Hello Earth Rocks, I distinctly get the impression from reading the comments, that you’re searching for a simpler explanation to tides? Well, how about this: The centrifugal force from *Orbital* motion around the barycentre, is not 6.5 times stronger. The Earth orbits, revolves, translates around the barycentre. Each point of mass on Earth, maps out a circle of equal radius _(4670km)_ as it orbits. The centrifugal force is equal across the whole earth. And when you subtract that (rω² ~ 3.3x10^-5 m/s²) from the gravitational attraction and factor in the earth’s own gravitation. You get an acceleration of +/- one ten millionth of a (g) on either side and *zero* at the centre, which is due to the *free-fall* motion experienced by all Planetary bodies, which are in motion in our solar system. Thank you for your support, ~ _wavy!_
The discussion for this video has rich with the math and details, and I appreciate that being provided for everyone. This video was designed for an introductory level course, and I specifically didn't want to get too deep into the details on barycenter (wanted to keep it simple), but thanks for adding it to the discussion. :)
@@EarthRocks You’re more than welcome, and I really appreciate the reply. Thank you. My only video was specifically designed, as an introductory course into dispelling the popular misconception, that the Earth remains static in space, when in fact it Rocks! …but, didn’t want to get too deep into the details about barycentre motion and apparent forces (wanted to keep it silly). Maybe we could work on something together in the future. :)
This is the best video I've seen about explaining tidal forces and movements. Thanks a lot. So many other videos do not have the faintest clue how the moon is related to Earth and where the plane of movement is located. You Rock!
The center of mass in the Earth-Moon system, also known as the barycenter, is located within Earth itself. It lies about 4,671 km (2,902 miles) from Earth's center, which is approximately 75% of Earth's radius on the moon side. While it seems like the Moon orbits Earth, both the Earth and Moon actually orbit this common center of mass. However, because the Earth is much more massive than the Moon, the barycenter is located closer to Earth's center, causing the Earth to wobble slightly as it orbits the barycenter. This means that the elliptical path we generally attribute to Earth's orbit around the Sun is actually the orbit of the barycenter of the Earth-Moon system. So you would observe the earth wobble about the barycenter the side opposite the moon is traveling faster around the barycenter and thus it is centrifugal force that is causing the bulge opposite the moon in about the same magnitude as the moon facing side.
Discussing with somebody else (who says only differential gravity counts), I think I found another simpler way to explain tides: Total force exerted on any given particle (because of moon-earth rotation/revolving), has to be equal to the centripetal force required for its revolving, to accomplish Newton´s 2nd Motion Law). The main force “supplier” is moon´s pull AT EACH LOCATION. The other possible forces only can be exerted on each particle by surrounding ones (internal stresses), or by the very gravitational pull from the rest of our planet. Closer to moon particles "feel" stronger pull than required centripetal force. THAT excess (and not any "differential gravity") has to be passed onto the rest of earth through their neighbors. On their turn, those neighbors exert opposite forces back onto considered particle (3rd Newton´s Motion Law). At farther hemisphere interactions are a kind of mirror image of what at closer hemisphere, because the farther the particle, the bigger the deficit of moon´s pull (compared to required centripetal force). That constitutes a complex "chain of transmission", where total force acting on each particle (leaving aside own weight) has to satisfy 2nd Newton´s Motion Law (F=mω²r) ... THAT FIELD OF FORCES, exerted DIRECTLY ON EACH PARTICLE, and the INERTIAL EFFECTS (centrifugal forces included) caused by the fact that all particles are being "forced" to revolve together, ARE ACTUALLY THE DIRECT CAUSE OF TIDES !! Basically solid earth is stretched, and water moves towards where the two bulges build up. None of those forces is "artificial", and all are directly felt by considered particle, what implies "natural" additions of forces, without any artificial intervention of our minds. That would be necessary to deduct moon´s gravitational pulls at locations very distant from each other (the so called “differential gravity”), what obviously material stuff can´t do !!
Minutes ago I sent what follows to "What Physics Teachers Get Wrong About Tides / Space Time / BPS Digital: "I have the "bad" habit of ruminating this type of things even at night in bed ... Last night I had a kind of eureka moment. Sublunar bulge has to be the result of both Lunar and own Earth pulls ... But NOT because water close to poles and previously low tide areas "pushes" closer contiguous water towards sublunar point and , "those tiny tangential sideways pushes on all the chunks of water added up over half the surface of the planet can produce a pretty decent increase in water pressure" as it is said at 5:35, for the reasons I exposed a couple of days ago. English is not my mother tongue, I type with only two fingers, and now I don´t have time enough to elaborate all relative to intervening forces properly, But just a short analogy, the one gave me (arguably?) the solution. What actually happens is similar to a tsunami, but with a relatively very short wave, not rectilinear but circular, that moves towards sublunar point. It´s what happens if we drop a stone vertically onto a flat surface of water. Initially the stone makes a "hole", but afterwards a wave of water moves in radially, and water forms a "peak" on the center, against water own weight. Displacements need not to be big, but the wave, not having space to form smaller and smaller circles, necessarily has to increase in hight, the closer to sublunar point the more. Kind of radial inward resonance ..."
My last line can be confusing, because it's listing three different things. "And if you’re looking for the most intense sun on the planet, be sure to travel to the equator at the equinoxes, 23.5 degrees North in June, and 23.5 degrees South in December." Equinoxes AND solstices. Sorry for the confusion.
I´ve frequently seen that most people don´t fully understand the "peculiar" type of movement of the earth, in its dance with the moon: its revolving around earth-moon common center of mass (barycenter). Its location is INSIDE our planet, app. 1/3 of its radius from the surface (but please note it is not a material point). The barycenter is the one which actually follows the elliptical orbit around the sun. But with that movement earth doesn´t properly orbits around the moon, as many people say: earth CM follows a circular path around the barycenter, and the rest of earth material points follow identical circles, but around different geometric points, and maintaining its orientation relative to distant stars ... LOGICALLY I´m considering ONLY earth-moon dance, disregarding earth daily spinning and the orbiting of the couple around the sun, because they have nothing to do with moon-earth dynamics. As it is certainly tough to grasp, a few days ago I managed to imagine an analogy, and sent it to another site. IT CAN BE VERY USEFUL TO MOST OF YOU: "I suggest you put your open hand horizontally on a table. Spot the table point app. under the middle of the first bone of your middle finger (B, from barycenter). And try slowly to make your hand “revolve”, with the first knuckle of same finger (C, app. at hand center) following a circle around B . If initially you had imagined a “moon” a couple of feet away in line with your middle finger, and that “moon” rotated around B simultaneously to hand revolving, ALWAYS in line with B and C and at side opposite to C, you would have a scenario quite similar to actual earth-moon dance. If you carefully observe the end of any of your fingers (or any other point of your hand), you will find that THEY ALL follow equal circumferences, ALWAYS keeping their location farthest from the moon (logically, within each own path). THAT´S, I think, A GOOD WAY to see what quoted from NOAA´s scientits work IS QUITE CORRECT: " ... since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”. The NOAA work is the one I referred to here a couple of weeks ago.
I have tried for my entire career as an Earth Science educator to get tides right. I have always taught the bulges as due to the differential forces of gravity (and hence acceleration) experienced on the near side versus the far side. Essentially the far side lags behind. I have seen so many contradictory and error-prone descriptions of tides in many different books (essentially problems with the vector diagrams). Then I saw this video: th-cam.com/video/pwChk4S99i4/w-d-xo.html, and I threw my hands up in the air. I'm not a physicist and can't decide which vector/force explanations are the best, so I decided to keep it very simple in the video and focus on the things we can understand easily and get right. Clearly I need to keep working on it. Thanks everyone for all the comments.
To clarify, you threw your hands up because you finally got it? Or because you still didn't understand it, or how to explain it? I guess what I'm asking is, is this video of yours providing a different explanation than that PBS Space Time one?
Wrong. Gravity between moon and earth cannot be felt, because centrifugal forces from earth and moon, both in orbit around their common baricenter, nullifies gravity same as the International Space Station, ISS.
On my post of yesterday only forces on sublunar and antipodal earth areas are referred to. There is where tide-generating dynamic imbalance is the greatest. On transverse section through earth CM, that imbalance doesn´t occur. On the rest of closer hemisphere moon´s gravity prevails, being net tide-generating force the closer to moon, the bigger. On the other hand, on farther hemisphere, where moon´s pull is smaller, centrifugal force prevails, being the imbalance the farther from moon the greater. Water from all parts of the oceans tend to move and pile up on both opposite areas (50/50) ... But VERY, VERY LITTLE through squeezing (as proposed by some people) if any !! The unique area where there are some inward vertical components of net tide-generating force imbalance is where this is null or close to null, and there almost the whole of the pull vector is tangential (horizontal) ...
A book on tides more than 300 page long, written by Dr. Bruce Parker, is linked below. He spent most of his career in NOAA and much of his time working on tide related problems as a specialty even while tackling jobs with a much broader scope. Positions he held at NOAA included: Chief Scientist of the National Ocean Service; Director of the Coast Survey Development Laboratory; Director of the World Data Center for Oceanography; Principal Investigator for the NOAA Global Sea Level Program; and head of the U.S. national tides and currents program (in a earlier organizational form before it became CO-OPS). Among his awards are the NOAA Bronze Medal, the Department of Commerce Silver and Gold Medals, and the Commodore Cooper Medal from the International Hydrographic Organization. Dr. Parker is presently a Visiting Professor at the Center for Maritime Systems at the Stevens Institute of Technology. Dr. Parker has written many papers on tidal subjects, some of which are included in the References section of this book, as well as many tidal analysis programs, some still being used in some form in CO-OPS. He also had the privilege of organizing the program for the International Conference On Tidal Hydrodynamics in 1988 and editing the book that resulted. Dr. Parker received his Ph.D. in physical oceanography from The Johns Hopkins University, and prior to that an M.S. in physical oceanography from the Massachusetts Institute of Technology, and a B.S./B.A.in biology and physics from Brown University ... AND HE SAYS: "At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point). At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system). On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction. In a hypothetical ocean covering the whole Earth with no continents (see Figure 2.8) THERE WILL BE TWO TIDAL BULGES RESULTING FROM THESE IMBALANCES OF GRAVITATIONAL AND CENTRIFUGAL FORCES, ONE FACING THE MOON (WHERE THE GRAVITATIONAL FORCE IS GREATER THAN THE CENTRIFUGAL FORCE), AND ONE FACING AWWAY FROM THE MOON (WHERE THE CENTRIFUGAL FORCE IS GREATER THAN THE GRAVITATIONAL FORCE). Logically, he also analyses thoroughly tidal local effects all over the world ... tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
My "discussion fellow" tried to improve his model as follows: "Let's have the spaceship apply an even force to the whole of the Earth and all the water on it with a tractor beam so that the Earth simply accelerates and the moon accelerates after it. In this case, there are two bulges in the water with one ahead and the other behind. The bulges in this case are not caused by any kind of rotation, but the mechanism for forming them is the same as it is in the real case of the Earth with the moon orbiting it - it's the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance" ... I answered: "Your model has really improved … Now both bulges would be identical to real, natural ones (disregarding effects due to sun and other celestial objects) … But in that scenario bulges wouldn´t be due to "the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance” as you say, because MASSIVE STUFF CAN REACT ONLY TO FORCES EXERTED ON IT, either gravity from more or less distant matter, or direct pulls or pushes (even share stresses) exerted by contiguous material. IT CANNOT DIRECTLY REACT TO GRAVITATIONAL ATTRACTION DIFFERENCES BETWEEN POSITIONS THOUSANDS OF KM APART !! How “on earth” could it ?? The behind bulge would actually be due to the fact that there backward pull from moon, at any given position, is bigger than forward pull from the spacecraft. And ahead bulge would be due to the fact that there moon backward pull is smaller than forward pull from the spaceship !! Your new model makes possible to also improve the parallel I drew: now we would be exerting on our planet an artificial, uniform and outward (relative to moon-earth couple) force with the spaceship, which is actually replacing the also uniform and outward force that, in the real case with earth revolving around the barycenter, is exerted by nature. How? … As I previously said, moon-earth dance is continuously bending the trajectory of each earth molecule (forcing it to be circular). ALL earth material points are continuously at farthest distance from moon (within its respective circular trajectory), what implies centripetal and centrifugal forces inherent to that revolving are all parallel to the straight line between earth and moon centers of mass" He also had said: "... Take away the spaceship and the moon will fall into the Earth, and as it does so there will be an amplification of the two bulges, again not caused by centripetal force". And I replied: "That is another prove of the impossibility of having the tides we have had for billions of years without mentioned inertial effects due to the revolving of earth and the rotation of moon, both logically around their common center of mass. What quoted would not happen if, exactly at the moment of disconnection of the spaceship, somehow we could give earth and moon the “tangential” real velocity vectors they got in nature. The barycenter would follow its previous straight trajectory, at the speed acquired until that moment, and earth-moon couple would dance again, the earth with the two bulges that in your model were artificially originated ..."
On the work: www.newenglandphysics.org/other/French_Tides.pdf one can read: "… the production of ocean tides is basically the CONSEQUENCE of the gravitational action of the moon- and, to a lesser extent, the sun …The analysis of the phenomenon is, however, considerably helped by introducing the concept of INERTIAL FORCES as developed in the present chapter. … With respect to the CM of the earth-moon system … the earth´s center of mass has an acceleration of magnitude Ac … every point in the earth receives this same acceleration from the moon´s attraction. … for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite. If, however, we consider a particle on the earth´s surface at the nearest point to the moon …, the gravitational force on it is greater than the CENTRIFUGAL FORCE by an amount that we shall call Fo … By an exactly similar calculation, we find that the tide-producing force on a particle of mass m at the farthest point from the moon … is equal to -Fo …” What I´ve been saying on many sites time and again !! The authors, as the NOAA scientists D.C. I referred to a week ago, neither say “the production of ocean tides is ONLY due to differential gravitational action" (or something with that meaning), nor “forget” an acting inertial force such as centrifugal force, as many people, physicists included, keeps doing !!
