Effect of lunar/solar gravity on shape of earth

[daniel6874]

In Gamow's short book 'Gravity' he describes the effect of lunar gravity on the earth, in particular the seeming paradox that the oceans, in response to the tug of the moon, swell both in the direction of the moon and on the opposite side as well. He resolves the paradox by explaining that Kepler's laws require that the greater angular velocity of water on the opposite side of the Earth send them further out, to satisfy the requirement that greater velocity implies a greater distance from the orbital focus.

My question is this: where does the extra water come from? Is water from the poles drawn down to form the tidal surplus? There is no net addition of water, and water does not "expand." Does this mean that sea level at the poles drops?

Thanks for any enlightenment.

 

[xts]

Water from low tide areas is moved to high tide areas, but only within every single ocean. That is why there are no visible tides on small closed seas, like Baltic. Inertia of that oceanic water flowing in/out shallow waters makes tides to be much higher at the shore than on open ocean.

 

[chrisbaird]

I agree with xts. If your position on Earth is facing the moon, you will have high tide, which means that ocean water has been drawn by the moon's gravity to your location. Points on Earth that are 90 degrees away from you, measured relative to the line connecting the Earth and the moon, will be experiencing low tide. The water was drawn by moon's gravity from those locations to your position. It's not an issue of the poles always loosing water, because the moon does not hover constantly over the equator. The Earth's rotation and the moon's orbit means that it is constantly located over different spots on earth, so that every ocean coast constantly cycles through low-tide and high-tide.

 

[xts]

As usually, the main problem is America ;) It almost completely separates oceans. Bering's Straits is narrow and shallow - so no significant amount of water may flow between Pacific and Arctic Ocean. On the South, Drake's Passage is not that narrow, but it still stops any significant flow between Atlantic and Pacific.
But Atlantic is wide enough to make its own pendulum-like waving: as you have high tide at European coast, America is just at low tide. And vice versa. You have even bigger gap between Americas and Japan/Phillipines/Australia.

 

[russ_watters]

Also, I don't know who Gamow is, but it is not correct to attribute the far side bulge to centripetal acceleration. The tidal force is a STATIC force, completely unrelated to any rotation.

 

[Bill_K]

Gamow was a leading physicist who did seminal work on alpha decay and big bang nucleosynthesis.

Laying blame for the tidal bulge on centrifugal force is an unusual way of putting it, but essentially correct. In the rotating frame there are two "forces" acting on the Earth, the centrifugal force and the lunar gravitational attraction. At the Earth's center they are in balance, while at the Earth's far side the gravitational attraction is less and the centrifugal force is more. Both effects combined are responsible for the tidal bulge.

 

[russ_watters]

The centrifugal forces on the near and far side cancel each other out. It is no more correct than adding a term for the degrees of Kevin Bacon to each side of the equation. Sure, the math still works out, but that's just because you haven't actually changed anything.

 

[K^2]

russ, it's a matter of coordinate system choice. You can take a CS that rotates with the planet-moon system and has its center at the center of mass of two bodies. In that CS, the far side is clearly experiencing more centrifugal force. The effective gravitational potential, on the other hand, changes shape, so the gravitational tidal force is reduced. The net effect ends up being exactly the same.

Naturally, taking an inertial frame of reference and getting bulge entirely due to gravitational tidal force is a far less cumbersome way to do it.

 

[olivermsun]

The centrifugal forces on the near and far side cancel each other out.

The centripetal accelerations on the opposite sides of the Earth do not cancel out, as the gravitational potential is not constant with distance from the orbital focus in the inertial frame.

In terms of Kepler's law, this is the balance between (orbital) angular velocity with and distance.

 

[russ_watters]

You're not hearing me: if the Earth and moon were not orbiting each other, there would be no rotating coordinate system and the tidal force would be the same. It is not a matter of picking one coordinate system over another. I understand that the math still works.

 

[russ_watters]

The centripetal accelerations on the opposite sides of the Earth do not cancel out, as the gravitational potential is not constant with distance from the orbital focus in the inertial frame.

There is no g in the centripetal force equation.

