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Smoke in a space station

  1. Feb 20, 2010 #1
    What do you make of the following explanation regarding the movement of smoke:

    Let's assume there is no gravitational force from the outside.
     
  2. jcsd
  3. Feb 20, 2010 #2

    Borek

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    We should define smoke first, discuss it later.
     
  4. Feb 20, 2010 #3
    I'm not sure about the smoke, but there is no "centrifugal force" pressing objects against the outside rim.
    The forces are
    a) the outside rim pressing on the objects; this provides the centripetal force to move them in a circle.
    b) the Newtons 3rd law reaction to this force; the objects pressing against the rim.
    There is no other force pressing on the objects.
     
    Last edited: Feb 20, 2010
  5. Feb 20, 2010 #4

    Andy Resnick

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    The quote totally neglects the Coriolus force, which is highly non-negligible in these scenarios. Also, the 'gravity' gradient between a crewperson's head and feet is large, large enough to cause disorientation, IIRC.
     
  6. Feb 20, 2010 #5
    The details of the smoke particles are of course irrelevant, but use this one from Wikipedia...


    so let's assume it's cigarette smoke.
     
  7. Feb 20, 2010 #6
    I don't think so..assume it's just enough spin for a 1g force.....
     
    Last edited: Feb 20, 2010
  8. Feb 20, 2010 #7
    Post # 3 :
    utterly incorrect.....and anyway irrelevent....

    http://en.wikipedia.org/wiki/Centrifugal_force


     
  9. Feb 20, 2010 #8
    If you had read beyond the first paragraph you would have found
     
  10. Feb 20, 2010 #9

    Borek

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    Yes and no, what happens to smoke depends on its density as compared to the density of the surrounding air.
     
  11. Feb 21, 2010 #10

    Cleonis

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    A clearer example is a balloon filled with helium. On Earth that balloon will rise to the ceiling of a room.
    Likewise, on a rotating space station a helium filled balloon will rise to the ceiling.

    Now, it may sound counter-intuitive that any object would move against the vector of gravity. But that is what buoyant objects do, they move against the vector of gravity, both on Earth and inside a rotating space station.


    Here, I use the expression 'vector of gravity' in the following meaning: the direction of inertial motion.
    - On Earth, if you release a object (a non-buoyant one) it will move in the direction of the Earth's center of gravity, which is towards the floor.
    - On a rotating space station if you release an object (a non-buoyant one), it will move towards the floor. (Of course, the actual motion of the object is in a straight line, it's the floor that is on a curved path wrt the inertial coordinate system, intercepting the object.)
     
  12. Feb 21, 2010 #11
    Setting aside the original quote's incorrect use of centrifugal force, I think the poster is referring to the fact that cigarette smoke rises because the air around the cigarette is heated and rises due to convection, carrying the smoke with it.
    What happens in the case of the rotating space station is interesting.
    If it is full of air, and the air is rotating with the ship, then there will be a pressure gradient such that the air pressure will be greater nearer the rim.
    Adjacent concentric "layers" of air, the further they are from the axis of rotation, will need a greater centripetal force (greater r) to keep them rotating.
    Would not this pressure gradient in the ship's atmosphere give rise to a buoyancy effect similar to that experienced in a gravitational field? If so, would it give rise to the same effect whereby warm air, and hence the smoke, rises?
     
  13. Feb 21, 2010 #12

    Cleonis

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    To simplify the picture, imagine a space station with a very large diameter (kilometers across), rotating at such an angular velocity that the rim is pulling 1 G.

    In a bathtub all the water will be pulling 1 G; difference in G-load between bottom and top is negligable.
    In the bathtub anything that is buoyant will pop to the surface if you submerge it.

    The buoyancy of a helium filled balloon in air is the same as buoyancy in water. In a column of air or water there is a pressure gradient. Note that the pressure gradient doesn't arise from gravity gradient. It's just that the lower in the column the more mass is above it, pressing down: that stacking up of mass is what gives rise to pressure gradient.


    Now, in a very small space station, rotating very fast, there will be an additional factor giving rise to pressure gradient; the effect that is described by Stonebridge in the previous post. The further away from the axis of rotation the higher the G-load. But that will be significant only if the space station is very small indeed.
     
