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Thread: Tides and the Moon

  1. #1
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    I understand how there are really high/low tides during a New Moon (because the sun and moon are both pulling on the Earth in the same direction), but why are there really high tides during a Full Moon? The sun and moon would be on different sides of the Earth, so I don't understand why there would be high tides. I would think there would be less drastic tides when the moon is full...

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    The moon makes high tides on the side of the earth closest to it, as well as on the side farthest away. See the BA page for more details.

    So, new moon or full moon, it doesn't matter. It's when the moon is first or last quarter that the sun's tidal effect cancels the moon's tidal effect, and the tides are not so extreme.

    Oops, didn't see the other thread on this same subject, same title.

    <font size=-1>[Added cross reference]</font>

    <font size=-1>[ This Message was edited by: GrapesOfWrath on 2002-01-15 18:35 ]</font>

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    the moon does not pull at the tides!

    because then the moon would be exactly above one of the tides (who knows what the other high tide is doing). in fact, i think the moon is directly above low tide. so the tides are sticking out like ears from the bit of the earth thats facing the moon.

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    While the source of tides - Sun and Moon - are clear, the mechanism is difficult to grasp. The BA himself (who does understand) has a bash at explaining here on this site: http://www.badastronomy.com/bad/misc/tides.html

    I find this unconvincing, but my own concept was shot down comprehensively, failing to explain the tidal effects on a body falling directly onto another, rather than being in orbit. So I must try to understand the Standard Model.

    John

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    Re: the moon does not pull at the tides!

    Quote Originally Posted by dutche
    because then the moon would be exactly above one of the tides (who knows what the other high tide is doing). in fact, i think the moon is directly above low tide. so the tides are sticking out like ears from the bit of the earth thats facing the moon.
    Why the three year thread necromancy (there should be some sort of law on this) when your thread is much newer? (Last post back in December.) Although that 24 to 1 poll result kinda is a damper.

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    Quote Originally Posted by JohnD
    I find this unconvincing, but my own concept was shot down comprehensively, failing to explain the tidal effects on a body falling directly onto another, rather than being in orbit. So I must try to understand the Standard Model.
    If your concept, JohnD, is the Different Orbits theory, then it wasn't shot down at all. I think it's a valid - and intuitively easier to grasp - explanation. But the math is a lot easier in the Standard (Differential Gravity) Model.

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    Quote Originally Posted by Eroica
    Quote Originally Posted by JohnD
    I find this unconvincing, but my own concept was shot down comprehensively, failing to explain the tidal effects on a body falling directly onto another, rather than being in orbit. So I must try to understand the Standard Model.
    If your concept, JohnD, is the Different Orbits theory, then it wasn't shot down at all. I think it's a valid - and intuitively easier to grasp - explanation. But the math is a lot easier in the Standard (Differential Gravity) Model.
    JohnD's point was that it couldn't be extended to other cases, so it was weaker in that sense.

    Worse, the intuitive notions are often extrapolated to conclusions that are just not true. So, it may be easier to grasp intuitively, but it also causes intuitive mistakes.

    Myself, I think "differential gravitation" is extremely intuitive--gravity is a force that varies as 1/r^2, and there is a stretching force associated with the changes as r increases--its derivative, in other words, also known as the tide-generating force. It is proportional to 1/r^3--the derivative of 1/r^2.

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    Gosh, thanks!
    But I agree - I would rather understand a concept that can survive extreme conditions, than one that failed at the extremes, for the reason you mention, ATP.
    JOhn

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    Quote Originally Posted by JohnD
    Gosh, thanks!
    I should have said "I think..."

    I've been involved in this discussion before, though. I don't really have a problem with the "different orbits" explanation, or the "centrifugal force" explanation, but I've noticed that the other methods always seem to result in wrong results, unless one makes a lot of careful adjustments--which makes them less intuitive, and less simple.

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    Quote Originally Posted by A Thousand Pardons
    JohnD's point was that it couldn't be extended to other cases, so it was weaker in that sense.
    Could you give an example where it doesn't work? Certainly it has no trouble explaining the tidal bulges that occur when a body is falling into the Sun (rather than orbiting it). In fact, it is the Differential Gravity theory that fails in extremis:

    http://www.badastronomy.com/phpBB/vi...83&amp;#156483


    Worse, the intuitive notions are often extrapolated to conclusions that are just not true. So, it may be easier to grasp intuitively, but it also causes intuitive mistakes.
    Example?