 

[K^2]

You're not hearing me: if the Earth and moon were not orbiting each other, there would be no rotating coordinate system and the tidal force would be the same. It is not a matter of picking one coordinate system over another. I understand that the math still works.

I'm well aware of that. Yet, whether or not the force is there is still a matter of coordinate system. That's kind of the deal with any fictitious force, gravity not excluded. To say that one coordinate system is the right one and the other is wrong is silly. Some just happen to be easier to work with. You don't need centrifugal force to explain tidal bulging. The most general explanation would certainly need to avoid it. But it's not wrong when dealing with specific system to say that the bulging is partially caused by centrifugal force. With suitable choice of coordinate systems, it is.

 

[russ_watters]

Let me try this another way: if one learns that the tidal force is a consequence of differential centripetal acceleration, how would they calculate the tidal force in a scenario where there is no rotation?

 

[K^2]

By accounting for ALL fictitious forces. Like I said, gravity is one of them. In one CS, it's just gravity. In another, it's gravity + centrifugal. It's a CS choice.

 

[russ_watters]

So you would invent a rotation if none existed? How would you calculate how fast the universe is rotating in order to select that rotating frame of reference?

 

[olivermsun]

The centripetal accelerations on the opposite sides of the Earth do not cancel out, as the gravitational potential is not constant with distance from the orbital focus in the inertial frame.
There is no g in the centripetal force equation.

I was talking about the potential due to the sun's gravitational field, which is the centripetal force toward the "orbital focus" in Gamow's words. I was not talking about centripetal force due to the Earth's rotation (I don't believe Gamow was, either).

 

[daniel6874]

Also, I don't know who Gamow is, but it is not correct to attribute the far side bulge to centripetal acceleration. The tidal force is a STATIC force, completely unrelated to any rotation.

I was surprised to find additional replies to this question. Russ, to what would you attribute the far-side bulge, then? If the answer to this appears in subsequent discussion, please alert me to the comment number. Thanks.

 

[K^2]

So you would invent a rotation if none existed? How would you calculate how fast the universe is rotating in order to select that rotating frame of reference?

It's a coordinate system transformation. It's not changing laws of physics. I'm not sure what your complaint here is.

 

[Hurkyl]

It's a coordinate system transformation. It's not changing laws of physics. I'm not sure what your complaint here is.

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.
4. The Earth's revolution around the Earth-moon center causes a bulge on the far side of the moon and flattens the side nearer to the moon
5. The Earth's revolution around the Sun is insignificant.

I lay 80% odds Russ thinks you are saying "#3 is true, and completely explains the tides", or possibly that you are saying "#1 and #3 are the same thing", or something similar. (my first time reading through the thread, it certainly sounded to me like that was what you were saying)

 

[K^2]

What I'm saying is that choice of center and rate of rotation for a rotating coordinate system are completely arbitrary. For any choice, I'll end up with some centrifugal force that I'll have to add to all massive objects. For some choices of the rotating frame, such as choosing Earth-Moon CoM as center of rotation, centrifugal force is certainly greater on far side, and therefore contributes to tidal bulging.

In general, a rotating frame of reference has its own tidal force associated with it. If you are working in a rotating frame of reference, the CS tidal force must be compounded with tidal force due to metric. (Well, if you transform the metric, that should take care of itself, but I hope you get what I mean.)

 

[Hurkyl]

I was surprised to find additional replies to this question. Russ, to what would you attribute the far-side bulge, then? If the answer to this appears in subsequent discussion, please alert me to the comment number. Thanks.

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.

 

[russ_watters]

What I'm saying is that choice of center and rate of rotation for a rotating coordinate system are completely arbitrary.

I am aware. What I asked was how would you choose one if none was obvious?

For example, if a dumbell shaped object with weights 1000kg, 10km apart is held stationary wrt the background of stars, 30,000km from Earth and is dropped, falling straight toward earth, with the bar pointing away from earth, how would you find the rotating frame of reference necessay to calculate the tidal force between the weights?

 

[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.

 

[willem2]

There is really only one way of explaining the tides in a simple way, and that is in a rotating reference frame, where the difference in gravity from the moon is the only cause of the bulges.

