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Thread: Why is there a tidal bulge on the far side of the Earth?

  1. #61
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    Quote Originally Posted by kilopi
    The tidal bulge is always pointed at the Sun, except for the slight offset that results in tidal slowing. If Mercury is in a resonance, the slowing on one part of the orbit must be offset by speeding up on another part.
    The Earth's tidal bulge is always slightly ahead of the Moon, isn't it? And the Moon pulling on it slows the Earth's rotation (Bad Astronomy pp 72-73).

    In the case of Mercury, though, the tidal bulge lags behind the Sun at perihelion (due to the increase in orbital velocity) and runs ahead of it at aphelion (due to the decrease in orbital velocity). So the Sun slows the planet's rotation during the slower half of the orbit, and speeds it up during the faster half. Is this what you mean?

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    Quote Originally Posted by Eroica
    The Earth's tidal bulge is always slightly ahead of the Moon, isn't it? And the Moon pulling on it slows the Earth's rotation (Bad Astronomy pp 72-73).
    Yes, that's what I meant.
    In the case of Mercury, though, the tidal bulge lags behind the Sun at perihelion (due to the increase in orbital velocity) and runs ahead of it at aphelion (due to the decrease in orbital velocity). So the Sun slows the planet's rotation during the slower half of the orbit, and speeds it up during the faster half. Is this what you mean?
    No, not necessarily. The tidal slowing should in fact be continuous, but if there is a true resonance, then the slowing would be offset by some sort of speed up. For instance, if there were a dipole moment in the density distribution (like the tidal distribution, but permanent) such that it was oriented to speed up the rotation at perihelion then one revolution later (and one and a half rotation later) it would again be oriented at perihelion to speed up the rotation. It's orientation at other parts of the orbit would both speed up and slow down the rotation, but the effects would either cancel or be smaller than that at perihelion.

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    Thanks for that, kilopi. I found this interesting PDF which explains the whole process in mathematical detail.

    They do confirm my suspicions about the tidal bulge:
    Quote Originally Posted by Andre Balogh and Giacomo Giampieri
    the bulge is carried forward with respect to the subsolar point, except at perihelion, where the converse happens (since during this short phase the Sun’s apparent motion is retrograde....
    However, they conclude that:
    the torque acting on the permanent deformations dominates [over the tidal torque], and causes the preferred orientation of Mercury.
    Which, I guess, is precisely the point you were making about the dipole moment in the density distribution.

    I might have known the reason would be far from simple.

    [Edited to provide better hyperlink]

  4. #64
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    Quote Originally Posted by Eroica
    Quote Originally Posted by Andre Balogh and Giacomo Giampieri
    the bulge is carried forward with respect to the subsolar point, except at perihelion, where the converse happens (since during this short phase the Sun’s apparent motion is retrograde....
    Interesting! I would not have suspected that, without actually computing it. I wonder how long the retrograde motion lasts (I seem to have a problem accessing that link to the pdf file right now).

    For a perfectly circular orbit, the apparent motion of the Sun is equal to one rotation, whereas the actual rotation is one and a half. Thus, it has to get close and speed up quite a bit to make the Sun appear retrograde.

    I suspect that if there were not a permanent dipole embedded in the material of Mercury, it would have progressed to a real tidal lock.

  5. #65
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    Sorry about the dodgy link. The article is titled Mercury: The Planet and its Orbit. It's by Andre Balogh and Giacomo Giampieri.

    You can also download it from Giacomo Giampieri Publications.

    The retrograde motion of the Sun only lasts about 6 or 7 days:
    ... at perihelion the orbital angular velocity becomes larger than the spin angular velocity, causing the Sun to move on a retrograde orbit for a short interval of time.

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    Quote Originally Posted by Eroica
    The retrograde motion of the Sun only lasts about 6 or 7 days:
    Which possibly means that the retrograde motion is not great--so the apparent rotation speed is small also. That would tend to reduce the effect of tidal slowing--I mean, speeding.

    Did they mention what the apparent maximum retrograde rotation was?

