you see it this way, you can see that jumping wouldn't really do anything in itself.
tl;dr
Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.
Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.
answered Dec 13 '18 at 22:01
Bill KBill K
1215
What is thetl;drdoing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?
– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
add a comment |
While you are standing on the floor of this rotating spacecraft you are actually moving sideways along with the floor.
If you continued sideways in a straight line you'd have to pass through the floor since the floor is rotating in a circle which crosses the straight line of your trajectory. This of course doesn't happen because you are pushing against the floor and the floor is pushing against you. This push is what give the impression of gravity and keeps you rotating along with the floor.
That push between you and the floor is perpendicular to your direction of motion.
Once you jump there will briefly be a greater push than needed to keep you in the circle where floor is moving, so you'll be moving away from the floor. Your acceleration until you leave the floor is still perpendicular to your sideways motion. So the two won't cancel out instead you will be moving in a "diagonal" direction which take you away from the floor.
Once you are no longer touching the floor you will continue along a straight line. However any straight line inside a circle will have to eventually meet the circle again. At that point you will fall to the floor.
If you wanted to experience weightlessness inside of this craft you'd have to cancel out your sideways motion. You don't do that by jumping but rather by walking or running against the direction of rotation until you have canceled out the sideways motion.
While walking or running at the exact right speed you can make a tiny jump to float away from the floor.
If you get it exactly right you'd end up floating inside of the craft which would still be rotating around you. Of course even your tiny jump still gave you a bit of speed in some direction, so eventually you'll hit the floor again which would still be rotating around you.
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
add a comment |
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Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it hits the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!
The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.
Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge tangential velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
|
show 10 more comments
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it hits the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!
The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.
Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge tangential velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
|
show 10 more comments
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it hits the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!
The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.
Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge tangential velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
Now, the question: if, while being inside a rotating space station, astronaut would jump really high, wouldn't he then experience zero gravity until he again will touch some part (wall or floor) of the station? Am I missing something in my understanding?
Well, here's a related question. Suppose you find yourself in an elevator at the top floor of a skyscraper when the cable suddenly snaps. As the elevator plummets down, you realize you'll die on impact when it hits the bottom. But then you think, what if I jump just before that happens? When you jump, you're moving up, not down, so there won't be any impact at all!
The mistake here is the same as the one you're made above. When you jump in the elevator, you indeed start moving upward relative to the elevator, but you're still moving at a tremendous speed downward relative to the ground, which is what matters.
Similarly, when you are at the rim of a large rotating space station, you have a large velocity relative to somebody standing still at the center. When you jump, it's true that you're going up relative to the piece of ground you jumped from, but you still have that huge tangential velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
edited Dec 14 '18 at 16:45
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
answered Dec 12 '18 at 18:26
knzhouknzhou
42.6k11117205
42.6k11117205
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
|
show 10 more comments
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
13
13
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
Another, related example: A hovering helicopter on earth does not see the earth move underneath it with 1000 km/h or so (due to earth rotation), simply because that same helicopter was moving at the same speed with the earth all along.
– user1583209
Dec 12 '18 at 18:36
2
2
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
Please do correct me if I am wrong, but I think here what happens in your explanation is that although you do land on the same spot of the disc after jumping off, it’s less due to the efffect of gravity and more due to the ring rotating to ‘catch’ you. Although you do have a horizontal velocity, when you jump off I think you don’t actually move in a circle, rather you move in a straight line until hitting the ring again. In that sense, OP is technically correct since you aren’t really(assumed gravity negligible) under the effect of some centrifugal force when you jump?
– EigenFunction
Dec 12 '18 at 19:20
10
10
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
@EigenFunction You definitely are under such a force in the rotating frame of reference of the rim of the station. True, an external observer would see you move in a straight line and intersect the rim again with no centrifugal force involved, but that observer never sees a centrifugal force--only the centripetal normal force of the station's rim on your feet.
– eyeballfrog
Dec 12 '18 at 20:28
11
11
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
@EigenFunction You could always say that, though. For example, you could say that gravity on Earth doesn't really exist: a thrown ball always goes in a straight line, but the floor is just constantly accelerating up to catch it. The fact that this is always just as good a description is the content of the equivalence principle, i.e. the foundation of general relativity. You just can't tell the two apart.
