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Space elevator

  1. Sep 4, 2012 #1
    i am not able to understand how, where and why stresses will act on the space elevator considering it to be a rope or ribbon,
    i mean to ask that if we make a free body diagram of the space elevator then where and in which direction forces will act on it.
    and why that forces will act as they are doing
     
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  3. Sep 4, 2012 #2

    mfb

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    It depends on your coordinate system:

    Non-rotating: Downward forces from gravity everywhere, towards the center of earth. An additional part is tension in the cable, which is a bit hard to show in a sketch as it is relevant everywhere. It has its largest value at geostationary orbit. Below, the reduction of tension (with lower altitude) gives a net force upwards for every cable element, and above, the reduction of tension (with higher altitude) gives a net force downwards for every cable element.
    Every cable element with mass m gets a total force of mωr^2 inwards, keeping in a circular orbit.

    Rotating: Downward forces from gravity everywhere, towards the center of earth. Additional, every mass element gets a force of mωr^2 outwards, and tension as described above. The sum of all forces is 0 for all mass elements.
     
  4. Sep 4, 2012 #3

    A.T.

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  5. Sep 10, 2012 #4
    Basically, every point on the ribbon, from the geosynchronous level (35,786) and below, must have high enough tensile strength to support all of the mass below that point.
     
  6. Sep 10, 2012 #5

    A.T.

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    It's not that simple. The stress at each point is not just a function of the mass below it, but also of the angular velocity:
    http://en.wikipedia.org/wiki/Space_elevator#Cable_section

    And above the geosynchronous level, there is stress too.
     
  7. Sep 10, 2012 #6
    A.T., you are correct. I oversimplified by not discussing the forces on that mass. If you assume that the counterweight just barely pulls the ribbon taut, so there is no upward force from it, then the forces acting on each section of mass of the ribbon would be the gravitational force down and the centrifugal force up. To determine the total force affecting any point on the ribbon, you would need to integrate the forces below and above that point. If the counterweight is creating some upward force, then this would need to be added.
     
  8. Sep 10, 2012 #7

    jbriggs444

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    If you assume a motionless cable (in the rotating frame) or a uniformly rotating cable (in the inertial frame) then it is sufficient to integrate the force contributed by each mass element from one end of the cable up to the chosen point.

    It does not matter which end you integrate from. The magnitude of the calculated force will be identical either way. The signs of the calculated forces will be opposite, of course.

    If you were to integrate both "below and above" and then add the results together you'd be sure to get the wrong answer, either by double-dipping (with a sign error) or cancellation (without the sign error).

    The wording of a counterweight "just barely pulling the ribbon taut" seems off. The ribbon is not going to go slack without a counterweight. It is taut under centrifugal force. The counterweight is there so that the ribbon's orbit neither falls from nor flies away from a circular orbit.
     
  9. Sep 10, 2012 #8
    If you use the point on the ribbon at the geosynchronous level as an example, the net force on the downward side will be toward Earth (gravitational force will be higher than centrifugal force), and on the upward side will be away from Earth (centrifugal force will be higher than gravitational force). Therefore, the forces will be pulling in opposite directions by an equal quantity. You must, of course, get the sign correct when you add them.

    The ribbon is only taut from centrifugal force when there is a counterweight or an incredibly long ribbon past the geosynchronous level. Without the counterweight (or the incredibly long ribbon), the net forces for the entire ribbon will be toward Earth and the ribbon will fall to Earth.
     
  10. Sep 10, 2012 #9

    jbriggs444

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    When you add them, the result should be zero. If your goal was to compute the tension in the ribbon then adding upward force to downward force is just plain wrong.

    Suppose that you have two teams playing tug-of-war, each pulling on a rope. Team A is pulling with 1000 pounds-force to the left. Team B is pulling with 1000 pounds-force to the right.

    Is the tension on the rope:

    A. 2000 pounds-force
    B. 1000 pounds-force
    C. 0 pounds-force
    D. None of the above

    My answer is B. It seems that your answer is A.

    The ribbon is taut regardless. It is rotating and not massless. To a first approximation, the ribbon will have an elliptical orbit such that [portions of] the ribbon may or may not intersect with the surface of the earth. Whether the ribbon falls "to the Earth" will depend on that detail.
     
