Hi Allen,Of course you know that we'd both agree that the equatorial bulge effect is "with respect to the distant stars." Superimposed on top of that are the ocean tidal cycles. If you're talking about the former, Regner's point is indeed "kinda irrelevant," but since this thread started focusing on the ocean tides, I limited my remarks to the superimposed effect.
Martin On May 9, 2008, at 4:39 PM, Allen Daves wrote:
I would like to point out however, that the tidal bulges are on both sides of the earth & earth's oceans at the same time, the earth is flatened around it's whole cercumfrence supposedly due to the centrifical force that is cused by inertia wrt the stars not just the tidal actions of the sun/moon.....Even that explination would not reslove the the fact that the acceleration is detectable but it does not address the overall inertial paradox that exist for both the tides as well as any given orbiting body in a eliptical orbit where the accelerations advance and slow down wrt the very same causes of inertial reaction when not in orbit. (the whole point of the diagram)Therfore that (sun only) explination is not only too simplistic but kinda irrelevant to the issue as a whole that I am trying to point out.----- Original Message ---- From: Martin G. Selbrede <mselbrede@xxxxxxxxxxxxx> To: geocentrism@xxxxxxxxxxxxx Sent: Friday, May 9, 2008 2:25:19 PM Subject: [geocentrism] Re: TidesI understood Regner to be mounting an argument based on relative differential force (the entire premise of tidal forces in the first place). It is this differential force that supposedly gave rise to the rings of Saturn, when a sufficiently large chunk of material was disrupted as it passed within the Roche limit (meaning the force between the near-side and far-side parts of the body exceeded the tensile strength of the material comprising that body). This tidal disruption causes the alleged shattering of the body into a myriad of small chunks, each finding a unique orbit around Saturn due to inter-body collisions over time.I believe Regner mounted a Sun-only explanation for the sake of didactic simplicity, and he even went so far as to point this out. I saw nothing faulty in his analysis given the limits he himself provided to frame his discussion.Martin On May 9, 2008, at 4:14 PM, Neville Jones wrote:Regner,Your analysis of the cause of the tides is invalid, because gravitational attraction between parts of extended objects being exactly equivalent to all mass of each object being concentrated into mathematical points is valid for these objects being rigid, not flexible. You cannot assume the World to be rigid on the one hand and then subtract the theoretical force on a rigid World from both sides of a flexible World in order to 'explain' the tides.Where would these forces on the oceans be acting? At the centre of mass of the ocean, or on individual water molecules, or at the centre of mass of the World?The pull would vary as the inverse-square of the distance, which would result in very serious tidal waves arising as the Moon-side ocean was pulled more than the other was 'left behind'.Rather than being<- | - > as depicted, the force would be proportional tox*x^(-1) (x*x+2xR+R*R)^(-1) (x*x+4xR+4R*R)^(-1)where x is the distance from the centre of mass of the Moon to the closest tip of the ocean nearest to the Moon and R is the radius of the World, and x >> R in the heliocentric model.Ocean swell would also vary with the Moon's orbit.The analysis is thus far too simplistic. As with other aspects of the so-called 'solar system', such forces would lead very quickly to disorder and destruction.Neville. -----Original Message----- From: art@xxxxxxxxxxxxxx Sent: Wed, 07 May 2008 15:18:40 +1000There has been a couple of references to tides, lately, and much confusionon the subject. I'll do what I can to clear it up a bit. Please take the time to read the whole post before commenting. The force of gravity moves objects The Moon and the Earth orbit each other due to their gravitational interaction. It can be shown that you can simplify this problem:Every single bit of the Moon interacting gravitationally with every singlebit of the Earth. is exactly equivalent toAll the mass of the Moon being located at the centre-of-mass of the Moon, and all the mass of the Earth being located at the centre-of- mass of theEarth.From such an analysis we find that the two objects will orbit their commoncentre-of-mass in elliptical orbits. Tidal forces deform objectsBut the Earth and the Moon are extended objects so what effect does that have? Well, the Moon is still close enough to the Earth, that the gravityfrom the Moon will change over the diameter of the Earth: Moon Near Centre Far side of Earth side O <-------- <------- <------Okay, if we now subtract the part that actually moves the Earth, that is we subtract <------- which is the same as adding -------> to all threevectors (arrows) above, and we get the tidal forces: Moon Near Centre Far side of Earth side O <- | -> So looking at the Moon-Earth system from the outside we see that theMoon-side of Earth experience a larger gravitational pull from the Moon than the centre of the Earth, and the opposite side of Earth experience experience even less gravitational pull from the Moon. Seeing it from the Earth point of view, we have a Moon-ward force on the Moon-side of Earth and an opposite force on the opposite side of Earth. This is what causes the tidal bulge, and this is why there are (approximately) two high- and two low-tides a day. The oceans, being a liquid, obviously responds much stronger than the bedrock, but the Earth's crust is flexing too, as has been measured. The Earth rotates under this tidal bulge, which makes it a dynamical phenomenon. The gravitational force is proportional to 1/r² where r is the distancebetween the Earth and the Moon. The tidal force is the difference ingravitational force over a certain distance, which means it can be expressed as the r-derivative of the gravitational force. The tidal force is therefore proportional to 1/r³. That means the tidal force falls off more rapidly with distance than the gravitational force. This is the reason that the gravitational force of the Sun dominates the orbit of the Earth, but the Moon's tidal forceson Earth are larger than the Sun's.The Sun also contributes a significant tide (about half that of the Moon)and when the two are aligned at new Moon or full Moon, we get morepowerful tides called spring tides. At either half Moon we get neap tides. The ellipticity (eccentricity) of the Moon-Earth and the Sun-Earth orbitsalso causes variations in the strength of tides on Earth. The Moon experience (about 22 times) stronger tides from the Earth because of the Earth/Moon mass ratio of 81 and the size ratio of 3.7.The Lunar tide from the Sun is 3.7 times weaker than here on Earth, dueto the 3.7 times smaller diameter of the Moon.The tides would of course be exactly the same in a geocentric Universe.Allen's concerns in the "Acceleration thread"To account for tidal influences, the centre-of-mass of the test- mass of the accelerometer, should coincide with the centre-of-mass of the free-falling laboratory. Tides would have NO effect on such a set-up.But then the tidal effects on such a lab are very small in the first place: Less than a billionth per meter of the gravitational force keeping the labAND the test-mass in orbit.That's all for now (sorry for such a long post), and I hope it clarified afew things. - Regner Receive Notifications of Incoming MessagesEasily monitor multiple email accounts & access them with a click. Visit www.inbox.com/notifier and check it out!-------- Martin G. Selbrede Chief Scientist Uni-Pixel Displays, Inc. 8708 Technology Forest Place, Suite 100 The Woodlands, TX 77381281-825-4500 main line (281) 825-4507 direct line (281) 825-4599 fax (512) 422-4919 cellmselbrede@xxxxxxxxxxxxx / martin.selbrede@xxxxxxxxxxxx
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