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    Quote Originally Posted by kilopi
    Did they mention what the apparent maximum retrograde rotation was?
    No. I've quoted all they have to say about the retrograde motion of the Sun. Among their conclusions are the following:
    Quote Originally Posted by Balogh and Giampieri
    (i) the 3:2 resonance is due to permanent deformations on the equatorial plane [of the planet], and to the non-zero eccentricity [of its orbit]
    (ii) tidal torque by itself is not able to produce a non-synchronous resonance. In contrast, it could prevent reaching or maintaining it if the permanent deformations were not large enough
    In an earlier post I wrote:
    Quote Originally Posted by Eroica
    Instead, [Mercury] does the best it can: it makes sure that it always has its tidal bulges lined up with the Sun when it is closest to the Sun and the force of gravity is strongest.
    It turns out that it is the permanent (non-tidal) bulge or deformation which points towards the Sun at perihelion, not the tidal bulges (which, as you said, point more-or-less towards the Sun at all times). A number of websites I have consulted make this mistake.

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    Quote Originally Posted by Eroica
    It turns out that it is the permanent (non-tidal) bulge or deformation which points towards the Sun at perihelion, not the tidal bulges (which, as you said, point more-or-less towards the Sun at all times). A number of websites I have consulted make this mistake.
    It is subtle, isn't it?

    Worse, if the permanent deformation truly did point towards the sun at perihelion, the effect would be more or less symmetric and wouldn't contribute the required speed up--so it probably does not point to towards the sun at perihelion!

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    Quote Originally Posted by Eroica
    Quote Originally Posted by Balogh and Giampieri
    (i) the 3:2 resonance is due to permanent deformations on the equatorial plane [of the planet], and to the non-zero eccentricity [of its orbit]
    (ii) tidal torque by itself is not able to produce a non-synchronous resonance. In contrast, it could prevent reaching or maintaining it if the permanent deformations were not large enough
    Ah ha! Apparently, at least one of my two cents worth of guessing was on the money.

    Does this deformation help explain the ecentricity of the orbit as well? Is the Sun "flingining" it around at perihelion with a little extra kick from the deformation tug to hold the eccentricity?
    We know time flies, we just can't see its wings.

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    Tides.

    This page is perhaps the best explanation for the tides that I have read so far. While gravity alone CAN produce tidal forces, the apparent negative movement of the far bulge can only really be explained by the centrifugal force of the Earth orbiting the Earth-Moon centre of mass.

    http://co-ops.nos.noaa.gov/restles3.html

  11. #71
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    Re: Tides.

    Quote Originally Posted by Ian R
    This page is perhaps the best explanation for the tides that I have read so far. While gravity alone CAN produce tidal forces, the apparent negative movement of the far bulge can only really be explained by the centrifugal force of the Earth orbiting the Earth-Moon centre of mass.

    http://co-ops.nos.noaa.gov/restles3.html
    I haven't looked at it lately, but we've discussed it before. Unless it has been updated, it can't be trusted. It was so full of errors, mistakes, and misconceptions, that I would consider it an embarrassment to NOAA.

    PS: The centrifictional force is the same at all points on the Earth--in magnitude and direction. So there is no differential because of it. That webpage was amended to say that years ago, possibly in response to our emails, but other statements were left in place that conflicted with that notion.

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    I have a problem with the Sawicki (BA) model for tides which goes like this: -

    Dunk a tennis ball in water, hold it up and look at it. See the water drain to the underside? That's a tidal bulge.

    Now, according to Sawicki's model this tide is due to 'differential' gravity and we can calculate what that is:-

    using Sawicki's equation (1) the gravitational pull of the earth a(e) at distance d(e) is: a(e) = G*M(e) / d(e)^2 (where d(e) is the distance to the centre of earth) (1)
    = 6.67259e-11 * 5.978e24 / 6.378e06^2
    = 9.81 ms^-2 (yes, that sounds familiar)

    then, using Sawicki's equation (4) for the difference between the earths pull on the closest/farthest point on the ball:

    Da(e) = a(e) * 2R / d(e) (where R is the radius of the ball, 0.05m, say) (4)
    = 9.81 * 0.1 / 6.378e06^2
    = 2.41e-14 ms^-2 (?????????)

    You can see the problem can't you - differential gravity is 2.41e-14 ms^-2 . This is vanishingly small, it isn't even pico-gravity, but according to Sawicki that's what causes the tide.

