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Thread: Tidal waves (again)

  1. #1

    Tidal waves (again)

    http://oceanlink.island.net/oinfo/tides/tides.html

    This explanation differs from the one on this site. Is this one wrong? Or are both correct?

  2. #2
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    This thought experiment is flawed:



    That's a centrifugal force. With gravity, the Earth should not experience a centrifugal force. What the bulging of the sea level on the other side than the moon is, is that the moon is attracting the earth more than it's attracting the water on the other side of the earth. So, the Earth displaces a bit, but the water stays where it is.

  3. #3
    Thinks about this... the moon and the earth revolve around a common point. There is a sentripal force working on the earth from the moon, pointing towards that point. That force is the moons gravity.

    That will create a sentrifugal force that is equal to the sentrifugal force...

    Where am I wrong?

  4. #4
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    To be honest, I am not sure, but I believe my explanation and the one on that page (the "usual" explanation) are in fact the same thing. Theirs uses a non-rotating frame of reference, as if you are outside the Earth-Moon system looking in, while I use a frame the rotates at the same rate the Moon goes around the Earth. The math should work out to be the same, but to be frank I have never done it. The math is pretty hard.

  5. #5
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    The BA's explanation is right.

    The oceanlink explanation has the Moons gravitational field only affecting the side of the Earth facing the moon while attributing the bulge on Earth's opposite side to centrifugal force.

    Gravitational "pull" from the moon (or any mass) is proportional to the square of the distance from the mass. The Earth is some 13000 km in diameter. That means the side of Earth facing the moon is roughly 3 to 3.5 percent closer to the moon than the side of Earth not facing the moon.

    This "streches" the Earth along a line passing through the center of the Earth and the center of the moon. Same thing happens to the moon. And the Earth & moon don't need to be rotating around each other for this to happen: they just need to be in the same vicinity.

    The oceanlink explanation seems to me to say the Earth is somehow stuck into the fabric of space with a stick and the moon acts like a magnet that deforms the Earth in the direction of the magnet (moon) around the stick. Nope. (Not trying to be condescending: just the way I read that explanation...)

    The Earth is free to move, and does indeed orbit around the center of mass between the Earth and the moon. And while the Earth/moon pair orbits this center of mass, they're both "elongated" a tiny bit by the difference in gravitational force present across thier relatively large diameters.

    Doug.

    PS: Still trying to understand this myself. For me, it's not intuitive at all...

  6. #6
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    OOPS! Just read the BA's response. Now I don't really know what to think... 8-[

    What I was trying to imagine is if the Earth & moon were not in orbit around each other; simply moving towards each other.

    I'd'a thought you'd still get buldges (tides)...

    Doug.

  7. #7
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    "OOPS! Just read the BA's response. Now I don't really know what to think...

    What I was trying to imagine is if the Earth & moon were not in orbit around each other; simply moving towards each other.

    I'd'a thought you'd still get buldges (tides)... "

    I think you would still get the bulges. Though I think you need the rotation of Earth to get the two bulges- that means that Moon shouldn't have two bulges, just one- if I'm thinking of this correctly.

    What would happen if Moon had a polar orbit?

  8. #8
    Ok, think about it this way. While revolving the only force you can "feel" is the sentrifugal force. So we only need to consider that one.

    The sentrifugal force will be strongest at the point farthest from the moon, weaker at the center of the earth, and weakest at the point closest to the moon.

    And because the sentrifugal force is equal to the moons gravitational pull this explanation is the same as the other one.

    Am I right?

    EDIT: But the drawings at the site I gave must be wrong then.

  9. #9
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    Quote Originally Posted by gbaikie
    I think you would still get the bulges. Though I think you need the rotation of Earth to get the two bulges- that means that Moon shouldn't have two bulges, just one- if I'm thinking of this correctly.
    You're not. The far bulge occurs because the Moon attracts that side of the Earth less than it does the core of the Earth. Rotation is irrelevant.

    What would happen if Moon had a polar orbit?
    The tidal bulges would go around the Earth south-north-south-north instead of around the equator. There would also be some interesting resonance effects with the solar element of the tides.
    Everything I need to know I learned through Googling.

  10. #10
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    I like the NOAA explanation...

    Turns out it's both the centrifugal force of the Earth orbiting the Earth/moon barycenter and Newton's Universal Gravity Law (which I badly mangled earlier...).

    There seems to be an error in Figure 1's caption:

    This revolution is responsible for a centrifugal force component (Fg) necessary to the production of the tides.

    should be:

    This revolution is responsible for a centrifugal force component (Fc) necessary to the production of the tides.

    hth, Grey...

    Doug.

