effect of lunar/solar gravity on shape of earth

D H

Better: We can say that centrifugal forces can be used to explain the tides. Saying that they cause the tides is going a bit too far IMO.

Consider a moon the same mass as ours that is in an orthogonal orbit about some planet the same size as the Earth. (Orthogonal orbit: An orbit with a zero angular momentum. When the moon hits the planet, the impact angle will be orthogonal to the ground.) At the instant that this hypothetical moon is one lunar distance from the planet, the tidal forces exerted by this moon on that planet will be exactly the same as the tidal forces exerted by our orbiting Moon on the Earth. Explaining that this is the case is a piece of cake if you use a gravity gradient based explanation. Now try from explaining using a centrifugal force based explanation.

You most certainly could explain the tides from the perspective of an Earth-fixed frame. Explanation follows my response to harrylin's post.

That's because oceanographers are elided. (Elided because of the old mantra, If you can't say something nice don't say anything at all.)

The reason that centrifugal forces due to the Earth's rotation must be disregarded is because those forces have already been taken into account. The starting point for an Earth-fixed based explanation of the tides is the geoid. The geoid is an equipotential surface of the gravitational plus centrifugal forces. Accounting for those centrifugal force again would be a double booking.

olivermsun

Hmm...what does this have to do with anything?

D H

What's wrong with it is that there is no math. Where people do use math in a centrifugal force-based explanation they inevitably do it wrong, and then hand-wave the mistakes away. I know that a correct explanation is possible, but I have yet to see one.

Compare that to a gravity gradient based approach. The desired outcome is the gravitational acceleration due to the Moon (or Sun) relative to the Earth as a whole. The Earth as a whole is accelerating toward the Moon by

[tex]{\boldsymbol a}_e = \frac {GM_m}{||{\boldsymbol R}||^3}\,{\boldsymbol R}[/tex]

where R is the vector from the center of the Earth to the center of the Moon. The gravitational acceleration toward the Moon at some point r away from the center of the Earth is

[tex]{\boldsymbol a}_p = \frac {GM_m}{||{\boldsymbol R}-{\boldsymbol r}||^3}\,({\boldsymbol R}-{\boldsymbol r})[/tex]

The relative acceleration is the difference between these two vectors:

[tex]{\boldsymbol a}_{\mbox{tidal}} = {\boldsymbol a}_p - {\boldsymbol a}_e[/tex]

The vector R can always be expressed in terms of a unit vector [itex]\hat{\boldsymbol r}[/itex] directed along [itex]\boldsymbol r[/itex] and some other unit vector [itex]\hat{\boldsymbol{\theta}}[/itex] normal to [itex]\hat{\boldsymbol r}[/itex]:

[tex]{\boldsymbol R} = R(\cos\theta \,\hat{\boldsymbol r} + \sin\theta\,\hat{\boldsymbol{\theta}})[/tex]

Using this notation and making a first order approximation for r<<R yields

[tex]{\boldsymbol a}_{\mbox{tidal}} \approx \frac {GM_m r}{R^3}
\left((3\cos^2\theta-1)\,\hat{\boldsymbol r} + 3\cos\theta\sin\theta\,\hat{\boldsymbol {\theta}}\right)[/tex]

Note that this covers the entire globe, not just the sublunar point. Try to do the same with a centrifugal-based approach. This is one of those cases where the inertial explanation is much easier, and much more general, than a rotating frame explanation.

D H

The places you are most likely to find very bad and/or mathematically incorrect explanations of the tides are textbooks and web sites written by oceanographers.

olivermsun

What is easy to explain in a few words as a "reason" for tides and what is easiest to compute for the surface of the earth are not necessarily the same thing. That doesn't necessarily make the explanation with "no math" less right, nor does it make the most expedient mathematical approach more right.

Personally, I found it kind of an interesting exercise to convince myself (mathematically) that several of the explanations offered earlier in the thread were indeed equivalent. As a side bonus, it tends to help you understand where people arguing from the various viewpoints are (mutually) confused.

Oh boy. I don't suppose you'd care to back this up with any math, or it it just going to be hand-waving?

russ_watters

Yes. So if we ignore the effect of the centrifugal force, we'll get a different result for an measured acceleration than if we include it -- a wrong prediction about reality. So:

1. Because including a centrifugal force component has a real/non-zero effect on the acceleration of a dropped object, we can say that both gravity and rotation play a role in causing the measured acceleration. One causes positive acceleration, one causes negative acceleration, but both together form the cause of the observed experimental result.

