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## effect of lunar/solar gravity on shape of earth

 Quote by A.T. The tidal force due to gravity gradient would be the same. But would the bulges be the same without rotation?
What bulges are you referring to? The earth has a doughnut shaped bulge of 23 km due to its rotation and two additional bulges due to the tidal force.

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 Quote by russ_watters Hurkyl, what I think I'm seeing for the near side doesn't sound like any of those: The tidal force creates a near-side bulge and flattens the far side. Then #4 would explain the far side bulge.
It just so happens someone posted a discussion of exactly what I'm referring to in another thread (the second diagram shows it):
 One of the few books that clearly defines "tide" at the outset is The Planetary System by Morrison and Owen [1966]: "A tide is a distortion in the shape of one body induced by the gravitational pull of another nearby object." This is definition (2) above. It clearly says that tides are the result of gravitation, without any mention of rotation effects....
http://www.lhup.edu/~dsimanek/scenario/tides.htm

Now here's where it gets really interesting:
 In this representation [the centrifugal force misconception] we can treat this system as if it were an inertial system, but only at the expense of introducing the concept of centrifugal force. It turns out that when this is done, the centrifugal force on a mass anywhere on or within the Earth is of constant size, and is therefore equal to the size of the gravitational force the moon exerts on the same amount of mass at the center of the Earth.... ... We are now focusing on the effects due only to the Earth-moon system. The motion of the Earth about the Earth-moon center of mass, causes every point on or within the Earth to move in an arc of the same radius. This is a geometric result most books totally ignore, or fail to illustrate properly. Therefore every point on or within the Earth experiences the same size centrifugal force. A force of constant size throughout a volume cannot give rise to tidal forces (as we explained above). The size of the centrifugal force is the same as the force the moon exerts at the Earth-moon center of mass (the barycenter), where these two forces are in equilibrium. [This barycenter is 3000 miles from the Earth center—within the Earth's volume.]
That's a twist I actually wasn't aware of, but of course it makes sense: The barycenter is not a fixed point that the earth rotates around, but a virtual point that moves as the moon moves around the earth.
 Mentor And just to be as clear as possible, here's the equation for tidal force: $$F=2G \frac{mM}{r^3}dr$$ http://burro.cwru.edu/Academics/Astr...ity/tides.html Does anyone have an equation for tidal force that includes a centrifugal force?

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 Quote by K^2 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.
 Recognitions: Science Advisor 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: $$ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} -1\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2 - 2 r \omega dt d\phi$$ I'm sure you know how to compute free-fall Reiman tensor.

 Quote by russ_watters 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 by russ_watters And just to be as clear as possible, here's the equation for tidal force: $$F=2G \frac{mM}{r^3}dr$$ 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?

 Quote by Hurkyl Agree or disagree: Tidal force from the moon causes a bulge on the near and far sides of the Earth, and flattens the places in-between. 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. 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.

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 Quote by A.T. 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.

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 Quote by K^2 2. Schwarzschild metric in the rotating frame: $$ds^2 = \left(r^2\omega^2 sin^2\theta + \frac{2M}{r} -1\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 sin\theta d\phi^2 - 2 r \omega dt d\phi$$ 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.

 Quote by russ_watters 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 by russ_watters 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 by russ_watters 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.

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 Quote by A.T. 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.

 Quote by Hurkyl 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.

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 Quote by daniel6874 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.

 Quote by Hurkyl 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.

 Quote by russ_watters 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.
 "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.

 Quote by A.T. 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?

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