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Moon's Tidal Forces puff up Earth

  1. Dec 18, 2008 #1
    It is easy to show*, that the Differential (Tidal) Force, produced by the Moon, around the surface (equator) of the Earth, is:
    [tex]\vec{\Lambda F} \approx \frac{G \, M_{moon} \, \delta m \, R_{earth}}{D^{3}} \times \left( 2 cos(\theta) \vec{x} - sin(\theta) \vec{y} \right)[/tex]​
    The radial component of this Tidal Force is:
    [tex]\Delta F_{r} \equiv \vec{\Lambda F} \bullet \vec{r} \approx \frac{G M_{moon} \delta m R_{earth}}{D^{3}} \times \left( 2 cos(\theta)^{2} - sin(\theta)^{2} \right)[/tex]
    [tex] \equiv \Phi \times \left( \frac{1}{2} + \frac{3 cos(2 \theta)}{2} \right) [/tex]​
    This does not vanish, when integrated over one full revolution. Instead, there is a net outward radial force, which would (on its own) rip off the Earth's crust, and propel it up & out into space:
    [tex] \Delta F_{r} \equiv \delta m \times \frac{\partial^{2}}{\partial t^{2}} r[/tex]
    [tex] = \delta m \times \omega^{2} \frac{\partial^{2}}{\partial \theta^{2}} r[/tex]
    [tex] = \Phi \times \left( \frac{1}{2} + \frac{3 cos(2 \theta)}{2} \right)[/tex]​
    This yields terms that are linearly, and quadratically, proportional to rotation angle ([tex]\theta[/tex]). Thus, the radius (r) would rapidly diverge with increasing rotation angle.
    * Carroll & Ostlie. Introduction to Modern Astrophysics, pg. 763.
    Thus, this net outward radial (tidal) force must be offset by some sort of Hookes' Law (cf. Stress-Strain):
    [tex] \Delta F_{r} \equiv \Phi \times \left( \frac{1}{2} + \frac{3 cos(2 \theta)}{2} \right) - K \times r[/tex]​
    The periodic solutions of this differential equation can be obtained, by Fourier decomposition:
    [tex] r(\theta) \equiv \sum a_{n} \times cos(n \, \theta)[/tex]​
    This yields:
    [tex] r(\theta) = \frac{\Phi}{2 K} + \frac{3 \Phi}{2} \frac{cos(2 \theta)}{K - 4 \, \delta m \, \omega^{2}}[/tex]​


    The constant term ([tex]\frac{\Phi}{2 K}[/tex]) represents the net average radial expansion of the Earth, induced by the Moon's Tidal Forces. The "spring force", resulting from this radial expansion, counteracts the Moon's tides. Indeed, the Earth "puffs up" until this "spring force" offsets said tides.


    To estimate the Earth's Spring Constant (K), consider constructing a (rigid) Dyson Sphere around the Earth. A large number (N) of hooks are lowered down from the inner surface of the Dyson Sphere, and anchored in the Earth. Equal tensions (T) are then applied to all of the anchor cables, generating a total expansive force (F) of N x T. Under this stress, the Earth (slightly) expands, by an amount [tex]\delta R[/tex]. The Spring Constant (K) can then be calculated from the Bulk Modulus (B):
    [tex] B \equiv \frac{Force / Area}{\Delta V / V}[/tex]
    [tex] = \frac{ F / 4 \pi R^{2} }{ (4 \pi R^{2} \delta R) / \frac{4 \pi R^{3}}{3} } [/tex] ​
    [tex]K \equiv \frac{F}{\delta R} = 12 \, \pi \, R \, B[/tex]
    [tex] \approx 2.4 \times 10^{19} N \, m^{-1}[/tex]​
    where we have used B = 1011 N m-2 for rock*.
    * D.C. Giancoli. Physics, pp. 310ff.
    Then, since K >> 4 [tex]\delta m \, \omega^{2}[/tex], we have:
    [tex] r(\theta) \approx \frac{\Phi}{2 K} + \frac{3 \Phi}{2 \, K} \times cos(2 \theta)}[/tex]​
    Let [tex]\lambda[/tex] represent the Rock Tide amplitude. Then,
    [tex]\lambda = r_{max} - r_{min} = 2 \times \frac{3 \Phi}{2 \, K}[/tex]
    [tex] = 6 \times \frac{\Phi}{2 \, K}[/tex]​
    Thus, if [tex]\lambda[/tex] ~ 30 cm, then the tidally induced average radial expansion is ~ 5 cm (ie., the Earth's crust oscillates from -25 to +35 cm). This result seems plausible, and may be Order-of-Magnitude accurate.