The corollary of this is what I am trying to convey in this thread:

2. Because including a centrifugal force component has no effect on the calculated tidal forces, we can say that the centrifugal force does not play any role in the cause the tidal force. Well if I was unclear before, that difference is the entire point I'm trying to make. But I would be more affirmative with it: Because the centrifugal force does not cause the tides, it should not be used in the explanation of the tides because using it in the explanation will fool people into thinking that that means it is the cause. Again, a corollary: explaining the far side bulge using a centrifugal force causes people to wrongly believe that static gravitational force cannot explain the far side bulge. The reason websites exist that are dedicated to dispelling that misconception is because explaining the tides using centrifugal force causes that misconception.

D H

What's wrong with it? Everything.

Starting with the first sentence.

The tide-raising forces at the earth's surface thus result from a combination of basic forces: (1) the force of gravitation exerted by the moon (and sun) upon the earth; and (2) centrifugal forces produced by the revolutions of the earth and moon (and earth and sun) around their common center-of-gravity (mass) or barycenter.

Right of the bat, this is just wrong. First off, what centrifugal force? To have a centrifugal force you have to have a rotating frame. There is no centrifugal force in an inertial frame in which the moon orbits the earth. There is a centrifugal force in a frame that rotates with the earth-moon system, but in that frame the earth and the moon are stationary. They are not orbiting one another in that frame.

A centrifugal force explanation of orbits can be useful at times. A geosynchronous satellite sits in a fixed position in the sky from our earthbound perspective. From our perspective, such a satellite indeed does have gravitational and centrifugal forces that are in balance, and per Newton's first law (extended to non-inertial frames), the satellite remains stationary. A rotating frame in which the earth and moon do not move can be useful at times. Such a frame is very useful for describing the Lagrange points, for example, and also is very nice for depicting the trajectory of a satellite in transit from the earth to the moon.

Continuing with the next paragraph,

With respect to the center of mass of the earth or the center of mass of the moon, the above two forces always remain in balance (i.e., equal and opposite). In consequence, the moon revolves in a closed orbit around the earth, without either escaping from, or falling into the earth - and the earth likewise does not collide with the moon.

That is the silly centrifugal force description of an orbit. And it is wrong. Look at it from the perspective of Newton's first law. If the two forces truly are in balance (i.e., equal and opposite), there is no closed orbit. There is only straight line motion or no motion at all.

Now let's look at the image that follows. You will see a similar image in most descriptions that invoke centrifugal force.



The moon and earth are clearly shown as orbiting one another here. There is no centrifugal force in this diagram.

Let's go on to the next diagram.



And this is flat out wrong. What's wrong is that this site (and almost every text that invokes centrifugal force) has the centrifugal force identical at the center of the earth, at the sublunar point, and at its antipode. The centrifugal force at a point removed from the axis of rotation by a distance r is rΩ². The sublunar point is only 1068 miles from the barycenter; the center of the earth, 2895 miles, and the antipodal point, 6858 miles. The centrifugal force is not the same at these three points. Do the math right and you do not get the tidal forces as shown in the diagram from the perspective of this rotating frame.

That's okay, though. Force is a frame dependent quantity when one allows fictitious forces to enter the picture. We don't want the tidal forces as observed from the perspective of this frame. We want the tidal forces as observed on the earth. To get a correct explanation we need to move the origin to the center of the earth -- and we need to get rid of that monthly rotation. The centrifugal forces that we added in need to be subtracted out.

What we're left with is a non-rotating with origin at the center of the Earth. This frame is the earth-centered inertial frame. Despite its name, this isn't really an inertial frame; it is accelerating toward the moon and the sun. This acceleration of the origin results in a fictitious force, and it is a uniform fictitious force. This fictitious force is the inertial force, not the centrifugal force.

There is little need to invoke the earth-moon barycenter here and there is absolutely no need to invoke the centrifugal force.

harrylin

Yes, I already mentioned that in post #41.

olivermsun

You seem to start with the conclusion that the explanation is wrong (or more accurately, not to your tastes), and then go on to interpret everything in the "most wrong" way possible. Since you do indeed get the same numerical answers for the tide-producing "forces" in either an inertial frame or a moving frame with "centrifugal" forces, let's try and see why this could happen despite "everything" being wrong.