    Please do tell me - where am I going wrong?

  13. #73
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    Dunk a tennis ball in water, hold it up and look at it. See the water drain to the underside? That's a tidal bulge.
    I think this is where you are going wrong. The movement of the water in this example is not due to tidal forces - the water moves downwards because of gravity, yes, but the ball is too small for tidal forces to have any appreciable effect upon it.

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    In Sawicki's paper, the Earth (or Moon, or whatever) is in freefall. If you hold the tennis ball up and look at it, it's not in freefall. You'd have to drop it first and then see how the water moves.

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    Ian R: The question is a little more subtle than that Ian, but thanks anyway - you might even be right.

    Eroica: I think the term 'free-fall' is extremely dodgy but I accept and will address your comment.

    Sawicki states that the earth as a whole is in free-fall about the sun. This is not correct. By any useful definition of 'free-fall', whilst point O (Sawicki's fig.1) is in free fall, but points F and C are not. This is because O is the only point at which centripetal acceleration (w^2*r) is equal to gravitational acceleration (G*M(s) / d(s)^2). At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.

    This analysis does not require any mysterious 'reverse' gravity and, if you care to do the sums, you'll see that the imbalances are exactly equal and opposite. In contrast, Sawicki's 'differential' model requires (or at least implies) a gravity reversal and it produces a 0.01% imbalance. This may not seem much but in the earth-moon system it's very significant - I've not calculated it recently but I think it's as much as 5%.

    Incidentaly, my original post contains an error:- 'differential' gravity should be 1.54e-07 not 2.41e-14. My mistake - I incorrectly squared d(e). Sorry about that.

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    Quote Originally Posted by RichardMB
    Dunk a tennis ball in water, hold it up and look at it. See the water drain to the underside? That's a tidal bulge.
    No, not really.
    You can see the problem can't you - differential gravity is 2.41e-14 ms^-2 . This is vanishingly small, it isn't even pico-gravity, but according to Sawicki that's what causes the tide.

    Please do tell me - where am I going wrong?
    The tidal bulge is also proportional to the radius of the body. A tennis ball is a lot smaller than the Earth.
    Quote Originally Posted by RichardMB
    Sawicki states that the earth as a whole is in free-fall about the sun. This is not correct. By any useful definition of 'free-fall', whilst point O (Sawicki's fig.1) is in free fall, but points F and C are not.
    Useful definition? By that definition of free fall, there are no "bodies" in free fall. That's not very useful--a term that applies to no body.

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    Quote Originally Posted by RichardMB
    At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.
    Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?

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    Quote Originally Posted by RichardMB
    Sawicki states that the earth as a whole is in free-fall about the sun. This is not correct. By any useful definition of 'free-fall' ... [snip]
    My definition of "free-fall" is "subject only to gravitational forces." What's yours?

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    Quote Originally Posted by Eroica
    Quote Originally Posted by RichardMB
    At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.
    Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?
    I'm pretty sure he is talking about the centrifictional force due to the body's revolution around the central body. Gravity is the centripetal force.

    One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body. That's why most explanations use the differential gravity, because it doesn't matter whether the body is in the free fall of an orbit, or falling "straight down."

    I don't have the trouble that many have in recognizing centrifugal effects as a force, but in this case, it is not necessary.

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    Eroica wrote:
    My definition of "free-fall" is "subject only to gravitational forces." What's yours?

    Oh, cheers, mine's a Boddingtons. Your definition looks fine to me, but are you absolutely certain this is what Sawicki means? because if he means something else you'll not be understanding his model. My definition isn't important, I gave it a stab and milli360 jumped on me so I've lost interest - I've really no idea what it means.

