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Widdekind
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It is easy to show*, that the Differential (Tidal) Force, produced by the Moon, around the surface (equator) of the Earth, is:
INTERPRETATION:
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.
ESTIMATION of EFFECT:
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):
APPENDIX:
We show that K >> 4 [tex]\delta m \, \omega^{2}[/tex]:
[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] \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]
[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.[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]
* 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]
INTERPRETATION:
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.
ESTIMATION of EFFECT:
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]
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]
[tex] \approx 2.4 \times 10^{19} N \, m^{-1}[/tex]
where we have used B = 1011 N m-2 for rock*.[tex] \approx 2.4 \times 10^{19} N \, m^{-1}[/tex]
* 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.[tex] = 6 \times \frac{\Phi}{2 \, K}[/tex]
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]
[tex]\approx 4.4 \times 10^{24} kg[/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]