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Effect of lunar/solar gravity on shape of earth

by daniel6874
Tags: earth, effect, gravity, lunar or solar, shape
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russ_watters
#37
Aug23-11, 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?
russ_watters
#38
Aug23-11, 05:32 PM
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Quote Quote by K^2 View Post
Are we looking at tidal force on the dumbbell? Ok. How about this. Ceter - Earth's center at t0, traveling at 30,000km/s in direction tangential to the line connecting Earth and dumbbell. Rotation - omega=1 with axis perpendicular to the direction of CS center's travel and the line connecting Earth and dumbbell. Dumbbell is stationary at t0 in this coordinate system, yet clearly experiences centrifugal tide.

Does that help?
Yes, thanks. So:

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.
K^2
#39
Aug23-11, 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 free-fall Reiman tensor.
A.T.
#40
Aug24-11, 01:59 AM
P: 3,912
Quote Quote by russ_watters View Post
What bulges are you referring to?
The bulges that cause tides. In your scenario without rotation, would the tides will still be the same, as in the case of Moon and Earth orbiting eachother?

Quote Quote by russ_watters View Post
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?
That is the tidal force due to gravity gradient alone, so it obviously doesn't include the effects of rotation. My question is if the effective tidal force for a rotating system would be different, and cause different tides than in a non rotating system.

For simplicity lets replace the Earth with two masses connected via a string :

O-------O 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?
harrylin
#41
Aug24-11, 02:23 AM
P: 3,181
Quote Quote by Hurkyl View Post
Agree or disagree:
  1. Tidal force from the moon causes a bulge on the near and far sides of the Earth, and flattens the places in-between.
  2. Tidal forces from the sun cause a bulge on the sides of the Earth near and far from the Sun, and flattens the places in-between.
  3. The Earth's rotation causes a bulge evenly distributed around the equator, and flattens the poles. [..]
The last one quoted (no.3) is misleading: due to the centrifugal action the land mass is bulged the same as the oceans, in a steady state (this is verified with atomic clocks: at mean sea level they have the same rate everywhere). Thus the earth's rotation cannot explain the tides.

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.
russ_watters
#42
Aug24-11, 05:54 AM
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Quote Quote by A.T. View Post
The bulges that cause tides. In your scenario without rotation, would the tides will still be the same, as in the case of Moon and Earth orbiting eachother?
The tides are the same size whether there is an orbit or not and whether there is a rotation or not.
That is the tidal force due to gravity gradient alone, so it obviously doesn't include the effects of rotation. My question is if the effective tidal force for a rotating system would be different, and cause different tides than in a non rotating system.
I'm sorry, but that's nonsense. By definition that's all the tidal force is.
For simplicity lets replace the Earth with two masses connected via a string :

O-------O 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?
If the system rotates, the force on the string will be greater, but that isn't a tidal force. What you are describing sounds to me like the constant height equatorial bulge, not the tidal force.
russ_watters
#43
Aug24-11, 05:57 AM
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Quote Quote by K^2 View Post
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 free-fall Reiman tensor.
You're claiming you can calculate a tidal force from that equation? Doesn't look like it to me, but I'd love to see it. I'd also like to see a reference for that being an equation for tidal force.
A.T.
#44
Aug24-11, 06:44 AM
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Quote Quote by russ_watters View Post
I'm sorry, but that's nonsense. By definition that's all the tidal force is.
I'm not asking about the definition of the term "tidal force". I'm asking if the tides on Earth are solely an effect of the "tidal force" due to gravity gradient, or if the effect is modified by the gradient in the centripetal force required for the orbit of the Earth's COM around the common COM of Earth & Moon. (In the co rotating frame it would be the gradient of the centrifugal force)

Quote Quote by russ_watters View Post
If the system rotates, the force on the string will be greater, but that isn't a tidal force.
I don't care how it's called. If the force is greater in a rotating system then it means the Earth's is stretched more, than it would be due to gravity gradient alone. That would mean that the rotation of the system does affect the magnitude of the tides on Earth.

