Your ellipsis covers the parenthetical "(or depression)" that apparently refers to that same acceleration. That depression is the low tide. The tidal acceleration is recorded by accelerometers around the world. It's in every book on the theory of tides that I've ever seen.
I have.

This sounds to me like another way of expressing the Different-Orbits theory that has already been mooted in this thread. What do you think? Do both explanations (yours and the DO theory) amount to the same thing?

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milli360 wrote:
That would explain why I’ve never heard of it, but I can confirm that I’ve recently found it from another source. Having said that, my position remains that, whilst such an acceleration may exist, it is not a pre-requisite. If it does exist then yes, a complete theory of tides should address it – mine included (if ever I should produce one).

The DO theory is correct as far as it goes and I have no problem with it. But I’m sure the author could have taken it further.

I’ve given more detail in particular why no outward force is necessary. I’ve shown how centripetal acceleration, solar gravity and terrestrial gravity interact, and I’ve given you the tools to check the results for yourself.

Does it help any or does it raise more problems?

Which outward force? I think we've discussed before, neither your explanation nor the BA's uses anything but gravity, right? As a centripetal force?

milli360 wrote:
I was making a general point that any explanation should show that no outward force is required, to be credible it must show this explicitly.

However, since you ask, here a snip from Sawicki:

'... there appears to be another force, the tidal force, pulling the mass m at points C and F up...' (Sawicki's emphasis),

and one from the BA:

'It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides. This is called vector addition. If we do that, our diagram will look like this:

<- X ->
far center near
side of Earth side '

Not what I’d call explicit (that didn't print properly did it? - take a look at the BA home page).

But I’m not much bothered about this, what concerns me most is that Sawicki’s paper totally ignores the centripetal acceleration of earths orbit. I’m frankly baffled that he should be so familiar with the suns gravitational gradient (D(a)s) yet (apparently) so ill informed about the complementary centripetal gradient.

Replacing centripetal acceleration with ‘The Earth is simply in free fall towards the Sun’ is staggeringly naïve – it produces in a wrong explanation and it produces incorrect results.

I assume you got that from the webpage. There may be more there, but the book seems explicit enough: "It seems paradoxical that gravity can act in such a way as to make something feel a force away from an object, but in this case it's because we are measuring that force relative to the center of the Earth."

That's fairly explicit, saying that gravity is the only force required.
As I've mentioned before, if you calculate the centripetal force required for each piece of a non-rotating Earth, the gradient is zero. In other words, the force is constant, in direction and magnitude, across the body of the Earth.
I disagree. The Earth is in free fall, after all, and the calculations match the data. The alternative you've suggested doesn't seem even to explain the low tide.

As someone said Einstein said, you want to make the theory as simple as possible, but no simpler.

OPPOSITE FACE LUNAR TIDAL BULGES ARE ELECTROSTATIC?

"Why is there a tidal bulge on the opposite side of the Earth?" Excellent question! But in reading the posts, I'm still not satisfied...

In fact I've never found a truly satisfactory reason in all my readings why on the opposite side of the globe facing away from the moon or sun, are nearly the same as the tidal bulge facing them. The idea that the gravitational pull on the opposite side of the planet is canceled by the planet does not satisfy me. 'Push' gravity does not explain it effectively either, nor does the "three point pull". So I would like to present an idea which I suspect is original, if not crazy, though another may have thought of it first: the risen tide on the planet's opposite face away from the moon or sun is due to water's electrostatic charge.

As outlandish as this may appear at first blush, consider the following suggestions:

1. There is a magnetic field from pole to pole, if off axis, that permeates the globe. This magnetic field gives a negative charge value to the northern hemisphere and a counter positive charge to the southern hemisphere, with a more neutral reading around the equator, where the positive and negative meet.

2. Tides appear to be greater towards the poles and lesser near the equator, which cannot be accounted by for planetary spin, since that would yield the opposite effect. So another cause must be found, which leads to the idea that like charges repel, even in the planet's ocean mass.

3. If like charges for large water mass, such as the oceans, repel, then the waters on the northern hemisphere, as they rise in response to the moon's or sun's gravity, will create a bulge towards the gravitational attraction and one opposite to it. This bulge creates a large negatively charged mass which, being of like charge, will be repelled on the opposite side of the planet, so that there too the water will bulge outwards. So due to like charge repelling like charge, the opposite side of the planet's mass is now counter-bulging in response to the mass bulge on the side facing the moon or sun. On the equator, where the positive-negative charge more or less cancels, there is less tidal opposite face response, so that there is less bulge than where the electrostatic charge is greater. On the southern hemisphere, the positively charged water mass likewise creates its own opposite bulge, same as in the northern.

