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Effect of lunar/solar gravity on shape of earth 
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#37
Aug2311, 05:27 PM

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And just to be as clear as possible, here's the equation for tidal force:
[tex]F=2G \frac{mM}{r^3}dr[/tex] http://burro.cwru.edu/Academics/Astr...ity/tides.html Does anyone have an equation for tidal force that includes a centrifugal force? 


#38
Aug2311, 05:32 PM

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1. How did you come up with this coordinate system? 2. Can you show me how you would compute the tidal force using your coordinate system? By that I mean, show me the equations you would plug that information into to compute the force. 


#39
Aug2311, 07:07 PM

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1. I didn't want to deal with Coriolis effect, so I picked one where dumbbell system is stationary. I also like to use heavy body as center for CM for simpler symmetry. All other numbers are arbitrary.
2. Schwarzschild metric in the rotating frame: [tex]ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} 1\right) dt^2 + \left(1  \frac{2M}{r}\right)^{1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2  2 r \omega dt d\phi[/tex] I'm sure you know how to compute freefall Reiman tensor. 


#40
Aug2411, 01:59 AM

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For simplicity lets replace the Earth with two masses connected via a string : OO to the moon > Will the force in the string will be the same, regardless if the system rotates or not, or will it be different? 


#41
Aug2411, 02:23 AM

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The common (and probably correct) explanation is that the acceleration of the water due to the moon's "pull" is higher than that of the earth on the side of the moon, and lower than that of the earth on the other side. 


#42
Aug2411, 05:54 AM

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#43
Aug2411, 05:57 AM

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#44
Aug2411, 06:44 AM

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#45
Aug2411, 08:42 AM

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#46
Aug2411, 02:51 PM

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You suggest that a *difference* in attraction might result in a rebellion of waters at the far side of the earth, but this is contrary to the idea that the moon's gfield is continuous. It may fall off quickly as the distance from the moon, but it will not change directions. While I agree with Russ, who only acknowledges me in the third person, that resort to tensor calculus is unhelpful, and I can be persuaded that the argument he cites (Siminak, who directly contradicts Gamow) regarding centrifugal forces is correct, I do not see in the cited page the calculation of the "gradient of the moon's gravitational force" resulting in "tidal forces" whose sense is precisely opposite to that of the force creating themthe moon's pull. It's counterintuitive and I would like a cite for the calculation. 


#47
Aug2411, 02:57 PM

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The waters on the far side move away from the center of the Earth for exactly the same reason that the center of the Earth moves away from the waters on the near side. 


#48
Aug2411, 03:31 PM

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You are proposing that the earth arranges itself along the axis of lunar pull, and that if we suddenly placed the moon on the opposite side, in a while the (tidal) shape of the earth would be essentially unchanged. 


#49
Aug2411, 03:47 PM

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http://www.vialattea.net/maree/eng/index.htm It comes to the same conclusion: The orbiting of the Earth's COM around the common COM of Moon & Earth has no effect on the tides. I have not checked the math in every detail. It goes into great detail explaining common errors, and it has the calculation in all possible reference frames. 


#50
Aug2411, 03:54 PM

P: 882

"Tensorial"? "Simanek's gradient"? "Schwarzschild's metric"? Guys, stick to common sense!
Tide is a phenomenon that on a full/new moon you may go to the beach in the afternoon to pick crabs, moules, and other tasty living beings, while on the quartermoon you could drown in the place you used to pick crabs a week before, rather than being able to pick something for a dinner. No tensorial analysis is able to explain why on Bretonian beach tides are 2 metres high, while on Lofoten they may be 8 meters high. And even Simanek won't help you hunting crabs. 


#51
Aug2411, 04:06 PM

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#52
Aug2411, 06:39 PM

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Edit: Ok, here we go. Corrected form of the metric, first. [tex]ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} 1\right) dt^2 + \left(1  \frac{2M}{r}\right)^{1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2  2 r^2 \omega sin^2\theta dt d\phi[/tex] Given that metric, all you have to do is compute the Rieman tensor. [tex]R^\alpha_{\beta\gamma\delta} = \frac{\partial \Gamma^\alpha_{\beta\delta}}{\partial x^\gamma}  \frac{\partial \Gamma^\alpha_{\beta\gamma}}{\partial x^\delta} + \Gamma^\alpha_{\gamma\rho}\Gamma^\rho_{\beta\delta}  \Gamma^\alpha_{\delta\sigma}\Gamma^\sigma_{\beta \gamma}[/tex] Where the Christoffel symbols are defined as follows. [tex]\Gamma^\alpha_{\beta\gamma} = \frac{1}{2}g^{\alpha\delta}\left(\frac{\partial g_{\beta\delta}}{\partial x^\gamma} + \frac{\partial g_{\gamma\delta}}{\partial x^\beta}  \frac{\partial g_{\beta\gamma}}{\partial x^\delta}\right)[/tex] And using the line element I posted earlier, the nonzero elements of the metric tensor are following. [tex]g_{tt} = r^2\omega^2 sin^2\theta + \frac{2M}{r}  1[/tex] [tex]g_{rr} = \left(1\frac{2M}{r}\right)^{1}[/tex] [tex]g_{\theta\theta} = r^2[/tex] [tex]g_{\phi\phi} = r^2 sin\theta[/tex] [tex]g_{t\phi} = r^2\omega sin^2\theta[/tex] The tidal acceleration from curvature is easy. [tex]a^\alpha =  R^\alpha_{\beta r \delta}u^\beta u^\delta L[/tex] Since object starts out at rest, u is trivial. [tex]u^\alpha = (\left(1\frac{2M}{r}\right)^{1/2}, 0, 0, \omega \left(1\frac{2M}{r}\right)^{1/2})[/tex] (Feel free to verify that u^{α}u_{α}=1) So all you really need are 3 vectors: R^{α}_{trt}, R^{α}_{φrφ}, and R^{α}_{trφ}. Because I'm obviously not going to do all that work, I threw it into a Mathematica package and crossed my fingers. [tex]a^\alpha=\left(0, \frac{2M}{r^3}L, 0, 0\right)[/tex] Which is exactly the same result as I should get without rotation, so that's a good sign. If I substitute in the numbers for M, L, and r, I get tidal acceleration of 0.00015m/s². At the mass of 1000kg, that's 0.15N of tidal force. You should get exactly the same from classical method. (Especially since the classical formula gives you 2GM/r³.) 


#53
Aug2411, 10:00 PM

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#54
Aug2511, 04:59 AM

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