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Artificial gravity in space

  1. May 9, 2013 #1
    Hi

    As I'm sure we're all aware, there are many planned missions to Mars.

    Given its distance (some 3 years away given today's tech) by the time any humans make it there, their muscles would be completely useless on account of wastage as a result of no gravity.

    Therefore, artificial gravity is proposed in the form of a rotating spaceship.

    Though that got me thinking; would it actually work?

    I think it wouldn't.

    Let's say a ship is moving at a constant along a vector (x axis) at 5000 m/s relative to the launch pad. Everyone in that ship, relative to the ship, is moving at 0 m/s. Then someone jumps along the same vector at 1 m/s. That person is then moving at 5001 m/s relative to the launch pad.

    The Captain decides to begin revolving the ship at 10rpm along the x axis.

    During that acceleration period from revolving at 0rpm to 10rpm, everyone is 'stuck to the floor' as the ship has greater velocity than they do.

    And this is where my thinking comes into play.

    The ship has reached its maximum revs/minute along the x-axis, at 10rpm, its velocity still at 5000 m/s.

    Surely anyone 'stuck to the floor' is now ALSO moving 10 m/s relative to the ship and thus if they were to jump at 1/ms would then be moving at 1 m/s relative to the ship thus cancel out the 'artificial-gravity'

    Or in other words, artificial gravity is impossible unless the ship is under constant acceleration which, as we know, is impossible given the effects of mass at great speeds.

    Thoughts?
     
  2. jcsd
  3. May 9, 2013 #2
    Hai Paul Wilson,
    I don't know much about artificial gravity but according to equivalence principle, acceleration is indistinguishable from gravitational field. So if the rocket were to accelerate at a constant rate of 10m/s^2,then the persons in the rocket will feel just like gravity on surface of the earth. So do you think " by the time any humans make it there(mars), their muscles would be completely useless"?
     
  4. May 9, 2013 #3
    Yes,keeping the acceleration constant is impossible for now because no fuel has been invented that can help the rocket reach higher velocity that is required to reach the mars.
     
  5. May 9, 2013 #4
    Indeed if a constant acceleration of 10m/s^2 were POSSIBLE then gravity would be felt. But we all know that that is not possible over such a great distance. The energy does not exist to constantly accelerate at that rate, therefore the ship must stop accelerating at some point in order to A) conserve energy or B) it has completely run out of it.

    The theory I saw proposed revolving at a constant rate to create it. Which I derived to be nonsense. That same theory I described.

    EDIT: Ash got there microseconds before me!
     
  6. May 9, 2013 #5
    How that cancel out artificial gravity? Acceleration is identifiable in every reference frame.
     
  7. May 9, 2013 #6
    How? Explain
     
  8. May 9, 2013 #7
    HAHA!! yeah!
     
  9. May 9, 2013 #8
    What should i explain? acceleration is identifiable in every reference frame?
     
  10. May 9, 2013 #9
    If you are moving at a constant 10m/s in a spaceship and jump at 1m/s you are then moving at 1m/s relative to the spaceship and are therefore no longer under acceleration.

    The theory I saw did not propose constant acceleration, merely a constant velocity.
     
  11. May 9, 2013 #10
    If you are asking how acceleration is identifiable in every reference frame,let me give you an example.

    Suppose an object is moving at 10 meters per second relative to me and 20 meters per second relative to you. IF object were to accelerate and increase its speed to 20 meters per second relative to me,then it would also accelerate and reach at 30 meters per second relative to you. So i know that object accelerated at 10 meters per second per second and you know that object accelerated at 10 meters per second per second. So in any reference frame,object will accelerate at 10 meters per second per second,SO we can say that acceleration is identifiable in every reference frame.

    And if rocket were to accelerate,person inside would feel same as gravity. say in earth that person jumped at 1m/s. wouldn't the person fall to the ground afterwards? Does the jumping cancel out earth's gravity? NO.right? since acceleration is identifiable and acceleration is indistinguishable from gravitational field,same would be the case if rocket were to accelerate.
     
  12. May 9, 2013 #11
    I agree with you on that.that person is moving 1m/s relative to spaceship.

    Sorry,i don't know much about the gravity that is created when an object rotates and i do know that Einstein's general theory of relativity does say that a rotating body can itself create a feeling of gravity. I think other posters would give answer to your post.
     
  13. May 9, 2013 #12

    Nugatory

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    Rotation IS constant acceleration, which is why it does produce a satisfactory artificial gravity.

    Google around for the carnival rides called "Rotor" and "Graviton", or try this article on wikipedia: http://en.wikipedia.org/wiki/Rotor_(ride). These are devices that use rotation to apply constant acceleration to people

    You may object that these devices are spinning but not moving.... but they are moving, at many miles a second, because they're attached to the earth which is orbiting the sun, which is rotating around the Milky Way galaxy, which is itself moving through space.
     