I´ll try and elaborate what I said yesterday. Logically there are many differences between actual formation of the sublunar bulge, and the analogy I exposed. On the one hand, bulge formation is the result of not just a single circular wave inching inwards. An "infinite" number of short circular "waves" form, all across the ocean hemisphere. On the other hand, those waves don´t work as normal waves, which consist only in the propagation of an "irregularity" somehow previously caused. In reality those irregularities are continuously being formed, thanks to the "joint venture" of own Earth´s pull on the water (weight), and Moon´s pull. I´ll try and expose some more ideas, for the sake of an easier understanding (hopefully). Let us imagine a solid sphere, and that we put on its surface a series of thin rings of decreasing size, in such a way that every ring is in horizontal position and touches contiguous rings, and also the sphere. If we tried to lift a ring with many imagined vertical strings, maintaining it horizontally, if in upper part it could be easily done, but if at lower part it would have to lift contiguous smaller ring, "pushing" it. But if also smaller ring is lifted with a little stronger pull, there would be no push at all. If now we imagine the same but with our planet as sphere, tring to lift a ring (as Moon would try to do, if far over), being the oppose force (own weight) so hugely bigger, the lifting would be impossible. But now this own weight is not aligned with lifting strings as in the first case: Earth´s vector pull on each ring form a cone towards our planet center of mass. The result of all those cone shaping vectors, and the "infinite" number of pulls from the Moon (practically with a cylindrical shape) gives an "infinite" number of inwards vectors, causing a tendency to decrease the size of the ring. If rings are solid as initially imagined, all that only can produce a compression in the circumferential sense, not producing any "push" on inner contiguous ring. If now we change to the real case, imagining a kind of ring of water drops, things change because water is freer to move But those ring forming drops, even if inner ring previous space had become "void", can´t "shrink" in order to get into that smaller space … The UNIQUE possible "outlet" is upwards (towards the Moon), at least for a small fraction of the water. And then rest of the ring water could close in, forming a smaller but kind of thicker ring (lifted drops are also pulled inwards). Apart from the opposite sense gradient of relative to Moon pull vector and angular factors (when going from 90º away from sublunar point to 0º) , this idea of circular waves shows that the pace of their size decreasing changes too, from close to null 90º away area, to a maximum at sublunar area … So many factors would require a scientifically accurate study, to see real distribution of forces over the hemisphere. But I guess that at most places only pulls happen, "suction" rather than squeezing. Though as previously said, a tiny "circumferential" compression happens, that doesn´t add up inwards and towards sublunar point, because all over the hemisphere "excess" of water rises a little bit, even against its own weight. No pushes accumulate: what accumulates is water lifting (and partially water tangential displacement). Somebody could wonder how that is possible, if water weight is so hugely bigger than Moon´s pull … But that is not difficult: even a child can move upwards a heavy ball, if on a sufficiently small slope. And, as previously said, the so called bulges have almost no slope at all: an average of not more than a few meters between places almost 10,000,000 meters apart (at the equator)
See all the posts from Rafael in the comments, and you'll find the answers. It can get quite complicated, and this video was intended to keep things simple (Tides 101).
There is a very, very interesting youtube video, "What if the moon did not existed - Neil F. Comins" . Perhaps too long, due to its interactivity (between NFC and the audience). NFC is a really eminent astronomer ... Just google his name and you´ll see. He talks about causes of tides during some 10 or 15 minutes, starting app. 06:00. I first watched it some months ago. I had seen he clearly considers both gravity and inertial forces (due to earth-moon "dance") is what actually causes tides (not just the so called "differential gravity"). But only recently I realized he clearly also says that to consider moon orbits the earth is WRONG. I´ve been long discussing tides on other sites, here among many others. As NFC says, there is a lot of confusion out there. One of the main roots of that confusion is precisely what mentioned. People consider centrifugal force, at least in the case of earth movement and tides (NFC says just "outward" force) is fictitious, because they say earth is in free fall in its orbit ... What I posted here three weeks ago, especially 1st and 2nd parts of my series "MY ULTIMATE GO?" is quite in line with what NFC says, though he drastically just says what mentioned, and I elaborated a three part analogy trying to explain it to anybody interested.
Very nice! --- Just one objection. The diagram at 2:33 creates the impression that there are gravitational forces creating a bulge on the Earth's side that is averted from the Moon, i.e., the arrows pointing to the left. Actually, there are forces that do cause this bulge, counteracting the Earth's gravitation: the centrifugal forces resulting from the Earth's eccentric rotation around the Earth-Moon barycentre, located two-thirds from the centre to the surface towards the Moon. I've found this diagram on de.wikipedia, wikimedia.org/wikipedia/commons/4/4e/GezeitenkräfteErdoberfläche.png I'm sure that you don't need a translation.
Quite right! But, you know, many scientists (perhaps most of them), consider centrifugal forces are not real (!!), only apparent or fictitious ... And they don´t even mention those words. The video is really nice, as you say. But with that objection, and a similar one when they refer to Sun related bulges. In this case, with the barycenter at the Sun (I presume), centrifugal forces are even more evident. But they only consider gravitational pull differences !!
Thank you for your comments. Your link doesn't work, but I'm assuming these images would be correct: w3.salemstate.edu/~lhanson/gls214/images2/gls214_tides3.gif, www.mt-oceanography.info/IntroOc/notes/figures/fig11a1.html. I chose in this video to simplify the forces and focus on their net result. I will reconsider this for the next version. Thanks again. :)
MY ULTIMATE GO ? (4th part) As said on 2nd part, I consider that rather than talking about “free fall” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were parts of an unique extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about their common center of mass, maintaining the distance between them due to their dynamic equilibrium. An ice skating couple can spin like a single dancer, because they are linked usually through their hands … If one of them had a pony-tail, when spinning, instead of hanging vertically, the pony-tail would move upwards, against its own weight … Clearly due to centrifugal force. The same would happen whatever the way they were linked to each other. If it were a kind of mutual “gravitational” pull sufficient not to need their hands (as earth-moon case), and they had suitable initial "tangential" speeds, the pony-tail would also “feel” that inertial force (why wouldn´t it?) … It would react to the TOTAL force acting DIRECTLY ON IT (WHATEVER the pull on other parts of the dancer, what the pony-tail “ignores" …), and raise where centrifugal force prevailed. Earth particles react to those inertial forces in a similar way to what quite clearly explained for the diminishing of our weight on the equator, and causes the equatorial bulge (though the author prefers not to mention the adjective “centrifugal” … ): “The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force (man´s weight) is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. IN FACT, THE ENTIRE BODY OF THE EARTH EXPANDS SLIGHTLY AND THE MAN AND SCALE MOVE OUTWARD FROM THE AXIS OF ROTATION SLIGHTLY, UNTIL FORCES COME INTO BALANCE WITH THE REQUIREMENTS OF ROTATIONAL STABILITY AT THE NEW RADIUS. THIS IS THE REASON FOR THE EQUATORIAL BULGE OF THE EARTH DUE TO ITS OWN AXIAL ROTATION" www.lockhaven.edu/~dsimanek/scenario/centrip.htm That´s why the NASA scientists I´ve referred to many times say: "At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point). At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system). On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction. In a hypothetical ocean covering the whole Earth with no continents there will be TWO TIDAL BULGES RESULTING FROM THESE IMBALANCES OF GRAVITATIONAL AND CENTRIFUGAL FORCES, ONE FACING THE MOON (WHERE THE GRAVITATIONAL FORCE IS GREATER THAN THE CENTRIFUGAL FORCE) AND ONE FACING AWAY FROM THE MOON (WHERE THE CENTRIFUGAL FORCE IS GREATER THAN THE GRAVITATIONAL FORCE" I DO KNOW many people (physicists included) will say: "But you are using a non-inertial frame of reference, where centrifugal forces are “added” by us, in order to keep Newton´s Motion Laws valid. But those forces are “fictitious”, they don´t actually exist" … (or something similar). This post is already rather long. I´ll leave my refuting of what usually is deduced from that question of frames of reference for another post.
MY ULTIMATE GO ? (3rd part) To diminish the risk of misinterpretations, before delivering the conclusions I referred to on 2nd part, I´ll repeat here the "breakdown" of earth complex movement into its simpler (and more easily graspable) components ... Earth´s movement (as a whole) is basically the addition of three simple, but quite different, movements: 1) Earth-moon common center of mass orbits the sun (once a year). Both celestial objects, as a "couple", are kind of linked to that CM (called barycenter), and move with it. That movement, and sun´s pull, originates what I call sun-related tidal effects (sea tides included). For the sake of simplicity, I usually disregard that movement, when analyzing main component of tides: moon-related tides. 2) The "couple", as I said yesterday on 2nd part, revolves/rotates around the barycenter (once every some 28 days). THAT IS THE MOVEMENT I ALWAYS REFER TO (unless clearly said the contrary), discussing and analyzing it AS IF ONLY THAT MOVEMENT WERE HAPPENING ... Moon´s pull, and dynamical effects originated by that movement´s features, causes two opposite bulges that continuously change position, logically with same periodicity of some 28 days (relative to the rest of the universe). Details of the singularity that earth revolves (instead of rotating like the moon) are explained on the post I sent here a week ago, with a "handy" analogy ... 3) Earth also has its daily spinning, which, apart from causing the permanent equatorial bulge (by the way, thanks to huge centrifugal forces inherent in such fast circular movement), makes us to perceive the period of the movement of the pair of bulges as if it were once a day, instead of once every some 28 days. Some time ago I posted the link to a video relative to that saying: "I suggest anybody interested to have a look at : th-cam.com/video/G-MOZId0FNw/w-d-xo.html where it´s clearly seen that daily movement of the bulges is only apparent, that they are almost still and it is the solid part of our planet (though also the bulk of ocean waters due to friction) what is actually spinning … The formation of the bulges is a rather slow process (some 28 days the complete cycle) … Nothing to do with all those daily local whirlpools, due to the much faster Earth spin, and with any other local singularity".
Another comment I´ve sent to the site I mentioned two days ago: "The idea is well orientated, but the explanation is FLAWED since its very first minute. At 0:38 it is said: "The moon's gravity is stronger at Point A and weaker at Point B than it is at Earth's center. The net effect of this differential of the Moon's gravity across the Earth is to stretch the oceans out like taffy, ergo why the oceans bulge out at opposite points along the Earth/Moon line. Now, that explanation sounds plausible and a lot of well known scientists give it, but as we'll see, it's not correct. … They aren't actually being lifted or stretched …" What do you mean with "lifted"? If "levitated", what nobody has said, logically that is quite impossible. But water in sublunar bulge (let us say less than 45º away from sublunar point) kind of weights a little bit less due to Moon´s pull, and due to that its level rises a little bit, though only a small portion of actual full high tide. And our planet DOES stretch due to opposite forces. Its solid parts experience tensile internal stresses to counter those forces, but as water is freer to move, its answer ALL ACROSS oceans is massive displacement of water towards sublunar and antipodal area ... And, by the way, that displacement happens, rather than through any "squeezing", in the way I explained a couple of days ago, as far as I can understand !!"
As a continuation to my post of yesterday, just to say I´ve already sent mentioned message, adding what follows: "By the way, apart from nature itself (and its own motion laws), mentioned tangential speeds could be considered the actual cause of tides, whatever the energy source that made those speeds possible. Without those initial speeds, and Moon-Earth "dance" originated by them, tides wouldn´t have been happening for billions of years as they have !"
Thank you. I suppose you also refer to my more recent posts, but you have put your comment on a post of mine of 1 month ago ... Perhaps because, I don´t know why, something is wrong with the order of the posts ! The same fellow of my discussion of last days, also told me that "first thing in the morning" ... And I replied: "I´m glad you say so, hopping it implies you´ve found things I´ve said are worth a book (though you exaggerate ...) ! But, you know, I type with only two fingers, and not being English my mother tong it would take me “eons” … By the way, he´s sent another long "refuting" post, but I need some time to read it more slowly, ruminate and refute it ...
My discussion fellow replied: "The removal of rotation does not remove the tidal bulges - they remain, but clearly the Earth has to be free to accelerate towards the moon for this to be the case. If you hold the Earth in position, then you have a different result where a single bulge will appear on the side nearest the moon" I answered: "There we are! It´s unbelievable that you consider absolutely necessary to let the earth free to get the rectilinear accelerated movement towards the moon (to have the second bulge), but time and again you say that (as it is in the real case) the curved accelerated movement of the earth (its revolving around the moon-earth barycenter), has nothing to do with the formation of the outer bulge, and that it is not necessary at all !! They are two different ways inertia manifests itself, but due to same basic Physics laws ... In the first case, outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge appears. In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth also stretches, and outer bulge builds too ..." Needless to say those outwards stresses could be classified as "centrifugal", at least in the broad sense of the term ...
MY ULTIMATE GO ? (2nd part) C) Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right … And not just as big as E´s (as in case B) but much, much bigger (see below how much). If with the correct values for masses, M-E distance, and initial speeds, we would have our real case, E being the Earth and M the Moon … We know the result is that, instead of E being orbiting around M as in case A, and somehow opposite to what in case B (when E got “fully free” to fall directly onto M), now E is “forced” to bend its trajectory much more, because M´s location is changing in a sense opposite to E´s speed … E and M revolves/rotates around their common center of mass. E´s trajectory curvature is much, much bigger than when orbiting around M (case A). For similar (but kind of opposite) reasons to case B, IT IS ERRONEOUS TO CONSIDER E “ORBITS” AROUND M, AND TO SAY THAT E IS IN A “ FREE FALL" … In cased A we had a “partially free" fall, and in B a “fully free” fall … But NOW I CONSIDER WE HAVE NO FREE FALL AT ALL, BECAUSE INITIAL SPEEDS AND MUTUAL PULLS ARE CAUSING A KIND OF DANCE OF THE COUPLE, KEEPING THEIR SEPARATION ACTUALLY CONSTANT, AS IF IT WERE A SINGLE TWO-PARTS OBJECT SPINNING AROUND ITS CENTER OF MASS … With this and last post I´ve already delivered what “promised”: a fresh explanation of one of the main roots of my stand … But I´ll leave for another post the elaboration of main consequences of that, as far as I can understand ...