    Myself, I think "differential gravitation" is extremely intuitive--gravity is a force that varies as 1/r^2, and there is a stretching force associated with the changes as r increases--its derivative, in other words, also known as the tide-generating force. It is proportional to 1/r^3--the derivative of 1/r^2.
    The maths is intuitive, I'll grant you that. But when you start talking about gravity generating "a negative force" (Bad Astronomy, p68), you can hardly claim to be dealing with intuitive ideas any more.

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    So what about Spring and Neap tides? :-? ops:

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    Quote Originally Posted by Sticks
    So what about Spring and Neap tides? :-? ops:
    What about them? Spring tides are tides of an exceptionally large range because the Sun and Moon are working together. They're called "spring" because the tide springs in and out (nothing to do with the season).

    Neap tides (neap = "nipped") are tides of unusually short range because the Sun and Moon are counteracting each other. The Sun is trying to produce a high tide while the Moon is trying to produce a low tide, and vice versa.

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    Quote Originally Posted by Eroica
    Quote Originally Posted by A Thousand Pardons
    JohnD's point was that it couldn't be extended to other cases, so it was weaker in that sense.
    Could you give an example where it doesn't work? Certainly it has no trouble explaining the tidal bulges that occur when a body is falling into the Sun (rather than orbiting it). In fact, it is the Differential Gravity theory that fails in extremis:
    If done correctly, and with attention to details, all points of view result in correct results, as I've said many times before. JohnD may have been commenting on his own personal intuitive use of the model.
    Worse, the intuitive notions are often extrapolated to conclusions that are just not true. So, it may be easier to grasp intuitively, but it also causes intuitive mistakes.
    Example?
    Here's my post from Oct 2003, in answer to one of your posts. It links {"Errors tend to creep in") to a discussion of just those sort of errors.
    Myself, I think "differential gravitation" is extremely intuitive--gravity is a force that varies as 1/r^2, and there is a stretching force associated with the changes as r increases--its derivative, in other words, also known as the tide-generating force. It is proportional to 1/r^3--the derivative of 1/r^2.
    The maths is intuitive, I'll grant you that. But when you start talking about gravity generating "a negative force" (Bad Astronomy, p68), you can hardly claim to be dealing with intuitive ideas any more.
    If any sort of math is intuitive, we're in pretty good shape.

    I'm not sure what the problem is with a negative force. If you have force in one direction, then a negative force would just be a force in the opposite direction. Force has a direction, as well as a magnitude. That seems intuitive enough, to me.

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    Quote Originally Posted by A Thousand Pardons
    I'm not sure what the problem is with a negative force. If you have force in one direction, then a negative force would just be a force in the opposite direction. Force has a direction, as well as a magnitude. That seems intuitive enough, to me.
    The problem is not that the force is negative. The problem is that it is generated by gravity, which is always attractive. How can gravity per se ever cause something to move away from the force generating the gravity?

    The water in the tidal bulge on the far side of the Earth is climbing away from the Moon in spite of - not because of - the Moon's gravity. That's how I see it now.

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    Quote Originally Posted by Eroica
    Quote Originally Posted by A Thousand Pardons
    I'm not sure what the problem is with a negative force. If you have force in one direction, then a negative force would just be a force in the opposite direction. Force has a direction, as well as a magnitude. That seems intuitive enough, to me.
    The problem is not that the force is negative. The problem is that it is generated by gravity, which is always attractive. How can gravity per se ever cause something to move away from the force generating the gravity?
    The water in the tidal bulge on the far side of the Earth is climbing away from the Moon in spite of - not because of - the Moon's gravity. That's how I see it now.
    If I know what you're talking about without reading the entire linked thread (certainly not guaranteed), it's not that the force is actually negative as such; it's that the net acceleration, relative to the net acceleration on another point, is the difference between the acceleration between point A and point B. Thus, depending which way you consider "up" (or "positive", if you want to be more precise about it), it can be negative, if the subtrahend (subtracted acceleration) is greater than the minuend (local acceleration).

