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Variation of gravity in a Rotating SpaceStation

  1. May 23, 2012 #1
    This is very simple question, and i just need a 2nd opinion.

    We have a Space Station (preferably a torus) with angular velocity ω and radius r. We have a car inside which OPPOSES the angular velocity and moves with the speed ωr . So, will the gravity felt in this car be Zero? Or will it be something else?

    If we replace the car with Usain Bolt, and ask him to run at the speed of ωr , opposing Angular velocity, then, what will be the change in this case?
  2. jcsd
  3. May 23, 2012 #2


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    Nothing. Both will have problems propelling themselves at such a speed, because they will loose traction at the outer wall.
  4. May 23, 2012 #3


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    in case you're wondering …

    in the rotating frame of the spaceship, there is still a centrifugal force on the car of mΩ2r outward, but there's also the velocity-dependent Coriolis force of 2mΩ2r inward, net force of mΩ2r inward, supplying the centripetal acceleration which explains why the car appears to be rotating with speed Ωr even though there's no reaction force from the "floor" of the spaceship! :biggrin:
  5. May 23, 2012 #4
    Assume that the station is rotating to re-create gravitational effects on Earth.
    ie the rotation is such that a 1kg mass placed on bathroom scales on the 'floor' of the station registers 10N
    In the context of this post can anyone explain what the reading on bathroom scales would be when
    a) a 1kg mass is placed on bathroom scales placed in the car when it is 'staionary'
    b) a 1kg mass is placed on bathroom scales placed in the car when it is being driven around the space station....what happens to the reading.... greater? smaller, .... the same?
  6. May 23, 2012 #5


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    That is answered in the question, isn't it? 10N
    Depends in which direction and how fast it drives, the centrifugal force will be modified by the Coriolis force.
  7. May 23, 2012 #6
    Absolutely correct for the first part.
    so... suppose the car is travelling at the same speed as the ring when it is simulating gravity.... what will the bathroom scales read???
    (in each direction)
  8. May 23, 2012 #7


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    0N for opposite direction and 20N for outrunning the rotation
  9. May 23, 2012 #8
    if you are standing in such a space station how can you tell which way it is rotating?
  10. May 23, 2012 #9


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    You just proposed the experiment that tells you the direction. If you mean just by standing still, you can't. But if you can move your arms, the Coriolis force will tell you how it rotates.
  11. May 23, 2012 #10
    So if I am standing 'still' in this space station I cannot tell which way it is rotating unless I use the coriolis force on my arms ?? How will the coriolis force reveal the direction of rotation.?
  12. May 23, 2012 #11


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    Just like in your scales example. You can do a similar test by swinging your arms and see which way they appear heavier. Maybe while holding a heavy object in your hand.
  13. May 23, 2012 #12
    That is not an acceptable answer !!!! wave your arms around.... come on
    HOW can you determine, 'in a physics way', which way the station is rotating?
  14. May 23, 2012 #13
    That is 'in a physics way.' If you mean in a controlled way, put to accelerometers (to measure gravity) into two cars, have them move in opposite directions, see which one measured the least acceleration.
  15. May 23, 2012 #14
    'Accelerometers to measure gravity'.... that is what I mean by bathroom scales !!!!
    How would they tell you the direction of rotation??....if they are on a moving car?
    Last edited: May 23, 2012
  16. May 23, 2012 #15


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    See post #7
  17. May 23, 2012 #16
    I could not fully understand post 3, does your post agree with post 3?
    I have to explain these things to 16 and 17 year old students.
    Any help in this are would be valuable.
  18. May 23, 2012 #17


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  19. May 23, 2012 #18
    If you hold a ball in your hand in this space station and let it go what will happen?
  20. May 23, 2012 #19


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    It will accelerate according to the sum of centrifugal and Coriolis forces.
  21. May 23, 2012 #20


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    I'm not sure of the precise scenario that your answer comes from but, if the car is doing the same 'speed' over the surface as the station peripheral speed then its resultant peripheral speed will be twice that of the station (?). The acceleration is given by v2/r so you would expect to measure 40N, not 20N.
  22. May 23, 2012 #21


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    In order to explain something, you need to understand it yourself at least one level higher, so that's a serious problem right there.

    So first, lets try to explain it to you on a level above of what you'll need to explain to students.

    Suppose, I have an object in an inertial frame of reference. For inertial frame of reference, we have Newton's Second Law.

    [tex]F = ma[/tex]

    Now, acceleration is the second derivative on position. Furthermore, it is a vector. For a rotation problem 2D is sufficient. So let me rewrite it.

    [tex]F_x = m \frac{d^2x}{dt^2}[/tex]
    [tex]F_y = m \frac{d^2y}{dt^2}[/tex]

    In a rotating reference frame, it's much easier to use polar coordinates.

    [tex]x = r cos(\theta)[/tex]
    [tex]y = r sin(\theta)[/tex]

    Lets rewrite the equations for forces again.

    [tex]F_x = m \frac{d^2}{dt^2}\left(r cos(\theta)\right)[/tex]
    [tex]F_y = m \frac{d^2}{dt^2}\left(r sin(\theta)\right)[/tex]

    Taking derivative once.

    [tex]F_x = m \frac{d}{dt}\left(\frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt}\right)[/tex]
    [tex]F_y = m \frac{d}{dt}\left(\frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt}\right)[/tex]

    Taking derivative twice.

