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Fundamental Frequency of the earth's crust

  1. Jun 3, 2006 #1
    Here's one for you. Is there a way of working out what the fundamental frequency of the Earth's crust is? Has anybody done this. And what would happen if we somehow matched this frequency, say by all dancing to a particularly banging dance remix of Electric Light Orchestra's "Mr Blue Sky" all at once? Came up in a conversation and we need to find out before we can concentrate on the more important matter of exams.
  2. jcsd
  3. Jun 3, 2006 #2
    I would pose that, since the earth is a dynamical, non-solid sphere(with regards to the moving molten material deep inside), that there is no "fundamental frequency"
  4. Jun 3, 2006 #3
    Damn! We couldn't decide if the crust may have. A look on google found some article that suggested it may have a fundamental frequency of 10Hz, but you know what the internet is like. Contains everything in the whole wide world ever, including all the nonsense people talk. Ahh well, I suppose it means I don't have to find someone to dancify ELO for me.
  5. Jun 3, 2006 #4
    I'd imagine you have very strong damping forces on this scale, you're looking at thousands of kilometers of goo...
  6. Jun 3, 2006 #5


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    The fundamental frequency of the Earth as a whole is about 1/54 minutes.

    See for instance https://www.physicsforums.com/showthread.php?t=119217 for some textbook references. (MTW's "Gravitation, pg 1036 - the problem shows up in figuring out the Earth's response as a gravity wave detector).

    The proposed Lagrangian for this problem is essentially

    L = A xdot^2 - B x^2

    The numerical factors of A and B, which are given in the text, would allow one to compute the amplitude of the vibration (i.e. height change at the equator) given the total amount of energy stored in that vibrational mode.

    I'll give a description of the form of the results for the factor B in terms of some well-known physical constants of the Earth.

    Total stored energy (the term proportional to x^2) is given by

    (stored energy) = (1/5) * (Gravitational binding energy of Earth) * ( x /r)^2

    where r is the radius of the Earth
    x is the distance that the equator moves
    and the Gravitational binding energy of the Earth is

    http://en.wikipedia.org/wiki/Gravitational_binding_energy gives

    2.24 ^10^32 joules

    The value for factor A can be estimated from the resonant frequency, if desired, but we don't really need it for what follws.

    Given the above relation, one can compute the total stored energy in a vibrational mode as a function of the height change at the equator (x).

    What isn't given is the appropriate "damping factor" to use. Basically the differential equation for the above needs to be extened to look just like a standard "resonant circuit", aka a "spring / mass / damper" problem. The value of the damping constant (the Q) isn't known (by me, someone somewhere probably knows it) but is probably not terribly large. (The Lagrangian I actually wrote down above doesn't contain the term for the damping factor, but we know that it is there).

    To calculate movement if we knew the value of Q, and the input power of the exciting vibration, we would use

    energy lost per cycle = (stored energy / Q) = (input power) * (54 minutes)

    where the equation for stored energy has been previously given.

    You can put this all together to solve for 'x' as a function of input power, and the presumed value for Q, but I haven't worked out the details.

    People jumping up and down won't be near the resonant frequency in any event (1/54 minutes) - you'd need some different mechanism to incite resonance (everyone hiking up the mountains for an hour?)
    Last edited: Jun 3, 2006
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