Recent days I´ve been really busy with previously commented discussion. The other party, defending the idea that ONLY differential gravity from moon intervenes on tides, said many absurd things about curved movement dynamics ... Eventually I had to tell him he was utterly wrong, living aside the bottom of the question, clearly in how he supports his stand. We had to discuss, not about the term "centrifugal" force: about "centripetal force", which he had said it is a "grey area", that people say about it things not fully rational (??). The day before yesterday he said: "A force attributed (?) to rotation which (the force) persists when the rotation is removed should not be given a name that implies that it is caused (!!) by rotation". In other paragraphs he also referred to "forces caused (or not) by rotation", or to "forces caused (or not) by certain movement", as if we could say some forces are caused by some movements, and others are not ... And I had to tell him: BUT THINGS ARE THE OTHER WAY AROUND !! MOVEMENTS CANNOT CAUSE ANY FORCE ... FORCES DO CAUSE MOVEMENTS (OR CHANGE THEM) !! His ideas are flawed even at the very "basics of basics" level ... Tens of not short posts ... to no avail. I´ve wasted a lot of time !!
The explanation I made yesterday (my first post) in relation with the imbalance between required centripetal forces and actual gravitational pull at each area (what -> internal stresses -> Earth stretch -> two opposite bulges) would match perfectly (if saying Earth where I had erroneously said Moon, as corrected yesterday in my second post) with the case of Sun related bulges: the farther (from Sun) the area, the smaller the gravitational pull, but the bigger the required centripetal force … As I also said, Moon related bulges are a little bit trickier to explain. Coming back to the drawing board, I can elaborate as follows. As Earth/Moon barycenter is situated app. 2/3 Earth´s radius from its center (logically, at Moon side), Earth movement in this "couple dance" is a kind of wobbling, similar to when a child plays with a hulla-hop (probably not spelled so). Though Earth daily spinning generates much, much stronger internal stresses and centrifugal forces (some 28/29 times higher angular speed …), these act always same way: trying and increasing Earth´s diameter at the equator and nearby. Earth wobbling due to its "dance" with the Moon generates internal stresses and deformations which we can perceive added to those mentioned stronger ones. Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation. Earth parts to the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration there. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: closer to the axis parts have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL … At Earth parts at right side of that axis, required centripetal forces are smaller, also the farther the bigger. BUT now gravitational pulls from the Moon are opposite to required centripetal forces, the closer to the Moon, the bigger. Only own gravitational Earth´s pull can supply required centripetal forces … As their distribution don´t match with Moon´s pull distribution, also internal forces occur to compensate the imbalance. And that spherical segment is similarly stretched too, and a bulge also occurs at that side (that bulge is usually explained, but partially, in a much simpler way: just because there Moon´s pull is higher ...) And due to Earth´s daily spinnig, those two opposite bulges seem to rotate around Earth, trying to keep in line with Moon (as actual causing forces). Result: our by all known, but not so well known, tides (high tides where the bulges, and the rest ...) Globally considered, as we also have, as a consequence of the same, quite an array of different local tides, VERY WELL explained in the video.
I´ll continue to refute what said on “Misconceptions about tides” (“lockhaven.edu-dsimanek” previously linked sites). They author of that article also says: “ … Once they launch into the rotating coordinate mode and start talking about centrifugal forces … They also forget that the non-uniformity of moon's gravitational field over the volume of the earth is alone sufficient to account for both tidal bulges, bulges that would be essentially the same if ... the earth and moon were not moving relative to each other”. HE is the one who apparently forgets that hypothetical case is NOT possible … unless some other external force were countering the gravitational attraction between both celestial bodies !! Let us “play” with our imagination. Some “possible” (?) cases could be: 1) A kind of “cosmic" superman were able to keep them separated their current distance, pushing them outwards on their inner sides. Earth would not get tidal bulges but quite the opposite: it would become slightly flatter (internal compression stresses increase) 2) Similar situation, but pulling outwards from their outer sides (where strong solid parts in the Earth side ...). Kind of bulges would build, but outer one would be quite different, depending on the way our superman could "grasp" the Earth … (internal tensile stresses increase) 3) Some “ miraculous” outward forces were applied directly on Moon and Earth respective centers of mass, countering mutual gravity attraction … Earth inner hemisphere would get an ovoid shape, kind of bulge (internal tensile stresses increase), but outer hemisphere would experience the opposite type of deformation: it would get slightly flater (internal compression stresses increase). THEREFORE to get our real bulges it is paramount the way Moon-Earth mutual attraction is countered to keep actual distance … And the real way is their rotation/revolving around their barycenter !! THAT EARTH REVOLVING implies inertia related outward forces, quite an intervening fact in bulges formation, as explained on my last post. So, to say “… bulges that would be essentially the same if ... the earth and moon were not moving relative to each other” is erroneous, or at least pointless, because it compares something real with something impossible ...
Tides are not binary, some places have one a day. Locations at same latitudes on same ocean don't experience tides when others do. Tidal bulges are out of phases with moon's location. Low tide 9/20/17 during new moon happens 1 hr 45 min after moon meridian (Sebastian, FL) Tides near equator are smaller than tides nearer poles.
Yes, you are correct! The last portion of this video explains that -- see the Amphidromic systems section at the end for more information. The majority of this video discusses the theory behind the causes of tides (tidal bulges). The last portion of the video explains the impact of enclosed tidal basins on the bulges (reality).
This very morning I sent what follows to another site, replying to a post trying to refute that centrifugal forces inherent to earth revolving count for tides ... I´m sure it could be interesting for many of you. Mentioned post said: "Imagine a case where there's a planet and a "moon", but there is no orbiting - the planet is being pulled along on a string in a straight line by a powerful spaceship and is accelerating, while the moon is being pulled along behind it by gravity such that it keeps up and maintains separation. The moon's gravity is causing two bulges in the planet's ocean, but there is no rotation involved in the system at all, so are the bulges driven by centripetal force in a case such as this one? No" I replied: "You yourself answered your question: No. But, I´d rather say "no" to the very question: if the powerful spaceship would be making the earth accelerate, and with the moon back at current distance, earth deformation would depend on the way the spaceship force were transmitted to earth: 1) If somehow it were transmitted pulling on earth c.g., we wouln´t have two bulges: forward hemisphere would get flatter. 2) If the spaceship pushed on earth back side, the whole earth would get flatter (if pushing through a sufficiently big flat surface): no bulges at all. 3) Only if pulling from forward earth parts, we would get two bulges. But forward bulge would not be due only to less gravitational pull from more distant moon, but also to the very spaceship direct pull ... We could even draw a parallel with our real case. In your imaginary case, the agent which causes moon-earth distance doesn´t diminish, is the mega-spaceship ... In the real case, similarly the agent is a kind of "hidden" force ("centrifugal" f., or at least "outward" force), due to inertia and the fact that all earth parts try to follow the tangent, and moon-earth dance, through invisible "gravity strings", is continuously "bending" the trajectories of all those earth parts ... Without mentioned dynamic effects of that earth-moon dance, we could not even be here discussing the issue, because they were paramount in the very history of our planet evolution and the beginning of life !"
The discussion I referred to a week ago has continued. Yesterday I sent him what follows, I hope self-explanatory. It is directly connected to what on this youtube video. "You´ve said a lot of times something like: if we “remove" the revolving movement of earth, the bulges remain … So, that movement cannot be causing the tides. I already did refute a similar idea of yours, in relation with your idea of a mega-spaceship causing bulges. But you didn´t get it, and said I had taken your case in the wrong way !! Let us forget the spaceship, and have just moon and earth. If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision. OK. Therefore, revolving was not causing them (YOUR deduction). IN A QUITE SIMILAR WAY, we could reason as follows: If, somehow (a cosmic “hiperman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …) Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !! Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong … Any guess?"
(Continuation of my post of 2 weeks ago) After other posts interchanged with my companion fellow, I had to tell him: "You stubbornly keep ignoring that, IN ALMOST A MONTH, apart from those 28/29 cycles of earth daily spinning, and apart from a little less than a 29/365 of the complete orbit around the sun, our planet does complete a cycle of its revolving around the barycenter ... Disregarding the daily spinning, which has nothing to do with moon-earth dynamics, we can look at the moon e.g. midnight (when visible), and write down its angular position in its apparent trajectory. And do the same next day ... We´ll see there has been a change towards the east: THAT is the real movement of the moon. And earth, in its dance with the moon, ALWAYS at the other side of the barycenter, will have followed the same angle in its revolving around the barycenter. By the way, I posted a link to a google image which shows that "dance" in three different positions, very interesting, but I didn´t see any further comment here ... Did you see it? I said: (#152 and #153) "There is an image: main-qimg-851d9284749b378191c8ae87e4e2e4c2 (copy and paste on search engine window) that can help guys understand better moon-earth dance, especially earth revolving around the barycenter. By the way: the image has a detail a little bit erroneous ... Any guess? A Google page appears. Please click on images …"
MY ULTIMATE GO ? Once again I´m going to try and convey a core idea of my stand, now in a “fresh” way … It will have three parts. I´ll post them neither on same post, nor on three different ones, but on two posts … (that way readers will have more time to “ruminate” what proposed). A) Not to need to include a figure, please kindly IMAGINE one with two celestial objects on same “vertical” line (on your screen). Let us imagine the lower one (M) on a fix location, and the upper one (E) moving “horizontally” from right to left, and let us consider M´s gravity starts to pull downwards E just when reaching M´s vertical (remember, M is considered somehow on a fix location). If within certain game of relation between masses, distance and speed of E, E will start “orbiting” around M. M´s gravity, acting as centripetal force, changes E´s speed vector direction, but it is not “able” to change its size. A “free fall” to many, but rather a “partially free fall” to me, because E´s inertia avoids a direct straight line fall. B) Let us now imagine M, at the very moment E gets at M´s vertical, starts moving also from right to left, with same velocity as E. Being now M´s pull on E always vertical, E will fall towards M, following a parabolic line. Now M´S PULL IS ABLE not only to bend E´s speed vector, but also TO INCREASE ITS VERTICAL SPEED WITH AN ACCELERATION PROPORTIONAL TO M´S PULL … same way as if neither of them had an initial horizontal speed … THAT´S WHAT I CONSIDER A "FULLY FREE” FALL ... And WE SHOULD NOT SAY E IS ORBITING AROUND M, because it will fall onto it !!
I´ll continue to refute what dealt with on my post of four days ago. One of first paragraphs of “Misconceptions about tides” (“lockhaven.edu-dsimanek” previously linked sites), says: "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 of the earth”. That last comment seems to show the author is biassed: he starts considering rotation (“revolving”, rather) has nothing to do with tides … With that fixed idea, his reasoning may be kind of distorted, reaching erroneous conclusions !! Further comments we can do about it: 1) If only "the gravitational pull of another nearby object” (the Moon in our case) counted, Moon and Earth would have fallen onto each other long, long ago … 2) I know standard Physics says: but they both rotate/revolve around the barycenter, and that prevent them to fall onto each other !! 3) Right. Therefore, without that rotation/revolving we wouldn´t have had the tides we´ve had for billions of years !! 4) And as Moon´s gravitational pulls on farthest hemisphere ocean water “units" are insufficient to cause required centripetal acceleration at their place, those water “units" tend to follow in the direction of their tangential speed (logically, due to the “inertia” phenomenon). Part of own Earth pull on that water has to be “used” to produce that centripetal acceleration deficit. 5) The author also himself says on www.lockhaven.edu/~dsimanek/scenario/centrip.htm that (explaining why on the equator we weigh less than at higher latitudes, being our weigh maximum at both poles): "The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation". EXACTLY THE SAME happens (why it wouldn´t? …) to water further from the Moon due to Earth revolving around Moon-Earth barycenter, because its weight there has same direction and sense as required centripetal force. And that water lightening there, together with the displacement of water from rest of further hemisphere (caused similarly, but with different angles between weight and required centripetal forces) is actually the reason of the antipodal bulge formation !! The varying Moon´s pull also intervienes, but only causing previously mentioned centripetal force “deficit” !! Another thing is that, as Earth revolves without rotation (in its “dance” with the Moon), required centripetal force is constant across it. And as at Earth c.g. it is in balance with Moon´s pull there, the “fields” of gravitational differences, and the one of differences at each place between required centripetal force and gravity, are equivalent !! That fact keeps kind of hidden to many the error of thinking Earth revolving has nothing to do with tides formation … The so called “standard model”, considering only gravitational gradient, would give erroneous results when analyzing e.g. Earth tidal effects on the Moon, or on any of the many billions of celestial objects tidal locked to another, because in those cases centripetal acceleration is NOT constant: it has to increase proportionally to distances to respective barycenter !!
I have to say now, despite my negative comments of two weeks and one month ago, that what said in the video is essentially correct. Why that change ...? If interested, see my last comments in: www.thenakedscientists.com/forum/index.php?, especially last one, of this very morning ...
