effect of lunar/solar gravity on shape of earth

russ_watters

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.

russ_watters

It just so happens someone posted a discussion of exactly what I'm referring to in another thread (the second diagram shows it): http://www.lhup.edu/~dsimanek/scenario/tides.htm

Now here's where it gets really interesting: 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.

russ_watters

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

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

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.

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?

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

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

The tides are the same size whether there is an orbit or not and whether there is a rotation or not. I'm sorry, but that's nonsense. By definition that's all the tidal force is. 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

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.

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)

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.

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

Yes. The tides are soley due to the tidal force.

daniel6874

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

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

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.

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

"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

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?