Quote Quote by russ_watters View Post
What you are describing sounds to me like the constant height equatorial bulge, not the tidal force.
No. The equatorial bulge is due to the spin of the Earth around it's own axis in 24h periods. I'm talking about the effects of the rotation of the Earth's COM around the common COM of Earth & Moon in 27d periods.
russ_watters
#45
Aug24-11, 08:42 AM
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Quote Quote by A.T. View Post
I'm not asking about the definition of the term "tidal force". I'm asking if the tides on Earth are solely an effect of the "tidal force" due to gravity gradient...
Yes. The tides are soley due to the tidal force.
daniel6874
#46
Aug24-11, 02:51 PM
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Quote Quote by Hurkyl View Post
The primary effect of the gravity of an object is to pull other objects towards it.

Tidal force is the difference in gravitational attraction. If you have an elastic object and you tug (along a line) on the opposite sides with different force, then you stretch the object out along that line. This is the secondary effect of gravity -- to stretch objects. To deform spheres into things like seen at this wiki page.
Your reply is given without reference to other posts. To suggest that a prolate spheroid results from the moon's gravitational pull on the earth begs the question. The question is, why would a pull in the direction of the moon cause the waters on the far side of the earth to move in the *opposite* direction? At worst we would expect a bulge at the far side that did not exceed the "equatorial bulge."

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 g-field 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 them--the moon's pull. It's counterintuitive and I would like a cite for the calculation.
Hurkyl
#47
Aug24-11, 02:57 PM
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Quote Quote by daniel6874 View Post
The question is, why would a pull in the direction of the moon cause the waters on the far side of the earth to move in the *opposite* direction?
The waters on the far side don't move away from the moon.

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.
daniel6874
#48
Aug24-11, 03:31 PM
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Quote Quote by Hurkyl View Post
The waters on the far side don't move away from the moon.

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.
If you are agreeing with Russ, can you explain Simanek's "gradient"? The moon's gravitational field is a vector field--he speaks of finding the "gradient of the moon's gravitational force upon...[the volume of the earth]." Well, then it would be tensorial.

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.
A.T.
#49
Aug24-11, 03:47 PM
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Quote Quote by russ_watters View Post
Yes. The tides are soley due to the tidal force.
Okay, thanks. I found this website:
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.
xts
#50
Aug24-11, 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.
daniel6874
#51
Aug24-11, 04:06 PM
P: 64
Quote Quote by A.T. View Post
Okay, thanks. I found this website:
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.
This appears helpful. I will be interested to see if the effect is not slightly attenuated on the far-side of the earth w/r to the moon. And perhaps I can get a refund from Dover for Gamow's book?
K^2
#52
Aug24-11, 06:39 PM
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Quote Quote by russ_watters View Post
You're claiming you can calculate a tidal force from that equation? Doesn't look like it to me, but I'd love to see it. I'd also like to see a reference for that being an equation for tidal force.
Yeah, except I messed up transform, and the metric I gave you isn't Ricci flat. I'll fix it and get back to you.

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 non-zero 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 uαuα=-1)

So all you really need are 3 vectors: Rαtrt, Rαφrφ, and Rαtrφ.

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/s. 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/r.)
olivermsun
#53
Aug24-11, 10:00 PM
P: 617
Quote Quote by daniel6874 View Post
...I can get a refund from Dover for Gamow's book?
Perhaps, but from what I could see from the OP, there was nothing wrong with Gamow's explanation.
daniel6874
#54
Aug25-11, 04:59 AM
P: 64
Quote Quote by K^2 View Post
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/s. 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/r.)
Since you have gone to the trouble, perhaps you could explain why this might imply a bulge on the far side of the earth? The page Russ cited says the author found the gradient of the force-field to arrive at his conclusions about the movement of water. The gradient field in his picture, if correct, answers the question. I have access to Mathematica, so if you outline the calculation, that would be sufficient. It is clear we must take into account the shape of the earth.


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