At this point, no frame has been specified, so I would tend to assume that the usage of centrifugal force is colloquial, or else that a rotating frame is about to be used.

I think it's fair to assume that a reader will understand that the centrifugal forces "observed" in the rotating frame are the result of orbital motion.

In the rotating frame, which as you pointed out is the only frame in which the centrifugal "force" is observed, the forces are balanced. Hence in the rotating frame, we do not observe the moon falling into or away from the earth. The reader probably realizes that such motion (or lack thereof) observed in the rotating frame corresponds to a closed orbit in the inertial frame.

Is centrifugal force really this confusing?

The orbiting satellite problem is commonly illustrated in physics textbooks using a schematic such as this. The orbital motion is shown in the inertial frame so that we understand the "big picture," but the solution includes defining (and sketching in) the proper moving frame axes. Here no particular coordinates have been defined (yet?).

I'll grant you, this figure is not really clear or rigorous regarding what quantities are really being represented by what "vectors." Amusingly, however, the centrifugal forces actually do become equal at points A and B when the math is "done right," and you do in fact get the correct tide-producing forces.

We didn't add in or take away anything at all. The earth is observed to be rotate even in the "inertial" frame, and we simply choose to decompose its total rotation into parts which are either convenient or intuitively useful. For the rotating frame, some may choose to keep the part of the rotation which locks the earth and moon faces together and then deal with the consequences later.

This just sounds like frame (or word) games to me.

And finally, to respond to this oft repeated point:
Which is why, at the end of the rotating frame computation, you subtract the "symmetric" component of centrifugal force which is due to rotation around the earth's center. Then, you get a nice final answer which is numerically correct, leaves the symmetric part in the geoid where it's already done for you, and shows you some interesting equivalences between frames rotating around different axes to boot.

A.T.

I think in general that cause/effect arguments tend to get too philosophical. But I think the point you trying to make is:

- The gravitational gradient is the same in every reference frame
- The tidal effect computed in every reference frame must be the the same

Therefore:

- The sum of all inertial forces (which might arise from the choice of the reference frame) must not affect the tidal effect.

A.T.

1) In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?

2) If the centrifugal force vector is the same for A & B & C (as shown in the picture) then it cannot cause/affect the tides, so why show/mention it in an explanation of tides?

olivermsun

The residual vectors really are equal, pointing in the same direction and everything, once the earth rotation at 1/lunar orbit is taken out. I suppose that's part of what's amusing about it. The vectors make sense since every point on the earth (if you take out all the rotation around its axis) translates in a circle with the same radius as the distance from the center of the earth to the barycenter.

It provides the constant "offset" that gives you a tide-producing force toward the moon on the near side, away on the far side, and with the whole system static in the moving frame (the moon and earth maintain their relationship).

A.T.

I have not idea what you mean by "residual vectors" and "taking out the earth rotation". The inertial centrifugal force in rotating frames points away from the frame's axis of rotation, and has the magnitude m [itex]\omega[/itex]² r, where [itex]\omega[/itex] is the angular velocity of the frame, and r is the distance to the frame's axis of rotation. So again:

In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)?


A uniform inertial force field causes uniform acceleration, so it cannot cause or contribute to any deformation (tides).

olivermsun

The moon's gravitational force is nonzero and pointed toward the moon (even if there is a gradient), yet the earth center stays still in the rotating frame, and the "tide-producing force" on the antipodal side of the earth even has opposite sign. Hence the centrifugal force is "necessary" to describe what is observed in the rotating frame.

Let r be the radius of the earth, x the distance from the barycenter to the center of the earth, and [itex]\omega[/itex] the orbital frequency. In the rotating frame, the "centrifugal acceleration" at the center of the earth is [itex]\omega^2 x[/itex] directed to the right, at A is [itex]-\omega^2 (r - x)[/itex], directed toward the left, and at B is [itex]\omega^2 (r + x)[/itex] toward the right again. Since people like to remove the component of centrifugal force which is due to rotation around the earth's axis and put that into the geoid, take out outward accelerations [itex]-\omega^2 r[/itex], [itex]+\omega^2 r[/itex] from the accelerations at A and B, respectively. The remaining centrifugal bits are what I called the "residual" vectors. What are they?

A.T.

So the Fc vectors shown in the picture are actually not the centrifugal force, but the uniform components of the centrifugal force, left after the radial component in respect to the Earth's center was removed.