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    Quote Originally Posted by Eroica
    Quote Originally Posted by A Thousand Pardons
    I'm not sure what the problem is with a negative force. If you have force in one direction, then a negative force would just be a force in the opposite direction. Force has a direction, as well as a magnitude. That seems intuitive enough, to me.
    The problem is not that the force is negative. The problem is that it is generated by gravity, which is always attractive. How can gravity per se ever cause something to move away from the force generating the gravity?
    Just looking at the situation from the standpoint of Newtonian mechanics, the earth would tend to move in a straight line--away, on a tangent, from the Sun. The Sun's gravity pulls it back to orbit. So, in that sense, the farside tidal bulge is closer to the Sun than it would have been were the Sun's gravity not acting upon it.
    The water in the tidal bulge on the far side of the Earth is climbing away from the Moon in spite of - not because of - the Moon's gravity. That's how I see it now.
    The tidal bulge due to the moon acts similarly, but on a different scale.

  17. #17
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    gravity tides

    i asked "would the moon point at a high tide" proving the netonian theory of tides or "i think the moon is directly above low tide. so the tides are sticking out like ears from the bit of the earth thats facing the moon."

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    Re: gravity tides

    Quote Originally Posted by dutche
    i asked "would the moon point at a high tide" proving the netonian theory of tides or "i think the moon is directly above low tide. so the tides are sticking out like ears from the bit of the earth thats facing the moon."
    The actual tide at a specific place on the globe is not just a function of the moon's position. The shape of the coast, and the depth of the ocean, and prevailing winds, and weather, all have an effect. It can completely change the resonance of the tide, at a particular place--so the tide at that place should not be generalized to a global theory. Tidal theories take tidal measurements all over the world into account.

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    Quote Originally Posted by Eroica
    The water in the tidal bulge on the far side of the Earth is climbing away from the Moon in spite of - not because of - the Moon's gravity. That's how I see it now.
    The water is thrown away from the Earth by centrifugal force, (I know, it doesn't exist, but it works like it does.) creating the tidal buldge on the side of the Earth opposite that of the Moon. Spring and neap tides involve the Sun. When the Sun, Earth, and Moon line up, then you get a greater tide. When they form a right angle with the Earth at the corner, then the tide is less pronounced. I'll let someone else tell me which is which; I've forgotten.

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    Quote Originally Posted by Maddad
    Quote Originally Posted by Eroica
    The water in the tidal bulge on the far side of the Earth is climbing away from the Moon in spite of - not because of - the Moon's gravity. That's how I see it now.
    The water is thrown away from the Earth by centrifugal force, (I know, it doesn't exist, but it works like it does.) creating the tidal buldge on the side of the Earth opposite that of the Moon. Spring and neap tides involve the Sun. When the Sun, Earth, and Moon line up, then you get a greater tide. When they form a right angle with the Earth at the corner, then the tide is less pronounced. I'll let someone else tell me which is which; I've forgotten.
    Eroica explained it in the post seven before yours.

    I guess that's a ToSeek.

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    I read this paper recently, so you might also find it interesting if it had not been shown here before: Myths about Gravity and Tides by Mikolaj Sawicki.

    There is one thing there that I am trying to understand, same as Eroica, that the far side bulge is in spite of, not because of. From the paper, bottom pg. 2 and top pg. 3, with illustration:

    "Consider the point C on Earth closest to the Sun and the point F on a far side of Earth. The Sun pulls harder on a unit of mass at the point C, not as hard on a unit of mass at Earth center O, and weaker yet on a unit of mass at point F. The acceleration a_s of Earth as a whole in free fall towards the Sun is determined by the gravitational pull of the Sun on Earth's center. Hence the unit of mass at C has a tendency to accelerate towards the Sun with acceleration a_s + (delta)a_s, i.e. more than the center of Earth, while a mass at the far side F has a tendency to accelerate towards the Sun with acceleration a_s - (delta)a_s, i.e. to lag behind the center of Earth."

    This reminds me of what I learned long ago in high school physics that it is okay to add and subtract the gravitational force this way, so it explains why there is a bulge on the far side, instead of a "dent". So it is "in spite of" rather than "because of" to logical thinking. But that is not what actually happens! So you have to add and subtract the Sun's gravitational force from the Earth's center, which seems to me to still be illogical, but that is how it is explained. :-? I'm not totally convinced however and would still get a "D" on that exam just like before!