    [b]Eroica wrote:
    Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?
    You've misunderstood me, I didn't mention earth's axis, I was referring to earth's annual rotation about the solar barycentre same as Sawicki was - I'll come back to it in a moment. What I was saying is that without centripetal acceleration (annual solar) the sun's gravity would produce a single tide, much the same as on the tennis ball and for the exactly same reason.

    milli360 wrote
    One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body.
    I once found a site once that explained this idea in detail. It's wrong. Look again at Sawicki's fig.1. The mistake is to think of points F, O and C as points on earth, this is wrong, they're points in space. Point O is 1AU from the sun so we can plug this into a(g) = GM/D^2 to determine the sun's gravitational acceleration at point O. By the same token we can use a(c) = w^2r to find it's centripetal acceleration. If we do this we'll find that they are exactly the same. None of this requires the presence of earth. You can remove earth altogether if you wish. You can replace it with jupiter. You can spin it backwards or forwards, fast or slow, pole over pole. It won't make the slightest bit of difference - at point O solar gravitation will equal centripetal acceleration. Point F is farther out, at 1AU+1 earth radius, gravitational acceleration (=GM/D^2) is therefore lower and centripetal acceleration (= w^2r) is greater. Again, I must stress, it has nothing whatever to do with earth. Once we've done our calculations and we know what a(g) and a(c) is at all three points, we can stick the earth back and ask what affect these accelerations might have - tides for instance.

    Don't be gulled by smart graphics. Get hold of a decent textbook.

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    Quote Originally Posted by RichardMB
    What I was saying is that without centripetal acceleration (annual solar) the sun's gravity would produce a single tide, much the same as on the tennis ball and for the exactly same reason.
    No, that is definitely not true, unless by "centripetal acceleration" you are actually referring to gravity itself. As I pointed out, if the body is falling straight down, a tidal bulge would be induced. No centrifugal force necessary.
    milli360 wrote
    One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body.
    I once found a site once that explained this idea in detail. It's wrong.
    The site was wrong?
    Look again at Sawicki's fig.1. The mistake is to think of points F, O and C as points on earth, this is wrong, they're points in space. Point O is 1AU from the sun so we can plug this into a(g) = GM/D^2 to determine the sun's gravitational acceleration at point O. By the same token we can use a(c) = w^2r to find it's centripetal acceleration.
    You mean centrifugal. Gravity is centripetal. However, F, O, and C are points on Earth--we are trying to calculate the tide on Earth, not in space.
    If we do this we'll find that they are exactly the same. None of this requires the presence of earth. You can remove earth altogether if you wish. You can replace it with jupiter. You can spin it backwards or forwards, fast or slow, pole over pole. It won't make the slightest bit of difference - at point O solar gravitation will equal centripetal acceleration. Point F is farther out, at 1AU+1 earth radius, gravitational acceleration (=GM/D^2) is therefore lower and centripetal acceleration (= w^2r) is greater.
    If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth. The reason for that is you have to remove all effects of the rotation of the Earth. Otherwise, the situation would be akin to the moon's, where there is one complete rotation in every revolution.

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    milli360 wrote:
    'The site was wrong?
    Yes. Where it says that centripetal acceleration, on a body in orbit, is the same at all points on the surface of the body (or some thing like that), the site is wrong.

    'If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth.'
    If you say so - but since I never mentioned either 'centrifugal' or 'force' I'd rather not comment. But I do agree that gravity is centripetal - by definition.

    Let's get back to Sawicki's model though, again referring to his fig.1.

    Point O is 1AU from the sun and it takes 365.3 days to complete an orbit so we can work out a(c) (centripetal acceleration), at O, using a(c) = w^2 r - it comes to 5.9303e-03 ms^-2. We also know the solar mass (1.989e30 kg) so we can work out a(g) (gravitational acceleration), at O, using a(g) = GM/d^2 - it also comes to 5.9303e-03 ms^-2. At point O therefore centripetal and gravitational acceleration are the same.

    Point F is farther from the sun at 1AU+ 1 earth radius but it's period is the same at 365.3 days. Repeating the above calculations, using this increased distance, gives a(c) = 5.9306e-03 ms^-2 and a(g) = 5.9298e-03 ms^-2. Solar gravity is less than the centripetal acceleration, which means that (at point F) the sun in not able to supply all the acceleration required - something else is making up the difference.

    Stop and think about this, we started looking for the force (acceleration) that pulls the tidal bulge at F up, but after doing 3 simple calculations we find that we really need a force (acceleration) to pull it down. Not an outward force, an inward force - the same as the sun's gravity.