– knzhou
Dec 12 '18 at 21:16
8
8
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
You don't have rotational velocity when you jump. You have linear velocity. For a large enough radius and a short enough jump, the difference is negligible but it is certainly not the same. The only thing providing you with artificial gravity is constantly having your mass accelerated @ a 90 degree angle to your velocity.
– Stian Yttervik
Dec 13 '18 at 9:08
|
show 10 more comments
If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
|
show 2 more comments
If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
|
show 2 more comments
If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
If you jump then you are in free fall, apart from air resistance, so you are weightless. This holds for any jump. For a brief moment you experience zero gravity!
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
answered Dec 12 '18 at 19:30
my2ctsmy2cts
4,7432618
4,7432618
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
|
show 2 more comments
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
6
6
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
You may add that unfortunately, your environment continues to move around you. (This is true even when on earth, because the gravity exerted by our planet is indistinguishable from an accelerated movement of our planet towards us. (Hello to all flat-earthers!)
– Peter A. Schneider
Dec 13 '18 at 12:42
1
1
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
On the other hand you could run opposite to the direction of the ships rotation, and find yourself in lower and lower gravity as you went faster. going at the same angular velocity you would indeed find yourself weightless.
– R. Rankin
Dec 13 '18 at 21:56
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 10:52
1
1
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
@gerrit: You should experience a Coriolis effect in any rotating reference frame, including the space station.
– Kevin
Dec 14 '18 at 20:38
1
1
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
@gerrit The test with throwing up a ball / shooting an arrow straight up would be the one I would be using to check whether I'm in a rotating space station, and in which direction it turns. Gravity does not have Coriolis forces, rotating frames of reference do. So, when you emulate gravity with rotation, you inevitably also get unnaturally high Coriolis forces.
– cmaster
Dec 15 '18 at 14:33
|
show 2 more comments
As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.
Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.
But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)
When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.
From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.
The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.
For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.
Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.
All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
|
show 2 more comments
As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.
Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.
But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)
When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.
From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.
The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.
For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.
Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.
All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
|
show 2 more comments
As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.
Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.
But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)
When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.
From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.
The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.
For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.
Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.
All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
As I understand it, in order for the centrifugal force (which is responsible for creating gravity, in this case) to work, object it works upon should be attached to the wheels 'spoke' or 'rim'.
Not exactly. Centrifugal "force" is what's known as a "pseudo-force". It's the result of analyzing events using a non-inertial reference frame. If you're on a merry-go-round, and you treat the merry-go-round as being stationary while the world revolves around it, you will find that objects have a tendency to go towards the outside of the merry-go-round. So within the "The merry-go-round is stationary" point view, you have to posit some force pushing the objects away from the center, which is the centrifugal force.
But this force doesn't "really" exist: when you analyze the situation from the point of view of someone not on the merry-go-round, the objects are traveling in a straight line. It's just that any straight line will necessarily go away from the center. (Draw a circle, then draw straight line next to it. Imagine traveling along that line. From the point of view of the circle, you're first getting closer, then traveling away.)
When something is rotating, its velocity is constantly changing: although its speed is constant, the direction is changing, so the velocity is changing. Changing velocity means acceleration, and acceleration means force. This force is directed towards the center. Imagine driving around a circle counterclockwise. If you were to let go of the wheel, you would fly off the circle. You have to constantly turn left to stay on the circle. So there is a force, but it's towards the center of the circle, and is called the centripetal force.
From an inertial reference frame pint of view, a force is needed to stay on the circle; the centripetal force. But from a circular motion reference frame, the object is stationary. So if there's a centripetal force pulling the object in, there must be another force, the centrifugal force, pushing it out. So if you're standing in a rotating space station, you'll going to feel a force of the floor pushing you "up" towards the center of the space station, and since it feels like you're at rest (the space station is moving with you), it's going to seem that there must be some force pushing you "down" into the floor.