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    Quote Originally Posted by nutant gene 71
    This reminds me of what I learned long ago in high school physics that it is okay to add and subtract the gravitational force this way, so it explains why there is a bulge on the far side, instead of a "dent". So it is "in spite of" rather than "because of" to logical thinking. But that is not what actually happens! So you have to add and subtract the Sun's gravitational force from the Earth's center, which seems to me to still be illogical, but that is how it is explained. :-? I'm not totally convinced however and would still get a "D" on that exam just like before!
    I don't know if this means anything to you or not, but keep in mind, gravitational acceleration (from the Newtonian point of view, without getting relativistic) is a vector quantity, not a scalar. Even though we often refer to gravity only in terms of its magnitude, it does have a direction (what we usually call "down", pardon the technical jargon :wink. But when you add vectors with opposite directions, the magnitude of the vector sum is the same as the difference between the two magnitudes.
    I'm not sure if this will mean anything to you or not, but I'm sure it will to others here. Perhaps someone else could grab the ball from here, and explain vector addition in terms a layperson can understand? Or is that a complete contradiction in terms?

  23. #23
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    Quote Originally Posted by JohnOwens
    Quote Originally Posted by nutant gene 71
    This reminds me of what I learned long ago in high school physics that it is okay to add and subtract the gravitational force this way, so it explains why there is a bulge on the far side, instead of a "dent". So it is "in spite of" rather than "because of" to logical thinking. But that is not what actually happens! So you have to add and subtract the Sun's gravitational force from the Earth's center, which seems to me to still be illogical, but that is how it is explained. :-? I'm not totally convinced however and would still get a "D" on that exam just like before!
    I don't know if this means anything to you or not, but keep in mind, gravitational acceleration (from the Newtonian point of view, without getting relativistic) is a vector quantity, not a scalar. Even though we often refer to gravity only in terms of its magnitude, it does have a direction (what we usually call "down", pardon the technical jargon :wink. But when you add vectors with opposite directions, the magnitude of the vector sum is the same as the difference between the two magnitudes.
    I'm not sure if this will mean anything to you or not, but I'm sure it will to others here. Perhaps someone else could grab the ball from here, and explain vector addition in terms a layperson can understand? Or is that a complete contradiction in terms?
    You said "But when you add vectors with opposite directions, the magnitude of the vector sum is the same as the difference between the two magnitudes." I could understand this if the vectors were away from the Earth, opposite centripetal force of gravity, but don't understand it if the force vectors are the other way, into the center of the globe. Maybe I should learn more on how vectors work? I thought the opposite tidal bulge was from shifting baryonic centers of the sun and moon... but was told this is not true. Many thanks anyway, for trying to explain it to a "hopeless student." I hope that's not a "contradiction in terms."

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    Quote Originally Posted by Maddad
    Spring and neap tides involve the Sun. When the Sun, Earth, and Moon line up, then you get a greater tide. When they form a right angle with the Earth at the corner, then the tide is less pronounced. I'll let someone else tell me which is which; I've forgotten.
    Spring tides are greater than normal and neap tides are less.

  25. #25
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    Quote Originally Posted by nutant gene 71
    Quote Originally Posted by JohnOwens
    I don't know if this means anything to you or not, but keep in mind, gravitational acceleration (from the Newtonian point of view, without getting relativistic) is a vector quantity, not a scalar. Even though we often refer to gravity only in terms of its magnitude, it does have a direction (what we usually call "down", pardon the technical jargon :wink. But when you add vectors with opposite directions, the magnitude of the vector sum is the same as the difference between the two magnitudes.
    I'm not sure if this will mean anything to you or not, but I'm sure it will to others here. Perhaps someone else could grab the ball from here, and explain vector addition in terms a layperson can understand? Or is that a complete contradiction in terms?
    You said "But when you add vectors with opposite directions, the magnitude of the vector sum is the same as the difference between the two magnitudes." I could understand this if the vectors were away from the Earth, opposite centripetal force of gravity, but don't understand it if the force vectors are the other way, into the center of the globe. Maybe I should learn more on how vectors work? I thought the opposite tidal bulge was from shifting baryonic centers of the sun and moon... but was told this is not true. Many thanks anyway, for trying to explain it to a "hopeless student." I hope that's not a "contradiction in terms."
    OK, you've got one gravity vector (perhaps I should have specified, the gravitational vector, not the tidal pseudo-vector) from the Sun, which always points towards the Sun of course, and then you've got the gravity vector from the Earth, which always points towards the Earth (or is magnitude 0 for Earth as a whole). On the side of the Earth facing the Sun, these vectors are in opposite directions, so the sum's magnitude is the difference of their individual magnitudes. For Earth as a whole, it's just the magnitude of the Sun's gravity's vector. But on the dark side of the Earth, the net magnitude is the sum of the individual magnitudes. On the other hand, at the terminator, it's a Pythagoran sum of the magnitudes: g_t = sqrt(g_S^s + g_E^2), pointing mostly Earthwards of course.
    So, using some arbitrary and fairly fictional numbers just for an example, assume Earth's surface acceleration were 10 m/s^2, and the acceleration from the Sun's gravity around Earth's orbital distance were 0.005 m/s^2 at Earth's center, 0.004 m/s^2 on the far side of Earth, and 0.006 m/s^2 on the near side of Earth. Then the subsolar point would be getting pulled towards the Earth/away from the Sun at 9.996 m/s^2, the Earth as a whole would be getting pulled towards the Sun at 0.005 m/s^2, and the dark side would be getting pulled towards the Earth (and towards the Sun) at 10.004 m/s^2. Now, if you subtract out the influence of Earth's gravity and the average acceleration of the Earth to get at just the tidal effects of the Sun, and the difference that causes tides, then you end up with the near side being accelerated away from the Earth's center/towards the Sun at 0.001 m/s^2, the Earth's center under no acceleration (of course), and the far side of the Earth under 0.001 m/s^2 acceleration away from both Earth and Sun.
    Hmm, I think I'd better stop for now, I'm starting to confuse myself, and forget what my point was. Maybe I'll post more after I re-read your post, but I've got stuff to do, so that won't be for a while....

