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Your weight on a Space Elevator ?

  1. Aug 25, 2004 #1
    Your "weight" on a "Space Elevator"?

    I'm not sure the term "Space Elevator" is widely used, and outside of those who've read Clarke's THE FOUNTAINS OF PARADISE I'm not sure the term would mean much to many, so by way of an overly simplified description of the term...

    You put a big rock in geosynchronous orbit, drop a rope from it, and tie off the rope to a spot on Earth's equator. Now you run an elevator car up and down the rope to get in and out of orbit, and you have your "Space Elevator".

    For a MUCH better explanation (not to mention a nifty "artist's concept" pic) check out this article over on FIRST SCIENCE.COM:
    http://www.firstscience.com/site/articles/elevator.asp

    For the sake of discussion, OUR "Space Elevator" consists of 4 parts:

    1) A "counter weight" (perhaps a large asteroid) slightly father out in its orbit than geosynchronous (so that the c.g. of the entire "Space Elevator" as a system is what orbits at the g.s.o. point as opposed to just the "counter weight").

    2) A "tether" that runs from the "counter weight" down to within 300 miles of Earth's surface where the "tether" is attached to...

    3) A Penn Station of New York sized "Railroad terminal" in orbit that remains stationary 300 miles above the Earth's equator, from which dangles...

    4) A large number of "tracks" that make their way down from the "Railroad Terminal" 300 miles up, to anchor points on Earth's equator.

    It's an old Idea, I know, but here's DA QUESTION...

    If you weighed 100pounds at Earth's sea level, how much would you weigh once you made your way up from Earth's surface to the "Railroad Terminal" 300 miles above Earth's surface?

    I can't imagine where you'd be weightless, even though you're in orbit. In fact the only place I'd imagine you'd be weightless is at the point, somewhere along the "tether", where you're actually at the g.s.o. point. Until you got that far out I'd expect you'd still have weight, and beyond the g.s.o. point (getting closer to the "counter weight" that's slightly farther out than g.s.o.) I'd expect your "weight" would start to go slightly negative (i.e. once past g.s.o. you'd have a tendency to start drifting toward the ceiling of your elevator car).
    First - Am I right about that.
    Secondly - how quickly would your weight drop as you ascended the elevator? Would it drop in a linear fashion from 100 pounds on Earth's surface, to zero at the g.s.o. point, to a slightly negative value past g.s.o.?

    How would you calculate your weight at 300 miles up under these conditions?
     
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  3. Aug 25, 2004 #2

    Jenab

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    This Railroad Terminal is not in orbit. It's being suspended by something that is in orbit. And that's got to be one strong tether. Is there any substance at all with cohesive strength enough to suspend a station that massive?

    The station would be moving around the Earth with the same angular velocity as a geosynchronous satellite, or 0.0000729212 radians per second. The radius of the station's path is 300 miles plus Earth's equatorial radius, which is about 6,860,800 meters. The centrifugal acceleration is the product of the square of the angular velocity and the radius, or about 0.036 meters per second squared.

    The "apparent" acceleration of gravity would be the local inertial acceleration minus this centrifugal acceleration. The local inertial acceleration is GM[earth]/(R[earth] + altitude)^2, or 8.472 meters per second squared. The apparent gravitational acceleration on the railroad station would be 8.436 meters per second squared. This is about 0.861 g, where 1.0 g is the inertial acceleration of gravity at the equator at sea level.

    Yes, but again that's one strong tether!

    The centrifugal force (directed outward) increases linearly with geocentric distance, assuming you stay in the equatorial plane. The inertial acceleration of gravity (directed inward) varies inversely with the square of the geocentric distance. If the angular velocity is held constant at 2 pi radians per sidereal day (86164 seconds), then these quantities are equal in magnitude and opposite in sign at g.s.o.

    weight = mass x acceleration
    = mass x (inertial gravitational acceleration - centrifugal acceleration)
    = mass x { GM[earth] / (R[earth] + altitude)^2 - w^2 (R[earth] + altitude) }

    where w is the angular speed in radians per second.

    Jerry Abbott
     
    Last edited: Aug 25, 2004
  4. Aug 25, 2004 #3

    LURCH

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    You are right about the decrease in weight being gradual from ground to weightlessness at GEO. As for the rate of that decrease, I'm unsure what you would weigh @300 miles (why did you choose that altitude, BTW?). I don't think the drop would be linear, since gravity's influence is dropping in a linear fashion, but your speed and the angle at which you turn in orbit are constantly changing inversely to another.
     
  5. Aug 25, 2004 #4
    To Jenab

    Huge, enormous, all but unbounded thanks Jenab!
    That answers not only my specific question, but also gives me a way of figuring out how much you'd weigh at any point along the line. "Teach a man to fish..." ya know?
    Thanks.

    As to a strong tether... :approve: posh! We all know that "adamantium" that mythical material used to secure the gates of Tartarus (and by most SF writers to do impossible jobs) is MORE than strong enough to stand the strain!
    :approve:

    On a more serious note...
    Tell ya what. Assuming somebody actually comes up with a material strong enough to build something like a "Space Elevator", I'm REALLY not sure I'd want to live anywhere NEAR the base of a structure like that. I mean think about it. If yer 'ol "adamantium" there had any elasticity to it at all, how many gazunka-tons of energy would be released if that puppy snapped?
    Prolly enough to make for some pretty dammed grizzly pictures on the six 'o clock news I'd imagine.
     
  6. Aug 25, 2004 #5
    To Lurch

    As to why that height...
    An honest answer?
    The view.
    I've got a desktop picture of the ISS from one of the shuttle's first visits on my 'puter, and GOD that's one HELL of a glorious view with Earth spread out beneath you and stretching clear out to nearly a third of where instinct tells you the horizon should be.