4. Why does the equatorial water not bulge as much as towards the poles? My best guess is that it is already bulging, because of the planet's spin, so that the additional gravitational pull of the moon or sun does not displace this water as much as in either hemisphere because the centrifugal force already displaces it. Thus, equatorial waters should experience lower tides. I would expect, by this reasoning, that the waters near the artic or antartic circles should rise the most, and then taper off gradually towards the poles, or flatten out towards the equator. Of course, this basic principle is further modified by ocean currents and land mass interferences.

So by this reasoning, the tidal bulges on the opposite side of the planet are actually electrostatic phenomena, where like charge repels like charge, even in the ocean waters. The same would be expected in the land mass, though to a much lower degree. In effect, the planet is pulsating with a gravitationally activated bulge which then triggers a like response on the opposite side of the planet through like electrostatic force repulsion. If this did not happen, the planet would wobble chaotically, which it does not.

Pardon my ignorance, but I am not familiar with the above mentioned electrostatic theory of tides from any known sources, but would love to find out this it is not so. To my thinking, this much better explains why the tides create a like bulge on the opposite side of the planet, than any theory now found in existing textbooks. Blame it on the hot sun, but I got this idea while hiking up and down the rocky peaks near Pena Springs, Anza Borrego desert, last weekend, watching out for snakes and other prickly things. New barbs won't hurt me, so all challenges are welcome. 8)

* * * * * (edited 3/18/04) * * *
... Having said all that, I must confess that this idea is really crazy, so ignore it... thanks. #-o

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No, it isn’t.

‘Gravity cannot act in such a way as to make something feel a force away from an object.’

That would be explicit.

O.k. Let me see if I understand you correctly here. You reckon the centripetal acceleration, of a 365.24 day orbit at Sawicki’s 3 points O, F and C, is exactly the same – there will be no difference in centripetal acceleration at F, C and O.
Is this what you are saying?
I use w^2r to calculate centripetal acceleration and I don’t actually get that result, so (bearing in mind your quote of Einstein’s) could you maybe show your calculations, just for points O and F say. That would be most useful.

Ah, yes. Free fall! When we’ve sorted out centripetal acceleration I’d like to get back to this.

But this is what you asked for: "I was making a general point that any explanation should show that no outward force is required, to be credible it must show this explicitly." And no "outward" force is necessary, gravity suffices.
And, as we've shown, wrong. Gravity is not causing the Earth to feel a force away from the moon, sure, but the difference in gravity causes the tides.
No problem. If the points F, C and O are not rotating (and we have to do the calculations with this in mind, otherwise we are factoring in an equitorial bulge that has nothing to do with tides), then as they orbit, all three describe a circle with the same radius. Halfway around the orbit, the point F is oriented like O was, so that the radius of both their orbits is the same as the radius of the orbit of O. That's why the w^2r computes to the same value--because they have the same r. And if you follow it through, even the direction is the same.

I am not impressed. This is a very silly idea and I’ve already explained it to you once. Until you can demonstrate a much firmer grip on centripetal acceleration there is really no point in my discussing it with you.

But it’s an interesting subject and I’m keen to explore it further – anybody out there up for some sensible discussion?

Here is an illustration that I drew a couple years ago. This is a larger version of the same.

The red dot is the center of the circle, and both orbit the red X. The circle does not rotate as it orbits, as can be seen from the orientation of the black dot at each step. The center of the black dot's orbit is the black X. The radius of the red dot's orbit is equal to the radius of the black dot's orbit.

Yes, they are equidistant from the points about which they are revolving, but they are not always equidistant from the Sun's centre of mass. So while their centrifugal accelerations are constant, the gravitational accelerations which the Sun imparts to them vary.

Take point F, for example. Sometimes it is closer to the centre of the Sun, so gravitational acceleration is greater than centrifugal acceleration - hence there is a tidal bulge towards the Sun. At other times it is further from the Sun, so centrifugal acceleration is the greater - hence the tidal bulge away from the Sun.

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That's true, but if at any single point in time the centrifugal acceleration is the same at all those points and gravitational acceleration is different, then it is still the difference in gravitational acceleration between the points that causes the bulge, ain't it?

And so even if there were zero centrifugal acceleration (not orbiting, just falling), there would still be differences in gravitational acceleration, and the net difference would be the same as when orbiting, so the tidal forces - and thus, the tidal bulges - would be the same.

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A point I made back on Oct 19. 8)

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I take it that you are still advocating for the differential orbits explanation, and you think it is similar to RichardMB's.

