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Artificial gravity ship: Floor?

  1. Jun 19, 2013 #1
    The other day, I was learning about creating "artificial" gravity by building a gigantic circularly shaped ship which would revolve at a frequency such that the centripetal acceleration coincided with the acceleration due to gravity [itex]g[/itex] (analogous to the shape and movement of earth).

    I can definitely buy this: Surely it makes sense that the ship revolves similar to the earth and thus gives rise to a centripetal acceleration toward the center of the ship which would be equal to [itex]g[/itex]. But then I saw a picture similar to the one below and immediately got confounded:

    http://www.school-for-champions.com/science/images/gravity_artificial.gif

    Since the revolution of that circular "ship" gives rise to the acceleration [itex]g[/itex] toward the center of the ship, wouldn't that person launch towards and "slam" into the center of the ship? According to me, it should look something like this:

    BrupV76.png

    Why? Well since the movement and shape of the ship is mimicing that of earth's, I figured the placement of the human beings ought to be similar. For comparison:

    HHnzc8C.jpg

    Is there something I am not seeing clearly?
     
    Last edited: Jun 19, 2013
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  3. Jun 19, 2013 #2

    mathman

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    The centripetal force is what holds him to the floor. What he is experiencing is centrifugal "force". If the floor wasn't there he would fly off on a tangent to the circle.
     
  4. Jun 19, 2013 #3

    Bandersnatch

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    In the first picture, if the floor suddenly dissapeared, the man would float away in a straight line, as per Newton's First law of motion. This means that for him to move in a curved(circular) path, some force has to constantly push him towards the centre, and that force is the reaction force of the floor on his feet. It does the job of the centripetal force, and in the reference frame glued to the man, he'd perceive it as a centrifugal force mimicking gravity(that is, in the rotating space station, as on Earth, the floor is pushing us up)

    If you were to use the second picture, there would be no force to keep the man glued to the sphere. He would indeed float away with constant velocity equal to the tangential velocity he had at that instant.

    The third picture does allow people to stand on the ground, because there exists a force, absent from the previous cases, i.e., gravity, which keeps the people glued to the floor.
     
  5. Jun 19, 2013 #4
    Ok, I can buy that but suppose he were to jump a little straight up: How is the centripetal acceleration then accelerating him back to the floor? I understand that he will continue in the circular motion due to inertia and therefore not change his position relative to the floor (right?), but I do not understand how the "centrifugal" force throws him back to the exact same position on the floor. Shouldn't he be heading towards the ship's center?

    Yes, sounds sensible.

    Ok so what we essentially need to make sure of is that there is some "surface" which can disable the human beings from escaping the circular motion, hence positioning people according to the first picture?

    I know that due to the earth's huge mass, its inherent gravitational attraction relative to us humans is superstrong and therefore keeps us "glued" to its surface. But is this gravitational force not considered as a centripetal one? Would not the ship similarly, due to its size, have this gravitational force attract the human beings towards its center? I mean, why don't we get thrown off tangentially off earth but would do that in the case of the ship?
     
  6. Jun 19, 2013 #5

    Bandersnatch

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    Sure, but in my opinion it's better to think of forces that deviate the object(or people) from what it does "naturally", that is, moving in a straight line at constant velocity.
    The curved surface is merely one of the ways of providing that force. There can be other ways, like gravity, or tension of a string, or magnets.
    The point is, that if you want to achieve circular motion, there has to be some force acting as the centripetal force.

    The Earth weighs so much more than any sensibly imagined space station ever could(six billion trillion metric tons versus e.g. ISS's 450 tons), that the force of gravity is completely negligible for the space station. If it weren't, then there would be no need to do the rotation trick to simulate gravity.
    I recommend doing the calculations yourself. You only need Newton's law of gravity here, and all values can be found out from wikipedia(for Earth) or made up(for space station; within reason).

    There is an interesting thing to note here. The force of gravity on people on Earth's surface is more than what is needed to keep us going in circular motion(and people on the poles do not move in circles anyway). If not for the surface pushing us up, we'd all fall down towards the centre of the planet.
    That's why for us to feel as if there was gravity, we need the floor to push us up, because that's what happens on Earth. This is exactly what the floor of the rotating space station does in the first picture.
     