  11. #11
    From the url DALeffler posted:

    "It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces."

    It seems that in this explanation the centrifugal force is just added to the gravitational force, even though it really makes no difference, because the difference between the force on oposite sides of the earth are neglible.

    The gravitational force drops by a factor of r^2, while the centrifugal force drops with a factor of r. That's why the latter can be neglected.

    Please correct me if I'm wrong.

  12. #12
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    Here's my take on what the NOAA/NOS site is saying:

    Start with a perect, rigid sphere of a planet covered with water that is not rotating and is not gravitationally bound to any other object. The water depth anywhere on the planet should be the same depth as any other point on the plant.

    Now spin the planet around any axis (again, with no other gravitationally bound object near it).

    My first assumption would be the centrifugal force of the planets own rotation would cause the water level to rise at the equator and fall at the poles, but would not cause tides: tides are a repeating change in water level at a given lat/lng from one time period to the next time period.

    Once our perfect rigid sphere attained a constant rotation rate, the depth of the water covering it would be different at different latitudes - being deeper at the equator and shallower at the poles - but would not have depth changes according to any period of the planets rotation.

    Now I know the coreolis effect has to come into play here somewhere. I don't have a clue as to how much that'd cause water levels to rise or fall, but I'm pretty sure it would. I'd also think coriolis would be another constant: once the planet was up and spinning at speed, you could still judge latitude soley by water depth.

    Throw in a massive satelite gravitationally bound to our perfect sphere.

    Now we have to account for the centrifugal force arising from both spheres orbiting a common center of gravity (the barycenter).

    That centrifugal force, from orbiting the barycenter - not the centrifugal force from the planets' rotation around it's own axis (which doesn't change water levels) - gives rise to 1/2 the force that causes "tides" (changing water levels) at points on our primary sphere according to the period of rotation around the barycenter (I think...).

    The other half of the force that causes tides is the gravitational attraction exerted by the massive satellite. That force has to be taken into account too.

    It's only when all the forces are added together it seems one force is responsible for tides on half the Earth and other forces are responsible for tides on the other half of the Earth, when in reality, it's BOTH forces, at the same time, that account for tides.

    Doug.

  13. #13
    I agree with everything you said except the last. Look at the site again. It says that the centrifugal force is the same everywhere on earth, and also that it doesn't have anything to do with the tides.

  14. #14
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    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.

    "It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces. "

    They're talking about two cetrifugal forces that combine: centrifugal force from the Earth's own rotation around it's axis (constant sea height @ lattitude); and centrifugal force from the Earth being in orbit with the moon (changing sea height according to lat/lng/moon orbit period)...

    The first is produced simply by the Earth spinning, the second is produced by the Eatrh & moon orbiting each other.

    The key here for me is "differential tides". You wouldn't have "tides" with an Earth and no moon. You would have different sea levels without the moon, but not "changing" sea levels - "tides"...

    Doug...

  15. #15
    In order for a tide to be created you need two things.

    1. A force acting differently on the near and the far side of the earth. Thereby stretching the earth/water.

    2. The force needs to change over time in order for us to notice a difference.

    As the centrifugal force doesn't fullfill the first requirment it doesn't create a tide.

    You don't need to look at the forces together, they can be looked at seperatly. And we know that the centrifugal component doesn't create any tides, so we can disregard it.

  16. #16
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    So, could you say that the bulges are a measurement/effect of the relative motions of Sun, Moon, and Earth. The spin of Earth is not needed to cause the bulges, but of course the Earth's spin complicates tides- giving two highs and lows a day. But without the Earth's spin, the bulges would be same magnitude, but would change bi-monthly. But tides would would not be of same magnitude, the non-smoothness of earth surface with the Earth's spin creates differences in tides.
    Now, what would happen if the Earth's Mass was say 1/2 of what it is now, but everything else was the same [including Earth's diameter, the Moon's orbital distance from Earth and Earth/Moon distance from Sun. The barycenter would further from the Earth center. The Moon would orbit slower. And basically the Earth would go around the Sun in around the same speed. I would guess that the bulges would be the same size. But I'm not sure. By reducing Earth's Mass you doing a similar thing as increasing the Moon's mass- moving the barycenter further way from the Earth.

  17. #17
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    From Grey Wanderer:
    In order for a tide to be created you need two things.

    1. A force acting differently on the near and the far side of the earth. Thereby stretching the earth/water.

    2. The force needs to change over time in order for us to notice a difference.

    As the centrifugal force doesn't fullfill the first requirment it doesn't create a tide.