How does the DO explanation deal with the low tide on the side of the Earth? That question came up earlier.

You’ve taken your eye off the ball. Milli360’s drawing is just a distraction that’s best ignored and a simple experiment will show why.

Remember the tennis ball? Tie a string to it, dunk it in water again and whirl it round your head – what do you see happens to the water? It goes round to the far side, isn’t that right? This is another tidal bulge – a centripetal acceleration tidal bulge.
Now ask yourself what possible difference will it make to this bulge if the ball were to rotate on its axis. Even if you’re unsure about centripetal acceleration you must surely realise that the balls own rotation doesn’t matter squat. The whole idea is just plain daft. So don’t analyse it – put it in the recycle bin.
And don’t wait 2 years.
----------------
In astronomy, things are only as complicated as you care to make them.

The Earth's own rotation causes a bulge, not of just a meter or two like the tidal bulge, but of twenty kilometers. That is how much difference there is between the equatorial radius and the polar radius. The flattening of Jupiter and Saturn is even more pronounced. But that flattening is definitely not a tidal bulge.

In order to study the tidal bulge, you have to ignore that bulge, which is due to the rotation of the planet. As far as the tide is concerned, that flattening bulge is "squat", and you have to remove it from the calculations somehow. You have to treat the case of a non-rotating planet.

I thought we had all agreed that the rotation of the Earth on its axis is irrelevant to tides? milli360's diagram is of an Earth that is not rotating! (By the way, I was defending you in the post you're now criticizing!)

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Yes.
Obviously the water that makes the tidal bulges on the near and far side's of the planet must come from somewhere. The tidal depressions are simply left behind when the water is displaced.

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That is the intuitive idea, but it's not enough. The data on the accelerometers worldwide, and the more involved theory, show an increase in acceleration, depressing the equipotential surface at the areas of low tide. It doesn't seem that either of you account for that in the explanation.

I am not arguing that there are not alternatives to ways of explaining the tides--different reference frames can be effective. However, it's not as simple as it seems, and the BA's approach is not incorrect.

PS:
Here's a simplified calculation.

At the center of the Earth, the acceleration due to the Sun's mass M is GM/R^2, where R is the distance to the Sun. At the far side of the Earth, the acceleration is GM/(R+r)^2, where r is the radius of the Earth. Subtracting the two gives GM(-2Rr-r^2)/(R^2 (R+r)^2), which is approximately -2GMr/R^3. That's in accord with the usual description of the tide-rasing force acceleration.

At the points of low tide, the distance to the Sun is very nearly the same as at the center of the Earth, GM/R^2. That's the problem I've mentioned. However, the force of the Sun's gravity is not directed parallel to that at the center of the Earth. There are two components, one directed parallel, and one perpendicular, towards the center of the Earth. The right triangle has sides r and R. When you subtract the two, as vectors, then, the difference is r/R times GM/R^2, or GM/R^3. That's half of the tide-raising accleration calculated above--and that is what is found in the accelerometer data.

The equipotential surface at low tide is depressed half as much as it is raised at high tide.

Please accept my apology’s, there was no criticism intended, none at all.
I just didn’t want you distracted by milli360’s diagram because it doesn’t correctly represent centripetal acceleration as it applies to tidal bulges.

You can easily change the diagram so that it does though. Simply draw two circles (centred on the red cross) a big one of radius 1AU+R (R=earth radius) and a small one of radius 1AU-R. Label the big circle F and the small circle C.

Circle F is the circle through which all the outer high tides pass, whilst circle C is the one through which all the inner high tides pass – they are, of course, coaxial.

For a given angular velocity, the centripetal acceleration for path F is greater than for path C. From which it then follows that the centripetal acceleration at the outer bulge is constant and larger than the small constant centripetal acceleration at the inner bulge.

To me this is so obvious but I’ve learned that it won’t be obvious to everyone.

Does it make sense to you or do you have problems with it?

I don't think anyone is arguing against the centripetal force calculation, but the question is whether it is relevant to actually computing the magnitude of the tides is another question. For instance, it doesn't seem to help much with the low tide--without introducing additional complications. The BA's explanation is a bit simpler.

You can ignore centripetal acceleration, as Sawicki & the BA have done, this gives you the simple, but incorrect, explanation that seems to be your preferred option.
For a correct explanation however, you need to include centripetal acceleration, and for my money it’s also the simplest.

No, it's correct.
That's just a backdoor way of including centrifictional force, and unnecessary.