    [tex]F_x = m \left(\frac{d^2r}{dt^2} cos(\theta) - 2 \frac{dr}{dt} sin(\theta)\frac{d\theta}{dt} - r cos(\theta) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta)\frac{d^2\theta}{dt^2}\right)[/tex]
    [tex]F_y = m \left(\frac{d^2r}{dt^2} sin(\theta) + 2 \frac{dr}{dt} cos(\theta)\frac{d\theta}{dt} - r sin(\theta) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta)\frac{d^2\theta}{dt^2}\right)[/tex]

    Suppose, now, that you are standing on the station rotating around r=0 with angular velocity ω. Suppose that you measure all angles and x/y relative to yourself. I'll call these θ', x', and y' coordinates. Clearly θ=θ' + ωt. (To within a choice of origin for θ) The equations for x' and y' are same polar equations as before.

    [tex]x' = r' cos(\theta ')[/tex]
    [tex]y' = r' sin(\theta ')[/tex]

    The r'=r, of course, since we chose the coordinate system with the same center, and the center of rotation does not move.

    Let us substitute θ=θ' + ωt into force equations.

    [tex]F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\frac{d\theta}{dt} - r cos(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 - r sin(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)[/tex]
    [tex]F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\frac{d\theta}{dt} - r sin(\theta' + \omega t) \left(\frac{d\theta}{dt}\right)^2 + r cos(\theta' + \omega t)\frac{d^2\theta}{dt^2}\right)[/tex]

    Of course, we can also expand dθ/dt the same way.

    [tex]F_x = m \left(\frac{d^2r}{dt^2} cos(\theta' + \omega t) - 2 \frac{dr}{dt} sin(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r cos(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 - r sin(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)[/tex]
    [tex]F_y = m \left(\frac{d^2r}{dt^2} sin(\theta' + \omega t) + 2 \frac{dr}{dt} cos(\theta' + \omega t)\left(\frac{d\theta'}{dt}+\omega\right) - r sin(\theta' + \omega t) \left(\frac{d\theta'}{dt} + \omega\right)^2 + r cos(\theta' + \omega t) \frac{d^2\theta'}{dt^2}\right)[/tex]

    Now, I'm going to simplify a bit, using the fact that x/r = cos(θ).

    [tex]F_x = m \left(\frac{d^2r}{dt^2}\frac{x}{r} - 2 \frac{dr}{dt}\frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y}{r} \frac{d^2\theta'}{dt^2}\right)[/tex]
    [tex]F_y = m \left(\frac{d^2r}{dt^2}\frac{y}{r} + 2 \frac{dr}{dt}\frac{x}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x}{r} \frac{d^2\theta'}{dt^2}\right)[/tex]

    Since x/r and y/r simply give us directions, we go over completely to primed coordinates.

    [tex]F_x' = m \left(\frac{d^2r}{dt^2}\frac{x'}{r} - 2 \frac{dr}{dt}\frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 - r^2 \frac{y'}{r} \frac{d^2\theta'}{dt^2}\right)[/tex]
    [tex]F_y' = m \left(\frac{d^2r}{dt^2}\frac{y'}{r} + 2 \frac{dr}{dt}\frac{x'}{r}\left(\frac{d\theta'}{dt}+\omega\right) - r \frac{y'}{r}\left(\frac{d\theta'}{dt}+\omega\right)^2 + r^2 \frac{x'}{r} \frac{d^2\theta'}{dt^2}\right)[/tex]

    And we extract from it the parts that come directly from differentiating x' and y'.

    [tex]F_x' = m \left(\frac{d^2 x'}{dt^2} - 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)[/tex]
    [tex]F_y' = m \left(\frac{d^2 x'}{dt^2} + 2 \frac{dr}{dt} \frac{y'}{r} \omega - 2 r \frac{x'}{r} \frac{d \theta '}{dt}\omega - r \frac{x'}{r}\omega^2\right)[/tex]

    The above already contains the centrifugal and Coriolis terms. In vector form, it can be equivalently written like this.

    [tex]F = m(a' + 2 \omega \times v - \omega^2 r \hat{r})[/tex]

    So long as you define vector ω as vector perpendicular to x/y plane.

    Finally, suppose you want to continue using Newton's 2nd in this coordinate system. Then you would have to define.

    [tex]F' = F - 2m\omega \times v + m\omega^2 r \hat{r}[/tex]

    Where F is the real forces acting on the object, and the two additional terms are Coriolis and centrifugal fictitious forces respectively.

    Obviously, this is not how you are going to explain the subject to high school kids. But this is level of understanding you should have before teaching centrifugal forces.
  23. May 24, 2012 #22


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    Yes you are right. In the station frame you have:
    centrifugal : 10N outwards
    Coriolis : 20N outwards
    But the net force must be : 10N inwards to provide the centripetal acceleration, so the wall must provide 40N inwards.
  24. May 24, 2012 #23


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    If you stand on the outer wall and try to drop the ball on your foot, it will deviate from the radial direction opposite to the station rotation.
  25. May 24, 2012 #24
    It depends on How fast. If we conside ωr as the speed. Then it will be 0N and 40N, since the V is doubled, or you can say ω is doubled, the acc. will become fourfold.

    Or you could throw a ball, and see how far it goes.

    Try and see, :). Or you could run really fast both ways, and see in which way you feel lighter.

    BUT, there is a problem, if we have ω<<<1 and r>>1km, we need accelerometers, or perhaps a window.


    Taking into consideration this post, and a previous post of mine, and some formulae deducted taking into these errors into account:

    We will be unable to replicate Earth-like conditions in space until and unless we sell mars and build a humongous Station.
    Last edited: May 24, 2012
  26. May 24, 2012 #25


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    This thread was started by Algren. Why has truesearch now taken it over?
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