I´ve just now sent what follows to the author of the included article link, after some pieces of email we have interchanged, especially in relation to his idea that ONLY differential gravity is the root cause of tides. I´M SURE it may interest many of you ... "On the one hand, I´m glad you find "my take" interesting. But on the other I´m kind of sad, because you found necessary to send me Newton´s 3rd Law "properly stated" ... Do you think I had used it wrongly ?. I was very careful with its application ... But in "my case" it cannot be applied exactly as originally stated, because three "bodies" interviene, rather than just two: Moon (A), a certain part of Earth (considered portion B), and the rest of our planet not included in B. When a cart is pulled by two horses, if accelerating and supposing no friction losses, 2nd and 3rd Newton´s Laws apply: - Added pull of the horses on the cart is equal to the mass of the cart multiplied by the acceleration it gets. - The cart is pulling back EACH horse with same size pull each horse is exerting on the cart. By the way, I´ve seen you yourself explain, and very well, a really similar case using EXACTLY same arguments as my take ones (your article www.lockhaven.edu/~dsimanek/scenario/centrip.htm Under the subtitle "Weight" you say: "Weight may be defined as the force required to keep an object at rest relative to its surroundings. This definition is consistent with most colloqual interpretations of the word (surprise!). The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation". TAHT CASE is quite similar to "mine", in wich Earth is revolving around the barycenter. A man on Earth surface where antipodal bulge builds would also experience ALL those changes, due to a lightening equally caused by its circular movement (though much smaller). Unique difference: now we have two "horses" pulling instead of only one ... IT´S ALSO CLEAR that one of the "horses" (the Moon) pulls Earth massive parts there (B) less than it does on closer parts, but what that actually causes is that the other "horse" (massive parts of Earth not being B) HAS TO pull more, and "supply" the difference. And corresponding reaction according to Newton´s 3rd Law, equal but opposite (therefore outward, "centrifugal" in its general sense) and exerted by B ON REST OF EARTH, is also bigger than at closer parts (every vector having opposite sense at closer hemisphere) And, as you say for equatorial bulge, I can say: "This is the reason for antipodal bulge of the Earth due to its revolving around Moon-Earth center of mass". You also told me: "I don't think it is helpful to think of tidal bulges as being caused by tensile forces in earth or water" ... I think it´s not a question about being "helpful" or not ... As far as I can understand, also in line with what you yourself say for the equatorial bulge case, root causes of tidal bulges are several: varying Moon´s pull, Earth own pull on each of its parts (gravitational or due to Earth rigidity), initial conditions that gave tangential speed to Earth (and Moon) ... and nature motion laws (basically due to the so called "inertia"), as Newton gave us. CHANGES in internal forces where solid Earth (together with corresponding stretching), and in pressure distribution where water (alongside its corresponding pilling up on opposite areas) are nature answers to above mentioned causes. They are actually the tides themselves, rather than their cause. By the way, apart from nature itself (and its own laws), mentioned tangential speeds could be considered the actual cause of tides, whatever the energy source that made those speeds possible. Without those initial speeds, and Moon-Earth "dance" originated by them, tides wouldn´t have been happening for billions of years as they have ! Any comments would be appreciated. But please kindly keep in mind quick replies may be affected by our "deep-rooted" ideas ... I would prefer a late but well "ruminated" reply.
MY ULTIMATE GO ? (5th part) Before trying and convey my stand on non-inertial frames issue, let us go back to the drawing board, as if we didn´t have any education on dynamics … Same image with Earth (E) on upper part, and vertically below the Moon (M). Let us suppose we only know how they move, with kind of the mind of a “smart” child. And then we learn that (as David Cooper said many times), if Earth and Moon tangential speeds somehow became null, they would begin to accelerate straightly towards each other, because of gravitational mutual attraction … If just stopping their circular movements, they get “free” to accelerate towards each other … THOSE CIRCULAR MOVEMENTS ARE WHAT, WHEN EXISTING, WERE SOMEHOW COUNTERING THOSE STRAIGHT ACCELERATIONS NOW THEY HAVE GOT !! No other logical possibility for a “clean” child mind … How come? Let us see … Curiously, with a rotating (so called "non-inertial”) frame of reference, same thing happens, but let us say “virtually” … As the frame of reference rotates with E and M, E and M don´t "rotate" anymore relatively to that frame of reference … Centripetal forces are not required whatsoever, and gravitational pull would make them fall onto each other … But that´s far from reality … Then physicists go on: "How can we do “real” maths with that “artificial” (the adjective is mine ...) reference system ? … We have to apply a “fictitious” force, which we´ll call centrifugal force". That way WE GET BACK TO THE REAL SCENARIO … as far as dynamics is concerned. Well, that force is certainly “fictitious”, because it has been introduced by us to work with that non-inertial frame of reference … BUT THET MEANS THAT THE REAL CIRCULAR MOVEMENTS, NOT NEEDING THE ADDITION OF ANY “ARTIFICIAL” FORCE, SOMEHOW PRODUCE AN INERTIAL EFFECT THAT KEEPS E AND M WITHOUT FALLING STRAIGHTLY ONTO EACH OTHER !! TO ME (AND NOT ONLY ME), THAT IS A REAL CENTRIFUGAL FORCE, QUITE REAL !! In case B (“MY ULTIMATE GO ?", posted 4 days ago), as E and M were moving horizontally at same speed, Moon´s pull was able not only to bend E´s trajectory, but also to accelerate E straightly towards M with its full strength, quite a “free” acceleration. That “freedom” was made possible by the movement of the moon towards the left, with same horizontal speed as E. The real case is dynamically quite the opposite: the movement of M in opposition to E´s speed, and with speeds which make them to revolve/rotate around the barycenter, makes impossible the decrease of E-M distance, and even any proper orbiting of E around M. It gives us a “rotating” scenario, where all forces have basically same direction as the line between centers of mass, and where the angular position doesn´t matter much (as far as Earth-Moon dynamics is concerned), as long as we accept centrifugal forces keep the separation constant … After all, if Earth and Moon were the only celestial objects in the universe, to talk about angular position of the system would have no sense at all … BUT, STILL, THE WOULD HAVE TO KEEP ROTATING. THAT WAY INHERENT INERTIAL FORCE, THE CENTRIFUGAL FORCE, WOULD KEEP THEM WITHOUT FOLLOWING STRAIGHTLY ONTO EACH OTHER !! Therefore, as I´ve said many times, I find quite correct what a NOAA scientist told me: "... to provide a basic description of the forces which create the tides. It's intended audience were the grade school children and adults of that time. It used terminology of science and forces which were common in the 1950s. Such as centrifugal force. Centrifugal force was always an "imaginary force" (not a real / measurable force). But that type of description made the concepts easier to understand and explain. That description and use of centrifugal force continued to be common practice until the 1970-80's. At that point, the terminology shifted and the TEXTBOOKS USED IN GRADE SCHOOLS WERE CHANGED TO USE A MORE MODERN TERMINOLOGY AND DESCRIPTION OF THIS “EFFECT”BEING A RESULT OF INERTIA RATHER THAN AN “IMAGINARY FORCE”". Initially I didn´t fully grasp his point ... But later I did. The problem is that many books, dictionaries included, keep following models with ideas which are several decades out of date …
When discussing with somebody else things said on the very interesting articles I´m lately referring to: www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip www.lockhaven.edu/~dsimanek/scenario/centrip.htm he asks me what below. I´ve seen many people somehow have rather confused ideas about daily manifested tides (does Earth daily spinning cause tides ??), when Moon-Earth "dance" movements have almost a month period ... That´s why I´ve decided to send this post. I answered what follows. "Your question: " ...(I had previously said) DAILY Earth spinning causes the "permanent" equatorial bulge, but it has NOTHING to do with Moon-Earth dynamics ... Does this mean that you also agree that the earth's spinning has nothing to do with tides?" My answer: I chose carefully my words. There is a trick is in the way you put the question ... The earth's spinning has nothing to do WITH THE ROOT CAUSE OF TIDES, but it does with the way they manifest ... When we say "tides", in its global sense (local changes of sea level can have other causes), we refer to the cyclical change of sea level related mainly to Moon position in its monthly rotation. To analyze their actual causes, we have to disregard the daily spinning of our planet (imaging we check the position of the bulges everyday at same time). That way we would have the bulges, always in line with the Moon, "moving" around our planet in an app. 28 days "two-parts" cycle ... THAT IS the basic phenomenon of tides. BUT, as Earth does spin daily, its meridian in line with the Moon changes continuously, and though bulges maintain its position relative to the Moon, they continuously change meridian too. And Moon related tides SEEM to have an app. 12+12 h. cycle ... BUT THAT SPINNING doesn´t cause the tides, though it causes the so called equatorial bulge, easily explainable as done on “lockhaven.edu-dsimanek” previously linked sites, when analyzing how we weigh less on lower latitudes, especially on the equator: "... Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation". ANOTHER THING is that the monthly revolving of the Earth in its "dance" with the Moon, also causes similar inertia related effects (not to use tricky concept "centrifugal forces"), which actually are one of the causes of tides, especially on the hemisphere further from the Moon, where antipodal bulge appears ... On closer hemisphere, though those inertial effects have also an outward direction, as Moon´s pull there is higher, the later prevails and the better understood bulge builds towards the Moon !!
As said in my answer of yesterday to W. L., I also find this video really good, but with same objection: centrifugal forces are "ignored" and they shouldn´t ! I´ll try and elaborate. At diagram at 2:23 "medium" gravitational pull (average as if acting on Moon´s C.G.) is rested both at nearer and farther sides, trying to explain the bulges ... The authors don´t consider that any centrifugal force, they´re just comparing to find kind of "relative bulges" ... But they are wrong, as far as I can understand. What´s causing "medium" black vector at C.G.? : the centripetal acceleration required for the rotation. But neither that REQUIRED (to orbit at each distance) acceleration is uniform across the Moon, from left to right. Farther parts of the Moon, to rotate at same angular speed, NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces. ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in our diagram, to farther parts ... But according to Newton´s 3rd Motion Law, if at any considered Moon´s section, right part is pulling (towards Earth) contiguous left part, this one is also pulling the former towards the left, same force but opposite direction: OUTWARDS. All those internal forces are centrifugal REAL forces, which, together with their "mirror" centripetal ones, stretch the Earth, its solid parts included. Logically, that deformation is much bigger where liquid parts, our oceans. And the TWO normal bulges ARE DUE TO THAT.
As a continuation to my comment of one month ago, this is just to say that afterwards I sent two other posts, #85 and #86, to: www.thenakedscientists.com/forum/index.php?topic=49715.new#lastPost They contain new ideas aiming to help understand the issue, and the reasons why I consider there is so much confusion out there ...
Sorry: I had recently been discussing why Moon has got tide locked to Earth, and in: "But neither that REQUIRED (to orbit at each distance) acceleration is uniform across the Moon, from left to right. Farther parts of the Moon, to rotate at same angular speed, NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces" and instead of saying "neither that ... acceleration is uniforme across the EARTH" and farther parts of the EARTH" I had a lapsus and said "Moon" !!! There´s another rather tricky detail: Earth´s rotation I´m referring to (some 28/29 day period, as Moon´s) is around Earth/Moon barycenter, app. 2/3 of Earth radius from its center, towards the Moon ... That would change the diagram: we have bigger centrifugal forces at Earth´s left far side, and smaller at right side, BUT this one in same sense as the pull from the Moon ... What said about internal stresses, centrifugal and centripetal, and about subsequent Earth stretch as cause of the two opposite high tides, keeps being valid !
Thank you for your honest reply. I´m not going to insist now on why tides actually happen, as far as I can understand. I suggest anybody interested to have a look at my post of two days ago, and at linked site. Regarding the linked youtube video (...4S99i4), utterly erroneous to me, last week I posted a couple of comments there, and less than an hour ago one more, as a continuation of the former. If anybody interested, there you can see my main reasons to say so.
As a continuation to my comment of four days ago, just to say that yesterday I sent another post to: www.thenakedscientists.com/forum/index.php?topic=49715.50 comment #84, with an analogy with a children game (the whip ?), which I´m sure could help understanding what I´ve been exposing about our topic ...
Hello Earth Rocks, I distinctly get the impression from reading the comments, that you’re searching for a simpler explanation to tides? Well, how about this:
The centrifugal force from *Orbital* motion around the barycentre, is not 6.5 times stronger. The Earth orbits, revolves, translates around the barycentre. Each point of mass on Earth, maps out a circle of equal radius _(4670km)_ as it orbits. The centrifugal force is equal across the whole earth. And when you subtract that (rω² ~ 3.3x10^-5 m/s²) from the gravitational attraction and factor in the earth’s own gravitation. You get an acceleration of +/- one ten millionth of a (g) on either side and *zero* at the centre, which is due to the *free-fall* motion experienced by all Planetary bodies, which are in motion in our solar system.
Thank you for your support, ~ _wavy!_
The discussion for this video has rich with the math and details, and I appreciate that being provided for everyone. This video was designed for an introductory level course, and I specifically didn't want to get too deep into the details on barycenter (wanted to keep it simple), but thanks for adding it to the discussion. :)
@@EarthRocks You’re more than welcome, and I really appreciate the reply. Thank you. My only video was specifically designed, as an introductory course into dispelling the popular misconception, that the Earth remains static in space, when in fact it Rocks! …but, didn’t want to get too deep into the details about barycentre motion and apparent forces (wanted to keep it silly). Maybe we could work on something together in the future. :)
This is an interesting video. It will take a while to think about these correlations. Thank you.
This is the best video I've seen about explaining tidal forces and movements. Thanks a lot.
So many other videos do not have the faintest clue how the moon is related to Earth and where the plane of movement is located.
You Rock!
The center of mass in the Earth-Moon system, also known as the barycenter, is located within Earth itself. It lies about 4,671 km (2,902 miles) from Earth's center, which is approximately 75% of Earth's radius on the moon side.
While it seems like the Moon orbits Earth, both the Earth and Moon actually orbit this common center of mass. However, because the Earth is much more massive than the Moon, the barycenter is located closer to Earth's center, causing the Earth to wobble slightly as it orbits the barycenter.