Just to follow up one of my previous posts. There is a derivation of the tidal potential that is presented in Stacey's Physics of the Earth (p115). It includes a once-per-revolution (once per year) rotation of the Earth, so it should satisfy your objections. After making the calculations and simplifying the algebra, Stacey comes up with a formula for the potential that includes three separate terms. The third term is - 1/2 w^2 r^2 sin^2 theta, where w is the rotation, r is the radius, and theta is the colatitude. It's the rotational potential due to the rotation, and it would contribute to the rotational potential if the rotation were much larger (once per day). If the problem were set up without that rotation, the algebra might be more difficult, but the answer would be the same, except this term would disappear. The first term is a constant, independent of position on the Earth, and, as Stacey says, "has no tidal effect." It's the gravitational potential at the center of the Earth due to the moon, with a small correction. The second term is the tidal potential, and is -Gmr^2/R^3 times (3/2cos^2 xi - 1/2), where xi is the angle off the line between the moon and Earth. That last factor is just a second order zonal harmonic, a prolate ellipsoid pointed at the moon.

The gradient of that tidal potential is the tidal force, and it is the same as that computed by the method of differential gravity that we've used above. It shows the depression of the low tide that seems to be missing from the differential orbit method.

So, as I've said, it is possible to include all the terms in a system which includes a rotating earth and come up with the right answer, but you have to be careful--for instance, you have to identify, and then ignore, the bulge induced by that rotation.

Exactly – and haven’t I been telling you that for weeks? Earth’s axial rotation is entirely irrelevant (I actually said it doesn’t matter squat”) so Stacey factors it out – brilliant. Now do you believe me.

My problem however isn’t with Stacey’s Physics of the Earth, it’s with Phil Plait’s Bad Astronomy, specifically this idea that the far tidal bulge is caused by differential gravity (attributed to Mik Sawicki). It isn’t – that’s Bad Physics.
The far bulge would exist even if you removed the sun and its gravity completely, so long as earth’s orbital path was maintained. In fact the far bulge would then be an awful lot bigger. Putting the sun and its gravity back would actually depress the far bulge – exactly the opposite of what Sawicki suggests.
Differential gravity can produce planet rupturing stresses, no doubt about that, but what differential gravity cannot do is induce any kind if push force. On it’s own solar gravity would pull all the oceans, and even the atmosphere, to the near side. To raise the far tide you need (and it is not optional) the centripetal acceleration of earth’s solar orbit.

That would depend on how the orbital path is being maintained in the absence of gravity, wouldn't it? Shooting a ball around on the inside of a circular track would cause it to flatten against the track, not stretch out . . .

Do you contend, then, that an object falling straight into a black hole would not be torn apart by tidal forces?

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Not the issue. Back in October, I said there is nothing wrong with the DO model per se, you have to be careful with it and apply it properly. And that doesn't make the BA's model not as good. Two weeks ago, I mentioned a treatment that I was partial to, and that's the one of Stacey's. Even it is still not complete though.

And the interesting thing about it is that the centripetal component is also factored out--it's irrelevant as well.
But then there would be no near bulge, right? It'd even be a "low", right? That's why it can be said to have nothing to do with the tides.
No, as SeanF points out, if the Earth is in freefall (and it is), there will be tides. Whether it is falling straight into the Sun or not.

That is not my contention. This is what I actually said:

So, yes, an object falling straight into a black hole would be torn apart. This is a valid application of differential gravity, which is not a problem. The earth however is not falling straight – it is following a curved path which requires centripetal acceleration, this is provided by solar gravitation and must be properly accounted for. I am not saying that differential gravity does not exist I am saying that it does not raise the far tide (nor does it raise the near tide side, that is primarily due to direct solar gravity).

Centripetal acceleration, w^2r, depends only on w (essentially the period) and r (the radius). If your scheme includes some kind of mechanical constraint then, yes, you can defeat the consequential bulge, but it will then manifest itself as some kind of distortion - flattening as you correctly surmise. Think of a 1AU tether instead or a (truly) massive bank of rocket motors.

Which centripetal component?

Right. The near bulge is produced by direct solar gravity (see answer to SeanF, above), so if we remove the sun and its gravity we loose the near bulge. This is the point of the two tennis ball experiments.


SeanF has not pointed this out to me – but I’m sure he will if he wants to. As for ‘freefall’ I really don’t know what it means. So far as I’m concerned earth is in orbit.

What I have not yet done is combine these two bulges. I don’t want to give the idea that you can simply stick these two bulges on the earth and get the high tides. It’s more complicated than that – but only a bit.

And what do you do for the low tides?

The point is, once you're done, you'll get the same answer that the BA gets--and his method is valid. It is not incorrect, as you claim.