    You don't need to look at the forces together, they can be looked at seperatly. And we know that the centrifugal component doesn't create any tides, so we can disregard it.
    The equation for centrifugal force is Fc = m*v^2/r, while the force due to gravity is Fg = GMm/r^2. Both of these forces depend on r, so both of these forces affect the near side, center, and far side of the earth differently. Because centrifugal force does this, it does fulfill your first requirement and help to create the tides. In all truth, we only have two options: both forces together cause the tide or either one will because they are both mathematically equivalent.

    Dax

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    From gbaikie:
    So, could you say that the bulges are a measurement/effect of the relative motions of Sun, Moon, and Earth. The spin of Earth is not needed to cause the bulges, but of course the Earth's spin complicates tides- giving two highs and lows a day. But without the Earth's spin, the bulges would be same magnitude, but would change bi-monthly. But tides would would not be of same magnitude, the non-smoothness of earth surface with the Earth's spin creates differences in tides.
    The earth's spin gives us the tides. If the earth didn't spin with respect to the moon, the tides would be much smaller, as they would only be due to the sun. Here, a high-tide to high-tide period would be twice a month since that's the length of a day . I don't completely understand what you mean by "differences in tides," but the smoothness (or not) of the earth's surface doesn't affect the height of the tides or the diurnal cycle. These come from the moon-sun-earth positions and from the earth's daily rotation, respectively.

    Now, what would happen if the Earth's Mass was say 1/2 of what it is now, but everything else was the same [including Earth's diameter, the Moon's orbital distance from Earth and Earth/Moon distance from Sun. The barycenter would further from the Earth center. The Moon would orbit slower. And basically the Earth would go around the Sun in around the same speed. I would guess that the bulges would be the same size. But I'm not sure. By reducing Earth's Mass you doing a similar thing as increasing the Moon's mass- moving the barycenter further way from the Earth.
    A back-of-the-envelope calculation (okay, I did it on a napkin) at midnight [meaning take it with a grain of salt] says that the height would decrease to half the current value in this case. This is consistent with a decreasing centrifugal force that would also be, although the decrease in each force may be different.[/quote]

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    Quote Originally Posted by DALeffler
    I like the NOAA explanation...
    I have some problems with the NOAA site, I've discussed it before on usenet and on the SDMB.

    The site in the OP, of this thread, is wrong. The BA's webpage is right. The NOAA site was horribly wrong, but they tried to fix it, but didn't spend enough time at it. At least, that's the case when I last combed through it in February (see the SDMB link for more comments).
    Quote Originally Posted by Tobin Dax
    The earth's spin gives us the tides.
    I disagree. The differing tides result from that fact that the moon is over different parts of the Earth, and the moon's gravity stretches out the Earth, as the BA's page says.

    Of course, one of the reasons that the moon appears over different parts of the Earth is because the Earth rotates once each day--but even if the Earth did not rotate, and the moon went around the Earth, we'd still experience tides. So, in that sense, the spin has nothing to do with it.

  20. #20
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    Maybe I should have been more specific, kilopi. The rotation of Earth with respect to another object gives us tides. That is what I was trying to say. This also creates a subtle point on which we may disagree: When Earth and the moon do become tidally locked, there will not be tides from the moon as we have defined them. Yet, in my opinion anyway, the moon still revolves around the earth in this case.

    In summary, spin has everything to do with. Although, it might also be important to note that I'm defining tides as the differing tides, not as the tidal bulges themselves (which do have nothing to do with spin).
    [/quote]

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    Quote Originally Posted by Tobin Dax
    This also creates a subtle point on which we may disagree: When Earth and the moon do become tidally locked, there will not be tides from the moon as we have defined them. Yet, in my opinion anyway, the moon still revolves around the earth in this case.
    No, I wouldn't necessarily disagree, but the definition of "tide" may be different, depending upon context.
    In summary, spin has everything to do with.
    Not spin, but that the moon is over varying parts of the Earth. As you point out, the Earth could still spin, but if the moon were geosynchronous, then there would not be any tides as you've defined them.
    Although, it might also be important to note that I'm defining tides as the differing tides, not as the tidal bulges themselves (which do have nothing to do with spin).
    We do agree then that the tidal bulges have nothing to do with spin.

    Back to the NOAA site. Here's a quote from the intro page:
    The Astronomical Tide-Producing Forces: General Considerations" my reading of it seems to be that the two tidal bulges are due to the moon and Sun, one each: "However, the gravitational forces of the moon and sun also act externally upon the earth's ocean waters. These external forces are exerted as tide-producing, or so-called 'tractive' forces. Their effects are superimposed upon the earth's gravitational force and act to draw the ocean waters to positions on the earth's surface directly beneath these respective celestial bodies (i.e., towards the 'sublunar' and 'subsolar' points). High tides are produced in the ocean waters by the 'heaping' action resulting from the horizontal flow of water toward two regions of the earth representing positions of maximum attraction of combined lunar and solar gravitational forces. Low tides are created by a compensating maximum withdrawal of water from regions around the earth midway between these two humps.
    That sure seems to me like they're claiming that there are only two bulges, one beneath the Sun and the other beneath the moon.

