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Jeff Morris wrote: OK, Nav, its clear you're not going to get this without some help. You keep claiming the centrifugal force varies across the Earth. However, that is not the case. Your assumption is that the Earth is rotating around the E-M barycenter, and that because that is offset from the Earth center, the centrifugal force is unbalanced. (Or more precisely, you claim the "r" in the centrifugal force equation is different on the near and far sides of the Earth.) I think it's you that does not understand that the rotation force is based on the lunar cycle -28 days! However, if we remove the daily rotation, Great idea -not based in reality of course. the Earth does not move around the barycenter quite like you think. Like I think? Only the center of the Earth describes a circle around the barycenter. So the Earth does wobble (now you are getting really close to the whole story -where harmonics of all the orbital periods give a complete answer). Now, don't all points on the surface move similarly around the barycenter and if they do, what is the difference in their orbital path to that of a circle? Now as I see it, from the math, the differential gravity model takes no account of this, exploiting the idea that the orbital motion of every point on the surface is perfectly circular around the moon (thereby allowing a cancellation of the centrigal component), a point that you seem to be having a bit of trouble grasping. By the way, the differential gravity idea first came from Newton. It's correct as far as it goes but the orbital mechanics of the Earth-Moon pair are more complicated (as far as I've been able to read, Newton only saw the free falling body aspect in his tidal proposal). The Devil _is_ in the details and you can't ignore the system rotation. This really is my last post on this. If you still haven't got the idea then I really can't make it any clearer and you'll just have to ponder why University Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON EARTH are due to the difference between centrifugal and gravity forces. Cheers |
"Nav" wrote in message ... Jeff Morris wrote: OK, Nav, its clear you're not going to get this without some help. You keep claiming the centrifugal force varies across the Earth. However, that is not the case. Your assumption is that the Earth is rotating around the E-M barycenter, and that because that is offset from the Earth center, the centrifugal force is unbalanced. (Or more precisely, you claim the "r" in the centrifugal force equation is different on the near and far sides of the Earth.) I think it's you that does not understand that the rotation force is based on the lunar cycle -28 days! However, if we remove the daily rotation, Great idea -not based in reality of course. We are looking for the dominant effect - the daily rotation is not a contributor to that. There are a variety of effects we're ignoring. the Earth does not move around the barycenter quite like you think. Like I think? OK, as you claim. Only the center of the Earth describes a circle around the barycenter. So the Earth does wobble (now you are getting really close to the whole story -where harmonics of all the orbital periods give a complete answer). Now, don't all points on the surface move similarly around the barycenter No. Only the center of the Earth revolves around the barycenter. Other points on the Earth revolve around other points nearby. Stop arguing and just work it out with a model. If you "wobble" a disk, all points wobble the same way. You're claiming that some points describe small circles, and some point describe large circles, but that clearly can't happen unless you rotate the disk. and if they do, what is the difference in their orbital path to that of a circle? All paths are the same. That's my point - stop talking and wobble a plate on the table without rotating it. All points on it trace the same circle. Hold a pencil on each side and look at the circles they trace. They all have the same radius, and therefore the same centrifugal force. Now as I see it, from the math, the differential gravity model takes no account of this, exploiting the idea that the orbital motion of every point on the surface is perfectly circular around the moon (thereby allowing a cancellation of the centrigal component), a point that you seem to be having a bit of trouble grasping. There is no need to grasp it. You entire argument is different points feel different rotation. But they don't - all points on the Earth feel the same centrifugal force. But your argument fails another test: The centrifugal force is "fictional" - it is just a convenience to simplify some problems. The only real force at play here is gravity, so any alternate approach must yield the same answer as a "gravity only" solution. By the way, the differential gravity idea first came from Newton. It's correct as far as it goes but the orbital mechanics of the Earth-Moon pair are more complicated (as far as I've been able to read, Newton only saw the free falling body aspect in his tidal proposal). The Devil _is_ in the details and you can't ignore the system rotation. This really is my last post on this. If you still haven't got the idea then I really can't make it any clearer and you'll just have to ponder why University Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON EARTH are due to the difference between centrifugal and gravity forces. As I've said, it's very easy to find numerous sites that scoff at the NOAA site you've mentioned, its regularly cited as "bad science." And I don't deny that there are a handful of sites that "handwave" that centrifugal force is the cause of the second bulge, but there are dozens that refute that in great detail. And virtually every published text supports my view. And you still haven't responded to the obvious flaws in the formula you proposed. If the centrifugal force from the Moon is as you claim, why does your math show that the Sun's contribution is only 1% of the Moon's? You when to great pains to show the math, but when I showed it was bogus you got very quiet on that front. |
"Nav" wrote in message
... .... If you still haven't got the idea then I really can't make it any clearer and you'll just have to ponder why University Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON EARTH are due to the difference between centrifugal and gravity forces. I have never claimed that you can't derive the tides by computing the difference between gravitational pull and the centrifugal force. My point has been that centrifugal force is constant across the Earth, and thus does not explain why the bulges are in opposite directions. The centrifugal force is exactly canceled by the average gravitational pull, and what is left over is the differential force. Your claim has been that the centrifugal force varies across the Earth, and that's what I've taken exception to. You keep citing the NOAA page, http://www.co-ops.nos.noaa.gov/restles3.html So I went back and read that in detail. In it is the "disclaimer": "While space does not permit here, it may be graphically demonstrated that, for such a case of revolution without rotation as above enumerated, any point on the earth will describe a circle which will have the same radius as the radius of revolution of the center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of the centrifugal force produced by the revolution of the earth and moon around their common center of mass (G) is the same at point A or B or any other point on or beneath the earth's surface. Any of these values is also equal to the centrifugal force produced at the center-of-mass (C) by its revolution around the barycenter." it goes on to develop differential gravity: "While the effect of this centrifugal force is constant for all positions on the earth, the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body." In other words, this site actually agrees with what I have been saying. Frankly I owe an apology to the site's author, since I maligned it without reading carefully. In fact, although it leads with a provocative line about a "little known aspect of the moon's orbital motion," and has a rather confusing diagram, its basic approach is correct and in full agreement with my claim. Apparently the site was actually changed at some point about two years ago because of complaints on another board. |