This means that the elliptical path we generally attribute to Earth's orbit around the Sun is actually the orbit of the barycenter of the Earth-Moon system. So you would observe the earth wobble about the barycenter the side opposite the moon is traveling faster around the barycenter and thus it is centrifugal force that is causing the bulge opposite the moon in about the same magnitude as the moon facing side.
Discussing with somebody else (who says only differential gravity counts), I think I found another simpler way to explain tides:
Total force exerted on any given particle (because of moon-earth rotation/revolving), has to be equal to the centripetal force required for its revolving, to accomplish Newton´s 2nd Motion Law). The main force “supplier” is moon´s pull AT EACH LOCATION. The other possible forces only can be exerted on each particle by surrounding ones (internal stresses), or by the very gravitational pull from the rest of our planet.
Closer to moon particles "feel" stronger pull than required centripetal force.
THAT excess (and not any "differential gravity") has to be passed onto the rest of earth through their neighbors.
On their turn, those neighbors exert opposite forces back onto considered particle (3rd Newton´s Motion Law).
At farther hemisphere interactions are a kind of mirror image of what at closer hemisphere, because the farther the particle, the bigger the deficit of moon´s pull (compared to required centripetal force).
That constitutes a complex "chain of transmission", where total force acting on each particle (leaving aside own weight) has to satisfy 2nd Newton´s Motion Law (F=mω²r) ...
THAT FIELD OF FORCES, exerted DIRECTLY ON EACH PARTICLE, and the INERTIAL EFFECTS (centrifugal forces included) caused by the fact that all particles are being "forced" to revolve together, ARE ACTUALLY THE DIRECT CAUSE OF TIDES !!
Basically solid earth is stretched, and water moves towards where the two bulges build up.
None of those forces is "artificial", and all are directly felt by considered particle, what implies "natural" additions of forces, without any artificial intervention of our minds. That would be necessary to deduct moon´s gravitational pulls at locations very distant from each other (the so called “differential gravity”), what obviously material stuff can´t do !!
Simpler way?
You are great.This video is great for concept clearing.
Minutes ago I sent what follows to "What Physics Teachers Get Wrong About Tides / Space Time / BPS Digital:
"I have the "bad" habit of ruminating this type of things even at night in bed ...
Last night I had a kind of eureka moment.
Sublunar bulge has to be the result of both Lunar and own Earth pulls ... But NOT because water close to poles and previously low tide areas "pushes" closer contiguous water towards sublunar point and , "those tiny tangential sideways pushes on all the chunks of water added up over half the surface of the planet can produce a pretty decent increase in water pressure" as it is said at 5:35, for the reasons I exposed a couple of days ago.
English is not my mother tongue, I type with only two fingers, and now I don´t have time enough to elaborate all relative to intervening forces properly, But just a short analogy, the one gave me (arguably?) the solution.
What actually happens is similar to a tsunami, but with a relatively very short wave, not rectilinear but circular, that moves towards sublunar point. It´s what happens if we drop a stone vertically onto a flat surface of water. Initially the stone makes a "hole", but afterwards a wave of water moves in radially, and water forms a "peak" on the center, against water own weight.
Displacements need not to be big, but the wave, not having space to form smaller and smaller circles, necessarily has to increase in hight, the closer to sublunar point the more.
Kind of radial inward resonance ..."
Thanks for making such a wonderful video. Your efforts are appreciated. A very good way of explaining the concept.
My last line can be confusing, because it's listing three different things. "And if you’re looking for the most intense sun on the planet, be sure to travel to the equator at the equinoxes, 23.5 degrees North in June, and 23.5 degrees South in December." Equinoxes AND solstices. Sorry for the confusion.
I´ve frequently seen that most people don´t fully understand the "peculiar" type of movement of the earth, in its dance with the moon: its revolving around earth-moon common center of mass (barycenter). Its location is INSIDE our planet, app. 1/3 of its radius from the surface (but please note it is not a material point).
The barycenter is the one which actually follows the elliptical orbit around the sun.
But with that movement earth doesn´t properly orbits around the moon, as many people say: earth CM follows a circular path around the barycenter, and the rest of earth material points follow identical circles, but around different geometric points, and maintaining its orientation relative to distant stars ...
LOGICALLY I´m considering ONLY earth-moon dance, disregarding earth daily spinning and the orbiting of the couple around the sun, because they have nothing to do with moon-earth dynamics.
As it is certainly tough to grasp, a few days ago I managed to imagine an analogy, and sent it to another site. IT CAN BE VERY USEFUL TO MOST OF YOU:
"I suggest you put your open hand horizontally on a table. Spot the table point app. under the middle of the first bone of your middle finger (B, from barycenter). And try slowly to make your hand “revolve”, with the first knuckle of same finger (C, app. at hand center) following a circle around B .
If initially you had imagined a “moon” a couple of feet away in line with your middle finger, and that “moon” rotated around B simultaneously to hand revolving, ALWAYS in line with B and C and at side opposite to C, you would have a scenario quite similar to actual earth-moon dance.
If you carefully observe the end of any of your fingers (or any other point of your hand), you will find that THEY ALL follow equal circumferences, ALWAYS keeping their location farthest from the moon (logically, within each own path).
THAT´S, I think, A GOOD WAY to see what quoted from NOAA´s scientits work IS QUITE CORRECT:
" ... since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”.
The NOAA work is the one I referred to here a couple of weeks ago.
I have tried for my entire career as an Earth Science educator to get tides right. I have always taught the bulges as due to the differential forces of gravity (and hence acceleration) experienced on the near side versus the far side. Essentially the far side lags behind. I have seen so many contradictory and error-prone descriptions of tides in many different books (essentially problems with the vector diagrams). Then I saw this video: th-cam.com/video/pwChk4S99i4/w-d-xo.html, and I threw my hands up in the air. I'm not a physicist and can't decide which vector/force explanations are the best, so I decided to keep it very simple in the video and focus on the things we can understand easily and get right. Clearly I need to keep working on it. Thanks everyone for all the comments.
Thank You!!
To clarify, you threw your hands up because you finally got it? Or because you still didn't understand it, or how to explain it?
I guess what I'm asking is, is this video of yours providing a different explanation than that PBS Space Time one?
Wrong. Gravity between moon and earth cannot be felt, because centrifugal forces from earth and moon, both in orbit around their common baricenter, nullifies gravity same as the International Space Station, ISS.
On my post of yesterday only forces on sublunar and antipodal earth areas are referred to.
There is where tide-generating dynamic imbalance is the greatest.
On transverse section through earth CM, that imbalance doesn´t occur.
On the rest of closer hemisphere moon´s gravity prevails, being net tide-generating force the closer to moon, the bigger.
On the other hand, on farther hemisphere, where moon´s pull is smaller, centrifugal force prevails, being the imbalance the farther from moon the greater.
Water from all parts of the oceans tend to move and pile up on both opposite areas (50/50) ...
But VERY, VERY LITTLE through squeezing (as proposed by some people) if any !! The unique area where there are some inward vertical components of net tide-generating force imbalance is where this is null or close to null, and there almost the whole of the pull vector is tangential (horizontal) ...
A book on tides more than 300 page long, written by Dr. Bruce Parker, is linked below.
He spent most of his career in NOAA and much of his time working on tide related problems as a specialty even while tackling jobs with a much broader scope. Positions he held at NOAA included: Chief Scientist of the National Ocean Service; Director of the Coast Survey Development Laboratory; Director of the World Data Center for Oceanography; Principal Investigator for the NOAA Global Sea Level Program; and head of the U.S. national tides and currents program (in a earlier organizational form before it became CO-OPS).
Among his awards are the NOAA Bronze Medal, the Department of Commerce Silver and Gold Medals, and the Commodore Cooper Medal from the International Hydrographic Organization.
Dr. Parker is presently a Visiting Professor at the Center for Maritime Systems at the Stevens Institute of Technology.
Dr. Parker has written many papers on tidal subjects, some of which are included in the References section of this book, as well as many tidal analysis programs, some still being used in some form in CO-OPS.
He also had the privilege of organizing the program for the International Conference On Tidal Hydrodynamics in 1988 and editing the book that resulted.
Dr. Parker received his Ph.D. in physical oceanography from The Johns Hopkins University, and prior to that an M.S. in physical oceanography from the Massachusetts Institute of Technology, and a B.S./B.A.in biology and physics from Brown University ...
AND HE SAYS:
"At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point).
At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system).
On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction. In a hypothetical ocean covering the whole Earth with no continents (see Figure 2.8) THERE WILL BE TWO TIDAL BULGES RESULTING FROM THESE IMBALANCES OF GRAVITATIONAL AND CENTRIFUGAL FORCES, ONE FACING THE MOON (WHERE THE GRAVITATIONAL FORCE IS GREATER THAN THE CENTRIFUGAL FORCE), AND ONE FACING AWWAY FROM THE MOON (WHERE THE CENTRIFUGAL FORCE IS GREATER THAN THE GRAVITATIONAL FORCE).
Logically, he also analyses thoroughly tidal local effects all over the world ...
tidesandcurrents.noaa.gov/publications/Tidal_Analysis_and_Predictions.pdf
My "discussion fellow" tried to improve his model as follows:
"Let's have the spaceship apply an even force to the whole of the Earth and all the water on it with a tractor beam so that the Earth simply accelerates and the moon accelerates after it. In this case, there are two bulges in the water with one ahead and the other behind. The bulges in this case are not caused by any kind of rotation, but the mechanism for forming them is the same as it is in the real case of the Earth with the moon orbiting it - it's the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance" ...
I answered:
"Your model has really improved … Now both bulges would be identical to real, natural ones (disregarding effects due to sun and other celestial objects) …
But in that scenario bulges wouldn´t be due to "the difference in the moon's gravitational attraction acting on the Earth as it reduces with distance” as you say, because MASSIVE STUFF CAN REACT ONLY TO FORCES EXERTED ON IT, either gravity from more or less distant matter, or direct pulls or pushes (even share stresses) exerted by contiguous material. IT CANNOT DIRECTLY REACT TO GRAVITATIONAL ATTRACTION DIFFERENCES BETWEEN POSITIONS THOUSANDS OF KM APART !! How “on earth” could it ??
The behind bulge would actually be due to the fact that there backward pull from moon, at any given position, is bigger than forward pull from the spacecraft. And ahead bulge would be due to the fact that there moon backward pull is smaller than forward pull from the spaceship !!
Your new model makes possible to also improve the parallel I drew: now we would be exerting on our planet an artificial, uniform and outward (relative to moon-earth couple) force with the spaceship, which is actually replacing the also uniform and outward force that, in the real case with earth revolving around the barycenter, is exerted by nature. How? … As I previously said, moon-earth dance is continuously bending the trajectory of each earth molecule (forcing it to be circular). ALL earth material points are continuously at farthest distance from moon (within its respective circular trajectory), what implies centripetal and centrifugal forces inherent to that revolving are all parallel to the straight line between earth and moon centers of mass"
He also had said:
"... Take away the spaceship and the moon will fall into the Earth, and as it does so there will be an amplification of the two bulges, again not caused by centripetal force".
And I replied:
"That is another prove of the impossibility of having the tides we have had for billions of years without mentioned inertial effects due to the revolving of earth and the rotation of moon, both logically around their common center of mass.
What quoted would not happen if, exactly at the moment of disconnection of the spaceship, somehow we could give earth and moon the “tangential” real velocity vectors they got in nature. The barycenter would follow its previous straight trajectory, at the speed acquired until that moment, and earth-moon couple would dance again, the earth with the two bulges that in your model were artificially originated ..."
On the work:
www.newenglandphysics.org/other/French_Tides.pdf
one can read:
"… the production of ocean tides is basically the CONSEQUENCE of the gravitational action of the moon- and, to a lesser extent, the sun …The analysis of the phenomenon is, however, considerably helped by introducing the concept of INERTIAL FORCES as developed in the present chapter.
… With respect to the CM of the earth-moon system … the earth´s center of mass has an acceleration of magnitude Ac … every point in the earth receives this same acceleration from the moon´s attraction.
… for a particle at the earth´s center, the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite. If, however, we consider a particle on the earth´s surface at the nearest point to the moon …, the gravitational force on it is greater than the CENTRIFUGAL FORCE by an amount that we shall call Fo …
By an exactly similar calculation, we find that the tide-producing force on a particle of mass m at the farthest point from the moon … is equal to -Fo …”
What I´ve been saying on many sites time and again !!
The authors, as the NOAA scientists D.C. I referred to a week ago, neither say “the production of ocean tides is ONLY due to differential gravitational action" (or something with that meaning), nor “forget” an acting inertial force such as centrifugal force, as many people, physicists included, keeps doing !!
thank you
I´ll try and elaborate what I said yesterday.
Logically there are many differences between actual formation of the sublunar bulge, and the analogy I exposed.
On the one hand, bulge formation is the result of not just a single circular wave inching inwards. An "infinite" number of short circular "waves" form, all across the ocean hemisphere.
On the other hand, those waves don´t work as normal waves, which consist only in the propagation of an "irregularity" somehow previously caused. In reality those irregularities are continuously being formed, thanks to the "joint venture" of own Earth´s pull on the water (weight), and Moon´s pull.
I´ll try and expose some more ideas, for the sake of an easier understanding (hopefully).
Let us imagine a solid sphere, and that we put on its surface a series of thin rings of decreasing size, in such a way that every ring is in horizontal position and touches contiguous rings, and also the sphere.
If we tried to lift a ring with many imagined vertical strings, maintaining it horizontally, if in upper part it could be easily done, but if at lower part it would have to lift contiguous smaller ring, "pushing" it.
But if also smaller ring is lifted with a little stronger pull, there would be no push at all.
If now we imagine the same but with our planet as sphere, tring to lift a ring (as Moon would try to do, if far over), being the oppose force (own weight) so hugely bigger, the lifting would be impossible.
But now this own weight is not aligned with lifting strings as in the first case: Earth´s vector pull on each ring form a cone towards our planet center of mass.
