Does artificial gravity based on centrifugal force stop working if you jump off the ground?











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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?










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  • 7




    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
    16 hours ago










  • What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
    – my2cts
    16 hours ago






  • 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
    12 hours ago






  • 2




    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
    11 hours ago






  • 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
    3 hours ago















up vote
12
down vote

favorite












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?










share|cite|improve this question




















  • 7




    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
    16 hours ago










  • What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
    – my2cts
    16 hours ago






  • 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
    12 hours ago






  • 2




    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
    11 hours ago






  • 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
    3 hours ago













up vote
12
down vote

favorite









up vote
12
down vote

favorite











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?










share|cite|improve this question















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






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edited 30 mins ago









knzhou

39.7k9110194




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asked 18 hours ago









Filipp W.

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2066








  • 7




    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
    16 hours ago










  • What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
    – my2cts
    16 hours ago






  • 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
    12 hours ago






  • 2




    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
    11 hours ago






  • 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
    3 hours ago














  • 7




    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
    16 hours ago










  • What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
    – my2cts
    16 hours ago






  • 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
    12 hours ago






  • 2




    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
    11 hours ago






  • 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
    3 hours ago








7




7




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
16 hours ago




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
16 hours ago












What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
– my2cts
16 hours ago




What you could worry about is the effect of tidal force. Even for a big wheel the variation of acceleration, $a=omega r$, is non negligible. You head will weigh less when standing up then when lying down.
– my2cts
16 hours ago




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
12 hours ago




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
12 hours ago




2




2




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
11 hours ago




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
11 hours ago




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
3 hours ago




@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
3 hours ago










5 Answers
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24
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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 reaches 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 rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.






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  • 4




    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
    17 hours ago






  • 1




    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
    17 hours ago






  • 3




    @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
    16 hours ago








  • 4




    @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
    15 hours ago








  • 1




    @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
    – candied_orange
    6 hours ago




















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10
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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!






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    up vote
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    down vote













    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.






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      up vote
      0
      down vote













      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.






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        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.






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          5 Answers
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          up vote
          24
          down vote














          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 reaches 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 rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.






          share|cite|improve this answer

















          • 4




            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
            17 hours ago






          • 1




            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
            17 hours ago






          • 3




            @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
            16 hours ago








          • 4




            @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
            15 hours ago








          • 1




            @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
            – candied_orange
            6 hours ago

















          up vote
          24
          down vote














          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 reaches 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 rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.






          share|cite|improve this answer

















          • 4




            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
            17 hours ago






          • 1




            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
            17 hours ago






          • 3




            @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
            16 hours ago








          • 4




            @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
            15 hours ago








          • 1




            @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
            – candied_orange
            6 hours ago















          up vote
          24
          down vote










          up vote
          24
          down vote










          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 reaches 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 rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.






          share|cite|improve this answer













          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 reaches 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 rotational velocity. You don't lose it just by losing contact with the ground, so nothing about the story changes.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 18 hours ago









          knzhou

          39.7k9110194




          39.7k9110194








          • 4




            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
            17 hours ago






          • 1




            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
            17 hours ago






          • 3




            @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
            16 hours ago








          • 4




            @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
            15 hours ago








          • 1




            @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
            – candied_orange
            6 hours ago
















          • 4




            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
            17 hours ago






          • 1




            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
            17 hours ago






          • 3




            @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
            16 hours ago








          • 4




            @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
            15 hours ago








          • 1




            @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
            – candied_orange
            6 hours ago










          4




          4




          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
          17 hours ago




          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
          17 hours ago




          1




          1




          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
          17 hours ago




          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
          17 hours ago




          3




          3




          @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
          16 hours ago






          @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
          16 hours ago






          4




          4




          @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
          15 hours ago






          @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
          15 hours ago






          1




          1




          @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
          – candied_orange
          6 hours ago






          @simple if you enter the spinning room from an axle having not touched the station you would indeed just float there while the station spins around you. If you enter that room from a rotating spoke you'll find you'll also float but you're body will spin as you float because you'll have the same angular momentum you had when you last touched the station.
          – candied_orange
          6 hours ago












          up vote
          10
          down vote













          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!






          share|cite|improve this answer

























            up vote
            10
            down vote













            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!






            share|cite|improve this answer























              up vote
              10
              down vote










              up vote
              10
              down vote









              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!






              share|cite|improve this answer












              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!







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 16 hours ago









              my2cts

              4,0992417




              4,0992417






















                  up vote
                  1
                  down vote













                  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.






                  share|cite|improve this answer








                  New contributor




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                    up vote
                    1
                    down vote













                    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.






                    share|cite|improve this answer








                    New contributor




                    SweepingsDemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.




















                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      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.






                      share|cite|improve this answer








                      New contributor




                      SweepingsDemon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      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.







                      share|cite|improve this answer








                      New contributor




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                      share|cite|improve this answer



                      share|cite|improve this answer






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                      answered 9 hours ago









                      SweepingsDemon

                      111




                      111




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                          up vote
                          0
                          down vote













                          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.






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote













                            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.






                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              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.






                              share|cite|improve this answer












                              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.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 7 hours ago









                              bobuhito

                              7341511




                              7341511






















                                  up vote
                                  0
                                  down vote













                                  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.






                                  share|cite|improve this answer








                                  New contributor




                                  Asuka Jr. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                    up vote
                                    0
                                    down vote













                                    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.






                                    share|cite|improve this answer








                                    New contributor




                                    Asuka Jr. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.




















                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      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.






                                      share|cite|improve this answer








                                      New contributor




                                      Asuka Jr. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      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.







                                      share|cite|improve this answer








                                      New contributor




                                      Asuka Jr. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      share|cite|improve this answer



                                      share|cite|improve this answer






                                      New contributor




                                      Asuka Jr. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                      answered 6 hours ago









                                      Asuka Jr.

                                      1




                                      1




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                                      New contributor





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