It is easy to show(adsbygoogle = window.adsbygoogle || []).push({}); *, that theDifferential (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 thisTidal Forceis:

[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]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] \equiv \Phi \times \left( \frac{1}{2} + \frac{3 cos(2 \theta)}{2} \right) [/tex]

[tex] \Delta F_{r} \equiv \delta m \times \frac{\partial^{2}}{\partial t^{2}} r[/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.

[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]

Thus, this net outward radial (tidal) force must be offset by some sort of*Carroll & Ostlie. Introduction to Modern Astrophysics, pg. 763.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, byFourierdecomposition:

[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]

INTERPRETATION:

The constant term ([tex]\frac{\Phi}{2 K}[/tex]) represents the net average radial expansion of the Earth, induced by the Moon'sTidal 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.

ESTIMATION of EFFECT:

To estimate the Earth'sSpring Constant(K), consider constructing a (rigid)Dyson Spherearound the Earth. A large number (N) of hooks are lowered down from the inner surface of theDyson Sphere, and anchored in the Earth. Equal tensions (T) are then applied to all of the anchor cables, generating a total expansive force (F) ofNx T. Under this stress, the Earth (slightly) expands, by an amount [tex]\delta R[/tex]. TheSpring Constant(K) can then be calculated from theBulk Modulus(B):

[tex] B \equiv \frac{Force / Area}{\Delta V / V}[/tex]Therefore:

[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]where we have used B = 10

[tex] \approx 2.4 \times 10^{19} N \, m^{-1}[/tex]^{11}N m^{-2}for rock*.

Then, since K >> 4 [tex]\delta m \, \omega^{2}[/tex], we have:*D.C. Giancoli. Physics, pp. 310ff.

[tex] r(\theta) \approx \frac{\Phi}{2 K} + \frac{3 \Phi}{2 \, K} \times cos(2 \theta)}[/tex]Let [tex]\lambda[/tex] represent theRock Tideamplitude. Then,

[tex]\lambda = r_{max} - r_{min} = 2 \times \frac{3 \Phi}{2 \, K}[/tex]Thus, if [tex]\lambda[/tex]

[tex] = 6 \times \frac{\Phi}{2 \, K}[/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 beOrder-of-Magnitudeaccurate.

APPENDIX:

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]So:

[tex] (K - 4 \, \delta m \, \omega^{2}) \, \lambda = 3 \frac{G \, M_{moon} \, \delta m \, R_{earth}}{D^{3}} [/tex]And:

[tex] \delta m = \frac{K}{(4 \, \omega^{2}) + \frac{3 \, G \, M_{moon} \, R_{earth}}{\lambda \, D^{3}} }[/tex]And:

[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]

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# Moon's Tidal Forces puff up Earth

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