Jeff Morris wrote: In other words, this site actually agrees with what I have been saying. Frankly I owe an apology to the site's author, since I maligned it without reading carefully. In fact, although it leads with a provocative line about a "little known aspect of the moon's orbital motion," and has a rather confusing diagram, its basic approach is correct and in full agreement with my claim. Apparently the site was actually changed at some point about two years ago because of complaints on another board. The site certainly agrees with me. Don't confuse rotation of the earth with rotation about the system center. Here is the exact quote and it's concurrance with my view is as clear as day: "1. The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, 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. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2." Note the "one of the two forces". Cheers |
Jeff Morris wrote: "Nav" wrote in message ... ... If you still haven't got the idea then I really can't make it any clearer and you'll just have to ponder why University Departments of Oceanographics (and NOAA etc.) all say that THE TIDES ON EARTH are due to the difference between centrifugal and gravity forces. I have never claimed that you can't derive the tides by computing the difference between gravitational pull and the centrifugal force. My point has been that centrifugal force is constant across the Earth, and thus does not explain why the bulges are in opposite directions. The centrifugal force is exactly canceled by the average gravitational pull, and what is left over is the differential force. Your claim has been that the centrifugal force varies across the Earth, and that's what I've taken exception to. F= mr omega^2. The distance from the barycenter to all points on earth is NOT the same. 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. There really is no point going on any more is there? For myself, I'm sick of repeating the same point over and over. EOT Cheers You keep citing the NOAA page, http://www.co-ops.nos.noaa.gov/restles3.html So I went back and read that in detail. In it is the "disclaimer": "While space does not permit here, it may be graphically demonstrated that, for such a case of revolution without rotation as above enumerated, any point on the earth will describe a circle which will have the same radius as the radius of revolution of the center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of the centrifugal force produced by the revolution of the earth and moon around their common center of mass (G) is the same at point A or B or any other point on or beneath the earth's surface. Any of these values is also equal to the centrifugal force produced at the center-of-mass (C) by its revolution around the barycenter." it goes on to develop differential gravity: "While the effect of this centrifugal force is constant for all positions on the earth, the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body." In other words, this site actually agrees with what I have been saying. Frankly I owe an apology to the site's author, since I maligned it without reading carefully. In fact, although it leads with a provocative line about a "little known aspect of the moon's orbital motion," and has a rather confusing diagram, its basic approach is correct and in full agreement with my claim. Apparently the site was actually changed at some point about two years ago because of complaints on another board. |
Holy Back Pedal!!! Now you're claiming you agreed with me all along??? Just
22 hours ago we had this exchange: Actually, applying it in this context is your problem. Centrifugal acceleration is constant, it doesn't vary across the surface of the Earth as you claim. Remember, it doesn't even exist, its actually a reference frame shift. It is a much larger force than differential gravity but you want to ignore it? You are wrong Jeff, it does vary across the surface of the earth Jeff -the Barycenter is at ~3/4 r! On the moon side it's ~1/7 as large. Finally, (repeating yet again) it is the ____DIFFERENCE______ between inertial and gravity forces that make the tides. To say it's only "differential gravity" (I shudder at that term) is clearly wrong - this was a simple proof. You actually provided the math that "proves" centrifugal force varies, and thus causes tides much larger predicted by the accepted formula. Now you're claiming you never meant that at all? Right, Navie. Its the Constellation all over again. "Nav" wrote in message ... Jeff Morris wrote: In other words, this site actually agrees with what I have been saying. Frankly I owe an apology to the site's author, since I maligned it without reading carefully. In fact, although it leads with a provocative line about a "little known aspect of the moon's orbital motion," and has a rather confusing diagram, its basic approach is correct and in full agreement with my claim. Apparently the site was actually changed at some point about two years ago because of complaints on another board. The site certainly agrees with me. Don't confuse rotation of the earth with rotation about the system center. Here is the exact quote and it's concurrance with my view is as clear as day: "1. The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, 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. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2." Note the "one of the two forces". Cheers |
"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. My issue has always been that since Centrifugal Force is constant, it doesn't directly explain the two bulges. It is useful for people who would have trouble with the seeming "negative gravity" of the far side bulge, but it is also possible to fully describe the tides quite simply without Centrifugal Force at all. There really is no point going on any more is there? For myself, I'm sick of repeating the same point over and over. But your point seems to change on each post. First you claim CF varies, then you claim to be in complete agreement with a site that says its constant. Now you're saying it varies again. You provide your formula and challange me to work it out, then never explain why it give bogus answers. Constitution Constitution CONSTITUTION!!!!! opps .... constellation EOT Cheers You keep citing the NOAA page, http://www.co-ops.nos.noaa.gov/restles3.html So I went back and read that in detail. In it is the "disclaimer": "While space does not permit here, it may be graphically demonstrated that, for such a case of revolution without rotation as above enumerated, any point on the earth will describe a circle which will have the same radius as the radius of revolution of the center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of the centrifugal force produced by the revolution of the earth and moon around their common center of mass (G) is the same at point A or B or any other point on or beneath the earth's surface. Any of these values is also equal to the centrifugal force produced at the center-of-mass (C) by its revolution around the barycenter." it goes on to develop differential gravity: "While the effect of this centrifugal force is constant for all positions on the earth, the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body." In other words, this site actually agrees with what I have been saying. Frankly I owe an apology to the site's author, since I maligned it without reading carefully. In fact, although it leads with a provocative line about a "little known aspect of the moon's orbital motion," and has a rather confusing diagram, its basic approach is correct and in full agreement with my claim. Apparently the site was actually changed at some point about two years ago because of complaints on another board. |
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!!!!!! Cheers |
"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." or, in other words, exactly what I just said above. 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. |
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. |
"Thom Stewart" wrote in message ... Donal, I hope you're satisfied! For a damned 1/4 of a point!? All this differential and centrifugally, how does this Knowledge help a sailor to ride the tides? That was the original question. Remember? IIRC, I gave him my honest opinion. I hope you're satisfied (g) I'll admit that I'm pleased. Now, I wonder if Scot did any Tide Riding while he has been on Vacation Cruise? It will be interesting to find out how his trip went. Does anybody know when we can expect to hear from him? Regards Donal -- |
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. |
"Jeff Morris" wrote in message ... "Donal" wrote in message ... "Martin Baxter" wrote in message Donal wrote: So why does the moon seem to have a greater impact on the tides? Well duh! Remember F=G*(m'*m")/(d^2), Emmm... Huh? What the hell does that mean in English? Did you not take physics in school? Sadly, I didn't! With hindsight, I suspect that I had a poor teacher who managed to make the subject appear much duller than it really is. My high school physics teacher was possibly the worst teacher I ever had - a true nut case who shouldn't have been left alone with children. Fortunately I found much better teachers in college. I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. Where's Gilligan when you need him? Regards Donal -- |
"Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Hence, they bulge out from the middle. That's actually all that is needed to explain the tides; its so simple a lot of people have trouble getting it. Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Well, you're right that there are low tides at right angles, but the way the math works out the far side high tides are virtually the same. The magnitude of the differing pull 4000 miles closer to the Moon is about the same as 4000 miles further out. (Though I'm curious now just how much they do differ from each other ...) Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. OK. Centrifugal force is the explanation for children. Its kind of like the Tooth Fairy of physics. The problem is that while CF can explain the "outward force" needed for the far bulge, its still the differing gravitational force that defines the size and shape of the tidal force. And CF isn't needed at all if you can accept the "free fall) argument. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. The Earth and Sun are also an orbiting pair. There is no qualitative difference, only quantitative. Where's Gilligan when you need him? Gilly is an educated man. I'm sure he agrees with me. |
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"Jeff Morris" wrote in message ... "Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Very neat! However, your view seems to be a little bit simplistic. Why should a solid fall more slowly than a fluid in a gravitational field? If your theory was correct, then there wouldn't be any tide at all. You seem to be ignoring momentum. Regards Donal -- |
Jeff,
Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout "Jeff Morris" wrote in message ... "Donal" wrote in message ... "Jeff Morris" wrote in message I've still got a suspicion that if we expand your equation, we will find that the sun has a greater gravitational influence on the earth than the moon does. Yes, its does. The direct gravitational pull of the Sun is enormous, much larger than the Moon's. However, the tides are caused by the difference in pull between the near side and the far side. Sorry! This doesn't make any sense at all. How does the water on the far side(of the earth) know that there is a different pull on the other side? It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Hence, they bulge out from the middle. That's actually all that is needed to explain the tides; its so simple a lot of people have trouble getting it. Even if you were correct, then there would be a high tide facing the moon, a low tide at right angles to the moon, and a much lower *high* tide opposite the moon. The reality is that the HW opposite the moon is only fractionally smaller. Well, you're right that there are low tides at right angles, but the way the math works out the far side high tides are virtually the same. The magnitude of the differing pull 4000 miles closer to the Moon is about the same as 4000 miles further out. (Though I'm curious now just how much they do differ from each other ...) Centrigugal force explains why there is a high tide on the oppisite side of the Earth from the moon - if you consider that the two bodies are rotating around a common centre. OK. Centrifugal force is the explanation for children. Its kind of like the Tooth Fairy of physics. The problem is that while CF can explain the "outward force" needed for the far bulge, its still the differing gravitational force that defines the size and shape of the tidal force. And CF isn't needed at all if you can accept the "free fall) argument. Since the Moon is a lot closer, that difference is more significant. If you remember any calculus, you'd know that differentiating a 1/r^2 function yeilds a 1/r^3 function. The inverse cube of the Sun's distance becomes a tiny number compared to the Moon's. The net result is that the Moon's effect on the tides is 2.2 stronger than the Sun's. Tsk...tsk. The moon only has a stronger effect on tides because the Earth and Moon are an orbiting pair. The Earth and Sun are also an orbiting pair. There is no qualitative difference, only quantitative. Where's Gilligan when you need him? Gilly is an educated man. I'm sure he agrees with me. |
"Scout" wrote in message
... Jeff, Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout I've always said that Centrifugal Force can be used as part of the explanation, as long as you end up with the same answer. There are several different ways of looking at this, all valid. (I hope I can get through this without mangling the terms too badly ...) The problem with Centrifugal Force is that it is a "fictional force." It is only needed if you work in a non-inertial, or accelerating reference frame. If you are in a car going around a curve, your reference frame is accelerating towards the center of the curve, and thus you feel a Centrifugal Force in the opposite direction. To an outside observer, the CF doesn't exist, the only force is the car pulling the passenger around the turn. The outside observer can analyze the situation completely without invoking CF. (The passenger feels CF push him outward, the observer sees the car pull the passenger inward.) In the Earth-Moon system there is gravity pulling both the Earth and Moon around curves. Because the gravity acts on all objects, we don't notice ourselves being pulled around. The magnitude of the Centrifugal force is to small to notice, but in that reference frame it exists. To the outside observer, we're just in freefall, being pulled inward by gravity. The problem with CF arises when you look carefully at the math. One pitfall Nav fell into was trying to calculate CF as a function that varies with the distance to the barycenter. However, all points on the Earth do not rotate around the barycenter, only the center does. Other points describe the same circle around nearby points, so that all points on Earth feel the same Centrifugal Force. (This is a tough concept to explain in words; its easier to do it graphically. Consider a plate wobbling around a point but with no rotation - each point on the plate describes the same circle.) BTW, Nav provided two commonly used formulas, one for gravity and the other for CF. Although they look quite different, you should appreciate that they are the same, since the angular velocity is determined by the gravitational force. The CF will be the same (with the opposite sign) as the gravitational pull at the Earth's center. Since the CF is a constant force, it can't describe the two bulges in opposite directions. It is gravity itself that varies with distance. The differential force can be derived either by subtracting the average gravitational force which causes the freefall at the center of the Earth, or it can be derived by adding the centrifugal force. Since the two are the same, except for the sign, the math is identical. So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! |
"Jeff Morris" wrote
[snip] So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! Yes, I can, as I've watched my physics teaching partner wince quite a bit this week as we discussed this thread. He was quick to cover our whiteboard with formulas and drawings. It's an interesting thread though, and notwithstanding my oversimplified analogies, I've learned a lot from it. By the way, I saw that same wince from a black history professor when I suggested that the Civil War was fought to free the slaves, and then again when I suggested to an ancient lit professor that The Odyssey has all the earmarks of an Arnold Schwarzenegger movie. Probably explains why I like a good fart joke. Scout |
"Scout" wrote in message
... "Jeff Morris" wrote [snip] So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! Yes, I can, as I've watched my physics teaching partner wince quite a bit this week as we discussed this thread. He was quick to cover our whiteboard with formulas and drawings. It's an interesting thread though, and notwithstanding my oversimplified analogies, I've learned a lot from it. By the way, I saw that same wince from a black history professor when I suggested that the Civil War was fought to free the slaves, and then again when I suggested to an ancient lit professor that The Odyssey has all the earmarks of an Arnold Schwarzenegger movie. Probably explains why I like a good fart joke. Sort of like the wince I got from an African-American/Cherokee friend when I asked him how he was going to celebrate Columbus Day? jeff ps So how does your partner rate my explanations? |
"Jeff Morris" wrote