  26. #26
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    Quote Originally Posted by JohnOwens
    ... I think I'd better stop for now, I'm starting to confuse myself, and forget what my point was. Maybe I'll post more after I re-read your post, but I've got stuff to do, so that won't be for a while....
    I think, I think I got it! Here's how it looks to me:

    Moon's (Sun's) gravitational force is ---->N--------->O-------> F

    where N is Earth's near side, and F is Earth's far side. Now,

    --->N ----->O -----&lt; F

    where N is near side of planet, O is center of planet, and F is far side. So what have we got? The Earth's gravity, which always pulls towards it center at O, is subracted from N, since Earth is pulling back on the Moon's gravitational force, but on the right at F the force is also subtracted, so you can subtract the vector force from F. But this can be true in only one case: that the Earth's mass is shielding the Moon's and Sun's gravitational force! Instead of adding the vector on the right, as I would do intuitively, not thinking the Earth is a gravity shield, it is more correct to subtract the gravitational force from the right, from F. Now it makes sense. But only if Earth is shielding the Moon's and Sun's gravity!

    So this is significant, but I suspect it would once again get me in trouble, since the Earth is not supposed to be "shielding" gravity, if I understand it correctly. This was debated at some length on the Allais Effect thread sometime ago. So now I got an "A" on the gravity vectors, but a "C-" for the reason why? :-?

    It took me long enough, but still not clear about that "shielding" stuff.

  27. #27
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    you're on the right track.

    Think about the force of the moon's gravity at three points on earth, the near side, the center, and the far side.

    It's big on the near side, medium in the middle, and weak on the far side.

    Now, we use the middle of the earth as the "stationary" center, to see how things move with respect to that (a common reference point).

    Subtract the force on the center from those of the near and far sides.

    This "weakens" the near side, but completely flips the "far side" to point the other way.

    This manipulation of the vectors basically shows how: The near side is pulled away from the center, but the center is pulled along too, catching up with the near side, and leaving behind the far side.

    I'm sleepy, so that last bit may not have made much sense.

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    Quote Originally Posted by Ricimer
    you're on the right track.
    Except for the shielding stuff

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    Quote Originally Posted by A Thousand Pardons
    Quote Originally Posted by Ricimer
    you're on the right track.
    Except for the shielding stuff
    The reason I said "shielding" was because I was thinkng of a "barycenter wobble" cause for far side tidal bulge. In my mind's eye, it's the old water bucket on a string trick, where Earth's mass is gravitationally tied to Moon and Sun. So oceanic fluid on the near side responds to direct gravity, the gravitational "tether", while the far side water responds ( if shielding effect is true) to barycenterwobble effect, because the direct pull from the Moon-Sun is canceled out. Hence, two bulges, but for two very different reasons, though both tied to gravity: (1) Near side=direct pull of gravity, and (2) far side=nixed pull of gravity (shielded) where only barycenter wobble pulls in opposite direction (of gravity tether), so both sides balance out. But now I am way over my head, so the tide is likely to engulf me if I go any deeper.

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