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least common denominator
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Poppycock! There is no gravity... THE EARTH SUCKS!
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Gary Fisher 29er 2019 Kia Stinger 2.0t gone ![]() 1995 Miata Sold 1984 944 Sold ![]() I am not lost for I know where I am, however where I am is lost. - Winnie the poo. |
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Correct, but not sure that the term much smaller is accurate. Due to the large mass of the sun, tidal forces with a sun-earth only system would be of the order 46% of the forces of the sun-moon-earth system.
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Alexander '75 911S Targa '86 951 SOLD |
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That's cute.
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I don't care what geology or oceanography told you, the fact is, what PHYSICS tells you is the truth here, not what geology or oceanography tells you.
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Mike Bradshaw 1980 911SC sunroof coupe, silver/black Putting the sick back into sycophant! |
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wow!
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How is what I said wrong? If the moon really did instantly disappear, the tides would not disappear instantly, because they are a wave. If you were to ignore the cause and just observe the tides, you would be able to imagine a wave with two crests and two troughs approximately every 25 hours. If you drop a pebble in a pond, the pebble instantly disappears, but the waves keep going for a little while longer. I don't know how long it would take for the moon tides to go away; maybe a few days, maybe longer. I don't quite know what you mean by the moon not orbiting. It sounds like you mean to say, "if the moon stayed at one point above the surface of the earth." In that case, the moon would be in geostationary orbit around the earth. Right now, it takes about 27.3 days for the moon to orbit earth. In your scenario, the moon would orbit the earth every 24 hours. Anyway, if the moon were in geostationary orbit above earth, then the high point of the ocean would be stuck in one place, like you say. And it would not be noticed until sophisticated surveying techniques discovered the phenomenon. But the sun's tides would be noticeable, superimposed on the stationary moon "tide".
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Charlie 1966 912 Polo Red 1950 VW Bug 1983 VW Westfalia; 1989 VW Syncro Tristar Doka |
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Are you a physicist or a troll? Please be more thoughtful in your childish responses. I am embarrassed by being associated with you.
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Alexander '75 911S Targa '86 951 SOLD |
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least common denominator
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Reading? That might lead to free thought... I'm against it!
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Gary Fisher 29er 2019 Kia Stinger 2.0t gone ![]() 1995 Miata Sold 1984 944 Sold ![]() I am not lost for I know where I am, however where I am is lost. - Winnie the poo. |
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Lots of truths and half truths and misunderstanding here
![]() Wikipedia, Tide has an excellent piece that explains all. For those that don't like to surf, here's the pith of it "The ocean's surface is closely approximated by an equipotential surface, (ignoring ocean currents) which is commonly referred to as the geoid. Since the gravitational force is equal to the gradient of the potential, there are no tangential forces on such a surface, and the ocean surface is thus in gravitational equilibrium. Now consider the effect of external, massive bodies such as the Moon and Sun. These bodies have strong gravitational fields that diminish with distance in space and which act to alter the shape of an equipotential surface on the Earth. Gravitational forces follow an inverse-square law (force is inversely proportional to the square of the distance), but tidal forces are inversely proportional to the cube of the distance. The ocean surface moves to adjust to changing tidal equipotential, tending to rise when the tidal potential is high, the part of the Earth nearest the Moon, and the farthest part. When the tidal equipotential changes, the ocean surface is no longer aligned with it, so that the apparent direction of the vertical shifts. The surface then experiences a down slope, in the direction that the equipotential has risen."
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Bill Verburg '76 Carrera 3.6RS(nee C3/hotrod), '95 993RS/CS(clone) | Pelican Home |Rennlist Wheels |Rennlist Brakes | |
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As for the size of tides if there were only a sun, it's sort of explained here on a NOAA website:
"First, the tidal force envelope produced by the moon's gravitational attraction is accompanied by a tidal force envelope of considerably smaller amplitude produced by the sun. The tidal force exerted by the sun is a composite of the sun's gravitational attraction and a centrifugal force component created by the revolution of the earth's center-of-mass around the center-of-mass of the earth-sun system, in an exactly analogous manner to the earth-moon relationship. The position of this force envelope shifts with the relative orbital position of the earth in respect to the sun. Because of the great differences between the average distances of the moon (238,855 miles) and sun (92,900,000 miles) from the earth, the tide producing force of the moon is approximately 2.5 times that of the sun."