[snip] Sort of like the wince I got from an African-American/Cherokee friend when I asked him how he was going to celebrate Columbus Day? Ouch! ps So how does your partner rate my explanations? The great mediator saw truth in both models. He was a bit more forgiving of my "impellor in a great centrifugal pump" analogy, but slapped my wrist for saying this about centrifugal force: if it feels real, mustn't it be real? By the time he was done, my head was spinning and yet somehow my brain seemed to be bulging (quite appropriately) out of both sides of my head. In the end he called me an English teacher, which is his way of slandering me, and told me my paltry general science achievements were no match for his superior physics and math skills. I told him he could forget about me taking him sailing again and he took it all back. He's never done the newsgroups though, and was quite impressed with the ease with which like minded folks could debate a worthy topic. Scout |
Well just to confuse things a bit mo
Even if we only focus on the tide generating potential, there is a cupple of things that we haven't discussed yet, and one of them has to do with rotation: "The Coriolis freqency". The other thing one could include is the "parallax". I mention this just to make clear that the two models discussed above both are incomplete. Peter S/Y Anicula "Jeff Morris" skrev i en meddelelse ... "Scout" wrote in message ... Jeff, Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout I've always said that Centrifugal Force can be used as part of the explanation, as long as you end up with the same answer. There are several different ways of looking at this, all valid. (I hope I can get through this without mangling the terms too badly ...) The problem with Centrifugal Force is that it is a "fictional force." It is only needed if you work in a non-inertial, or accelerating reference frame. If you are in a car going around a curve, your reference frame is accelerating towards the center of the curve, and thus you feel a Centrifugal Force in the opposite direction. To an outside observer, the CF doesn't exist, the only force is the car pulling the passenger around the turn. The outside observer can analyze the situation completely without invoking CF. (The passenger feels CF push him outward, the observer sees the car pull the passenger inward.) In the Earth-Moon system there is gravity pulling both the Earth and Moon around curves. Because the gravity acts on all objects, we don't notice ourselves being pulled around. The magnitude of the Centrifugal force is to small to notice, but in that reference frame it exists. To the outside observer, we're just in freefall, being pulled inward by gravity. The problem with CF arises when you look carefully at the math. One pitfall Nav fell into was trying to calculate CF as a function that varies with the distance to the barycenter. However, all points on the Earth do not rotate around the barycenter, only the center does. Other points describe the same circle around nearby points, so that all points on Earth feel the same Centrifugal Force. (This is a tough concept to explain in words; its easier to do it graphically. Consider a plate wobbling around a point but with no rotation - each point on the plate describes the same circle.) BTW, Nav provided two commonly used formulas, one for gravity and the other for CF. Although they look quite different, you should appreciate that they are the same, since the angular velocity is determined by the gravitational force. The CF will be the same (with the opposite sign) as the gravitational pull at the Earth's center. Since the CF is a constant force, it can't describe the two bulges in opposite directions. It is gravity itself that varies with distance. The differential force can be derived either by subtracting the average gravitational force which causes the freefall at the center of the Earth, or it can be derived by adding the centrifugal force. Since the two are the same, except for the sign, the math is identical. So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! |
mission accomplished! lol
Scout "Peter S/Y Anicula" wrote Well just to confuse things a bit mo Even if we only focus on the tide generating potential, there is a cupple of things that we haven't discussed yet, and one of them has to do with rotation: "The Coriolis freqency". The other thing one could include is the "parallax". I mention this just to make clear that the two models discussed above both are incomplete. |
Scout,
I was determined not to re-enter this discussion, except to needle Donal about the 1/4 point. Your post about your physic teacher gave me enough enjoyment to venture another post. Your experience brought back memoirs of mine own days in High School with my own Physic Teacher about Tides and a Commercial Clammer who help me learn sailing. By the time I got finished, I had more faith in the Clammer. He lived in a shed on a float in the middle of the harbor. He was fond of the booze an when he know enough about any subject to satisfy himself, then he was satisfied. His explanation of the Tide, given to me willingly. was; and I quote. The Moon causes the Tides. It cause high tide because of its pull on earth, which screws up the pull between the earth and the Sun. Since the Tides are made up of water, the higher water, under the Moon creates less water on the other side of the Earth or low Tide. Since the Tide are water and the pull is less the farther away from the Moon they are not as high on the side of the earth facing the moon but higher than the water on the side of the Earth facing away away from the Moon. People call this difference 1/2 tide. The height of High Tide and Low Tide along with 1/2 Tides are affected by the Phase of the Moon. The difference of the location and of the time of tides are caused by the differences of the rotation of the Earth and the time of revolution and direction of the Moon This is explanation has served me well for over 60 years. Ole Thom |
Scout,
I forgot to mention my old friend. Frank, also told me he had Tide Tables and Almanics for the heigth and times of the tides and the position on the Moon. Ole Thom |
You're absolutely correct. There are numerous effects we're not considering. We've
only attempted to understand the primary cause of two tides a day. Even then, the math is a bit more complex than the simple formulas we've used. The parallax effects are certainly significant - they are caused non-circular orbits. And then there's Lunar declination to fold in. Of course, spring and neap must be considered - does everyone know when Syzygy is? And these are just the global effect - there's a whole litany of local effects to consider. Or you can keep a copy of Eldridge or Reed's handy. "Peter S/Y Anicula" wrote in message ... Well just to confuse things a bit mo Even if we only focus on the tide generating potential, there is a cupple of things that we haven't discussed yet, and one of them has to do with rotation: "The Coriolis freqency". The other thing one could include is the "parallax". I mention this just to make clear that the two models discussed above both are incomplete. Peter S/Y Anicula "Jeff Morris" skrev i en meddelelse ... "Scout" wrote in message ... Jeff, Remember that I first posted that very same sentiment, and even provided a graphic. I still believe that to be true, but have modified my internal model, giving allowance for the centrifugal force. I'm not a physicist, but the way I'm seeing it, there is a middle ground in this discussion. I'm curious to know if you're discounting centrifugal force as a contributor to the far bulge. Scout I've always said that Centrifugal Force can be used as part of the explanation, as long as you end up with the same answer. There are several different ways of looking at this, all valid. (I hope I can get through this without mangling the terms too badly ...) The problem with Centrifugal Force is that it is a "fictional force." It is only needed if you work in a non-inertial, or accelerating reference frame. If you are in a car going around a curve, your reference frame is accelerating towards the center of the curve, and thus you feel a Centrifugal Force in the opposite direction. To an outside observer, the CF doesn't exist, the only force is the car pulling the passenger around the turn. The outside observer can analyze the situation completely without invoking CF. (The passenger feels CF push him outward, the observer sees the car pull the passenger inward.) In the Earth-Moon system there is gravity pulling both the Earth and Moon around curves. Because the gravity acts on all objects, we don't notice ourselves being pulled around. The magnitude of the Centrifugal force is to small to notice, but in that reference frame it exists. To the outside observer, we're just in freefall, being pulled inward by gravity. The problem with CF arises when you look carefully at the math. One