    Good isn't it? and it makes a lot more sense - the earth itself can provide this, no trouble at all. The tiddly little 2.59e-05 ms^-2 needed comes from earths mighty 9.81 ms^-2, as a consequence the weight force at F is reduced by the same amount - and a reduction in the weight force will result in a bulge - as required.

    This explanation does away with the troublesome outward force, it does away with doubtful vector manipulations, it does away with asymmetric tidal forces.

    Still happy with 'differential gravity' are we ?

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    Quote Originally Posted by RichardMB
    Where it says that centripetal acceleration, on a body in orbit, is the same at all points on the surface of the body (or some thing like that), the site is wrong.
    I'm not sure why you keep using "centripetal" where you should be using centrifugal, but with that in mind, that it's the same at all points is not wrong. It is correct, assuming a sphere, once you completely subtract the effect of the rotation of the Earth.
    'If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth.'
    If you say so - but since I never mentioned either 'centrifugal' or 'force' I'd rather not comment. But I do agree that gravity is centripetal - by definition.
    If gravity is centripetal, why do you differentiate between gravitational and centripetal acceleration, below?
    Let's get back to Sawicki's model though, again referring to his fig.1.

    Point O is 1AU from the sun and it takes 365.3 days to complete an orbit so we can work out a(c) (centripetal acceleration), at O, using a(c) = w^2 r - it comes to 5.9303e-03 ms^-2. We also know the solar mass (1.989e30 kg) so we can work out a(g) (gravitational acceleration), at O, using a(g) = GM/d^2 - it also comes to 5.9303e-03 ms^-2. At point O therefore centripetal and gravitational acceleration are the same.

    Point F is farther from the sun at 1AU+ 1 earth radius but it's period is the same at 365.3 days. Repeating the above calculations, using this increased distance, gives a(c) = 5.9306e-03 ms^-2 and a(g) = 5.9298e-03 ms^-2. Solar gravity is less than the centripetal acceleration, which means that (at point F) the sun in not able to supply all the acceleration required - something else is making up the difference.

    Stop and think about this, we started looking for the force (acceleration) that pulls the tidal bulge at F up, but after doing 3 simple calculations we find that we really need a force (acceleration) to pull it down. Not an outward force, an inward force - the same as the sun's gravity.

    Good isn't it? and it makes a lot more sense
    It makes an intuitive sense, but it's wrong. Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force.
    This explanation does away with the troublesome outward force, it does away with doubtful vector manipulations, it does away with asymmetric tidal forces.
    All that troublesome math?
    Still happy with 'differential gravity' are we ?
    More than ever. I'm understanding better where the alternative leads.

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    milli360 wrote:
    'I'm not sure why you keep using "centripetal" where you should be using centrifugal'
    I use centripetal because it means towards a central point (inward) and that is precisely what I mean. Centrifugal means away from a central point (outward), my previous mail explained that there is no need for an outward acceleration - which is just as well because none exists. If you came across a model that so much as hints at an outward force be assured that it is wrong.
    'If gravity is centripetal, why do you differentiate between gravitational and centripetal acceleration .... ?'
    A good question: Gravitational acceleration a(g) is an acceleration produced by a mass. Although gravitation is centripetal, the term ‘centripetal acceleration’, a(c), is generally used to mean the acceleration required by a mass to cause it to move in a circular (elliptical) path. So in our present discussion a(c) is the centripetal acceleration required to move earth in a 365.3 day circle and a(g) is the centripetal acceleration provided by the sun (as a consequence of its mass). This is similar to the relationship between a 12v battery and a 12v bulb, both are 12v devices but one supplies it whilst the other requires it. Yes?

    'Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force. '
    Point L in Sawicki's fig 1 is at 90 degrees is this what you mean? Point L is 1AU from the sun so a(g) = a(c) (same as point O), there is no need for a contribution from earths gravity, which is just as well because there is no component of earths gravity in the direction of the sun anyway. The weight force is at a maximum at L so there is no bulge - this is low tide. That's what we want isn't it?

    ‘All that troublesome math?’
    The maths is really very simple (just + - ^ / and *) and anyway I haven’t actually given you any maths – I’ve given you the equations and their results. This is the minimum I think you need to test my model – and I hope you are testing it.
    I don’t think you’ve tested Sawicki’s maths though, have you? Take a look at his equation 3. It’s o.k. for Da(s) at point F but then he says it the same as Da(s) at point C – and that’s wrong. I suspect Mik knows it’s wrong too, but if they’re not the same it undermines the whole model.

