Guess at slowing of Earth spin

Discussion in 'Physics & Math' started by Billy T, Mar 11, 2006.

?

How big is "SLIGHTLY" (in first post of Billy T if not here)

Poll closed Mar 25, 2006.
  1. Less than 1.0E-4Meters

    58.8%
  2. Less than 1.0E-2Meters

    5.9%
  3. Less than 1Meter

    17.6%
  4. Less than 100Meters

    5.9%
  5. Less than 1000Meters

    0 vote(s)
    0.0%
  6. More than 1000Meters

    11.8%
  1. CANGAS Registered Senior Member

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    1,612
    To depart from science and indulge in beloved science fiction; the famous JOLT that the Enterprise so often enjoyed on our tv screens would be much more so if it could catch as good as Yogi Berra did.
     
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  3. Tortise Registered Senior Member

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    I agree with Pete. I think it might be easier to believe that if teleportation were possible, it would be done in a way that would concerve our known laws, including concervation of angular momentum.
     
    Last edited: Mar 26, 2006
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  5. CANGAS Registered Senior Member

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    Toretise:

    You do not believe in the Prime Directive of godunkems; anything is possible if postulated?

    Besides, as Pete pointedly pointed out, the Enterprise could conserve momentum by catching a good jolt.

    However, the point of the posts is not how teleportation might hypothetically function, but, rather, to rigorously examine the conservation of the main part of of the Earth's angular momentum if a very, very small part of it were "magically" removed.
     
    Last edited: Mar 26, 2006
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  7. Tortise Registered Senior Member

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    I was agreeing with him and just making a comment.
     
  8. CANGAS Registered Senior Member

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    Live long and prosper.
     
  9. Pete It's not rocket surgery Moderator

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    Long life and prosperity.
     
  10. Tortise Registered Senior Member

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    Live long and prosperkinesis
     
  11. D H Some other guy Valued Senior Member

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    2,257
    It seems to me that at least one of the conservation and/or laws of thermodynamics would have to be thrown out with teleportation. The teleportee is deep in a gravity well. Teleporting from the surface of the Earth to geosynchronous orbit represents a big increase in potential energy. Conserving energy would require a change in kinetic energy or temperature. A kinetic energy change would violate conservation of linear and/or angular momentum. A temperature change would (I think) violate one or more of the laws of thermodynamics.
     
  12. DaleSpam TANSTAAFL Registered Senior Member

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    The change in potential energy could be supplied by the energy required to do the teleportation even using "perfectly efficient" teleportation equipment.

    Hmm, maybe the Klingons could power their own ships by teleporting bombs down the gravity well and capturing the energy with such equipment.

    -Dale
     
  13. DaleSpam TANSTAAFL Registered Senior Member

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    D_H, BillyT

    Sorry about my short replies yesterday, I didn't have enough time to give a detailed response to your points. I think the best way to explain the various variables is to describe my thought process in a little more depth.

    After making the estimation when I was thinking about how to actually work this problem I realized that the beginning and the end points could both be described by a single rigid-body with changing moment of inertia by using the typical "massles rod" idea. Since everything involved in the problem is conservative it does not matter how the satellite gets to where it is going, only the beginning and end points matter for the result. So instead of thinking about launching the satellite in the traditional manner I thought about hoisting it aloft on this massless (telescoping) rod. Now, as the rod lengthens (r) the total moment of inertia increases (I = Ie + m r²) and so by conservation of angular momentum (L) the sidereal day (T) lengthens and the geosynchronous orbital radius (a) increases. Therefore a, the geo orbital radius, is a function of r, the radius to which we have hoisted our satellite on its massless rod.

    So, when I initially solved for a as a function of r the first thing I tried to do was to simply evaluate a(a)-a(re). Unfortunately this quantity turned out to be numerically equal to zero. Basically the problem is that the moment of inertial of the earth (Ie = 8.07E37 kg m²) is so large that, as expected, it completely obscures the contribution of the satellite.

    I tried several ways of directly simplifying the expression a(a)-a(re) in order to pull out Ie, but since it involves square roots and cubes of square roots it is annoyingly difficult to simplify. However, the expression for √(a³) is easy to separate out Ie. Unfortunately you cannot take differences in √() to answer the question about differences in a. That is when I decided to differentiate the expression for √(). This removed the Ie term and can easily be solved for da. Integrating then gives the desired quantities without requiring any precision-losing steps. The only subtraction involved is a²-re² which involves numbers that are different enough to be reasonably precise.

    The one thing that concerns me is why my results are 7 or 8 orders of magnitude off from superluminal's. The only quantity I see that is on that order is the geosynchronous radius itself (a = 4.2E7 m). Perhaps, superluminal, your number represents the multiplicitave factor rather than the distance itself?

    -Dale
     
    Last edited: Mar 26, 2006
  14. D H Some other guy Valued Senior Member

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    2,257
    Dale,

    You can take differences in a^(1/3) if you do a first-order approximation. Since the change in a is very small, the first order approximation is very accurate (19 decimal places), which is many orders of magnitude smaller than other error sources.

    BTW, your answer agrees with mine (you said 6e-13 m, I said 5.8e-13 m).
     