pitfall Nav fell into was trying to calculate CF as a function that varies with the distance to the barycenter. However, all points on the Earth do not rotate around the barycenter, only the center does. Other points describe the same circle around nearby points, so that all points on Earth feel the same Centrifugal Force. (This is a tough concept to explain in words; its easier to do it graphically. Consider a plate wobbling around a point but with no rotation - each point on the plate describes the same circle.) BTW, Nav provided two commonly used formulas, one for gravity and the other for CF. Although they look quite different, you should appreciate that they are the same, since the angular velocity is determined by the gravitational force. The CF will be the same (with the opposite sign) as the gravitational pull at the Earth's center. Since the CF is a constant force, it can't describe the two bulges in opposite directions. It is gravity itself that varies with distance. The differential force can be derived either by subtracting the average gravitational force which causes the freefall at the center of the Earth, or it can be derived by adding the centrifugal force. Since the two are the same, except for the sign, the math is identical. So take your pick, either explanation works, and I'm sure there are others. However, I hope you can appreciate that explanations like "gravity creates the inner bulge, centrifugal force creates the outer bulge" makes physicists wince! |
Thanks Thom,
thank God for those ol' salty guys ~ they always manage to keep it real! Scout "Thom Stewart" wrote [snip] Your post about your physic teacher gave me enough enjoyment to venture another post. Your experience brought back memoirs of mine own days in High School with my own Physic Teacher about Tides and a Commercial Clammer who help me learn sailing. By the time I got finished, I had more faith in the Clammer. He lived in a shed on a float in the middle of the harbor. He was fond of the booze an when he know enough about any subject to satisfy himself, then he was satisfied. |
"Donal" wrote in message
... It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Very neat! However, your view seems to be a little bit simplistic. It is simple. That's because there really isn't that much going on (at this level). Just the pull of gravity, which varies with distance. Everything else is red herring. Why should a solid fall more slowly than a fluid in a gravitational field? Why is there any difference? They both feel the same force. And, the land surfaces are distorted by the tides, roughly a meter, IIRC. To be honest, I could never figure out if the water is distorted more for some reason (its lighter?), or is in simply free to move, and thus get involved in the local shoreline effects. (That is, is the tide in the middle of the ocean the same as in the middle of a continent?) If your theory was correct, then there wouldn't be any tide at all. No, the force distorts both the land and the water. These distortions are the two bulges. In fact, because there is a difference in force, there must be some distortion - how much is a detail for the engineers! You seem to be ignoring momentum. Nope. If you use the "free fall" approach, momentum isn't really a factor in computing the force, though I suppose it gets involved when you calculate the actual motion. You sort of consider momentum with the centrifugal force approach, but you don't calculate it because the CF gets cancelled out. Consider an astronaut space walking outside a space station. They both float together, feeling no force, although they are both in freefall in their orbit. If the astronaut moves to a lower orbit, he will feel a stronger pull and be drawn in, unless he speeds up to compensate. If the astronaut moves to a higher orbit, the force is reduced. As I said, the force can be calculated without consideration of momentum. A purist might say momentum is considered because the mass and velocity of the every object in the system is folded together. And, the pure way force is defined is by how it changes momentum. But I don't think this is what you're talking about. |
"Jeff Morris" wrote in message ... "Donal" wrote in message ... It doesn't "know" anything. Because the Earth and Moon are an "orbiting pair," as you say, they are falling towards each other. Because the gravitational field varies, the near side falls faster than the middle; and the far side falls slower. Very neat! However, your view seems to be a little bit simplistic. It is simple. That's because there really isn't that much going on (at this level). Just the pull of gravity, which varies with distance. Everything else is red herring. Why should a solid fall more slowly than a fluid in a gravitational field? Why is there any difference? They both feel the same force. And, the land surfaces are distorted by the tides, roughly a meter, IIRC. To be honest, I could never figure out if the water is distorted more for some reason (its lighter?), or is in simply free to move, and thus get involved in the local shoreline effects. (That is, is the tide in the middle of the ocean the same as in the middle of a continent?) If your theory was correct, then there wouldn't be any tide at all. No, the force distorts both the land and the water. These distortions are the two bulges. In fact, because there is a difference in force, there must be some distortion - how much is a detail for the engineers! You seem to be ignoring momentum. Nope. If you use the "free fall" approach, momentum isn't really a factor in computing the force, though I suppose it gets involved when you calculate the actual motion. You sort of consider momentum with the centrifugal force approach, but you don't calculate it because the CF gets cancelled out. Consider an astronaut space walking outside a space station. They both float together, feeling no force, although they are both in freefall in their orbit. If the astronaut moves to a lower orbit, he will feel a stronger pull and be drawn in, unless he speeds up to compensate. If the astronaut moves to a higher orbit, the force is reduced. As I said, the force can be calculated without consideration of momentum. I don't understant this. In orbit, momentum is the force that balances the effect of gravity. Without momentum, your astronaut wouldn't "float" - he would crash straight into the Earth. In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? Furthermore, if he slowed down, then he would still feel like he was floating -- apart from the temperature, and perhaps the braking effect of the atmosphere. This makes me think that the orbiting "free-fall" astronaut doesn't feel that he is floating at all. He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? What does your physics friend say about this? A purist might say momentum is considered because the mass and velocity of the every object in the system is folded together. And, the pure way force is defined is by how it changes momentum. But I don't think this is what you're talking about. I'm not sure. I'm certainly *not* a purist. Regards Donal -- |
"Donal" wrote in message ... "Jeff Morris" wrote in message Consider an astronaut space walking outside a space station. They both float together, feeling no force, although they are both in freefall in their orbit. If the astronaut moves to a lower orbit, he will feel a stronger pull and be drawn in, unless he speeds up to compensate. If the astronaut moves to a higher orbit, the force is reduced. As I said, the force can be calculated without consideration of momentum. I don't understant this. In orbit, momentum is the force that balances the effect of gravity. Without momentum, your astronaut wouldn't "float" - he would crash straight into the Earth. Momentum is not a force. You're right that the astronaut has momentum, and that the force of gravity alters his momentum. In fact, Force is defined by how much it changes momentum. (Many people learn F=ma; in physics that is normally written as F=dp/dt, or Force equals the rate of change of momentum with respect to time.) My point is that you can determine the force on the astronaut without considering his momentum. To figure out how the force would alter his orbit, you would probably take momentum into account. Remember, I'm not trying to calculate the tides, only to show how gravity can cause two equal size bulges on the Earth. In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? So tell us, what is the difference? Floating in a space station is call "free fall" because it feels the same as jumping off a cliff. Furthermore, if he slowed down, then he would still feel like he was floating -- apart from the temperature, and perhaps the braking effect of the atmosphere. Yes, but that's not the point. The point is, if he is in a lower orbit, he experiences more gravity; in a higher orbit, less gravity. If his speed is not adjusted to compensate, he will drift further away from the space station. Just like the tides. This makes me think that the orbiting "free-fall" astronaut doesn't feel that he is floating at all. Haven't you ever seen astronauts floating? He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? http://science.howstuffworks.com/weightlessness1.htm What does your physics friend say about this? He would probably deplore the lack of education in your country. A purist might say momentum is considered because the mass and velocity of the every object in the system is folded together. And, the pure way force is defined is by how it changes momentum. But I don't think this is what you're talking about. I'm not sure. I'm certainly *not* a purist. Regards Donal -- |