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Why did you quote a source that used less precision than me?
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Alexander '75 911S Targa '86 951 SOLD |
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It's not the size of the tides but how you use them.
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Gary Fisher 29er 2019 Kia Stinger 2.0t gone ![]() 1995 Miata Sold 1984 944 Sold ![]() I am not lost for I know where I am, however where I am is lost. - Winnie the poo. |
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Also, the difference in distance between the moon or sun and one side of earth or the other plays a roll in tides. This difference is significant with the earth-moon system, insignificant with the earth-sun system. These factors together all combine to control the tides, so the 2.5 figure is more precise.
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1) I was quite kind in my first post, and told him that he was wrong, then explained the truth. I'm sorry if some of you thought I was being mean
but it's a fact that his oceanography class was telling him something wrong, and highly simplified.2) Wiki?? That link not only doesn't really say anything, it also ignores the most important part of tidal effects, which is the tangential compression of an object. It is as much of an effect as the radial extension. 3) By "Moon stop orbiting" I mean it locks into a geostationary orbit. If there was a large wave traveling around the planet, it would continue to do so, since water waves are a transfer of energy. Tidal motion is NOT a transfer of energy. It's water being pulled and twisted by gravity, nothing more. So, if magically the Moon locked into a geosync orbit right now, instantly, the tides would also stop traveling around the planet instantly, since they are integrally locked in the gravitational web of the Moon. 4) Troll? For answering the question and correcting someone so they don't continue to have the wrong idea of something for the rest of their life? Is that a troll in your mind?
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Mike Bradshaw 1980 911SC sunroof coupe, silver/black Putting the sick back into sycophant! |
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Yeah, troll! Lousy, *******, commie, pinko, socialista troll! Did I forget something? Yeah!, I'm talkin' ta you! You do strike me as a competent physicist, however.
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I think that you left out two letters "in"
Any competent physicist would understand that gravity is universally attractive. It does one thing and only one thing, it is a force of attraction between two masses. The magnitude of the attraction depends solely on the masses of the bodies and the distance between them, the direction is always attractive along the axis between them. Differential attractive forces due to differential distances will lead in the extreme case to 'spaghettification' of an object. The gravitational force on the near sides of the objects being stronger than the gravitational force on the far sides leads to a stretching in an axial direction, as the objects stretch due to this differential in gravitational forces they become longer and narrower but there is no tangential comperessive force other than the electronic and atomic bonds holding the molecules and atoms of the objects together. In the most extreme case even these bonds acan be overcome.
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http://www.lhup.edu/~dsimanek/scenario/tides.htm I'm willing to believe this website and am heading down to Safeway for a package of crow. Give Mike a Kewpie doll! As a scientist, I only want the truth, even if it dispels what I learned from my august professors at the University of California. In my defense, the website does say that most textbooks have it wrong, even post grad books and physics books. I nothing else, I've learned here that the tides are damned complicated. (After all that, the author of that website better be right.)
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![]() In fact, looking at your link, I see the same force diagram, which clearly shows the tangential effects. They are manifestations of the small differential tidal effects on the parts of the Earth that are off axis. Finally, all of this talk about the Moon orbiting the Earth is also wrong, since it actually orbits the Sun The Moon and the Earth co-orbit the Sun, because the direct gravitational force on the Moon from the Sun is stronger than what is on the Moon from the Earth...
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Mike Bradshaw 1980 911SC sunroof coupe, silver/black Putting the sick back into sycophant! |
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Pithy stuff, this.
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Mike Bradshaw 1980 911SC sunroof coupe, silver/black Putting the sick back into sycophant! |
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to a first approximation, yes.
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