The important point here is that the contact with the floor provides the centripetal force, but the centrifugal force exists in your reference frame regardless of whether you have contact with the floor. Go back to the example of driving in a circle. Suppose you drop a ball in the car. Before you dropped it, it was moving with the car, and so just as the car had a centripetal force keeping in circular motion, the ball had a centripetal force on it. But for the fraction of a second that it's in the air, it doesn't have the centripetal force.
For an outside observer, the car is turning left, while the ball is moving in a straight line. The car accelerates to the left into the ball, and when the ball lands, it is to the right of where it was dropped. For someone in the car, however, it seems like the car is stationary, and ball is accelerating to the right.
Similarly, if you were to jump in the space station, then since you are perceiving things in the space station's reference frame, it will seem like you are accelerating towards the floor. This apparent acceleration exists regardless of whether you're touching the space station. That you are accelerating in the station's reference frame doesn't require physical contact with the station because it's not a physical phenomenon. It's simply an attribute of the coordinate system.
All of this applies locally: if you jump up, your motion will, in the station's reference frame, on small scales be the same as if you were being pulled down by gravity. This is an approximation that breaks down as you go to larger scales. These deviations from the approximation show up as other pseudo-forces, such as the Coriolis force. So being in contact with the floor does matter in that it keeps you moving with the station and reduces these deviations.
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
answered Dec 13 '18 at 17:34
AcccumulationAcccumulation
1,886210
1,886210
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
|
show 2 more comments
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
Maybe worth emphasising the Coriolis force a bit more, and describing the case where someone jumps backwards fast enough to cancel the rotation and so the centrifugal force would still exist but would be exactly balanced by coriolis so they would appear to fly above the floor without falling. (until they hit a wall)
– bdsl
Dec 13 '18 at 17:42
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:01
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
@gerrit As I said, treating the effect as acting like gravity is a local approximation. Things like that are a higher order effect than the ball "accelerating" towards the "ground". For a large space station and human-sized heights, the difference will be rather small. But if we take it to an extreme, obviously if we "drop" a ball from the center of the space station, it won't go anywhere. And I think it's the other way around: for the ball and the station to be at the same angular velocity, the floor must be at a higher linear velocity, so it will travel farther than the ball.
– Acccumulation
Dec 14 '18 at 15:35
2
2
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
+1 This is the only answer which captures the essential fact that answers the question: rotating effects like centrifugal acceleration are caused by the choice of analyzing the problem in a rotating frame. If you analyze it in a non-rotating frame, you get the same actual motion, you just notate it differently.
– Cort Ammon
Dec 15 '18 at 2:19
1
1
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
just a note on 'pseudo' force: When you are on a stationary merry-go-round and the rest of the universe spins around it, the relativistic mass of the galaxies increases (in concentric rings), and since your distance to opposite sides of those rings differs (by your distance from the center) you fall outwards -- it's real gravity then, too.
– amI
Dec 15 '18 at 18:05
|
show 2 more comments
This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.
The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.
You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
add a comment |
This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.
The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.
You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
add a comment |
This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.
The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.
You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
This is an inertia problem. The phrasing of your question presupposes that the astronaut loses inertia when jumping, resulting in their upward trajectory being the only major acting force. At any point while rotating in a stationary position, the inertia of the astronaut is outwards from the rotational path at a 90 degree angle from the line connecting the astronaut and the axis of rotation. When the astronaut jumps, this inertia is maintained and two new forces are generated, namely the force the astronaut enacts on the space station (resulting in negligible acceleration on the part of the space station) and the force the space station enacts on the astronaut (resulting in non-negligible acceleration on the part of the astronaut towards the axis of rotation). The astronaut's relative velocity is related linearly to the vector sum of the inertial force acting on him as well as the force from his jump. This is why the "gravity" felt by the astronaut is determined by the angular velocity of the station floor; greater angular velocity means greater inertia as well as greater centripetal (and centrifugal) force, which means greater force (applied by jumping) is required to accomplish the same (relatively) vertical displacement.
The feeling of being pressed to the ground in an artificial gravity chamber is due to the rotating floor pushing the astronaut towards the rotational axis, when the astronaut (from a physics perspective) wants to keep going along the inertial path. This force resulting in this central acceleration is the centripetal force, while the opposing force enacted by the astronaut on the space station floor is the centrifugal force.