Jeff Morris wrote: 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!" Yes that's right for that site. The differential of the gavity equation does not give a force it gives the rate of chnage of force with distance. That's basic calculus. 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. Nothing bogus abot F=m r omaga^2. It's high school physics. 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. Jeff, stop a moment and put what you just said into an equation. According to you subtraction of a constant from both sides of an inequality makes an equality? 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. That's because you've forgetten that the tiode is due to the difference between forces -how many times do I have to repeat this???? 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. No it doen't. It's your miscomprehension about the force balance. As to your claim that the rotation around the barycenter "powers" the tides, well, that's not true. Jeff, don't be hysterical. Of course it's true. Furthermore, the barycenter is not the same distance from all points on earth so the centrifugal force varies across the earth. Here is the equation for the force on a mass m of water on the far side for a non-rotating earth (to keep it clear): GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2 On the near side: GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2 M is the moon mass, E the earth mass, s the distance from the center of the earth to the barycenter, omega the angular velocity of the earth-moon pair G the gravitational constant. Now look very carefully at the three terms in each equation. The first two are gravity, the third centrifugal. Two terms are different in both cases. The gravity term is smaller on the far side and the centrifugal term is bigger. On the near side the gavity is bigger and the cenbtrifugal is smaller. That is the proof of my argument. Cheers |
Jeff Morris wrote: Remember, I'm not trying to calculate the tides, only to show how gravity can cause two equal size bulges on the Earth. Equal? That'll be hard when it's non-linear. Think about it. Take as long as you like. Cheers |
"Nav" wrote in message ... Jeff Morris wrote: 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!" Yes that's right for that site. The differential of the gavity equation does not give a force it gives the rate of chnage of force with distance. That's basic calculus. Simply differentiating the gravitational force is not enough. But the equation normally given for differential gravity is dimensionally correct. 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. Nothing bogus abot F=m r omaga^2. It's high school physics. That formula is correct. Applying it with a varying "r" is bogus. That's the issue here, but you haven't noticed it. 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. Jeff, stop a moment and put what you just said into an equation. According to you subtraction of a constant from both sides of an inequality makes an equality? I have no idea what you're saying. If the force at the center is X, and on the near side is X+D, and on the far side is X-D, then if you subtract the central force, the near side is D and the far side is -D. 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. That's because you've forgetten that the tiode is due to the difference between forces -how many times do I have to repeat this???? I just tried to follow all your instructions. You challenged me to work the math, I did, it produced bogus answers. The difference in the centrifugal force that your equation predicts is 65 times more that the difference in the gravitational force, so subtracting still won't help much. Why don't you work out the math? And don't forget to do it for the Sun also. 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. No it doen't. It's your miscomprehension about the force balance. I'm just reporting the numbers your equation predicts. As to your claim that the rotation around the barycenter "powers" the tides, well, that's not true. Jeff, don't be hysterical. Of course it's true. Furthermore, the barycenter is not the same distance from all points on earth so the centrifugal force varies across the earth. This is your basic mistake. Even the web site you pointed to for support explicitly says this is not true. Let me repeat it again http://www.co-ops.nos.noaa.gov/restles3.html "While space does not permit here, it may be graphically demonstrated that, for such a case of revolution without rotation as above enumerated, any point on the earth will describe a circle which will have the same radius as the radius of revolution of the center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of the centrifugal force produced by the revolution of the earth and moon around their common center of mass (G) is the same at point A or B or any other point on or beneath the earth's surface. Any of these values is also equal to the centrifugal force produced at the center-of-mass (C) by its revolution around the barycenter. This fact is indicated in Fig. 2 by the equal lengths of the thin arrows (representing the centrifugal force Fc) at points A, C, and B, respectively." it continues with: "While the effect of this centrifugal force is constant for all positions on the earth, the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body." Go on. Read it, think about it. You were quick to cite this page when you thought it supported you. Here is the equation for the force on a mass m of water on the far side for a non-rotating earth (to keep it clear): GmM/(R+r)^2 + GmE/r^2 - m(r+s)omega^2 On the near side: GmM/(R-r)^2 + GmE/r^2 - m(r-s) omega^2 M is the moon mass, E the earth mass, s the distance from the center of the earth to the barycenter, omega the angular velocity of the earth-moon pair G the gravitational constant. Nope. This is incorrect. The proper equations a Farside: GmM/(R+r)^2 + GmE/r^2 - m s omega^2 On the near side: GmM/(R-r)^2 + GmE/r^2 - m s omega^2 Further, since GmE/r^2 = m s omega^2, we are left simply with: Farside: GmM/(R+r)^2 On the near side: GmM/(R-r)^2 This leads to the traditional differential equation 2GmMr/R^3 as shown in http://mb-soft.com/public/tides.html Now look very carefully at the three terms in each equation. The first two are gravity, the third centrifugal. Two terms are different in both cases. Nope. Only one term is different. The gravity term is smaller on the far side and the centrifugal term is bigger. On the near side the gavity is bigger and the cenbtrifugal is smaller. That is the proof of my argument. The proof fails because of a faulty assumption. Sorry Nav, Centrifugal Force is constant. You can use it if you chose, but it doesn't really change the math and isn't particularly interesting. The part of the equation that actually produces the two tides is the differential gravity. |
Yes, I should have said "roughly equal." But don't your equations show a huge
difference between the near and far side tides caused by the barycenter being closer to the near side? Shouldn't that be a clue that something is amiss in your theory? "Nav" wrote in message ... Jeff Morris wrote: Remember, I'm not trying to calculate the tides, only to show how gravity can cause two equal size bulges on the Earth. Equal? That'll be hard when it's non-linear. Think about it. Take as long as you like. Cheers |
