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Question about gravity

  1. Aug 7, 2005 #1
    To put an object into orbit just above the Earth's surface, it would need to tavel at roughly 8000 meters per second tangent to the surface of the earth - (escape velocity). Assume that there is no atmosphere or any obstructions to slow the object down or get in the way.

    The direction around the earth does not matter. It could travel around the earth following the equator or it could go over the poles. As long as it goes 8000 meters per second, it will go into orbit. Satellites do this every day.

    Suppose that you had a gyroscope with a 1 meter radius and you spun it at 1274 RPS. The outer edge of the gyroscope would be traveling at 8000 meters per second tangent to the surface of the earth.
    (2*Pi*R*1274 revolutions per second=8000 meters per second)

    Can someone explain why such an aparatus would not hover? Also, if you were to speed it up, why would it not accelerate away from the earth?

    Forget the fact that it would be close to impossible to build such an aparatus and spin it that fast without it flying apart. I just want to know why it would not work.
     
  2. jcsd
  3. Aug 7, 2005 #2

    Danger

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    At the stated RPM, the bloody thing might hover due to aerodynamic effects alone! As for the 'escape velocity' part, the device isn't moving in a constant direction away from the Earth; the rotor is under constant acceleration, but only because it's confined to an 'orbit' of sorts around its own axis. The mechanism as a whole isn't moving. If a chunk of the rotor was to break loose, it could achieve orbit (ignoring air resistance and whatnot). I wouldn't want to be near it when it happened. I can't offer any force vectors or formulae or anything; this is just off the top of my head.
     
  4. Aug 7, 2005 #3

    rbj

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    actually, Danger: it really is incumbant upon just curious to say, in the first place, why the heck he would think it would hover. perhaps if a little paint got splattered on that spinning disk some droplets would fly off and never touch the ground on his atmosphereless planet. but there is no ostensible reason for the spinning disk to hover.
     
    Last edited: Aug 7, 2005
  5. Aug 7, 2005 #4

    Danger

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    True. I didn't think of that because I'm not a teacher. My inclination is to try to answer if I can, whereas the proper approach is to get the student to work it out with guidance. I'll try to keep that in mind for future instances. Thanks for pointing it out.
     
  6. Aug 8, 2005 #5

    russ_watters

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    It's pretty simple: rotation and translation are two different things.
     
  7. Aug 8, 2005 #6
    In my mind, the matter in the outer ring of the gyroscope is moving in translation tangent to the surface of the earth. It also happens to be changing direction constantly, however the direction of the translation should not matter. You can put an object into a circular orbit in any direction as long as it it is moving tangent to the surface of the earth. Why should it matter if the direction of the translation changes?

    Part of the reason why I am asking this question is because I had read about an experiment performed by some researchers in Japan where they dropped a gyroscope spinning at 18000 RPM in a vacuum and found it to fall slightly slower when it was spinning than when it was not. See the following article for more details: http://ascension2000.com/ConvergenceIII/c306.htm
    Unfortunately the article does not say what the radius or the mass distribution of the gyroscope disk was therefore I cannot calculate if their observed "weight loss" is in line with what I am proposing with my question. In their experiment, they did not spin their gyroscope anywhere close to the speed necessary to fit in with my simple calculations and I also would be certain that their gyroscope was much smaller in radius than 1 meter. (I would need to know this information in order to calculate the instantaneous linear velocity of the outer edge of the gyroscope ring and also factor in the mass distribution of their ring).
    This means to me that their observed "weight loss" of 1 part in 7000 could possibly be in line with what I am proposing.
    What I am hoping to get from this forum is a more detailed analysis of my question. Can someone provide mathematical proof that my thought process if flawed?
     
    Last edited: Aug 8, 2005
  8. Aug 8, 2005 #7

    Danger

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    I don't have time to look at the link right now, and I certainly don't know any math. In keeping with rbj's observation about how the forums work, I'll just point out one thing. No matter how fast or in what direction some part of the rotor is moving at any given instant, some other part is going the opposite way. Net movement relative to Earth is zero.
     
  9. Aug 8, 2005 #8

    russ_watters

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    It is, but since it is a lot of matter moving in a lot of different directions, it doesn't work the same as one lump of matter moving in one direction.
    To expand on what Danger said, if you add up all the velocities of all the points on the edge of the circle, the net velocity is zero. The math for that is fairly complicated (and I'm lazy too), but you can easily see that for every point on the circle, there is a point directly across from it moving in the opposite direction. Adding those two velocities gives you zero.
     
  10. Aug 8, 2005 #9

    rbj

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    because something else, other than the planet's gravity, is pulling on the matter of the outer ring.

    here's the deal from a classical Newtonian POV: the matter that flies over the surface of this atmosphereless planet (let's say at a height of a meter) is not hovering. in fact, it is falling. it is being diverted downward in the same manner that would a stone if you had dropped it from a meter in height. it's just that horizontal velocity has translated the object so far in the time that the object has fallen that the planet's curvature has sorta dropped out from underneath the object, thus maintaining the height above the planet. it speed is just barely slow enough that the object doesn't fly off into space. the planet's gravity is tethering to one meter above the planet's surface.

    but with your disk, it ain't gravity that is tethering the matter in the outer ring from flying off but the molecular bonds of the matter inside of that outer ring. it's spinning around so fast but gravity is doing nothing to hold it together. this gyroscope could be mounted on a pole and, spinning at that radical speed, it would likely stand up topheavy for months or years, for all i know. but it would still need the pole to hold it up.
     
  11. Aug 11, 2005 #10

    LeonhardEuler

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    This is not gravity. Nuetrons have gravitational mass even though they are nuetral. It is easy to show that the electric force resulting from a positive charge and a negative one close to one another varies as the inverse of the cube of the distance, not the square as is the case with gravity. This force is not necisarrily attractive, the force could be attractive, repulsive, or niether(i.e. tangential). This theory of gravity would suggest that there is an increase in gravitational attraction when dipole moments are increased. One would expect the wieght to increase in chemical reactions where the products have greater dipole moments than the reactants (for example: [itex]H_2 + Cl_2 \rightarrow 2HCl[/itex]) This is not observed.
     
  12. Aug 11, 2005 #11

    rbj

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    and that is where you're mistaken.
    because that is what we learn in physics. it's about vectors, particularly two vectors:

    [tex] \mathbf{F} = m \mathbf{a} [/tex]

    in your situation the two vectors are not equal because they point in different directions.
     
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