The result of all those cone shaping vectors, and the "infinite" number of pulls from the Moon (practically with a cylindrical shape) gives an "infinite" number of inwards vectors, causing a tendency to decrease the size of the ring.
If rings are solid as initially imagined, all that only can produce a compression in the circumferential sense, not producing any "push" on inner contiguous ring.
If now we change to the real case, imagining a kind of ring of water drops, things change because water is freer to move But those ring forming drops, even if inner ring previous space had become "void", can´t "shrink" in order to get into that smaller space … The UNIQUE possible "outlet" is upwards (towards the Moon), at least for a small fraction of the water. And then rest of the ring water could close in, forming a smaller but kind of thicker ring (lifted drops are also pulled inwards).
Apart from the opposite sense gradient of relative to Moon pull vector and angular factors (when going from 90º away from sublunar point to 0º) , this idea of circular waves shows that the pace of their size decreasing changes too, from close to null 90º away area, to a maximum at sublunar area …
So many factors would require a scientifically accurate study, to see real distribution of forces over the hemisphere. But I guess that at most places only pulls happen, "suction" rather than squeezing. Though as previously said, a tiny "circumferential" compression happens, that doesn´t add up inwards and towards sublunar point, because all over the hemisphere "excess" of water rises a little bit, even against its own weight. No pushes accumulate: what accumulates is water lifting (and partially water tangential displacement).
Somebody could wonder how that is possible, if water weight is so hugely bigger than Moon´s pull … But that is not difficult: even a child can move upwards a heavy ball, if on a sufficiently small slope. And, as previously said, the so called bulges have almost no slope at all: an average of not more than a few meters between places almost 10,000,000 meters apart (at the equator)
2.22 what are all the "other forces"??
See all the posts from Rafael in the comments, and you'll find the answers. It can get quite complicated, and this video was intended to keep things simple (Tides 101).
There is a very, very interesting youtube video, "What if the moon did not existed - Neil F. Comins" . Perhaps too long, due to its interactivity (between NFC and the audience).
NFC is a really eminent astronomer ... Just google his name and you´ll see.
He talks about causes of tides during some 10 or 15 minutes, starting app. 06:00.
I first watched it some months ago. I had seen he clearly considers both gravity and inertial forces (due to earth-moon "dance") is what actually causes tides (not just the so called "differential gravity"). But only recently I realized he clearly also says that to consider moon orbits the earth is WRONG.
I´ve been long discussing tides on other sites, here among many others. As NFC says, there is a lot of confusion out there. One of the main roots of that confusion is precisely what mentioned. People consider centrifugal force, at least in the case of earth movement and tides (NFC says just "outward" force) is fictitious, because they say earth is in free fall in its orbit ...
What I posted here three weeks ago, especially 1st and 2nd parts of my series "MY ULTIMATE GO?" is quite in line with what NFC says, though he drastically just says what mentioned, and I elaborated a three part analogy trying to explain it to anybody interested.
Very nice! --- Just one objection. The diagram at 2:33 creates the impression that there are gravitational forces creating a bulge on the Earth's side that is averted from the Moon, i.e., the arrows pointing to the left. Actually, there are forces that do cause this bulge, counteracting the Earth's gravitation: the centrifugal forces resulting from the Earth's eccentric rotation around the Earth-Moon barycentre, located two-thirds from the centre to the surface towards the Moon. I've found this diagram on de.wikipedia, wikimedia.org/wikipedia/commons/4/4e/GezeitenkräfteErdoberfläche.png I'm sure that you don't need a translation.
Quite right! But, you know, many scientists (perhaps most of them), consider centrifugal forces are not real (!!), only apparent or fictitious ... And they don´t even mention those words.
The video is really nice, as you say. But with that objection, and a similar one when they refer to Sun related bulges. In this case, with the barycenter at the Sun (I presume), centrifugal forces are even more evident. But they only consider gravitational pull differences !!
Thank you for your comments. Your link doesn't work, but I'm assuming these images would be correct: w3.salemstate.edu/~lhanson/gls214/images2/gls214_tides3.gif, www.mt-oceanography.info/IntroOc/notes/figures/fig11a1.html. I chose in this video to simplify the forces and focus on their net result. I will reconsider this for the next version. Thanks again. :)
I suggest you to have a look at the comment I´ved finished posting just now.
wow! Just wow!
MY ULTIMATE GO ? (4th part)
As said on 2nd part, I consider that rather than talking about “free fall” within a given gravitational field, it is more realistic to consider earth-moon interactions as if they were parts of an unique extended object, with no massive parts apart from where both celestial objects actually are, and rotating/revolving about their common center of mass, maintaining the distance between them due to their dynamic equilibrium.
An ice skating couple can spin like a single dancer, because they are linked usually through their hands … If one of them had a pony-tail, when spinning, instead of hanging vertically, the pony-tail would move upwards, against its own weight … Clearly due to centrifugal force.
The same would happen whatever the way they were linked to each other. If it were a kind of mutual “gravitational” pull sufficient not to need their hands (as earth-moon case), and they had suitable initial "tangential" speeds, the pony-tail would also “feel” that inertial force (why wouldn´t it?) … It would react to the TOTAL force acting DIRECTLY ON IT (WHATEVER the pull on other parts of the dancer, what the pony-tail “ignores" …), and raise where centrifugal force prevailed.
Earth particles react to those inertial forces in a similar way to what quite clearly explained for the diminishing of our weight on the equator, and causes the equatorial bulge (though the author prefers not to mention the adjective “centrifugal” … ):
“The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force (man´s weight) is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. IN FACT, THE ENTIRE BODY OF THE EARTH EXPANDS SLIGHTLY AND THE MAN AND SCALE MOVE OUTWARD FROM THE AXIS OF ROTATION SLIGHTLY, UNTIL FORCES COME INTO BALANCE WITH THE REQUIREMENTS OF ROTATIONAL STABILITY AT THE NEW RADIUS. THIS IS THE REASON FOR THE EQUATORIAL BULGE OF THE EARTH DUE TO ITS OWN AXIAL ROTATION"
www.lockhaven.edu/~dsimanek/scenario/centrip.htm
That´s why the NASA scientists I´ve referred to many times say:
"At the center of the Earth there is a balance between gravitational attraction (trying to pull the Earth and moon together) and centrifugal force (trying to push the Earth and moon apart as they revolve around that common point).
At a location on the Earth’s surface closest to the moon, the gravitational attraction of the moon is greater than the centrifugal force of the Earth (moving around the center of the revolving Earth-moon system).
On the opposite side of the Earth, facing away from the moon, the centrifugal force is greater than the moon’s gravitational attraction.
In a hypothetical ocean covering the whole Earth with no continents there will be TWO TIDAL BULGES RESULTING FROM THESE IMBALANCES OF GRAVITATIONAL AND CENTRIFUGAL FORCES, ONE FACING THE MOON (WHERE THE GRAVITATIONAL FORCE IS GREATER THAN THE CENTRIFUGAL FORCE) AND ONE FACING AWAY FROM THE MOON (WHERE THE CENTRIFUGAL FORCE IS GREATER THAN THE GRAVITATIONAL FORCE"
I DO KNOW many people (physicists included) will say: "But you are using a non-inertial frame of reference, where centrifugal forces are “added” by us, in order to keep Newton´s Motion Laws valid. But those forces are “fictitious”, they don´t actually exist" … (or something similar).
This post is already rather long. I´ll leave my refuting of what usually is deduced from that question of frames of reference for another post.
MY ULTIMATE GO ? (3rd part)
To diminish the risk of misinterpretations, before delivering the conclusions I referred to on 2nd part, I´ll repeat here the "breakdown" of earth complex movement into its simpler (and more easily graspable) components ...
Earth´s movement (as a whole) is basically the addition of three simple, but quite different, movements:
1) Earth-moon common center of mass orbits the sun (once a year). Both celestial objects, as a "couple", are kind of linked to that CM (called barycenter), and move with it.
That movement, and sun´s pull, originates what I call sun-related tidal effects (sea tides included).
For the sake of simplicity, I usually disregard that movement, when analyzing main component of tides: moon-related tides.
2) The "couple", as I said yesterday on 2nd part, revolves/rotates around the barycenter (once every some 28 days).
THAT IS THE MOVEMENT I ALWAYS REFER TO (unless clearly said the contrary), discussing and analyzing it AS IF ONLY THAT MOVEMENT WERE HAPPENING ... Moon´s pull, and dynamical effects originated by that movement´s features, causes two opposite bulges that continuously change position, logically with same periodicity of some 28 days (relative to the rest of the universe).
Details of the singularity that earth revolves (instead of rotating like the moon) are explained on the post I sent here a week ago, with a "handy" analogy ...
3) Earth also has its daily spinning, which, apart from causing the permanent equatorial bulge (by the way, thanks to huge centrifugal forces inherent in such fast circular movement), makes us to perceive the period of the movement of the pair of bulges as if it were once a day, instead of once every some 28 days.
Some time ago I posted the link to a video relative to that saying:
"I suggest anybody interested to have a look at :
th-cam.com/video/G-MOZId0FNw/w-d-xo.html
where it´s clearly seen that daily movement of the bulges is only apparent, that they are almost still and it is the solid part of our planet (though also the bulk of ocean waters due to friction) what is actually spinning …
The formation of the bulges is a rather slow process (some 28 days the complete cycle) … Nothing to do with all those daily local whirlpools, due to the much faster Earth spin, and with any other local singularity".
Another comment I´ve sent to the site I mentioned two days ago:
"The idea is well orientated, but the explanation is FLAWED since its very first minute.
At 0:38 it is said:
"The moon's gravity is stronger at Point A and weaker at Point B than it is at Earth's center. The net effect of this differential of the Moon's gravity across the Earth is to stretch the oceans out like taffy, ergo why the oceans bulge out at opposite points along the Earth/Moon line. Now, that explanation sounds plausible and a lot of well known scientists give it, but as we'll see, it's not correct.
… They aren't actually being lifted or stretched …"
What do you mean with "lifted"? If "levitated", what nobody has said, logically that is quite impossible. But water in sublunar bulge (let us say less than 45º away from sublunar point) kind of weights a little bit less due to Moon´s pull, and due to that its level rises a little bit, though only a small portion of actual full high tide.
And our planet DOES stretch due to opposite forces. Its solid parts experience tensile internal stresses to counter those forces, but as water is freer to move, its answer ALL ACROSS oceans is massive displacement of water towards sublunar and antipodal area ...
And, by the way, that displacement happens, rather than through any "squeezing", in the way I explained a couple of days ago, as far as I can understand !!"
As a continuation to my post of yesterday, just to say I´ve already sent mentioned message, adding what follows:
"By the way, apart from nature itself (and its own motion laws), mentioned
tangential speeds could be considered the actual cause of tides,
whatever the energy source that made those speeds possible.
Without those initial speeds, and Moon-Earth "dance" originated by
them, tides wouldn´t have been happening for billions of years as they
have !"
Rafael Molina Navas this is enough information for a short book. You should publish it.
Thank you. I suppose you also refer to my more recent posts, but you have put your comment on a post of mine of 1 month ago ... Perhaps because, I don´t know why, something is wrong with the order of the posts !
The same fellow of my discussion of last days, also told me that "first thing in the morning" ... And I replied:
"I´m glad you say so, hopping it implies you´ve found things I´ve said are worth a book (though you exaggerate ...) !
But, you know, I type with only two fingers, and not being English my mother tong it would take me “eons” …
By the way, he´s sent another long "refuting" post, but I need some time to read it more slowly, ruminate and refute it ...
My discussion fellow replied:
"The removal of rotation does not remove the tidal bulges - they remain, but clearly the Earth has to be free to accelerate towards the moon for this to be the case. If you hold the Earth in position, then you have a different result where a single bulge will appear on the side nearest the moon"
I answered:
"There we are!
It´s unbelievable that you consider absolutely necessary to let the earth free to get the rectilinear accelerated movement towards the moon (to have the second bulge), but time and again you say that (as it is in the real case) the curved accelerated movement of the earth (its revolving around the moon-earth barycenter), has nothing to do with the formation of the outer bulge, and that it is not necessary at all !!
They are two different ways inertia manifests itself, but due to same basic Physics laws ...
In the first case, outer hemisphere massive stuff kind of "falls behind", because it is not sufficiently pulled by the moon to get the rectilinear acceleration of the whole earth. The whole earth stretches, and behind bulge appears.
In the real case, outer bulge equally requires the revolving movement, because outer hemisphere massive stuff is not sufficiently pulled by the moon to get the required centripetal acceleration for the revolving movement. Internal stresses (in both senses, inwards and outwards) appear, earth also stretches, and outer bulge builds too ..."
Needless to say those outwards stresses could be classified as "centrifugal", at least in the broad sense of the term ...
If it affects gravity, why doesn't our weight change?
MY ULTIMATE GO ? (2nd part)
C) Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right … And not just as big as E´s (as in case B) but much, much bigger (see below how much).
If with the correct values for masses, M-E distance, and initial speeds, we would have our real case, E being the Earth and M the Moon …
We know the result is that, instead of E being orbiting around M as in case A, and somehow opposite to what in case B (when E got “fully free” to fall directly onto M), now E is “forced” to bend its trajectory much more, because M´s location is changing in a sense opposite to E´s speed … E and M revolves/rotates around their common center of mass. E´s trajectory curvature is much, much bigger than when orbiting around M (case A).