    We show that K >> 4 [tex]\delta m \, \omega^{2}[/tex]:
    [tex]\lambda = r_{max} - r_{min} = 2 \times \frac{3 \Phi}{2} \frac{1}{K - 4 \, \delta m \, \omega^{2}}[/tex]​
    [tex] (K - 4 \, \delta m \, \omega^{2}) \, \lambda = 3 \frac{G \, M_{moon} \, \delta m \, R_{earth}}{D^{3}} [/tex]​
    [tex] \delta m = \frac{K}{(4 \, \omega^{2}) + \frac{3 \, G \, M_{moon} \, R_{earth}}{\lambda \, D^{3}} }[/tex]
    [tex]\approx 4.4 \times 10^{24} kg[/tex]​
    [tex] 4 \, \delta m \, \omega^{2} \approx 9.2 \times 10^{16} N \, m^{-1}[/tex]​
  2. jcsd
  3. Jan 1, 2009 #2
    Planetary Compressions from Atmospheres

    From the definition of Bulk Modulus (B) (above), we have that:
    [tex]\frac{\Delta V}{V} = B^{-1} \times Pressure[/tex]​
    [tex]\frac{4 \, \pi \, R^{2} \, \delta R}{\frac{4 \, \pi \, R^{3}}{3}} = B^{-1} \times P_{atm}[/tex]​
    [tex]\frac{3 \, \delta R}{R} = B^{-1} \times P_{a}[/tex]​
    Again assuming that B ~ 1011 N m-2 (for rock), we can estimate the atmospherically induced Planetary Compressions of Venus, Earth, & Mars:
    Code (Text):

    Planet     Pressure (Atm)     Compression (m)
    Venus       90                       180
    Earth        1                         2.1
    Mars         0.01                      0.011
    This simple analysis suggests, that, on Earth, atmospherically induced Planetary Compression swamps out the tidally induced Planetary Expansion from the Moon.
  4. Jan 1, 2009 #3
    Consider a spherical Asteroid of diameter D, at radius R from the center of a planet. Its opposite ends are pulled apart by Tidal Forces equal to:
    [tex]\Delta F_{tidal} = \frac{2 \, G \, M \, m}{R^{3}} \times D[/tex]​
    This Tidal Force puts the Asteroid under a Tensile Stress equal to the differential force divided by the cross-sectional area:
    [tex]Tensile \, Stress = \frac{\Delta F}{\frac{\pi \, D^{2}}{4}} = \frac{2 \, G \, M \, m}{R^{3}} \times \frac{4}{\pi \, D}[/tex]​
    But the Asteroid's mass (m) is equal to:
    [tex]m = \frac{\pi}{6} \, D^{3} \times \rho[/tex]​
    [tex]Tensile \, Stress = \frac{4 \, G \, M \, \rho \, D^{2}}{3 \, R^{3}} [/tex]​
    Now, the Ultimate Tensile Strength (U) of rock is roughly 2 x 106 N m-2*. Thus, for a given Planetary Mass (M), and radial distance (R), the Ultimate Tensile Strength of rock limits the maximum size of an impactor before it's ripped appart by Tidal Forces:
    [tex]D^{2} \leq U \times \frac{3 \, R^{3}}{4 \, G \, M \, \rho}[/tex]​
    As the impactor approaches (ie., as R decreases), progressively smaller bodies are ripped appart.
    * Giancoli. Physics, pg. 313.
    For the Earth, at the "top" of its atmosphere (altitude ~100 km), we have that:
    [tex]D \leq 19 \, km[/tex]​
    where we have used a density of 2.96 g cm-3 for the impactor*. For comets, w/ density of 1.6 g cm-3, and using the Ultimate Tensile Strength of ice (~600 kPa**), we find:
    [tex]D \leq 14 \, km[/tex]​
    * The impactor, having negligible Surface Gravity, is "gravitationally uncompressed". See: https://www.physicsforums.com/showthread.php?t=278022
    ** http://www.crrel.usace.army.mil/sid/hopkins_files/Publications/hopkins_iutam.pdf
    CONCLUSION: This simple analysis suggests, that rocky impactors larger than about 19 km, and icy impactors larger than about 14 km, will be tidally disrupted before reaching the surface.
    Last edited: Jan 1, 2009
  5. Jan 1, 2009 #4