  14. Feb 21, 2010 #13
    I'm not so interested in how we name centrifugal force..or centripetal..or whatever....unless it affec ts the question and answer, simply in what the smoke does. I was thinking according to Cleonis post #10:

    because increased air density is experienced as one moves towards the gravity source...due to induced artifical "gravity"....

    which led me to think
    So I am still thinking,but am not positive, that the lighter smoke moves towards the hub of the rotating space station...I guess that's what the author calls the "ceiling"...so I'm tending to agree with the author I quoted....

    but this is confusing:
    No??? Then why is the air in my office room more dense than air at 1,000ft altitude off the earth's surface?? no gravity, nor air at all..it would all drift off into space, no???
     
  15. Feb 21, 2010 #14
    I think the pressure gradient in earth's atmosphere does depend partly on differences in g at different heights, but that it's the variation with height of the mass of air above the place where you're measuring the pressure that's more significant. On a small rotating space-station, though, differences in the strength of the simulated gravity could be significant too.
     
  16. Feb 21, 2010 #15

    Cleonis

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    Let me discuss another matter first:
    Why is there a pressure gradient (in air pressure) between the floor and the ceiling of a room?
    Imagine a stack of weighing scales, the type of scale that people have in their house to weigh themselves. Let's say the stack is 20 pieces high, and each scale weighs 1 kg. Then the bottommost scale will read 19 kg, the weight of all the scales above it, and the second scale from the bottom up will read 18 kg.

    Likewise, the air at the floor of a room has the entire atmospheric column of air above it, while the air near the ceiling has the entire column of air minus the heigth of the room above it. That difference in mass-above-it is what gives rise to the pressure gradient. The same reasoning extends all the way into the upper atmosphere.


    The radius of the solid Earth is about 6400 kilometers. The Earth's atmosphere is a very, very thin layer. Several 10's of kilometers above the surface the air density is only fraction of what it is at the surface. There is very, very little gravity gradient between the Earth's surface and the upper regions of the atmosphere. Gravity is inversely proportional to the square of the distance to the center of gravitational attraction. The difference between Earth surface and upper atmosphere is the difference between 6380 km. and, say, 6400 km.: that's a tiny difference.
     
  17. Feb 21, 2010 #16
    This is true in an inertial frame (unless, of course, there does happen to be another force pressing on the objects!), but before you accuse Naty1 of not reading the article carefully enough, have a look at the section quoted in #7, which mentions your force (b) as one of two concepts to which the name "centrifugal force" is traditionally applied: "a reaction force corresponding to a centripetal force".

    Whether the other kind of "centrifugal force" is acting on an object depends on our choice of reference frame.
     
  18. Feb 21, 2010 #17

    Andy Resnick

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    a 1g force *at what radius*?

    ps- that means other radii have different pseudo-gravity forces.
     
  19. Feb 21, 2010 #18

    Andy Resnick

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    This is the essential point- the fluid dynamics of the internal atmosphere will most likely dominate. I wonder- if the spacecraft had no walls (i.e. a torus, or a large hollow cylinder rotating axisymmetrically), what would the equilibrium pressure distribution be?

    It's clear if the interior were vacuum, and the crewperson jumped to the rotation axis, they would experience a large 'gravity' gradient.
     
  20. Feb 21, 2010 #19
    Rasal...thanks but no need to defend against accusations.,.they are usually irrelevent, like here, and happen all the time just like in the world of science when someone's argument turns out to be incorrect,...,like man made global warming is "settled science" for example...no need to get sidetracked...

    no,no,no......there is no reason there!! what would happen if there were no gravity??

    Hint...W = ma= mg

    I believe I now recall an analogy, but I don't recall the source....a car at stready speed in a straight line suddenly suddenly swerves right. What happens to a tethered balloon inside??
    Correct answer: balloon swerves LEFT.... it shows, I think, the false nature of the logic above.

    Anyway, I am still ever slightly so unsure about coriolus...because the air is contained inside.....I think it's a non issue in the space station???.
     
  21. Feb 21, 2010 #20
    And Rasalhague is correct, pressure gradient is due to the weight of the air. That is the thing causing pressure. If you want to verify this, take an old milk jug, fill it with water, and poke holes down the side. The top hole will have little or no pressure, while the bottom hole has the most. Obviously this cannot be caused by gravity varying inside the jug, just the weight of the water.
     
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