    But I do have more practical reasons.
    Consider:
    Most people have a tendency to puke in micro gee, at least until they get used to it, but who has time to get used to micro gee if they're only there as long as you'd expect to be in a railroad station or airport?
    Personally I expect that any "interplanetary airlines" with "flights" into and out of an orbital station like that, is probably going to do just about ANYTHING they can to avoid having large numbers of passengers exposed to micro gee for any length of time.
    From what I've read the airplanes used to simulate micro gee for just a few minutes (by way of following a ballistic trajectory) are nick named the "Vomit Comets" by the crews that run them.
    Can you imagine dragging two or three kids along for the trip? Ya GOTTA feed 'em or they're gonna be cranky as all hell, and if they're fed, well,.. fed becomes just another term for "loaded, cocked, and ready to fire". :yuck:
    Then there's the problem of walking in low gravity environments. Wanna bet learning how to do that "Kangaroo Hop" we saw astronaughts on the moon do is tougher to get good at than it looks? Especially if you've got baggage like most people do on trips. And even once you DO get the hang of "The Hop", once you get going, making turns (to avoid other people not to mention walls) is likely to throw you a few very unwelcome curve balls to have to deal with, especially if you're dog tired and in a hurry to catch the last red-eye "down the rails" ya know? And I'll also bet that a few hours of doin "The Hop", when you're not used to it, leaves you sore as all HELL in muscle groups you never knew you had.
    How long is O.S.H.A. or whoever's in charge of regulating work environments going to put up with businesses forcing employees to risk developing bone problems due to working all day in too low a gravity environment before laws are passed that restrict exposure to micro gee?
    Bottom line...
    As I'm not a real big believer in the idea of Star Trek like "gravity plating" being invented any time soon, if "Space Elevators" are ever built, I strongly suspect that areas of them that you're "average Joe" is going to travel through, or work in for extended periods of time, is also going to be an area where there's a decent amount of gravity still present.
    Trying to get hoards of people from areas that spin, through connecting areas that don't, and back into spinning sections is always going to be expensive, or at least more expensive than just restricting the crowds to areas that don't have to spin, so I wouldn't be at all surprised to find that a "railroad station" would be located much lower down on a "Space Elevator" than industrial sections farther up.
     
  7. Aug 25, 2004 #6

    enigma

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    One potential material for the space elevators is carbon nanotube. In theory, it has the tensile strength to serve the purpose. The engineering details still need to be worked out, the least of which is: "how do we make these tubes more than a few inches long".

    As for how heavy you are:

    The acceleration of gravity at any radial distance 'r' is

    [tex]\frac{G*M}{r^2}[/tex]

    If r is in km and G*M=398,600,400

    then the result you get will be in m/s^2. You can plug in the Earth's surface (6378 km) and you'll see a familiar number.

    Additionally, the centrifugal acceleration acting on a person (in m/s^2) is:
    [tex]1000 \times \omega^2 \times r[/tex]

    [itex]\omega[/itex] is the rotational rate of the Earth, or 7.29 * 10^-5 radians/sec, and r is in kilometers from the center of the Earth.

    Your actual acceleration toward the Earth is the gravitational acceleration minus the centrifugal acceleration. If you plot the two graphs, the intersection is the distance from the center of the Earth to GEO.

    Your weight at any altitude on the elevator would be your mass times the net acceleration.
     
    Last edited: Aug 25, 2004
  8. Aug 26, 2004 #7

    LURCH

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    unfortunately, I think that launching ships from 300 miles would far outway the savings of keeping passengers "well-grounded". It would almost completely eliminate the advantages of having the space ellevator in the first place.

    But maybe passengers could be wisked past GEO altitude in an ellevator car that continually accelerates (for the g-forces), to a spot where there's at least some simulated gravity caused by cetrifugal force. Launching from above GEO would be the ideal; simply "drop" the ship away from the Earth.

    Regarding living close to the base:

    I wouldn't want to live anywhere near the Equator, I don't think. The Earth is about 24,000 miles around, and GEO is almost 24,000 miles up (close enough, once you add a little to put your counterweight outside GEO). So if that big tetherball at the end breaks apart from the strain or something, the huge strong band of unobtainium will start falling, and it's not going to stop falling untill it goes all the way 'round the planet!
     
  9. Aug 27, 2004 #8

    HallsofIvy

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    Actually there is a problem with the "space elevator" that is seldom mentioned: with one end at the surface of the earth, the other end CAN'T be in geo-synchronous orbit (and, if I recall correctly, Clarke never claimed it was). The whole point of geo-synchronous orbit is that an object in geo-synchronous orbit both remains above a given point on the equator of the earth AND is in orbit. The fact that it is in orbit means that there is no net force up or down. An object below the height of geosynchronous orbit has a net force downward, an object above, a net force upward.

    If an object in geo-synchronous orbit were attached to a cable to the earth, the entire cable below it would have net force downward and would pull the object down with it. What you need is for the cable to extend out BEYOND geo-synchronous orbit so that the the total force upward on the part of the cable beyond geosynchronous orbit balances the total force downward on the part of the cable that is below geo-synchronous orbit. The "end" of the cable would have to be much much higher than geo-sychronous orbit (since the force falls off as r2, to balance, the section higher than geo-synchronous orbit must be many times as long as the section below. The end, of course, would not, technically, be in "orbit" since it would be held in position by the force of the cable.
     
  10. Aug 27, 2004 #9

    enigma

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    The gravitational force falls of as r-2. It's not the gravitational force which holds it up though.

    It's the centrifugal acceleration which holds it, and that force increases proportionally with 'r'.

    You do need to go up past GEO, but you can also put a tether a bit outside to keep the line taut.
     
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