  14. May 9, 2013 #13
    Indeed, I agree, to quote your last sentence, if rocket were to accelerate but I reiterate myself AGAIN, the theory I saw did not propose constant acceleration, it theorised a constant velocity. It was not under constant acceleration but instead a constant velocity.

    Therefore, there would be no artificial gravity to those on board.

    Please re-read; I made it quite clear the ship was not under acceleration relative to the launchpad once it had finished its revolution sequence but was instead moving at a constant rate. Thus there would be no apparent effects of gravity to anyone onboard.
     
  15. May 9, 2013 #14

    PeterDonis

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    This is not correct. Spinning the ship can indeed create an effect that is similar to the effect gravity creates when you stand on the surface of the Earth.

    Suppose the ship is moving at a constant velocity in the x direction (so that there is no linear acceleration present). Suppose the ship is disc-shaped, and is oriented perpendicular to its linear motion (i.e., the outer rim of the ship is a circle in the y-z plane). And suppose the ship is spinning about the x-axis (meaning the x-axis goes through the center of the disc and is perpendicular to it), so it is rotating in the y-z plane.

    Then a person inside the ship can stand on the inner surface of its outer rim just as they could on the surface of the Earth, and feel the same weight there, if the ship's diameter and rotation rate are set appropriately. This is because, relative to someone sitting at the center of the disc (i.e., on the x-axis), a person standing on the outer rim will be accelerating inward, towards the center, because the outer rim of the ship is pushing on them (just as the surface of the Earth pushes up on you when you're standing on it). Note that a person at the x-axis will *not* feel weight; they will be in free fall.
     
  16. May 9, 2013 #15
    Indeed. I didn't take into account the effects of relativity from the Sun, Solar System, etc.

    But do those carnival rides not rely upon the gravity of earth? Which of course, once we leave the Earth's gravitational influence, the Sun's, assuming we are in orbit around it, becomes relatively little if in the same carnival ride...?

    PS, I'm off to bed now. No replies until tomorrow from me.
     
  17. May 9, 2013 #16

    Nugatory

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    Only to the extent that if gravity didn't hold them on the surface of the earth they'd float off into space and then the operators wouldn't be able to make money by selling rides to people on the surface of the earth. The force that pins you to the walls of a Rotor is completely independent of the earth's gravity; it just depends on the diameter and rotational speed of the machine.
     
  18. May 9, 2013 #17

    Wes Tausend

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    Paul,

    Suppose that there was a merry-go-round (round-about) attached in this spaceship. On earth, passengers could ride around on it and experience an addition to the real gravity they were feeling as the ship was parked. They would literally seem to weigh more than their real weight that same morning standing "still" upon a bathroom scale.

    If you have not experienced an aggressive merry-go-round ride as a child (example, someone did not push it so fast you nearly slung off) then you are missing an important part of lifes experiences, one that would namely let you see how rotational centrifugal force emulates that of gravity. The artificial gravity may be stronger or weaker than real gravity (depending on rpms), but there will be the essence just the same.

    On this spaceship merry-go-round, a ride may be taken in outer space when the passenger is normally totally weightless. But when he is spun around, he will be tend to be thrown to the outside, and suddenly the sensation will be familiar; that of gravity. If the merry-go-round is spun at exactly the right rpm, he will seem to weigh exactly what he weighed on earth, only outward rather than downward. But why use a small merry-go-round when a large portion of the ship may be spun. Then everyone can enjoy the artificial gravitational benefits at once.

    The spin is everything in space; the steady forward motion towards Mars makes no difference at all, just like when the ship was steadily parked before take-off. Granted, any ship acceleration will briefly interfere, but in most cases it will simply add to the total artificial gravity. This no different than when we sit in a car and it accelerates. We weigh both the same as real gravity, plus weigh a little more, because of the additional acceleration.

    Were you aware that you seemed to weigh more in an accelerating car? It's true. You could put a swing in the back of a pick-up and stand on a scale placed on the seating board while parked. You will weigh the same as in the house. But when the pick-up takes off, the swing will swing back, and you will measure an additional weight during a period of acceleration (or braking, deceleration!). The same thing would occur if you swung around in a circle from a maypole and weighed yourself with a scale because it is just like a merry-go-round. I hope you can see this now. Good luck in solving this for yourself.

    Wes
    ...
     
    Last edited: May 9, 2013
  19. May 10, 2013 #18
    But rotation is itself constant acceleration. it is indistinguishable from gravity.
     
  20. May 10, 2013 #19
    This is a satisfactory answer. But i have a doubt.

    if we are present in a body(for example a big disk like shaped one) and if it were to rotate,won't the person in that body feel a force pushing outwards(Centrifugal force)?

    isn't that effect opposite of gravity? Isn't the same case as that of introduced by poster?
     
  21. May 10, 2013 #20
    i couldn't understand this point.if the outer rim pushed the person outward how can be accelerating inwards?( isn't Centrifugal force taking here when a body rotates?)
     
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