For similar (but kind of opposite) reasons to case B, IT IS ERRONEOUS TO CONSIDER E “ORBITS” AROUND M, AND TO SAY THAT E IS IN A “ FREE FALL" … In cased A we had a “partially free" fall, and in B a “fully free” fall … But NOW I CONSIDER WE HAVE NO FREE FALL AT ALL, BECAUSE INITIAL SPEEDS AND MUTUAL PULLS ARE CAUSING A KIND OF DANCE OF THE COUPLE, KEEPING THEIR SEPARATION ACTUALLY CONSTANT, AS IF IT WERE A SINGLE TWO-PARTS OBJECT SPINNING AROUND ITS CENTER OF MASS …
With this and last post I´ve already delivered what “promised”: a fresh explanation of one of the main roots of my stand …
But I´ll leave for another post the elaboration of main consequences of that, as far as I can understand ...
Recent days I´ve been really busy with previously commented discussion. The other party, defending the idea that ONLY differential gravity from moon intervenes on tides, said many absurd things about curved movement dynamics ... Eventually I had to tell him he was utterly wrong, living aside the bottom of the question, clearly in how he supports his stand.
We had to discuss, not about the term "centrifugal" force: about "centripetal force", which he had said it is a "grey area", that people say about it things not fully rational (??).
The day before yesterday he said:
"A force attributed (?) to rotation which (the force) persists when the rotation is removed should not be given a name that implies that it is caused (!!) by rotation".
In other paragraphs he also referred to "forces caused (or not) by rotation", or to "forces caused (or not) by certain movement", as if we could say some forces are caused by some movements, and others are not ...
And I had to tell him:
BUT THINGS ARE THE OTHER WAY AROUND !! MOVEMENTS CANNOT CAUSE ANY FORCE ... FORCES DO CAUSE MOVEMENTS (OR CHANGE THEM) !!
His ideas are flawed even at the very "basics of basics" level ...
Tens of not short posts ... to no avail. I´ve wasted a lot of time !!
Thank you
The explanation I made yesterday (my first post) in relation with the imbalance between required centripetal forces and actual gravitational pull at each area (what -> internal stresses -> Earth stretch -> two opposite bulges) would match perfectly (if saying Earth where I had erroneously said Moon, as corrected yesterday in my second post) with the case of Sun related bulges: the farther (from Sun) the area, the smaller the gravitational pull, but the bigger the required centripetal force …
As I also said, Moon related bulges are a little bit trickier to explain.
Coming back to the drawing board, I can elaborate as follows.
As Earth/Moon barycenter is situated app. 2/3 Earth´s radius from its center (logically, at Moon side), Earth movement in this "couple dance" is a kind of wobbling, similar to when a child plays with a hulla-hop (probably not spelled so).
Though Earth daily spinning generates much, much stronger internal stresses and centrifugal forces (some 28/29 times higher angular speed …), these act always same way: trying and increasing Earth´s diameter at the equator and nearby.
Earth wobbling due to its "dance" with the Moon generates internal stresses and deformations which we can perceive added to those mentioned stronger ones.
Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation.
Earth parts to the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration there. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: closer to the axis parts have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL …
At Earth parts at right side of that axis, required centripetal forces are smaller, also the farther the bigger. BUT now gravitational pulls from the Moon are opposite to required centripetal forces, the closer to the Moon, the bigger. Only own gravitational Earth´s pull can supply required centripetal forces … As their distribution don´t match with Moon´s pull distribution, also internal forces occur to compensate the imbalance. And that spherical segment is similarly stretched too, and a bulge also occurs at that side (that bulge is usually explained, but partially, in a much simpler way: just because there Moon´s pull is higher ...)
And due to Earth´s daily spinnig, those two opposite bulges seem to rotate around Earth, trying to keep in line with Moon (as actual causing forces). Result: our by all known, but not so well known, tides (high tides where the bulges, and the rest ...) Globally considered, as we also have, as a consequence of the same, quite an array of different local tides, VERY WELL explained in the video.
I´ll continue to refute what said on “Misconceptions about tides” (“lockhaven.edu-dsimanek” previously linked sites).
They author of that article also says:
“ … Once they launch into the rotating coordinate mode and start talking about centrifugal forces … They also forget that the non-uniformity of moon's gravitational field over the volume of the earth is alone sufficient to account for both tidal bulges, bulges that would be essentially the same if ... the earth and moon were not moving relative to each other”.
HE is the one who apparently forgets that hypothetical case is NOT possible … unless some other external force were countering the gravitational attraction between both celestial bodies !!
Let us “play” with our imagination. Some “possible” (?) cases could be:
1) A kind of “cosmic" superman were able to keep them separated their current distance, pushing them outwards on their inner sides. Earth would not get tidal bulges but quite the opposite: it would become slightly flatter (internal compression stresses increase)
2) Similar situation, but pulling outwards from their outer sides (where strong solid parts in the Earth side ...). Kind of bulges would build, but outer one would be quite different, depending on the way our superman could "grasp" the Earth … (internal tensile stresses increase)
3) Some “ miraculous” outward forces were applied directly on Moon and Earth respective centers of mass, countering mutual gravity attraction … Earth inner hemisphere would get an ovoid shape, kind of bulge (internal tensile stresses increase), but outer hemisphere would experience the opposite type of deformation: it would get slightly flater (internal compression stresses increase).
THEREFORE to get our real bulges it is paramount the way Moon-Earth mutual attraction is countered to keep actual distance … And the real way is their rotation/revolving around their barycenter !! THAT EARTH REVOLVING implies inertia related outward forces, quite an intervening fact in bulges formation, as explained on my last post.
So, to say “… bulges that would be essentially the same if ... the earth and moon were not moving relative to each other” is erroneous, or at least pointless, because it compares something real with something impossible ...
thank you very much for the lesson
Excellent
Tides are not binary, some places have one a day. Locations at same latitudes on same ocean don't experience tides when others do. Tidal bulges are out of phases with moon's location. Low tide 9/20/17 during new moon happens 1 hr 45 min after moon meridian (Sebastian, FL) Tides near equator are smaller than tides nearer poles.
Yes, you are correct! The last portion of this video explains that -- see the Amphidromic systems section at the end for more information. The majority of this video discusses the theory behind the causes of tides (tidal bulges). The last portion of the video explains the impact of enclosed tidal basins on the bulges (reality).
This very morning I sent what follows to another site, replying to a post trying to refute that centrifugal forces inherent to earth revolving count for tides ...
I´m sure it could be interesting for many of you.
Mentioned post said:
"Imagine a case where there's a planet and a "moon", but there is no orbiting - the planet is being pulled along on a string in a straight line by a powerful spaceship and is accelerating, while the moon is being pulled along behind it by gravity such that it keeps up and maintains separation. The moon's gravity is causing two bulges in the planet's ocean, but there is no rotation involved in the system at all, so are the bulges driven by centripetal force in a case such as this one? No"
I replied:
"You yourself answered your question: No.
But, I´d rather say "no" to the very question: if the powerful spaceship would be making the earth accelerate, and with the moon back at current distance, earth deformation would depend on the way the spaceship force were transmitted to earth:
1) If somehow it were transmitted pulling on earth c.g., we wouln´t have two bulges: forward hemisphere would get flatter.
2) If the spaceship pushed on earth back side, the whole earth would get flatter (if pushing through a sufficiently big flat surface): no bulges at all.
3) Only if pulling from forward earth parts, we would get two bulges. But forward bulge would not be due only to less gravitational pull from more distant moon, but also to the very spaceship direct pull ...
We could even draw a parallel with our real case. In your imaginary case, the agent which causes moon-earth distance doesn´t diminish, is the mega-spaceship ... In the real case, similarly the agent is a kind of "hidden" force ("centrifugal" f., or at least "outward" force), due to inertia and the fact that all earth parts try to follow the tangent, and moon-earth dance, through invisible "gravity strings", is continuously "bending" the trajectories of all those earth parts ...
Without mentioned dynamic effects of that earth-moon dance, we could not even be here discussing the issue, because they were paramount in the very history of our planet evolution and the beginning of life !"
The discussion I referred to a week ago has continued. Yesterday I sent him what follows, I hope self-explanatory. It is directly connected to what on this youtube video.
"You´ve said a lot of times something like: if we “remove" the revolving movement of earth, the bulges remain … So, that movement cannot be causing the tides.
I already did refute a similar idea of yours, in relation with your idea of a mega-spaceship causing bulges. But you didn´t get it, and said I had taken your case in the wrong way !!
Let us forget the spaceship, and have just moon and earth.
If, somehow, we "remove" the revolving, both earth and moon will accelerate towards each other … Though only a transient movement, bulges would remain (and increase) until the collision. OK. Therefore, revolving was not causing them (YOUR deduction).
IN A QUITE SIMILAR WAY, we could reason as follows:
If, somehow (a cosmic “hiperman” grasping the earth at its c.g…?), we “removed” that rectilinear, accelerated movement of earth towards the moon, outer bulge would not only disappear, but even change its sign, because outer earth hemisphere would get additional internal compressive stresses (before they were tensile s.), due to moon´s pull (though smaller than tensile stresses within inner hemisphere, closer to the moon …)
Therefore, if with the differential moon´s gravity maintained, outer positive bulge doesn´t exist, the moon differential gravity cannot be the cause of outer bulge !!
Same way you´ve “reasoned” lot of times … Either both are right, or both are wrong …
Any guess?"
(Continuation of my post of 2 weeks ago)
After other posts interchanged with my companion fellow, I had to tell him:
"You stubbornly keep ignoring that, IN ALMOST A MONTH, apart from those 28/29 cycles of earth daily spinning, and apart from a little less than a 29/365 of the complete orbit around the sun, our planet does complete a cycle of its revolving around the barycenter ...
Disregarding the daily spinning, which has nothing to do with moon-earth dynamics, we can look at the moon e.g. midnight (when visible), and write down its angular position in its apparent trajectory.
And do the same next day ... We´ll see there has been a change towards the east: THAT is the real movement of the moon. And earth, in its dance with the moon, ALWAYS at the other side of the barycenter, will have followed the same angle in its revolving around the barycenter.
By the way, I posted a link to a google image which shows that "dance" in three different positions, very interesting, but I didn´t see any further comment here ... Did you see it? I said:
(#152 and #153)
"There is an image:
main-qimg-851d9284749b378191c8ae87e4e2e4c2
(copy and paste on search engine window)
that can help guys understand better moon-earth dance, especially earth revolving around the barycenter.
By the way: the image has a detail a little bit erroneous ... Any guess?
A Google page appears. Please click on images …"
MY ULTIMATE GO ?
Once again I´m going to try and convey a core idea of my stand, now in a “fresh” way …
It will have three parts. I´ll post them neither on same post, nor on three different ones, but on two posts … (that way readers will have more time to “ruminate” what proposed).
A) Not to need to include a figure, please kindly IMAGINE one with two celestial objects on same “vertical” line (on your screen).
Let us imagine the lower one (M) on a fix location, and the upper one (E) moving “horizontally” from right to left, and let us consider M´s gravity starts to pull downwards E just when reaching M´s vertical (remember, M is considered somehow on a fix location).
If within certain game of relation between masses, distance and speed of E, E will start “orbiting” around M.
M´s gravity, acting as centripetal force, changes E´s speed vector direction, but it is not “able” to change its size. A “free fall” to many, but rather a “partially free fall” to me, because E´s inertia avoids a direct straight line fall.
B) Let us now imagine M, at the very moment E gets at M´s vertical, starts moving also from right to left, with same velocity as E.
Being now M´s pull on E always vertical, E will fall towards M, following a parabolic line.
Now M´S PULL IS ABLE not only to bend E´s speed vector, but also TO INCREASE ITS VERTICAL SPEED WITH AN ACCELERATION PROPORTIONAL TO M´S PULL … same way as if neither of them had an initial horizontal speed …
THAT´S WHAT I CONSIDER A "FULLY FREE” FALL ...
And WE SHOULD NOT SAY E IS ORBITING AROUND M, because it will fall onto it !!
thanks
I´ll continue to refute what dealt with on my post of four days ago.
One of first paragraphs of “Misconceptions about tides” (“lockhaven.edu-dsimanek” previously linked sites), says:
"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 of the earth”.
That last comment seems to show the author is biassed: he starts considering rotation (“revolving”, rather) has nothing to do with tides …
With that fixed idea, his reasoning may be kind of distorted, reaching erroneous conclusions !!
Further comments we can do about it:
1) If only "the gravitational pull of another nearby object” (the Moon in our case) counted, Moon and Earth would have fallen onto each other long, long ago …
2) I know standard Physics says: but they both rotate/revolve around the barycenter, and that prevent them to fall onto each other !!
3) Right. Therefore, without that rotation/revolving we wouldn´t have had the tides we´ve had for billions of years !!
4) And as Moon´s gravitational pulls on farthest hemisphere ocean water “units" are insufficient to cause required centripetal acceleration at their place, those water “units" tend to follow in the direction of their tangential speed (logically, due to the “inertia” phenomenon). Part of own Earth pull on that water has to be “used” to produce that centripetal acceleration deficit.
5) The author also himself says on www.lockhaven.edu/~dsimanek/scenario/centrip.htm
that (explaining why on the equator we weigh less than at higher latitudes, being our weigh maximum at both poles):
"The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation".
EXACTLY THE SAME happens (why it wouldn´t? …) to water further from the Moon due to Earth revolving around Moon-Earth barycenter, because its weight there has same direction and sense as required centripetal force.
And that water lightening there, together with the displacement of water from rest of further hemisphere (caused similarly, but with different angles between weight and required centripetal forces) is actually the reason of the antipodal bulge formation !!
The varying Moon´s pull also intervienes, but only causing previously mentioned centripetal force “deficit” !!
Another thing is that, as Earth revolves without rotation (in its “dance” with the Moon), required centripetal force is constant across it. And as at Earth c.g. it is in balance with Moon´s pull there, the “fields” of gravitational differences, and the one of differences at each place between required centripetal force and gravity, are equivalent !!
That fact keeps kind of hidden to many the error of thinking Earth revolving has nothing to do with tides formation … The so called “standard model”, considering only gravitational gradient, would give erroneous results when analyzing e.g. Earth tidal effects on the Moon, or on any of the many billions of celestial objects tidal locked to another, because in those cases centripetal acceleration is NOT constant: it has to increase proportionally to distances to respective barycenter !!