  15. DaleSpam TANSTAAFL Registered Senior Member

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    Your approach is very good, much more consice than mine. I didn't do a Taylor's expansion because I was thinking of a as a function of r not as a function of L. Although L doesn't change much r does. So your approach makes a first-order approximation reasonable, while I thought mine didn't.

    I think such close agreement with such different methods lends some credibility to the results. I suspect that our small difference probably comes mostly from using slightly different numbers. The obvious one is that you used the 24-hour day and I used the sidereal day, a difference of about 0.3%, but most of my other numbers (e.g. re, M, G) came from Mathematica and probably similarly had a few tenths of a percent difference each from the numbers you used.

    -Dale
     
    Last edited: Mar 26, 2006
  16. superluminal . Registered Senior Member

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    10,717
    My number:

    0.99999999999999999998213...

    is just the change in ω. I didn't calculate the actual distance but assumed the change would be on the same order of magnitude. The GSO radius is:

    r = <sup>3</sup>√(G x Me / ω<sup>2</sup>)

    Perhaps my assumption was wrong. Would one of you gents who have mathematica care to calculate it? If it does not agree with your answers, at least to within an order of magnitude, I'd be very interested in learning where I went wrong... I followed the simple method in my previous post, using the numbers in my other post.

    Thanks.

    P.S. This has been one of the more enjoyable exercises/threads. Good one BillyT.
     
  17. D H Some other guy Valued Senior Member

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    2,257
    You started with
    I init = (2/5MeRe^2 + MsRe^2)
    Ifinal = (2/5MeRe^2 + Ms(Re+Ds)^2)

    and used these to calculate the ratio

    wfinal/winit = Iinit / Ifinal

    That number is very close to one; in fact, it is one unless one resorts to extended precision. If you had calculated things a little differently you would not have needed extended precision.

    In many cases, it is preferable to work with the difference between pairs of numbers that are nearly the same. This is certainly the case here.

    Introduce
    delta I = Ifinal - Iinit
    delta w = wfinal - winit

    Starting with
    wfinal/winit = Iinit / Ifinal
    then
    delta w/winit = delta I/Ifinal =~ delta I/Iinit

    Since the change in moment of inertia is very, very small, the error in the approximation is very, very small.

    The MeRe^2 terms in delta I = Ifinal-Init cancel:
    delta I = Ifinal - Iinit = Ms((Re+Ds)^2-Re^2)
    Thus
    delta w/winit = -delta I/Init = -1.8e-20

    Expanding a = k * w^(-2/3) (Kepler's 3rd Law) about w=winit gives
    a(wfinal) = k*winit^(-2/3) -2/3k*winit^(-5/3)*(wfinal-winit) + ...
    or
    delta a/a(winit) = -2/3*delta w/winit

    Using your result with a(winit) = 4.2e7 meters,

    delta a = 2/3*1.8e-20*4.2e7 = 5e-13

    Your number differs from mine (5.8e-13) because you used 2/5MeRe^2 as the Earth's inertia. The Earth is not a uniform solid (its core is more dense than the mantle) and thus the solid body approximation is a bit high.
     
  18. superluminal . Registered Senior Member

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    Interesting. Thanks DH.
     
  19. DaleSpam TANSTAAFL Registered Senior Member

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    I plugged your formula above in using my numbers for G and Me. I took my number for ω and multiplied it by 1 for the beginning situation and multiplied it by your number above for the final situation. The difference between the two was 0 to standard numerical precision. I then repeated the procedure telling Mathematica that all of my input data had 100 decimal digits of precision. The result was 5E-13 m, so your result agrees with ours as well.

    The precision of the result was 79 digits indicating that 21 decimal digits of precision were lost. Machine precision on my computer is 15.95 decimal digits, so I guess it is not surprising that the original answer was 0.

    -Dale

    PS I see D_H already did the calculation and came out with the same result.
     
    Last edited: Mar 26, 2006
  20. superluminal . Registered Senior Member

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    10,717
    Cool! I did it originally using a web-based arbitrary precision calculator. The precision on my PC is also 15 digits as I mentioned earlier.
     
  21. Pete It's not rocket surgery Moderator

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    10,166
    I posted a thread on a simlar topic a while ago, thinking about whether energy is conserved when moving things through wormholes, but no one responded. [thread=46423]Here it is.[/thread]

    In the same vein as my initial thoughts in that dead-end thread, perhaps there might effectively be a very strong gravity field across a teleporter... ie pushing something through a teleporter "gate" (I'm thinking Stargate rather than Star Trek now) might be very very difficult if the gravitational potential is different on each side of the gate.
     
  22. Tortise Registered Senior Member

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    227
    I might be reading this wrong - I just woke up, but isn't that the crux of the problem? I mean if you were transporting someone from the earth to the moon, the higher gravity on earth might be the problem if you were to take them by rocket or if you were to transport them without violating one of our laws? Would it be naive to assume that such a tremendous amount of would could be done for nothing?
     
    Last edited: Mar 27, 2006
  23. Pete It's not rocket surgery Moderator

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    I think it would be naive... but that's exactly what happens in the popular idea of teleportation, right?
     

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