You can find this information in any basic university level physics book; pay special attention to chapters covering angular momentum and gravitational/circular forces. Fundamentals of Physics by David Halliday and Robert Resnick was the one I used in college.
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
answered Dec 13 '18 at 3:03
SweepingsDemonSweepingsDemon
711
711
add a comment |
add a comment |
Here's a really simple experiment you can perform right now.
You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.
Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.
Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.
But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.
This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
add a comment |
Here's a really simple experiment you can perform right now.
You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.
Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.
Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.
But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.
This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
add a comment |
Here's a really simple experiment you can perform right now.
You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.
Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.
Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.
But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.
This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
Here's a really simple experiment you can perform right now.
You are standing on the surface of the Earth which is rotating at, depending on your latitude, somewhere between 1000 and 1600 km/h. If you dare, stand up and jump. Did you suddenly hurdle sideways at a great speed? No. Similarly a lamp hanging from the ceiling isn't thrown violently to one the side despite not touching the ground.
Your momentum is conserved. And momentum is mass x velocity. Your mass doesn't change when you jump, so we can just talk about conserving velocity.
Velocity is a vector meaning it has a magnitude and a direction. Standing still you're traveling at, say, 1200 km/h going sideways, but so is the surface of the Earth and your house and the air so you don't perceive it. When you jump you retain that 1200 km/h sideways velocity and add about 2m/s upward velocity. Since everything else is moving sideways with you, all you perceive is the vertical jump.
But only your linear momentum is conserved. Gravity is keeping you stuck to the rotating Earth. If you were to somehow nullify Earth's gravitational pull you'd find yourself slowly, and then more and more rapidly, seeming to rise up from the surface of the Earth. Your linear momentum is carrying you along in a straight line, but the surface of the Earth is curved. It seems like the surface is dropping away, but without gravity sticking you to the surface you'll travel in a straight line over a curved surface.
This difference between the linear and angular momentum is the Coriolis force and has practical effects on Earth and on a space station. On Earth it's why storms swirl. The much smaller scale on a space station can have subtle effects if you throw a ball, shoot a gun, or even on your inner ear.
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
answered Dec 13 '18 at 21:49
SchwernSchwern
3,16921022
3,16921022
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
add a comment |
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
When I drop a ball on Earth, gravity accelerates it to the ground. When I drop a ball in artificial gravity, the ball conserves momentum from when I release it, whereas the floor of the space station is constantly accelerating toward the centre, leading to circular motion. I would expect my dropped object in artificial gravity to experience linear motion along a chord, and hit the floor/wall where the chord intersects it. But the distance along the chord is a shorter than along the rim, so shouldn't the observer in artificial gravity expect to see the ball fall forward?
– gerrit
Dec 14 '18 at 11:04
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
@gerrit The ball will still fall to the floor, because it is also experiencing centrifugal force when it is released and that is conserved, but it will appear to curve away from the observer. But this is an illusion (or the observer's rotating reference frame). The ball is continuing in a straight line, its linear momentum is conserved. It's the observer who is continuing to rotate away from the ball with the station. The larger the radius of the station the smaller the effect. wikiwand.com/en/Coriolis_force#Simple_cases
– Schwern
Dec 14 '18 at 17:15
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
Right, I didn't immediately make the link between jumping on a space station and storms hemisphere-dependent swirling. But the significance of the former does mean that the Coriolis force is observable for a jumper in artificial gravity, unless the station is truly immense.
– gerrit
Dec 14 '18 at 17:43
add a comment |
You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.
So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.
This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
add a comment |
You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.
So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.
This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
add a comment |
You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.
So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.
This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
You're not making any mistake except for thinking "artificial gravity" could be
nearly constant over a region as big as the structure itself. It really always
applies only to a region "much smaller" than the structure itself.
So, yes, a big jump (up and backwards on the wheel) could send you through the
middle of the wheel, where you would just float. Or, more simply, running fast
enough (backwards on the wheel) will cause you to levitate.