    Last edited: Jun 19, 2013
  7. Jun 19, 2013 #6

    Nugatory

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    No, he will continue traveling in a straight line, because of inertia, and this straight line will carry him back to the "floor".

    Look at your first picture, the one where the guy is standing on the "floor" at about the 11:00 position. At the exact moment that picture shows, he is moving up and to the right because that's the direction that the but of "floor" he's standing on is moving. His jump lifts him off the "floor" but he's still moving up and to the right, so his path carries him back to the "floor" again. In fact, if you do the math (assuming that the force of the jump is small compared with the centrifugal force) he'll come down at the same place - the ring rotates at just the speed that makes that spot come around to where he lands.
     
  8. Jun 20, 2013 #7
    What would happen if the man were to run in the opposite direction of rotation, at the same speed to which the circular ship was rotateing, then jumped towards the centre.
     
  9. Jun 20, 2013 #8
    He would see the floor moving under him while he 'floats' towards the centre.
    It would be the same situation as if an object was just resting ( from the point of view of an observer removed from the ship ) just above ( in the direction towards the centre ) the floor and then moved towards the centre.
     
  10. Jun 22, 2013 #9
    Understand that this is not a perfect means of producing artificial gravity. Under certain conditions objects will behave differently then then they do under the influence of real gravity. You have cited one scenario. With a little imagination you can come up with several more.
     
  11. Jun 22, 2013 #10
    Does he not have to move in a parabolic trajectory relative to the "floor" in order for him to land back on the same spot? Or were you meaning something like this:
    ukFJzBy.png
    He has the tangential speed represented by the green arrow, which is the same speed he has mid-air when he jumps straight up (the first transparent clone of him), and begins travelling in a straight-line fashion in the direction of the arrow.

    In the meantime, the spot from which he jumped up (the red spot) keeps moving in the circular path and meets up with the him once he comes back in contact with the "floor" of the ship. I wonder why he ought to land with his legs perpendicular to the floor? After all he was travelling in a straight-line, what could possibly have rotated him and enabled him land perfectly on his feet, as if he had jumped straight up and landed back down again on a ground here on earth?

    Hm, I am not quite sure how the math ought to look like, care to show how? We can find an expression for his tangential speed with which we can model his straight-line motion after the jump. However I am not sure how I, in terms of equations, can express the red spot and somehow through algebraic manipulations show that he would land back on it (given the premises of the scenario).

    So he would not land back on the same position from which he jumped off? As an analogue: It would be like you were running in the opposite direction on a very long treadmill; you would obviously not land back on the same position from which you jumped off. Or am I just confused?

    Thanks a bunch everybody for the replies!
     
  12. Jun 22, 2013 #11

    Nugatory

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    That's right. The artificial gravity produced by rotation only works as long as the objects in the space station are moving around at speeds that are small compared to the rotational velocity of the rim, and running fast enough to cancel out the velocity completely certainly doesn't meet that criterion. As soon as you consider very high jumps, or the trajectories of objects thrown across the station so that they hit the other side, or fast-moving objects, the artificial gravity stops looking so much like the real thing.
     
    Last edited: Jun 22, 2013
  13. Jun 22, 2013 #12

    Nugatory

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    Assume for the sake of argument that he starts his jump at exactly the 12:00 position (there's nothing special about that position except that it's easy to write down his speed at that moment so the math is easier).

    At that moment, before he jumps, his velocity vector is ##\widehat{v}=v_0\widehat{x}##, and after he jumps straight "up" with speed ##v_j## it is ##\widehat{v}=v_0\widehat{x}-v_j\widehat{y}## where ##v_0## is the speed of the rim of the space station and ##\widehat{x}## and ##\widehat{y}## are unit vectors in the positive x and y directions.

    Therefore, if he jumps at time t=0, his position at time t will be:
    [tex]x(t)=v_0{t}[/tex][tex]y(t)=R-v_j{t}[/tex]
    where ##R## is the radius of the space station.

    Now consider the trajectory of the red dot. It's going to be:
    [tex]x_{dot}=sin(v_0{t})\approx{v_0{t}}[/tex][tex]y_{dot}=R-cos(v_0{t})\approx{R({1-(v_0{t})^3})}[/tex]
    where the approximations come from the observation in post #11 that this entire artificial gravity thing only works when the speed and force of the jump is small. Because ##x_{dot}(t)=x(t)## we see that the jumper and the dot always have the same x coordinate - when the jumper meets the rim the dot will still be right under his feet.