I have to say now, despite my negative comments of two weeks and one month ago, that what said in the video is essentially correct.
Why that change ...?
If interested, see my last comments in:
www.thenakedscientists.com/forum/index.php?,
especially last one, of this very morning ...
The best
I´ve just now sent what follows to the author of the included article link, after some pieces of email we have interchanged, especially in relation to his idea that ONLY differential gravity is the root cause of tides. I´M SURE it may interest many of you ...
"On the one hand, I´m glad you find "my take" interesting. But on the
other I´m kind of sad, because you found necessary to send me Newton´s
3rd Law "properly stated" ... Do you think I had used it wrongly ?.
I was very careful with its application ... But in "my case" it cannot
be applied exactly as originally stated, because three "bodies"
interviene, rather than just two: Moon (A), a certain part of Earth
(considered portion B), and the rest of our planet not included in B.
When a cart is pulled by two horses, if accelerating and supposing no
friction losses, 2nd and 3rd Newton´s Laws apply:
- Added pull of the horses on the cart is equal to the mass of the
cart multiplied by the acceleration it gets.
- The cart is pulling back EACH horse with same size pull each horse
is exerting on the cart.
By the way, I´ve seen you yourself explain, and very well, a really
similar case using EXACTLY same arguments as my take ones (your
article www.lockhaven.edu/~dsimanek/scenario/centrip.htm
Under the subtitle "Weight" you say:
"Weight may be defined as the force required to keep an object at rest
relative to its surroundings. This definition is consistent with most
colloqual interpretations of the word (surprise!).
The figure shows the force vectors W and mg, which are the only forces
acting on the man. The vector F is their sum. W is directed along the
radius of the Earth. Being the radial component of the net force (it
is the net force in this case), its size is a = v2/R (the centripetal
force). Now compare these two cases. On the non-rotating Earth the
man's weight was of size mg. Remember, the weight of an object is the
force required to support it, i.e., the force exerted upward by the
weighing scale. With the Earth rotating, that force is smaller than
before. The contact force between the man's feet and the scale is
reduced. But all other such stress forces are reduced as well, within
the man, within the scale's springs, within the body of the Earth
itself. This causes a slight decompression of these materials, a
relaxation of the spring in the scales. In fact, the entire body of
the earth expands slightly and the man and scale move outward from the
axis of rotation slightly, until forces come into balance with the
requirements of rotational stability at the new radius. This is the
reason for the equatorial bulge of the Earth due to its own axial
rotation".
TAHT CASE is quite similar to "mine", in wich Earth is revolving
around the barycenter. A man on Earth surface where antipodal bulge
builds would also experience ALL those changes, due to a lightening
equally caused by its circular movement (though much smaller).
Unique difference: now we have two "horses" pulling instead of only one ...
IT´S ALSO CLEAR that one of the "horses" (the Moon) pulls Earth
massive parts there (B) less than it does on closer parts, but what
that actually causes is that the other "horse" (massive parts of Earth
not being B) HAS TO pull more, and "supply" the difference. And
corresponding reaction according to Newton´s 3rd Law, equal but
opposite (therefore outward, "centrifugal" in its general sense) and
exerted by B ON REST OF EARTH, is also bigger than at closer parts
(every vector having opposite sense at closer hemisphere)
And, as you say for equatorial bulge, I can say:
"This is the reason for antipodal bulge of the Earth due to its
revolving around Moon-Earth center of mass".
You also told me:
"I don't think it is helpful to think of tidal bulges as being caused
by tensile forces in earth or water" ...
I think it´s not a question about being "helpful" or not ... As far as
I can understand, also in line with what you yourself say for the
equatorial bulge case, root causes of tidal bulges are several:
varying Moon´s pull, Earth own pull on each of its parts
(gravitational or due to Earth rigidity), initial conditions that gave
tangential speed to Earth (and Moon) ... and nature motion laws
(basically due to the so called "inertia"), as Newton gave us.
CHANGES in internal forces where solid Earth (together with
corresponding stretching), and in pressure distribution where water
(alongside its corresponding pilling up on opposite areas) are nature
answers to above mentioned causes. They are actually the tides
themselves, rather than their cause.
By the way, apart from nature itself (and its own laws), mentioned
tangential speeds could be considered the actual cause of tides,
whatever the energy source that made those speeds possible.
Without those initial speeds, and Moon-Earth "dance" originated by
them, tides wouldn´t have been happening for billions of years as they
have !
Any comments would be appreciated. But please kindly keep in mind
quick replies may be affected by our "deep-rooted" ideas ... I would
prefer a late but well "ruminated" reply.
MY ULTIMATE GO ? (5th part)
Before trying and convey my stand on non-inertial frames issue, let us go back to the drawing board, as if we didn´t have any education on dynamics …
Same image with Earth (E) on upper part, and vertically below the Moon (M).
Let us suppose we only know how they move, with kind of the mind of a “smart” child.
And then we learn that (as David Cooper said many times), if Earth and Moon tangential speeds somehow became null, they would begin to accelerate straightly towards each other, because of gravitational mutual attraction …
If just stopping their circular movements, they get “free” to accelerate towards each other … THOSE CIRCULAR MOVEMENTS ARE WHAT, WHEN EXISTING, WERE SOMEHOW COUNTERING THOSE STRAIGHT ACCELERATIONS NOW THEY HAVE GOT !! No other logical possibility for a “clean” child mind …
How come? Let us see …
Curiously, with a rotating (so called "non-inertial”) frame of reference, same thing happens, but let us say “virtually” …
As the frame of reference rotates with E and M, E and M don´t "rotate" anymore
relatively to that frame of reference … Centripetal forces are not required whatsoever, and gravitational pull would make them fall onto each other … But that´s far from reality …
Then physicists go on: "How can we do “real” maths with that “artificial” (the adjective is mine ...) reference system ? … We have to apply a “fictitious” force, which we´ll call centrifugal force". That way WE GET BACK TO THE REAL SCENARIO … as far as dynamics is concerned.
Well, that force is certainly “fictitious”, because it has been introduced by us to work with that non-inertial frame of reference … BUT THET MEANS THAT THE REAL CIRCULAR MOVEMENTS, NOT NEEDING THE ADDITION OF ANY “ARTIFICIAL” FORCE, SOMEHOW PRODUCE AN INERTIAL EFFECT THAT KEEPS E AND M WITHOUT FALLING STRAIGHTLY ONTO EACH OTHER !! TO ME (AND NOT ONLY ME), THAT IS A REAL CENTRIFUGAL FORCE, QUITE REAL !!
In case B (“MY ULTIMATE GO ?", posted 4 days ago), as E and M were moving horizontally at same speed, Moon´s pull was able not only to bend E´s trajectory, but also to accelerate E straightly towards M with its full strength, quite a “free” acceleration. That “freedom” was made possible by the movement of the moon towards the left, with same horizontal speed as E.
The real case is dynamically quite the opposite: the movement of M in opposition to E´s speed, and with speeds which make them to revolve/rotate around the barycenter, makes impossible the decrease of E-M distance, and even any proper orbiting of E around M. It gives us a “rotating” scenario, where all forces have basically same direction as the line between centers of mass, and where the angular position doesn´t matter much (as far as Earth-Moon dynamics is concerned), as long as we accept centrifugal forces keep the separation constant …
After all, if Earth and Moon were the only celestial objects in the universe, to talk about angular position of the system would have no sense at all …
BUT, STILL, THE WOULD HAVE TO KEEP ROTATING. THAT WAY INHERENT INERTIAL FORCE, THE CENTRIFUGAL FORCE, WOULD KEEP THEM WITHOUT FOLLOWING STRAIGHTLY ONTO EACH OTHER !!
Therefore, as I´ve said many times, I find quite correct what a NOAA scientist told me:
"... to provide a basic description of the forces which create the tides. It's intended audience were the grade school children and adults of that time. It used terminology of science and forces which were common in the 1950s. Such as centrifugal force. Centrifugal force was always an "imaginary force" (not a real / measurable force). But that type of description made the concepts easier to understand and explain. That description and use of centrifugal force continued to be common practice until the 1970-80's. At that point, the terminology shifted and the TEXTBOOKS USED IN GRADE SCHOOLS WERE CHANGED TO USE A MORE MODERN TERMINOLOGY AND DESCRIPTION OF THIS “EFFECT”BEING A RESULT OF INERTIA RATHER THAN AN “IMAGINARY FORCE”".
Initially I didn´t fully grasp his point ... But later I did.
The problem is that many books, dictionaries included, keep following models with ideas which are several decades out of date …
When discussing with somebody else things said on the very interesting articles I´m lately referring to:
www.lockhaven.edu/~dsimanek/scenario/tides.htm#centrip
www.lockhaven.edu/~dsimanek/scenario/centrip.htm
he asks me what below.
I´ve seen many people somehow have rather confused ideas about daily manifested tides (does Earth daily spinning cause tides ??), when Moon-Earth "dance" movements have almost a month period ...
That´s why I´ve decided to send this post.
I answered what follows.
"Your question:
" ...(I had previously said) DAILY Earth spinning causes the "permanent" equatorial bulge, but it has NOTHING to do with Moon-Earth dynamics ... Does this mean that you also agree that the earth's spinning has nothing to do with tides?"
My answer:
I chose carefully my words. There is a trick is in the way you put the question ... The earth's spinning has nothing to do WITH THE ROOT CAUSE OF TIDES, but it does with the way they manifest ...
When we say "tides", in its global sense (local changes of sea level can have other causes), we refer to the cyclical change of sea level related mainly to Moon position in its monthly rotation. To analyze their actual causes, we have to disregard the daily spinning of our planet (imaging we check the position of the bulges everyday at same time). That way we would have the bulges, always in line with the Moon, "moving" around our planet in an app. 28 days "two-parts" cycle ...
THAT IS the basic phenomenon of tides.
BUT, as Earth does spin daily, its meridian in line with the Moon changes continuously, and though bulges maintain its position relative to the Moon, they continuously change meridian too. And Moon related tides SEEM to have an app. 12+12 h. cycle ...
BUT THAT SPINNING doesn´t cause the tides, though it causes the so called equatorial bulge, easily explainable as done on “lockhaven.edu-dsimanek” previously linked sites, when analyzing how we weigh less on lower latitudes, especially on the equator:
"... Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation".
ANOTHER THING is that the monthly revolving of the Earth in its "dance" with the Moon, also causes similar inertia related effects (not to use tricky concept "centrifugal forces"), which actually are one of the causes of tides, especially on the hemisphere further from the Moon, where antipodal bulge appears ...
On closer hemisphere, though those inertial effects have also an outward direction, as Moon´s pull there is higher, the later prevails and the better understood bulge builds towards the Moon !!
As said in my answer of yesterday to W. L., I also find this video really good, but with same objection: centrifugal forces are "ignored" and they shouldn´t ! I´ll try and elaborate.
At diagram at 2:23 "medium" gravitational pull (average as if acting on Moon´s C.G.) is rested both at nearer and farther sides, trying to explain the bulges ...
The authors don´t consider that any centrifugal force, they´re just comparing to find kind of "relative bulges" ...
But they are wrong, as far as I can understand.
What´s causing "medium" black vector at C.G.? : the centripetal acceleration required for the rotation. But neither that REQUIRED (to orbit at each distance) acceleration is uniform across the Moon, from left to right. Farther parts of the Moon, to rotate at same angular speed, NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces.
ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in our diagram, to farther parts ...
But according to Newton´s 3rd Motion Law, if at any considered Moon´s section, right part is pulling (towards Earth) contiguous left part, this one is also pulling the former towards the left, same force but opposite direction: OUTWARDS.
All those internal forces are centrifugal REAL forces, which, together with their "mirror" centripetal ones, stretch the Earth, its solid parts included.
Logically, that deformation is much bigger where liquid parts, our oceans. And the TWO normal bulges ARE DUE TO THAT.
As a continuation to my comment of one month ago, this is just to say that afterwards I sent two other posts, #85 and #86, to:
www.thenakedscientists.com/forum/index.php?topic=49715.new#lastPost
They contain new ideas aiming to help understand the issue, and the reasons why I consider there is so much confusion out there ...
Moon phrase never look like that . It always the whole chunk of the moon missing . I’ve seen it everyday
Nic one 2022
Sorry: I had recently been discussing why Moon has got tide locked to Earth, and in:
"But neither that REQUIRED (to orbit at each distance) acceleration is uniform across the Moon, from left to right. Farther parts of the Moon, to rotate at same angular speed, NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces"
and instead of saying "neither that ... acceleration is uniforme across the EARTH" and farther parts of the EARTH" I had a lapsus and said "Moon" !!!
There´s another rather tricky detail: Earth´s rotation I´m referring to (some 28/29 day period, as Moon´s) is around Earth/Moon barycenter, app. 2/3 of Earth radius from its center, towards the Moon ... That would change the diagram: we have bigger centrifugal forces at Earth´s left far side, and smaller at right side, BUT this one in same sense as the pull from the Moon ...
What said about internal stresses, centrifugal and centripetal, and about subsequent Earth stretch as cause of the two opposite high tides, keeps being valid !
Thank you for your honest reply.
I´m not going to insist now on why tides actually happen, as far as I can understand. I suggest anybody interested to have a look at my post of two days ago, and at linked site.
Regarding the linked youtube video (...4S99i4), utterly erroneous to me, last week I posted a couple of comments there, and less than an hour ago one more, as a continuation of the former. If anybody interested, there you can see my main reasons to say so.
As a continuation to my comment of four days ago, just to say that yesterday I sent another post to:
www.thenakedscientists.com/forum/index.php?topic=49715.50
comment #84, with an analogy with a children game (the whip ?), which I´m sure could help understanding what I´ve been exposing about our topic ...