This non-constant variation of your "artificial gravity" in spacetime was already
explained as a "tidal force" in my2cts's comment.
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
answered Dec 13 '18 at 4:43
bobuhitobobuhito
7441511
7441511
add a comment |
add a comment |
Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.
So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.
However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
add a comment |
Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.
So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.
However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
add a comment |
Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.
So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.
However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
Of course, the biggest problem with the supposition of becoming 'weightless' (returning to a null-G state) by making a 'big' jump is this: Assuming the OP was talking about not a created micro-gravity (.2G or less) but closer to one full gravity (Earth standard), is the size of the 'room' required to create that much tidal force. The smaller it is, the faster it would have to be spinning to create sufficient centripetal force to give the impression of that 'gravity', and subsequently the thinner the band of that 'gravity' as the further from the zone of that 'gravity' the less the force.
So unless it were some silly, tiny spinning room, the size of the 'station' required to create a 'natural gravitational' feel would far exceed the astronaut's ability to jump without outside propulsion. In that case, of course, the ability to then travel outside the functional 'gravity' into decreasingly accelerated zones (by removing his own laterally imparted velocity) he could then eventually reach the 0G center of the station or room.
However, with the additional concept of scale, the same could be said of the Earth itself. If you could 'jump' to a 'high enough' height, you could reach 0G (not really, but sufficient micro gravity as to be indistinguishable from 0G). This does give pause for thought, as not only would such a station need to be QUITE large, but that you'd better be wearing a low-pressure suit at the same time, as the atmosphere should be quite thin, since the Nitrogen/Oxygen atmosphere would ALSO be affected by the centripetal force of the station's rotation, thus not only be thin, but definitely not made up of the proper gasses for breathing (lighter gasses rising into the center of spin)... At least that's what I'd expect.
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
answered Dec 13 '18 at 6:21
Asuka Jr.Asuka Jr.
111
111
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
add a comment |
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
Although this is true, I think it answers a different question. I don't think the OP was asking about jumping from the 1G zone to the 0G zone, but rather that as soon as OP jumps, s/he loses contact with the ground and would start to "float" — except that OP is still moving linearly along a chord and will meet the rotating station again soon enough.
– gerrit
Dec 14 '18 at 10:58
add a comment |
Just a simplified example scenario depicting what the angular velocity answers say:
Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.
Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.
When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.
This also means, intuitively, that if you ran with the spin of the station you would get heavier.
Once you see it this way, you can see that jumping wouldn't really do anything in itself.
tl;dr
Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.
Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.
answered Dec 13 '18 at 22:01
Bill KBill K
1215
What is thetl;drdoing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?
– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
add a comment |
Just a simplified example scenario depicting what the angular velocity answers say:
Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.
Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.
When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.
This also means, intuitively, that if you ran with the spin of the station you would get heavier.
Once you see it this way, you can see that jumping wouldn't really do anything in itself.
tl;dr
Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.
Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.
answered Dec 13 '18 at 22:01
Bill KBill K
1215
What is thetl;drdoing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?
– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
add a comment |
Just a simplified example scenario depicting what the angular velocity answers say:
Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.
Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.
When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.
This also means, intuitively, that if you ran with the spin of the station you would get heavier.
Once you see it this way, you can see that jumping wouldn't really do anything in itself.
tl;dr
Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.
Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.
answered Dec 13 '18 at 22:01
Bill KBill K
1215
Just a simplified example scenario depicting what the angular velocity answers say:
Assume a large open cylinder set up as a jogging track that is an entire rotating "Ring" of the a space station.
Instead of jumping, run as fast as the track section is rotating but in the opposite direction. You will appear to lose weight as you accelerate since you will no longer have the angular momentum of the floor but something less than that.
When you reach the speed the station is spinning, you will be weightless and lift off the ground (if you don't lose traction first). While above the ground in this way you will be weightless.
This also means, intuitively, that if you ran with the spin of the station you would get heavier.