    Play with this some more, and you'll be able to show that the distance between the jumper and the rim describes the parabola that you'd expect.
     
    Last edited: Jun 22, 2013
  14. Jun 23, 2013 #13
    Ok! Interesting, the concept is not as flawless as one initially imagined.

    Thank you for that mathematical exhibition, I'll have a thorough look at it later. However, you write "you'll be able to show that the distance between the jumper and the rim describes the parabola that you'd expect": Is it really parabolic? As the person jumps off, he would travel in a straight-line motion right? So how does it end up being parabolic? See my post, #10, for further details.
     
  15. Jun 23, 2013 #14

    Nugatory

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    You're right, I was trying to say that you'll get the right quadratic distance from jumper and rim, the exact same result as you'd get for a person jumping off the ground in the earth's gravity. That formula for the distance leads to a parabolic trajectory if he jumps off at angle, just as an object thrown on earth will.

    BTW, I misplaced a factor of R in my equation for ##y_{dot}##. You can fix it for yourself if you try playing with the math - the mistake will be pretty obvious.
     
  16. Jun 24, 2013 #15
    my opinion about this

    i think the key to all this is the friction,which everybody ignores...
    so being earth or artificial ship,the bodies exert a friction which,acc.to the action-reaction Newton law is directed against the spinning;
    consequently,a reactive force appears to bodies,due to the conservation of the rotational moment of earth or ship;this force is opposite to friction and PERPENDICULAR on the spinning plane,acc.to the Noeter law...
    now you could compose yourself the spin dragging force and this force and you`ll obtain a force directed to the center of the earth,which is born from the rotation;
    another imaginary force(=gravity)is no longer necessarily,we manage with those;
    conclusion:gravitation doesn`t exists,mechanics is enough...
     
  17. Jun 24, 2013 #16

    sophiecentaur

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    Not "exactly". The reason for a parabolic trajectory under gravity is that there is constant acceleration. In the frame of the jumper, on earth, the ground will also be following a parabolic course. For someone jumping off the floor on the rotating spacecraft, the sums are different. They will be travelling at uniform velocity (no acceleration) during the flight but the floor will be moving on a circular path with constant magnitude acceleration towards the axis and constant angular velocity. The coriolis force is (one way) how the effect of jumping in the spacecraft is modelled.
    One noticeable difference in trajectories will be that jumping 'radially' ln the spacecraft will cause you to land at a different spot, whereas jumping vertically on Earth will bring you down where you started.
     
  18. Jun 24, 2013 #17

    sophiecentaur

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    Re: Coriolis force.
    Our balance mechanisms are very sensitive and I wonder whether the coriolis force would be sufficient to upset the balance of a passenger, getting up out of a seat, for instance, as the radius of rotation would be different. How big would the craft need to be to eliminate this effect, I wonder.
     
  19. Jun 24, 2013 #18
    Maybe you could try going from a crouched position to standing on a fairground ride.The type of ride that starts off in horizontal and when going fast enougth ends up vertical with all the occupants pinned to the side.I know Earth's gravity will still be present but it might give some indication.It makes me sick just to look at one.:surprised
     
  20. Jun 24, 2013 #19

    Nugatory

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    It's exact in the solution I posted, where I made the small-angle approximation that ##sin\theta\approx\theta## and ##cos\theta\approx{1-\frac{\theta^2}{2}}## where ##\theta=\omega{t}## is the angular change during the jump and ##\omega## is the angular speed of the station.

    If you can't make that small-angle approximation, the artificial gravity illusion becomes a lot less convincing - for example, you can jump high enough to "fall" into "the sky", and local pseudo-tidal effects become significant.
     
  21. Jun 24, 2013 #20

    Nugatory

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    Fairly big, I'd guess.

    At a radius of 100 meters, we'd need the rim to be moving at about 30 m/sec to produce 1g; that's about 2 rpm and I'm already starting to feel seasick. Stand up, moving your inner ear one meter closer to the center, and the acceleration would change by about 1%.... I'd be amazed if that were not both noticeable and nausea-inducing... But I will happily defer to anyone who has real data about the sensitivity of the inner ear (or who notices that I dropped a few factors of ten because it's late and I'm tired).
     
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