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No, you never said it wasn't needed, but you did minimize its importance, claiming
that it can't be used alone to predict the tides. You said: "Even worse, the site then goes on to use the _differential_ expression to calculate the ratio of forces between the moon and sun!" You even produced a bogus formula to support your claim that centrifugal force varies across the Earth. This would predict a tidal force 65 times stronger than differential gravity. If that isn't minimalizing differential gravity, I don't know what is. You demonstrated a further lack of understanding with: If the moon stopped its rotation around the earth and the earth and the moon was "falling" toward each other, there would still be two bulges. "YES but how big would they be (Hint: Smaller than the tides?)" Actually, the tides would be the same. One fundamental difference that we have is that you insist that taking centrifugal force into account is the *only* way to look at the problem. As I've said a number of times, centrifugal force is a "fictional force" that is only needed if you wish to work in the accelerating Earth-centric reference frame. In fact, it is required that there must be an alternate approach that does not use "fictional" forces. It is quite easy to explain the tides without Centrifugal force - I mentioned the approach in one of my first posts. There is a gravitational force pulling the Earth and Moon together. This creates an acceleration such that the Earth is in free fall, and no net force is felt at the Earth's center. However, there is a larger force on the Moon side, and a smaller force on the far side. Since the average force has been accounted for it must be subtracted from these two forces, and the smaller force ends up being the same magnitude as the larger, but in the opposite direction. Hence, two tidal bulges, both caused solely by gravity. Another fundamental difference we have is that I agree with the traditional value for the tidal force. Ignoring minor effects, the result predicted by Differential Gravity (whether or not you use centrifugal force as part of the explanation) that is about 2 feet for both the near and far side bulges. (The Sun's contribution is about half of the Moon's.) The land masses and shallow water tends to "pile up" the water to create tides that are somewhat higher. You, however, have claimed that the variation of the centrifugal force from the near side to the far creates a force that dominates the tides. Your formula predicts a force 65 times greater than the differential formula, which would seem to create tides 100 feet or more. Your explanation that land masses dampen these tides runs counter to common experience. And, your formula predicts the Sun's contribution is only 1% of the Moon's, which is clearly not the case. As to your claim that the rotation around the barycenter "powers" the tides, well, that's not true. In fact, even if the Earth and Moon were not rotating around the barycenter, the tides would be the same, at least, until the Earth and Moon collided. Frankly, it clear that you still do not accept the fact that CF is constant, exactly equals the average gravitational force, and thus has no interesting contribution to the tides. "Nav" wrote in message ... Now why try to distort the truth Jeff? I never ever said differential gravity was not needed. I always said that it's the difference between gravity and centrifugal forces. You do understand the connotations of the DIFFERENCE between forces don't you? It does not mean that either component is zero and actually implies that both are important. Shesh! Still it's nice to see that you now agree that centrifugal forces should not be ignored (as they are in the gravity only model). As I've said so many times, the key to understanding is that the system rotates about the barycenter and it is not just a gravity field problem. The rotation actually provides the energy needed to power the daily tides -think about it OK? Cheers Jeff Morris wrote: "Nav" wrote in message ... Jeff Morris wrote: "Nav" wrote in message ... F= mr omega^2. The distance from the barycenter to all points on earth is NOT the same. As the site expalins in the next paragraph, only the center of the Earth rotates around the barycenter. Other points rotate around neighboring points. Anyway, the site clearly shows in Fig. 2 that it is the _DIFFERENCE_ between gravity and centrifugal force that makes the tides, not gravity alone. We never disagreed on this point. HOLY BACKPEDAL!!!!!! I'm not sure you really want to go back over this thread - your record is rather shaky. Mine, however, has been quite consistent. Remember, I started by posting sites with differing approaches to show that this problem can be looked at in different ways. I then made my first comment about Centrifugal force with: "Remember that Centrifugal Force may be a handy explanation, but it is a "fictional force" that only appears real to an observer in an accelerating frame of reference. Therefore, whenever it is used to explain something, there must be another explanation that works in a non-accelerating frame." but then you started claiming that differential gravity wasn't needed, I responded with: "Before I thought you were just arguing philosophically how much we should credit centrifugal force, but now it appears you haven't really looked at the math at all. The reason why "differential gravity" is invoked is because it represents the differing pull of the Moon on differing parts of the Earth. Although this force is all obviously towards the Moon, when you subtract off the centrifugal force this is what is left. It is this differing pull that causes the two tides." a few posts later: "Given that, your argument falls apart. The centrifugal force is exactly the same on all points of the Earth, and (not by coincidence) is exactly opposite the net gravitational force. What is left over is the differential gravity." The bottom line here is that the tides are properly described by the differential gravity equation. Centrifugal force can be used to explain how an outward force can be generated, but it is not needed, and it does not yield the equation that describes the tides. Frankly, your the one who started this by claiming that the traditional explanation of tides is fundamentally flawed, and that the differential force normally cited is not what causes the tides. You really haven't produced any coherent evidence to support this claim. |
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