Once you see it this way, you can see that jumping wouldn't really do anything in itself.
tl;dr
Problem 1:
After you become weightless the air is probably rotating with the floor of the track, so it will push your body giving it some angular momentum again at which point you will be "Pulled" back to the surface. If being able to alter your weight was a partial goal of this exercise room they could make the floor and walls fairly smooth so that the air stayed (mostly) in place and ignored the spin altogether. If this were the case you'd start with a wind at your back, but the faster you ran the weaker the wind would become and when you became weightless there would be no air pushing you.
Problem 2: In order to run the speed of the station the starting gravity may have to be less than 1g, Not sure of the math at all, but I am sure with the right starting speed you could run until you were weightless.
answered Dec 13 '18 at 22:01
Bill KBill K
1215
edited Dec 13 '18 at 22:15
answered Dec 13 '18 at 22:01
Bill KBill K
1215
answered Dec 13 '18 at 22:01
Bill KBill K
1215
answered Dec 13 '18 at 22:01
Bill KBill K
1215
1215
What is thetl;drdoing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?
– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
add a comment |
What is thetl;drdoing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?
– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
What is the
tl;dr doing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?– gerrit
Dec 14 '18 at 11:02
What is the
tl;dr doing in the middle of your answer? The parts before and after are almost equally long. Which part is meant to be the summary?– gerrit
Dec 14 '18 at 11:02
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
The second part is more "issues" that could be brought up--arguing with myself as I was thinking but not really needed except for ancillary support.
– Bill K
Dec 14 '18 at 17:05
add a comment |
While you are standing on the floor of this rotating spacecraft you are actually moving sideways along with the floor.
If you continued sideways in a straight line you'd have to pass through the floor since the floor is rotating in a circle which crosses the straight line of your trajectory. This of course doesn't happen because you are pushing against the floor and the floor is pushing against you. This push is what give the impression of gravity and keeps you rotating along with the floor.
That push between you and the floor is perpendicular to your direction of motion.
Once you jump there will briefly be a greater push than needed to keep you in the circle where floor is moving, so you'll be moving away from the floor. Your acceleration until you leave the floor is still perpendicular to your sideways motion. So the two won't cancel out instead you will be moving in a "diagonal" direction which take you away from the floor.
Once you are no longer touching the floor you will continue along a straight line. However any straight line inside a circle will have to eventually meet the circle again. At that point you will fall to the floor.
If you wanted to experience weightlessness inside of this craft you'd have to cancel out your sideways motion. You don't do that by jumping but rather by walking or running against the direction of rotation until you have canceled out the sideways motion.
While walking or running at the exact right speed you can make a tiny jump to float away from the floor.
If you get it exactly right you'd end up floating inside of the craft which would still be rotating around you. Of course even your tiny jump still gave you a bit of speed in some direction, so eventually you'll hit the floor again which would still be rotating around you.
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
add a comment |
While you are standing on the floor of this rotating spacecraft you are actually moving sideways along with the floor.
If you continued sideways in a straight line you'd have to pass through the floor since the floor is rotating in a circle which crosses the straight line of your trajectory. This of course doesn't happen because you are pushing against the floor and the floor is pushing against you. This push is what give the impression of gravity and keeps you rotating along with the floor.
That push between you and the floor is perpendicular to your direction of motion.
Once you jump there will briefly be a greater push than needed to keep you in the circle where floor is moving, so you'll be moving away from the floor. Your acceleration until you leave the floor is still perpendicular to your sideways motion. So the two won't cancel out instead you will be moving in a "diagonal" direction which take you away from the floor.
Once you are no longer touching the floor you will continue along a straight line. However any straight line inside a circle will have to eventually meet the circle again. At that point you will fall to the floor.
If you wanted to experience weightlessness inside of this craft you'd have to cancel out your sideways motion. You don't do that by jumping but rather by walking or running against the direction of rotation until you have canceled out the sideways motion.
While walking or running at the exact right speed you can make a tiny jump to float away from the floor.
If you get it exactly right you'd end up floating inside of the craft which would still be rotating around you. Of course even your tiny jump still gave you a bit of speed in some direction, so eventually you'll hit the floor again which would still be rotating around you.
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
add a comment |
While you are standing on the floor of this rotating spacecraft you are actually moving sideways along with the floor.