If you really want to do it in a rotating reference frame, it can of course be done, but you need to account for:

1. The centrifugal force from the rotating reference frame. \omega^2 r where r is the distance from the barycenter.

2. The acceleration because of the rotation of the earth. Take into account that the Earth will rotate a bit slower, because of the reference frame also rotates in the same direction.

3. The coriolis force, because of the rotation of the earth.

4. The differences in gravity from the moon. Since these are equal to what you get in
a non-rotating frame, it follows that the first 3 effects must completely cancel.

 

[K^2]

I am aware. What I asked was how would you choose one if none was obvious?

For example, if a dumbell shaped object with weights 1000kg, 10km apart is held stationary wrt the background of stars, 30,000km from Earth and is dropped, falling straight toward earth, with the bar pointing away from earth, how would you find the rotating frame of reference necessay to calculate the tidal force between the weights?

It. Is. Arbitrary. You can chose ANY coordinate system. In any coordinate system, the results will be the same. In some systems, part of the tidal bulging is due to centrifugal force. In other systems, there is no centrifugal force.

You can choose ANY system you like and do the computations. You'll get the SAME answer.

 

[willem2]

It. Is. Arbitrary. You can chose ANY coordinate system. In any coordinate system, the results will be the same. In some systems, part of the tidal bulging is due to centrifugal force. In other systems, there is no centrifugal force.

You can choose ANY system you like and do the computations. You'll get the SAME answer.

But the differences in gravity are present in ALL coordinate systems. since the tides are also the same in ALL the coordinate systems, all other effects must cancel. Doing it in a rotating frame is also hopelessly complicated, and if you'd really bothered to compute the centrifugal forces, you'd see that you'd get the wrong numbers, because there would be several other effects you overlooked.

 

[russ_watters]

It. Is. Arbitrary. You can chose ANY coordinate system. In any coordinate system, the results will be the same. In some systems, part of the tidal bulging is due to centrifugal force. In other systems, there is no centrifugal force.

You can choose ANY system you like and do the computations. You'll get the SAME answer.

Yikes. Again, I am aware it is arbitrary. I am asking you to tell me what your choice is and how you made it! Ie:

For the example I gave, I would choose a frame that is Earth centered and not rotating. What would yours look like? Earth centered and rotating? How fast would you like it to rotate?

 

[olivermsun]

Since this thread persists: where did the OP ever mention rotating coordinates?

 

[russ_watters]

He doesn't specifically, he just talks about centrifugal forces. It's the meat of the explanation - the center third of the post.

 

[K^2]

For the example I gave, I would choose a frame that is Earth centered and not rotating. What would yours look like? Earth centered and rotating? How fast would you like it to rotate?

For sake of example, take center at Earth-Moon CoM, rotating with the same period as the system, so both objects are roughly stationary in that coordinate system. Clearly, there is centrifugal tide in this setup.

 

[russ_watters]

For sake of example, take center at Earth-Moon CoM, rotating with the same period as the system, so both objects are roughly stationary in that coordinate system. Clearly, there is centrifugal tide in this setup.

I didn't ask about the moon - I already knew what you chose for that case. I asked about the example I gave, where there is no obvious choice of rotation rate. What would you choose in that case?

 

[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?

 

[olivermsun]

He doesn't specifically, he just talks about centrifugal forces. It's the meat of the explanation - the center third of the post.

The explanation invokes angular momentum around the orbital focus, by Kepler's convention.

 

[A.T.]

if the Earth and moon were not orbiting each other, ... the tidal force would be the same.

The tidal force due to gravity gradient would be the same. But would the bulges be the same without rotation?

 

[russ_watters]

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.

 

[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.

 

[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/Astr221/Gravity/tides.html

Does anyone have an equation for tidal force that includes a centrifugal force?

 

[russ_watters]

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]

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.

 

[A.T.]

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 each other?

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/Astr221/Gravity/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 let's 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]

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]

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 each other?

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 let's 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]

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.

 

[A.T.]

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)

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.

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]

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]

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]

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]

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.]

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]

"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.