"Jeff Morris" wrote in message ... "Donal" wrote in message ... "Jeff Morris" wrote in message Consider an astronaut space walking outside a space station. They both float together, feeling no force, although they are both in freefall in their orbit. If the astronaut moves to a lower orbit, he will feel a stronger pull and be drawn in, unless he speeds up to compensate. If the astronaut moves to a higher orbit, the force is reduced. As I said, the force can be calculated without consideration of momentum. I don't understant this. In orbit, momentum is the force that balances the effect of gravity. Without momentum, your astronaut wouldn't "float" - he would crash straight into the Earth. Momentum is not a force. You're right that the astronaut has momentum, and that the force of gravity alters his momentum. In fact, Force is defined by how much it changes momentum. (Many people learn F=ma; in physics that is normally written as F=dp/dt, or Force equals the rate of change of momentum with respect to time.) Ok. What force opposes gravity so that a body may remain in orbit? My point is that you can determine the force on the astronaut without considering his momentum. In which case, there must be a "force" that is counteracting the effect of gravity. After all, gravity is trying to pull the orbiting Astronaut straight towards Earth. There must be another force that is opposing gravity. To figure out how the force would alter his orbit, you would probably take momentum into account. Remember, I'm not trying to calculate the tides, only to show how gravity can cause two equal size bulges on the Earth. Do you think that centrifugal force plays any part? If so, what do you think the ratio is between the centrifugal and differential gravity forces? In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? So tell us, what is the difference? Acceleration. Floating in a space station is call "free fall" because it feels the same as jumping off a cliff. Furthermore, if he slowed down, then he would still feel like he was floating -- apart from the temperature, and perhaps the braking effect of the atmosphere. Yes, but that's not the point. The point is, if he is in a lower orbit, he experiences more gravity; in a higher orbit, less gravity. If his speed is not adjusted to compensate, he will drift further away from the space station. Just like the tides. This makes me think that the orbiting "free-fall" astronaut doesn't feel that he is floating at all. Haven't you ever seen astronauts floating? Yes.... but they are constantly changing direction.... and therefor they should be aware of the effects of acceleration. He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? http://science.howstuffworks.com/weightlessness1.htm That is a very simplistic explanation. It refers to the fact that the astronauts will feel the acceleration at take-off, and yet it doesn't seem to understand that a change of direction is also acceleration. We humans can detect acceleration. If you sit in an automobile with your eyes closed, then you can feel an increase or decrease in speed .... or a change of direction!! As the astronauts are subjected to a constant change of direction, I suspect that they might not feel that they are completely free-floating. What does your physics friend say about this? He would probably deplore the lack of education in your country. Ask him anyway! Perhaps, if you allowed him to read the thread, he might be amazed at your lack of reading ability. After all, I've already explained that I gave up Physics at an early stage. Regards Donal -- |
"Donal" wrote in message ... I don't understant this. In orbit, momentum is the force that balances the effect of gravity. Without momentum, your astronaut wouldn't "float" - he would crash straight into the Earth. Momentum is not a force. You're right that the astronaut has momentum, and that the force of gravity alters his momentum. In fact, Force is defined by how much it changes momentum. (Many people learn F=ma; in physics that is normally written as F=dp/dt, or Force equals the rate of change of momentum with respect to time.) Ok. What force opposes gravity so that a body may remain in orbit? Nothing. The body remains in orbit because it has enough forward velocity (momentum also, but its the velocity that counts) so that while it "falls" into the Earth, the forward velocity keeps the body from hitting the Earth. For a low orbit, we normally thing of circular orbits, but for high orbits they can be quite eccentic. If the Earth were a "point source" it would only take a small velocity to stay in orbit. Of course, there are many ways to calculate this - somtimes its done in terms of "energy," other times as "delta V" so if you want to say the momentum of the body keeps it from falling into the Earth, that's OK. My point is that you can determine the force on the astronaut without considering his momentum. In which case, there must be a "force" that is counteracting the effect of gravity. After all, gravity is trying to pull the orbiting Astronaut straight towards Earth. There must be another force that is opposing gravity. See above. If the body were motionless (WRT Earth), it would fall directly in. But if it has any velocity, its has a chance of missing it. The reason why I say velocity is important, and not momentum, is that two bodies of different mass (and hence, different momentum) will float together in orbit. Of course, if you want to calculate the amount of fuel to burn, momentum becomes important. On Earth, we always have air resistance, and other forms of friction, so momentum is more significant. To figure out how the force would alter his orbit, you would probably take momentum into account. Remember, I'm not trying to calculate the tides, only to show how gravity can cause two equal size bulges on the Earth. Do you think that centrifugal force plays any part? If so, what do you think the ratio is between the centrifugal and differential gravity forces? It is possible to look at the problem without considering Centrifugal Force. However, even if you use CF, it is a constant, and equal to the net gravitation force. So if it helps to expain how there can be a force away from the moon, that's OK, but you must remember that it is the gravitation force that varies, so that's where the interesting math comes from. In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? So tell us, what is the difference? Acceleration. No, the acceleration is the same, more or less. (Not counting the difference in distance from the Earth, or air resistance, etc.) The only real difference is that the astronaut has enough velocity (hopefully) to miss the Earth as he falls. Floating in a space station is call "free fall" because it feels the same as jumping off a cliff. Furthermore, if he slowed down, then he would still feel like he was floating -- apart from the temperature, and perhaps the braking effect of the atmosphere. Yes, but that's not the point. The point is, if he is in a lower orbit, he experiences more gravity; in a higher orbit, less gravity. If his speed is not adjusted to compensate, he will drift further away from the space station. Just like the tides. This makes me think that the orbiting "free-fall" astronaut doesn't feel that he is floating at all. Haven't you ever seen astronauts floating? Yes.... but they are constantly changing direction.... and therefor they should be aware of the effects of acceleration. I must admit this subtlety has perplexed me - clearly the don't feel the G force, since it the same as a car in a tight turn. But I keep thinking it should be detectable, if only because the path is curving. He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? http://science.howstuffworks.com/weightlessness1.htm That is a very simplistic explanation. It refers to the fact that the astronauts will feel the acceleration at take-off, and yet it doesn't seem to understand that a change of direction is also acceleration. We humans can detect acceleration. If you sit in an automobile with your eyes closed, then you can feel an increase or decrease in speed .... or a change of direction!! As the astronauts are subjected to a constant change of direction, I suspect that they might not feel that they are completely free-floating. Of course, from a General Relativity, Gravity Well point of view, the obital path is a straight line in curved space. I should know the answer here - let me cogitate ... What does your physics friend say about this? He would probably deplore the lack of education in your country. Ask him anyway! Actually, it was Scout's friend. However, you should remember I majored in physics and worked for NASA doing spacecraft navigation. I may be rusty now, but 25 years ago I really knew this stuf! Perhaps, if you allowed him to read the thread, he might be amazed at your lack of reading ability. After all, I've already explained that I gave up Physics at an early stage. I haven't forgotton that. It was just a little dig since usually you Brits complain about our sorry education. |
Jeff Morris wrote: I must admit this subtlety has perplexed me - clearly the don't feel the G force, since it the same as a car in a tight turn. But I keep thinking it should be detectable, if only because the path is curving. (1) How tight is the turn and (2) how big is the imbalance between centrifugal and garvitational forces for such a small body? Of course, from a General Relativity, Gravity Well point of view, the obital path is a straight line in curved space. I should know the answer here - let me cogitate ... Eat some bran? :-P Cheers |