If you continued sideways in a straight line you'd have to pass through the floor since the floor is rotating in a circle which crosses the straight line of your trajectory. This of course doesn't happen because you are pushing against the floor and the floor is pushing against you. This push is what give the impression of gravity and keeps you rotating along with the floor.
That push between you and the floor is perpendicular to your direction of motion.
Once you jump there will briefly be a greater push than needed to keep you in the circle where floor is moving, so you'll be moving away from the floor. Your acceleration until you leave the floor is still perpendicular to your sideways motion. So the two won't cancel out instead you will be moving in a "diagonal" direction which take you away from the floor.
Once you are no longer touching the floor you will continue along a straight line. However any straight line inside a circle will have to eventually meet the circle again. At that point you will fall to the floor.
If you wanted to experience weightlessness inside of this craft you'd have to cancel out your sideways motion. You don't do that by jumping but rather by walking or running against the direction of rotation until you have canceled out the sideways motion.
While walking or running at the exact right speed you can make a tiny jump to float away from the floor.
If you get it exactly right you'd end up floating inside of the craft which would still be rotating around you. Of course even your tiny jump still gave you a bit of speed in some direction, so eventually you'll hit the floor again which would still be rotating around you.
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
While you are standing on the floor of this rotating spacecraft you are actually moving sideways along with the floor.
If you continued sideways in a straight line you'd have to pass through the floor since the floor is rotating in a circle which crosses the straight line of your trajectory. This of course doesn't happen because you are pushing against the floor and the floor is pushing against you. This push is what give the impression of gravity and keeps you rotating along with the floor.
That push between you and the floor is perpendicular to your direction of motion.
Once you jump there will briefly be a greater push than needed to keep you in the circle where floor is moving, so you'll be moving away from the floor. Your acceleration until you leave the floor is still perpendicular to your sideways motion. So the two won't cancel out instead you will be moving in a "diagonal" direction which take you away from the floor.
Once you are no longer touching the floor you will continue along a straight line. However any straight line inside a circle will have to eventually meet the circle again. At that point you will fall to the floor.
If you wanted to experience weightlessness inside of this craft you'd have to cancel out your sideways motion. You don't do that by jumping but rather by walking or running against the direction of rotation until you have canceled out the sideways motion.
While walking or running at the exact right speed you can make a tiny jump to float away from the floor.
If you get it exactly right you'd end up floating inside of the craft which would still be rotating around you. Of course even your tiny jump still gave you a bit of speed in some direction, so eventually you'll hit the floor again which would still be rotating around you.
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
answered Dec 15 '18 at 16:25
kasperdkasperd
101115
101115
add a comment |
add a comment |
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31
For an experimental exploration of answers to this question, take three friends and a dodgeball to your nearest playground with a merry-go-round. Get the merry-go-round up to speed, then have the folks who are riding it try to play "catch" with the dodgeball. Take turns observing from off the merry-go-round to see the difference between the rotating frame and the inertial frame. What happens is quite surprising, even if you make a prediction on paper beforehand, because the intuition you use to play "catch" is not well-adapted to rotating reference frames.
– rob♦
Dec 12 '18 at 20:02
2
It seems to work just fine. Space Station Centrifuge Gravity Simulation 196x NASA color 3min The word jump needs to be in the title.
– Mazura
Dec 13 '18 at 0:27
9
While jumping may remove you from the force imparted by the "floor" of the rotating drum, consider that you have momentum and will continue to move. The direction of this movement will be tangent to the curve of the rotating drum in the direction of rotation. So while you will not "fall" straight back down to the floor, you will drift forwards with your current momentum and... bump right back into the floor.
– Blackhawk
Dec 13 '18 at 0:56
1
@Anoplexian if the acceleration is temporarily zero, the velocity will be temporarily constant. I think you're speaking of the moment at which the velocity is temporarily zero, which is not the same as the acceleration being temporarily zero.
– phoog
Dec 13 '18 at 9:26
4
@DavidRicherby: During any jump, no matter how high (and neglecting air resistance), you're in free-fall, which is what we colloquially call "zero gravity".
– Dave Tweed
Dec 13 '18 at 15:13