    Still happy with 'differential gravity' are we ?

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    Quote Originally Posted by RichardMB
    A good question: Gravitational acceleration a(g) is an acceleration produced by a mass. Although gravitation is centripetal, the term ‘centripetal acceleration’, a(c), is generally used to mean the acceleration required by a mass to cause it to move in a circular (elliptical) path.
    Ah.
    'Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force. '
    Point L in Sawicki's fig 1 is at 90 degrees is this what you mean? Point L is 1AU from the sun so a(g) = a(c) (same as point O), there is no need for a contribution from earths gravity, which is just as well because there is no component of earths gravity in the direction of the sun anyway. The weight force is at a maximum at L so there is no bulge - this is low tide. That's what we want isn't it?
    Approximately at 90 degrees, but at the many points where a(g_=a(c), is what I mean. Low tide is a depression, the opposite of a bulge. That's my point, your theory shows no effect. Whereas, the "differential gravity" theory is in accord with observation.

    I'm partial to another treatment, which involves a calculation of the potential. To me, it's more comprehensive, and easier to understand--and I can use it to judge the other explanations, too.
    This is the minimum I think you need to test my model – and I hope you are testing it.
    It seems to fail.
    I don’t think you’ve tested Sawicki’s maths though, have you? Take a look at his equation 3. It’s o.k. for Da(s) at point F but then he says it the same as Da(s) at point C – and that’s wrong. I suspect Mik knows it’s wrong too, but if they’re not the same it undermines the whole model.
    Not sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?
    Still happy with 'differential gravity' are we ?
    Yep, except as noted.

    PS: Link to Sawicki's paper

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    milli306 wrote:
    ‘I'm partial to another treatment, which involves a calculation of the potential’
    Excellent – whatever you find helpful, I was discussing Sawicki’s paper.

    ‘Low tide is a depression, the opposite of a bulge. That's my point, your theory shows no effect. Whereas, the "differential gravity" theory is in accord with observation.’
    There is no need to postulate an acceleration at point L. Earth is elastic so think of Sawicki’s fig.1 as a balloon, pulling at point F and O will raise bulges at these points. Point(s) L will naturally depress as the balloon accommodates the changed internal stresses. Where’s the problem?

    But let’s have a look at what Sawicki says. According to Sawicki there is a sin alpha component of the sun’s pull at point L operating towards O (downwards), and he more or less just states that this amounts to D(a)s/2. By my calculations sin alpha is 4.25e-05 that’s 0.00000425 not a half so, even if we use the biggest D(a)s, the additional acceleration at L is 2.15e-11 ms^-2. That’s in accord with observation is it?

    (I must confess I’ve only given this a quick look - because you raised it - so if I’ve misunderstood it in someway do please explain it to me.)

    ‘Not sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?’
    Sawicki uses equ.3 to calculate Da(s) on the far side, that is point F at a distance d(s)+R (1AU+radius of earth). Using my data it comes to 5.0476e-07 ms^-2. On the near side, point C, the distance is d(s)-R (1AU-earth radius) and I make this –5.0483e-07 ms^-2. According to Sawicki these are the same. Well, what do you reckon, they don’t look the same to me.

    So what? Well, Sawicki says the acceleration at point C is a(s) + D(a)s, but since D(a)s is negative the acceleration at C is less than at O, so like F it lags behind the centre – the maths is telling us that there is a low tide at point C!
    That’s not all either. You now have a choice of positive or negative D(s)a so at point L you can have a bulge or a depression as you wish. You must watch these ‘higher order terms’ – they’re a bit like the ‘the curl of e’.

    You’re still not convinced are you?

