Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Space Elevator and Centrifugal Force

  1. Sep 24, 2008 #1

    LURCH

    User Avatar
    Science Advisor

    I was thinking about the Space Elevator idea, and I think most of us can agree that as one goes higher up the the tower, the pull of gravity gets weaker and "centrifugal force" gets stronger (compared to one another), untill the altitude of geosynchronous orbit, where the two become equal. So then, above 22,365 miles of altitude, centrifugal force would become stronger than gravitation, and anyone riding the Space Elevator to a point above this height would experience a net force holding them to the poart of the tower away from the Earth.

    But, if my thinking on this is clear, centrifgal force is only getting stronger when compared to gravity. In truth, centrifugal force gets weaker as one gets further from the center of rotation (if rotational frequency remains constant). So it would seem that, above geosynch, a person could experience some "artificial gravity" caused by the rotation of the tower, but this gravity could never be anything aproaching Earth Normal gravity.

    Do you all agree or diagree?
     
  2. jcsd
  3. Sep 24, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi LURCH! :smile:

    Surely "centrifugal force" = ω²r, and so increases with r, for a fixed ω? :smile:
     
  4. Sep 24, 2008 #3

    rbj

    User Avatar

    i think that if the elevator extended out another several thousand miles, one would experience a force of G in the opposite direction. i hope you don't make me calculate it.
     
  5. Sep 25, 2008 #4

    LURCH

    User Avatar
    Science Advisor

    LOL! As soon as I had posted this and walked away from my computer, I realized that I had been applying the wrong principal. Centrifugal force decreases with an increase in radius as velocity remains constant, not rotational period. Of course, as rotational period remains constant and radius increases, velocity must increase (as you have quite correctly pointed out, TT).

    I'm going to try to calculate the distance from the center of earth that would be necessary for people at the end of the tether to experience Earth-normal gravity. I'll get back to this sometime this weekend.
     
  6. Sep 25, 2008 #5
    I won't calculate it and spoil your fun, but I think that is going to be one very looong tether! How will you keep it from getting wrapped around the moon?:smile:
     
  7. Sep 26, 2008 #6

    LURCH

    User Avatar
    Science Advisor

    Do you really think it will come out to more than 10-times the length to geosynch?! Well, at any rate, tomorrow is Saturday and I plan to crunch my numbers then.
     
  8. Sep 26, 2008 #7
    I don't think civilizations that are much more advanced than or civilization will build such structures. I think that the reason why we think about these things is simply because we think humans will always stay in the pesent form. I think that's unrealistic.

    Humans are machines who carry their central processor inside their heads and need to carry it with them wherever they go. Clearly, if we need to invent things such as space elevators to accomodate for this, the space elevator won't be build, instead humans will be replaced by machines that are more flexible.


    The most efficient way to travel to a space station is simply to upload the data in your (electronic) brain to another machine on board the space station.
     
  9. Sep 26, 2008 #8

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Easy way to calculate the minimum length. Find the length needed to create 1g at the end without taking Earth's gravity into account.

    1g = 9.8m/s
    Earth's angular velocity = .000073 rad/sec

    centripetal acceleration = [itex] \omega^2 r[/itex]

    Solve for r

    I think you'll be surprised at the answer.
     
  10. Sep 27, 2008 #9
    Clearly, it is this principle upon which the space elevator depends. Its counter weight is just dead weight which is orbiting the planet in geo-sync but well above the geo-sync altitude (i.e. it is not in freefall, quite the opposite). The necessary tension in the cable is consequently fantastic, and the reason we need some super exotic material to construct it.

    If you were on the counterweight, you would be looking 'up' at Earth, the cable would appear to be hanging from the planet. The strength of this simulated gravity would depend on the mass of the counter-weight (the heavier, the less long the cable need be, depending on the weight of the cable itself and assorted equipment along it).

    The orbital insertion station/platform would be at 'proper' geosync altitude, this is where you would dock orbiting space ships etc in zero G. Possibly they could be moved farther out along the cable for assistance leaving orbit?

    Rather a nice dream, I particularly like this idea, as it appears reasonably feasible. The right materials engineering, and the robots to build the thing, and... well, read Arthur C Clarke, or the Mars trilogy for a more updated approach.
     
    Last edited: Sep 27, 2008
  11. Sep 27, 2008 #10

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your best advantage comes from placing your station at the counterweight for both departure and arrival. For departure because you are traveling at above orbital velocity, and for arrival since your ship will have to spend less fuel slowing to orbital velocity. For instance, if you put your counterweight at an altitude of 46786 km(compared to the 35926 for geosync), its linear velocity equals that of escape velocity for that altitude.
     
  12. Sep 27, 2008 #11
    Yes but clearly that is not practical, how do you propose to safely dock an incoming ship in a high G environment. The ship will have to not only slow but execute a sustained thrust to match the centripetal acceleration provided by the tension in the cable for the counter weight. That'd be for the duration of the docking maneuver from the moment it deviated from a natural orbit, not to mention it would have to be cradled or otherwise physically supported by the station when docked. Seems as though much less energy would be required to drop down into a lower orbit, even if you don't consider atmospheric braking as a possibility to assist this.
     
  13. Sep 27, 2008 #12

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I wouldn't consider 0.015 g a high G environment. The centripetal acceleration on a cable at an altitude of 46786 km would equal to angular velocity(.000073 rad/sec) squared times the radius (53164000 meters) this gives an answer of .28 m/s (0.029g). this will be lessened by the centripetal acceleration already provided by gravity at that altitude of 0.14 m/s (0.014g)
    A spacecraft traveling in a parabolic orbit with a perigee at an altitude of 46786 km will be traveling exactly at the same velocity as the counterweight when it reaches it. It is then just a matter of docking and cradling. There is no need for it to provide a thrust to match centripetal acceleration except maybe briefly during the actual docking, and then at just 15/1000 of a g.

    It's just a matter of trajectory matching and timing, something you would have to do in order to dock at a lower station anyway.
     
    Last edited: Sep 27, 2008
  14. Sep 27, 2008 #13

    LURCH

    User Avatar
    Science Advisor

    Finally got around to doing calculations, and I came up with something more than 1,000,000 mi.! In fact, it was right in the neighborhood of 1,150,000. Also according to my calculations, the increase in centrifygal force with increasing radius and constant orbital period was a linear, not an exponential, increase. This was contrary to what I had anticipated. I thought that since velocity increases in a linear fashion with the increasing radius, and centrifugal force is a function of the velocity squared, I would get an exponential acceleration curve.

    Would I be correct in assuming that the factor that flattens the curve is the division by the radius? Since the velocity increases linearly with the radius and the final answer involves the velocity squared divided by the radius, this makes intuitive sense. Or have I graphed my calculations wrongly, and the acceleration is curved?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Space Elevator and Centrifugal Force
  1. Space elevator (Replies: 3)

  2. Space elevator (Replies: 16)

Loading...