They form a uniform field, which cannot create tidal deformation, so they are not a good "reason" for the tides. But I see how in that particular frame they can be used to "explain" the opposite bulge on a static Earth:

<- Moon's gravity is pulling to the left with a r^-2 fall off.
-> Uniform centrifugal component pulls to the right

- In the Earth's center they cancel
- In A Moon's gravity wins and creates a bulge to the left
- In B Uniform centrifugal component wins and creates a bulge to the right

D H

What they are is invalid. That is exactly the hand-waving stuff I was talking earlier. It is invalid.

Here's a web site where an oceanographer does it right: http://oceanworld.tamu.edu/resources...apter17_04.htm. Emphasis mine:

Note that many oceanographic books state that the tide is produced by two processes: i) the centripetal acceleration at earth's surface as earth and moon circle around a common center of mass, and ii) the gravitational attraction of mass on earth and moon. However, the derivation of the tidal potential does not involve centripatal acceleration, and the concept is not used by the astronomical or geodetic communities.

This is a very nice description of the tides. Stewart starts off in terms of potential, develops the horizontal component of the tidal force, and then proceeds to dive deep, eventually reaching the concept of Doodson numbers.


Let's see what that constant [itex]r\omega^2[/itex] term of yours really is. The angular velocity of a pair of objects orbiting one another circularly due to gravity is given by

[tex]\omega^2= \frac{G(m_1+m_2)}{R^3}[/tex]
where ω is the angular velocity, G is the universal gravitational constant, m₁ and m₂ are the masses of the two bodies, and R is the distance between the two bodies. (This generalizes very nicely to elliptical orbits; it is Kepler's third law.)

Multiply both sides of the above by r₁, the distance between the center of mass and m₁, and we have the desired rω² term on the left hand side:

[tex]r_1\omega^2 = \frac{G(m_1+m_2)r_1}{R^3}[/tex]
The relationship between r₁ and R is given by m₁r₁=m₂r₂=m₂(R-r₁) or (m₁+m₂)r₁=m₂R. Substituting this into the right hand side of the equation for r₁ω² yields

[tex]r_1\omega^2 = \frac{Gm_2}{R^2}[/tex]
Recognizing that our generic body 1 is the Earth and that body 2 is the Moon, we have

[tex]r\omega^2 = \frac{Gm_{\text{M}}}{R^2}[/tex]
The right hand side is the translational acceleration of the Earth toward the Moon. This is what you really want because the answer you really want is in a frame with origin at the center of the Earth and rotating with the Earth.

You can use a centrifugal force concept to develop those forces in that frame with origin at the barycenter and rotating at the Earth-Moon orbital rate. Do it right and you will get some goofy answers. That's okay though because those answers are to a question other than the question we are trying to answer. To answer the question that was asked, you are going to have to transfer those forces from that frame to the Earth-centered, Earth-fixed frame. All of those centrifugal terms you so painstakingly added in have to be painstakingly subtracted out. A barycentric POV misses one fictitious force, that due to the translational acceleration of the Earth toward the Moon. This missed term also needs to be taken into account when transforming those forces from the rotating barycentric frame to an Earth-centered frame. Do the math right and you will get the same answer as you would have gotten by starting in an Earth-centered frame in the first place. So why not start in an Earth-centered frame in the first place? For one thing, the math is a lot easier, and for another, it avoids all this going around your elbow to get to your thumb business.

Just because the fictitious centrifugal force in the rotating barycentric frame at the center of the Earth has the same value as the fictitious inertial force in an Earth-centered at the center of the Earth does not mean these are the same fictitious force. They are very different fictitious forces.

A.T.

I don't think it misses that inertial force. I think the uniform component of the inertial centrifugal force in the barycentric frame (that olivermsun called "residual vectors", and you called "invalid", and is called Fc in the picture) is exactly that inertial force.

The decomposition of the centrifugal force in the barycentric rest frame is explained here in Fig 12
http://www.vialattea.net/maree/eng/index.htm

I agree. And in the Earth-centered frame you will always have that uniform inertial force field, pointing away from the Moon. This is even true without any rotation, just linear fall as described in Case 1 of:
http://www.vialattea.net/maree/eng/index.htm

This uniform inertial force field turns this:



into this:



So I think it is OK to use that uniform inertial force field to explain why the water on the opposite side is accelerated away from the fixed Earth's center.

What is misleading, is calling that uniform inertial force field "centrifugal force".