  27. #87
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    Quote Originally Posted by RichardMB
    There is no need to postulate an acceleration at point L.
    It's not a postulate, it's pretty much a fact. And, as you admit, your theory ignores it completely. It is about half the effect at the points of the tidal bulge.
    According to Sawicki there is a sin alpha component of the sun’s pull at point L operating towards O (downwards), and he more or less just states that this amounts to D(a)s/2. By my calculations sin alpha is 4.25e-05 that’s 0.00000425 not a half so, even if we use the biggest D(a)s, the additional acceleration at L is 2.15e-11 ms^-2. That’s in accord with observation is it?
    Alpha is the angle at the Earth's center, so at L, alpha is 90 degrees and the sine of alpha is one, not .00000425.
    ‘Not sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?’
    Sawicki uses equ.3 to calculate Da(s) on the far side, that is point F at a distance d(s)+R (1AU+radius of earth). Using my data it comes to 5.0476e-07 ms^-2. On the near side, point C, the distance is d(s)-R (1AU-earth radius) and I make this –5.0483e-07 ms^-2. According to Sawicki these are the same. Well, what do you reckon, they don’t look the same to me.
    As I said, he has ignored the higher order terms. They're a lot smaller than the main semidiurnal tide terms, but they contribute to the diurnal, and even sidereal tides.
    You’re still not convinced are you?
    I'm never convinced.

  28. #88
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    milli360 wrote:

    RichardMB wrote:
    There is no need to postulate an acceleration at point L.

    ’It's not a postulate, it's pretty much a fact. And, as you admit, your theory ignores it completely’.
    Gosh, a theory. No I don’t have a theory. For a simple problem like this I find Newton’s theory does admirably.

    The origin of tides is an orbital mechanics problem and there are two basic, and complementary, equations in orbital mechanics: GM/D^2 characterising the gravitational component and w^2r characterising the (elliptical) motion. Anybody reading these pages should know this by now.

    Where Sawicki goes wrong is to ignore the w^2r (centripetal acceleration) component and, by replacing it with ‘free-fall’ (whatever that means), he fatally cripples his model, and all that follows is corrupted.

    Whether acceleration at L is needed or not isn’t something I could get exited about (I suspect it isn’t), the point is it’s that Sawicki’s explanation of it has to be wrong.

    If you want a more plausible explanation for this acceleration (or depression) tell me where I can find out more about it and I’ll get onto it. Is it mentioned in Encarta, for instance or Webster’s, or Britannica? Where have you heard about this ‘fact’– other than in Sawicki’s paper?

  29. #89
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    Quote Originally Posted by RichardMB
    Gosh, a theory. No I don’t have a theory. For a simple problem like this I find Newton’s theory does admirably.
    Your explanation, then.

    Where Sawicki goes wrong is to ignore the w^2r (centripetal acceleration) component and, by replacing it with ‘free-fall’ (whatever that means), he fatally cripples his model, and all that follows is corrupted.
    That's wrong. As you point out above, the only centripetal acceleration component is due to gravity, and Sawicki definitely does not ignore gravity.
    Whether acceleration at L is needed or not isn’t something I could get exited about (I suspect it isn’t), the point is it’s that Sawicki’s explanation of it has to be wrong.
    I'm not so sure, and certainly not for the reasons that you've given.
    If you want a more plausible explanation for this acceleration (or depression) tell me where I can find out more about it and I’ll get onto it. Is it mentioned in Encarta, for instance or Webster’s, or Britannica? Where have you heard about this ‘fact’– other than in Sawicki’s paper?
    Are you asking how I know that there is a low tide??

  30. #90
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    milli360 wrote:

    Are you asking how I know that there is a low tide??
    No, I didn't mention low tide. What I said was:

    If you want a more plausible explanation for this acceleration .... Where have you heard about (it) other than in Sawicki’s paper?
    The question is perfectly clear - where have you heard about this acceleration (D(a)s/2 at point L) - other than in Sawicki's paper.

    That's wrong. As you point out above, the only centripetal acceleration component is due to gravity, and Sawicki definitely does not ignore gravity.
    No, I did not say that. What I said was that gravity is a centripetal acceleration. In fact there are at least 5 centripetal accelerations to consider with this problem.
    Yes, I do agree that 'Sawicki definitely does not ignore gravity' - the trouble is that's allhe considers.

    Now, I do appreciate your contribution to this discussion but in future you're going to have to try harder. I'm not going to bother to answer your questions if it's obvious that you haven't read my contribution.

    Please, is there anybody else out there with views or questions on this subject? See if we can't get it going again.

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