  22. #22
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    Quote Originally Posted by Tobin Dax
    Maybe I should have been more specific, kilopi. The rotation of Earth with respect to another object gives us tides. That is what I was trying to say. This also creates a subtle point on which we may disagree: When Earth and the moon do become tidally locked, there will not be tides from the moon as we have defined them. Yet, in my opinion anyway, the moon still revolves around the earth in this case.
    Well, you'd have a permanent high tide beneath and opposite the Moon, and a permanent low tide at the 90-degree points. They just wouldn't change other than for solar effects.
    Everything I need to know I learned through Googling.

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    Quote Originally Posted by ToSeek
    Well, you'd have a permanent high tide beneath and opposite the Moon, and a permanent low tide at the 90-degree points. They just wouldn't change other than for solar effects.
    In Tobin Dax's concept of tides, the "permanent tide" would not be considered a tide. Geophysists calculate and study the permanent tide, though.

  24. #24
    Quote Originally Posted by Tobin Dax
    The equation for centrifugal force is Fc = m*v^2/r, while the force due to gravity is Fg = GMm/r^2. Both of these forces depend on r, so both of these forces affect the near side, center, and far side of the earth differently. Because centrifugal force does this, it does fulfill your first requirement and help to create the tides. In all truth, we only have two options: both forces together cause the tide or either one will because they are both mathematically equivalent.
    With Fc you divide by r, and with Fg you divide by r^2. That should make a pretty big difference, shouldn't it? Enough to make us ignore the effect of the centrifugal force.

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    Quote Originally Posted by GreyWanderer
    With Fc you divide by r, and with Fg you divide by r^2. That should make a pretty big difference, shouldn't it? Enough to make us ignore the effect of the centrifugal force.
    Perhaps the other way around. Dividing by radius squared is going to make the quantity smaller.

    But, still, there is no effect of the centrifugal force. It's zero, basically, so it is safe to ignore.

  26. #26
    Quote Originally Posted by kilopi
    Quote Originally Posted by GreyWanderer
    With Fc you divide by r, and with Fg you divide by r^2. That should make a pretty big difference, shouldn't it? Enough to make us ignore the effect of the centrifugal force.
    Perhaps the other way around. Dividing by radius squared is going to make the quantity smaller.

    But, still, there is no effect of the centrifugal force. It's zero, basically, so it is safe to ignore.
    The quantity doesn't matter. Only the difference.

    Fg and Fc are equally strong at the center of the earth. But Fg differs more between the far side and the near side than Fc.

  27. #27
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    Quote Originally Posted by GreyWanderer
    The quantity doesn't matter. Only the difference.

    Fg and Fc are equally strong at the center of the earth. But Fg differs more between the far side and the near side than Fc.
    The difference in Fc between the far side and the near side is zero.

  28. #28
    Quote Originally Posted by kilopi
    Quote Originally Posted by GreyWanderer
    The quantity doesn't matter. Only the difference.

    Fg and Fc are equally strong at the center of the earth. But Fg differs more between the far side and the near side than Fc.
    The difference in Fc between the far side and the near side is zero.
    No, it isn't. Don't you agree that the centripetal force is stronger the farter you get away from the center of rotation? As long as you're rotating with the same rate.

  29. #29
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    Quote Originally Posted by GreyWanderer
    No, it isn't.
    It is, and that NOAA webpage points that out--of course, it disagrees with itself too, but that's beside the point.
    Don't you agree that the centripetal force is stronger the farter you get away from the center of rotation? As long as you're rotating with the same rate.
    The centripetal force is gravity, and it gets weaker the farther that you get from the center.

    I imagine you meant "centrifugual" (which I like to call centrifictional) anyway, but it's still constant across the Earth and moon. That's because you have to ignore the effect of the Earth's rotation in calculating it. Once you back that out, it's easy to show that each point experiences the same centrifictional force--and in the same direction. That's because each point, on a non-rotating Earth, follows a circle around a center that differs from the other points--but the radius of each circle is the same.

  30. #30
    Quote Originally Posted by kilopi
    That's because each point, on a non-rotating Earth, follows a circle around a center that differs from the other points--but the radius of each circle is the same.
    Thanks, I understand now.

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