  11. Sep 10, 2012 #10
    I agree with your analogy and that the tension in the rope would be 1000 pounds-force, and, therefore, I agree that it is not correct to add the opposing forces at the geosynchronous point for the space elevator ribbon. The tension on the space elevator ribbon at the geosynchronous level (or any point below that point) will be equal to the net force below that point. The closer to the Earth, the lower the tension.

    When I used the word "taut," I meant from the point the ribbon is connected to the Earth and upward. If the net forces are downward at the geosynchronous level, the entire ribbon will gradually be pulled toward the Earth until it completely falls to the ground, becoming slack at points close to the Earth as it falls. For example, if the counterweight of a space elevator were released from the ribbon, the net forces at the geosynchronous level would change from being net zero or upward to being downward and the ribbon would be pulled toward the Earth.
     
  12. Sep 10, 2012 #11

    jbriggs444

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    Ahh, that makes sense. I was automatically assuming that the tension at the bottom end of the elevator would be zero (and even thinking in terms of an unanchored elevator) and you were carefully arranging things so that the tension at the anchor would be zero.
     
  13. Sep 10, 2012 #12
    I was thinking about it as a "real" space elevator in which the bottom end of the ribbon would be anchored so that cargo could be loaded onto its climber (from the ground), and that the ribbon would need to be taut so that the climber, with its gripping wheels, can pull itself up. Also, the ribbon would need to be just barely taut to avoid excess tensional stress on it.

    An additional complexity in the analysis for a real space elevator ribbon, which is usually incorporated into conceptual designs, is that the ribbon would be tapered. Since the tension decreases going from the geosynchronous level downward, the ribbon does not need to be as wide either. This reduces the mass, and downward gravitational force, for the entire ribbon, and reduces the width (and mass) all along it. This tapering is also applied to the ribbon section between the geosynchronous level and the counterweight.
     
  14. Sep 10, 2012 #13
    I would think in a real space elevator the bottom end would be under compression instead of tension. It would be built like a tower supporting its weight from the ground up. There would be a transition section which would be under minimal tension or compression, and an upper section under tension as you were supposing. Building the elevator this way it is supported from both ends, allowing for a stronger structure.
     
  15. Sep 11, 2012 #14

    mfb

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    You cannot support any significant length with compression. The tallest construction in the world is less than 1 km high. While there are plans of 4km-constructions and concepts how 10-15 km might be achieved, this is completely negligible compared to the 36000 km of a space elevator.
    Tension is way easier to handle - you do not need additional support structures everywhere, and materials can handle more tension than compression forces.
     
  16. Sep 11, 2012 #15
    It would be nice if using a tower under compression would work (unfortunately, physics doesn't care about being nice) because that would reduce the tensile strength required for the section of the space elevator shaft (or ribbon) under tension. Currently, there are no materials available that have high enough tensile strength for an Earth-based Space Elevator. Carbon Nanotubes theoretically could work, but no one has been able to make any strands more that a few millimeters or centimeters long, which is a far cry from thousands of kilometers.

    However, a Lunar Space Elevator could be built using existing materials, because of the moon's lower gravity. LiftPort Group is currently pursuing this. http://liftport.com/
     
  17. Sep 12, 2012 #16
    I'm not sure I understand the compression problem.

    If you build something to 1km high then the weight of the top floor will be transferred down to the foundations. But if you build it to 36km, the weight of the top floor is entirely compensated for by the centripetal force keeping it in circular motion. All floors above this floor then exhibit a centrifugal force upwards, which, when transferred down through the building, acts to nullify the weight of one of the floors below 36km. So theoretically could we not keep building until we reach a point where the upward centrifugal forces of all the floors above 36km serve to cumulatively nullify the weights of all the floors below 36km?

    If we stop at this point, should the tension at the point of contact with the Earth not be zero? And then the only limiting factor is the tensile strength of the cable at some point (maybe 36km?) where it is maximised?
     
  18. Sep 12, 2012 #17

    jbriggs444

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    Yes, all of this is correct, except that it's 36,000 km, not 36 km.
     
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