"Jeff Morris" wrote in message ... "Donal" wrote in message ... In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? So tell us, what is the difference? Acceleration. No, the acceleration is the same, more or less. (Not counting the difference in distance from the Earth, or air resistance, etc.) The only real difference is that the astronaut has enough velocity (hopefully) to miss the Earth as he falls. That was a very revealing answer. Earlier in the thread you were confident enough about your position to question my lack of education. Now you seem to think that an object travelling at "x" miles an hour in a straight line ("headlong into space") has the same acceleration as a body travelling at "x" miles an hour in orbit. It's probably time that you consulted your physics partner. Before you let him read what you have written, you should make sure that there is a cloakroom near the PC. Otherwise, have a potty close at hand - because he is really going to **** himself when he reads your words. Haven't you ever seen astronauts floating? Yes.... but they are constantly changing direction.... and therefor they should be aware of the effects of acceleration. I must admit this subtlety has perplexed me - clearly the don't feel the G force, since it the same as a car in a tight turn. But I keep thinking it should be detectable, if only because the path is curving. Yeah. It's called acceleration - a bit like G-force. Please note that "a bit" = "exactly" in European understatement. He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? http://science.howstuffworks.com/weightlessness1.htm That is a very simplistic explanation. It refers to the fact that the astronauts will feel the acceleration at take-off, and yet it doesn't seem to understand that a change of direction is also acceleration. We humans can detect acceleration. If you sit in an automobile with your eyes closed, then you can feel an increase or decrease in speed .... or a change of direction!! As the astronauts are subjected to a constant change of direction, I suspect that they might not feel that they are completely free-floating. Of course, from a General Relativity, Gravity Well point of view, the obital path is a straight line in curved space. I should know the answer here - let me cogitate ... No, you don't know the answer. Trust me. Despite my dreadful education, I am confident that someone who doesn't understand the basic principles of acceleration is incapable of getting their head around the TGR. Cogitation would be a complete waste of your time. What does your physics friend say about this? He would probably deplore the lack of education in your country. Ask him anyway! Actually, it was Scout's friend. However, you should remember I majored in physics and worked for NASA doing spacecraft navigation. I may be rusty now, but 25 years ago I really knew this stuf! GULP!!! So why do you not seem to understand the difference between "velocity" and "speed"? Perhaps, if you allowed him to read the thread, he might be amazed at your lack of reading ability. After all, I've already explained that I gave up Physics at an early stage. I haven't forgotton that. It was just a little dig since usually you Brits complain about our sorry education. Little digs are very welcome. I appreciate a good insult - and calling me "you Brits" is definitelay a reasonable insult. You're not lagging too far behind Joe! He called me a Brit about six months ago. Regards Donal -- |
"Donal" wrote in message
... "Jeff Morris" wrote in message ... "Donal" wrote in message ... In fact, I think that your use of the word "float" reveals that you don't understand the situation at all. Your astronaut wouldn't feel any difference between a free fall orbit and a headlong race into deepest space, - would he? So tell us, what is the difference? Acceleration. No, the acceleration is the same, more or less. (Not counting the difference in distance from the Earth, or air resistance, etc.) The only real difference is that the astronaut has enough velocity (hopefully) to miss the Earth as he falls. That was a very revealing answer. Earlier in the thread you were confident enough about your position to question my lack of education. Now you seem to think that an object travelling at "x" miles an hour in a straight line ("headlong into space") has the same acceleration as a body travelling at "x" miles an hour in orbit. Actually, I was saying that any body under the influence of Earth's gravity will "feel" the same force, more or less. It doesn't matter whether its falling off a cliff, in orbit, or leaving orbit into space. The pull of the Earth is the same, adjusting of course, for the distance. It turns out that most non-technically inclined people think that these are three completely different situations (as they are, in some respects) but in all three the body is being accelerated by gravity exactly the same. Sorry Donal, that's physics. If you had taken a physics course, you might understand that. Your continued rant is making you look rather silly. It's probably time that you consulted your physics partner. Before you let him read what you have written, you should make sure that there is a cloakroom near the PC. Otherwise, have a potty close at hand - because he is really going to **** himself when he reads your words. Actually, my partner was given a PhD in Astrophysics only if he promised to continue as a programmer. How about my former boss, he won the Nobel Prize in Physics a few years ago. Haven't you ever seen astronauts floating? Yes.... but they are constantly changing direction.... and therefor they should be aware of the effects of acceleration. I must admit this subtlety has perplexed me - clearly the don't feel the G force, since it the same as a car in a tight turn. But I keep thinking it should be detectable, if only because the path is curving. Yeah. It's called acceleration - a bit like G-force. Please note that "a bit" = "exactly" in European understatement. He must feel a constant force as his direction of travel changes. I wonder if this has been documented on the Internet? http://science.howstuffworks.com/weightlessness1.htm That is a very simplistic explanation. It refers to the fact that the astronauts will feel the acceleration at take-off, and yet it doesn't seem to understand that a change of direction is also acceleration. We humans can detect acceleration. If you sit in an automobile with your eyes closed, then you can feel an increase or decrease in speed .... or a change of direction!! As the astronauts are subjected to a constant change of direction, I suspect that they might not feel that they are completely free-floating. Of course, from a General Relativity, Gravity Well point of view, the obital path is a straight line in curved space. I should know the answer here - let me cogitate ... No, you don't know the answer. Trust me. Despite my dreadful education, I am confident that someone who doesn't understand the basic principles of acceleration is incapable of getting their head around the TGR. Cogitation would be a complete waste of your time. OK, why don't you explain to us? How does someone in orbit determine they are travelling a curved path. No fair using outside references. No fair looking for effects over time. What sort of device will instantaneously tell you the local gravitational field? I'm not saying its impossible, but its not obvious. What does your physics friend say about this? He would probably deplore the lack of education in your country. Ask him anyway! Actually, it was Scout's friend. However, you should remember I majored in physics and worked for NASA doing spacecraft navigation. I may be rusty now, but 25 years ago I really knew this stuf! GULP!!! So why do you not seem to understand the difference between "velocity" and "speed"? Where did I talk about that??? You're losing it Donal. So, please explain to all of us: what is the difference between velocity and speed, and how was that relevent to anything I said? Perhaps, if you allowed him to read the thread, he might be amazed at your lack of reading ability. After all, I've already explained that I gave up Physics at an early stage. I haven't forgotton that. It was just a little dig since usually you Brits complain about our sorry education. Little digs are very welcome. I appreciate a good insult - and calling me "you Brits" is definitelay a reasonable insult. You're not lagging too far behind Joe! He called me a Brit about six months ago. Oh, sorry, I forgot. That explains your lack of education. |
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