    D H

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    What are you going on about here, Widdekind? Tidal theory has been well-known for some time. Start with A.E.H. Love. This forum is not the place to regurgitate known physics and it particularly is not the place to stamp your misinterpretations of known physic (i.e., this: "atmospherically induced Planetary Compression swamps out the tidally induced Planetary Expansion from the Moon").
  6. Jan 1, 2009 #5

    Vanadium 50

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    I'm afraid your planetary compression is nonsense.

    If 15 psi of atmosphere causes the earth to "sag" by 2.1 meters, another 15 psi will surely cause it to sag by another 2.1 meters. So a car, which might push down on the 4 tire contact points with a pressure of 32 psi, should sink about 14 or 15 feet into the ground.

    That should make it evident that your model is oversimplified to the point of being completely wrong.
  7. Jan 1, 2009 #6
    That is not correct. The Bulk Modulus only applies, when the whole surface of the object, is subjected to uniform pressure.

    Thus, this simple model would predict, that:
    1. If you put such cars across the whole surface of the Earth
    2. Then it would "sag" another 2 m

    There is a difference between squeezing a rock in your hands (my model), and poking a single point on the surface w/ a pencil (your suggestion). Presumably, in the latter case, some sort of "surface tension" like forces apply.

    A better example, is the Isostatic Compression of the planet, from the massive Glaciers, of the Last Ice Age. ~1 km of ice compressed the crust around 100 m*. The column mass of said ice created base pressures of about 107 Pa, or roughly 100 Atm. My model would predict, then, a ~200 m "sag".
    CONCLUSION: My analysis is Order-of-Magnitude accurate. By way of comparison, Carroll & Ostlie often describe the underlying physics of Astronomical situations w/ simple models that are Order-of-Magnitude accurate, to w/in factors of 3-4. My model matches that standard.

    ADDENDUM: Areas in Montana have experienced "about 328 meters of isostatic uplift"*. And, around Hudson's Bay, ~300 m of uplift have been recorded**. Assuming a 2 km column height, for the Laurentide Ice Sheet over Montana, my model predicts 400 m of "sag". And, Isostatic Uplift is ongoing, even today. Again, this analysis is Order-of-Magnitude accurate, as advertised. I have not "pressed" the Significant Digits.
    * aprn.org/wp-content/uploads/2008/03/bathymetry.ppt
    ** https://www.dmr.nd.gov/ndgs/ndnotes/Rebound/Glacial Rebound.htm
    Last edited: Jan 1, 2009
  8. Jan 1, 2009 #7
    Apparently, then, ~1 km of ice compresses the crust ~300 m. My model predicts ~200 m. This is Order-of-Magnitude accurate. The essential Physics has been captured.
  9. Jan 2, 2009 #8

    D H

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    Exactly, and it is exactly why your analysis is wrong.

    You are cherry picking data, for one thing. The ice sheets were up to 3 times thicker than that, particularly in the areas where post-glacial rebound is greatest (see for example http://www.cig.ensmp.fr/~iahs/redbooks/a079/079019.pdf, and the wikipedia article you yourself cited). This already puts your analysis off by close to an order of magnitude.

    That is just the start of the problem. "There is a difference between squeezing a rock in your hands and poking a single point on the surface with a pencil." The ice sheets did not cover the globe. Your analogy is a false one by your own words. While the immense ice sheets did indeed compress the earth beneath them somewhat, the primary effect was to force material toward the equator. See, for example, http://www.homepage.montana.edu/~geol445/hyperglac/isostasy1. Your analogy is valid only if the Earth were fully covered with ice. It wasn't.
    Last edited by a moderator: Jan 2, 2009
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