why are there no tides in a fish tank?

Discussion in 'Physics & Math' started by DrZygote214, Jul 18, 2014.

  1. DaveC426913 Valued Senior Member

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    Argh. I could not sleep until I solved this. But I'm confident I have the answer.

    It requires both gravitational gradient and rotation to produce the correct tides. If you look at the force vectors of each individually, neither produces the correct result, but the sum of the two vectors at each point balances perfectly.

    When I get in to work tomorrow, I will create and upload a diagram that will make it clear. If I do it tonight, I'll be asleep on my kb tomorrow. For now, it will have to live in dry-erase on the whiteboard in the kitchen. I'm only posting now (in my scivvies) to give my brain a passing chance at sleeping.
     
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  3. DrZygote214 Registered Member

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    Okay im back with some new (and old) stuff. I read everyone's posts but who knows if i interpreted it all correctly.

    First of all, remember this:

    for the Moon pulling an object on Earth:
    a = G*7.347e22 / 378,028,000^2 = 3.431 e-5 m/s/s // object on Earth's surface *closest* to Moon
    a = G*7.347e22 / 390,770,000^2 = 3.211 e-5 m/s/s // object on Earth's surface *farthest* from Moon

    For the Sun pulling an object on Earth:
    a = G*1.989e30 / 149,591,890,000^2 = 5.932051 e-3 m/s/s // object on Earth's surface *closest* to Sun
    a = G*1.989e30 / 149,604,632,000^2 = 5.931041 e-3 m/s/s // object on Earth's surface *farthest* from Sun

    This confused the hell out of me because clearly the Sun has a stronger pull almost by a factor of 100. But what I forgot to calculate is the DIFFERENTIAL as a percentage...

    For the Moon: 3.431/3.211 ~= 1.0685
    For the Sun: 5.932051/5.931041 ~= 1.0002

    So the Moon has a much greater differential, and tidal forces are defined as internal stresses an object experiences because of the difference of gravity at one side versus another. Therefore i can safely say that the Moon causes the tides, much more so than the Sun. So i have to disagree with Nomadd22 when he says...



    OKAY now about this dispute with russ_waters and bvillion. Let me first say that there are ***2*** centrifugal forces on Earth: one comes from Earth's rotation, the other comes from Earth's orbit about the Moon.

    In case you didn't know, the Earth orbits the Moon just as the Moon orbits the Earth. They orbit their common center of mass, known as the "Barycenter" (god only knows where that terminology came from, since Center of Mass is also a very common term; it confused the hell out of me when i was 8 years old, back when astronomy was my favorite subject).

    The Barycenter exists not at Earth's center, but more like 4,600 km from its center. In other words, Earth is rotating AND it's revolving around a point that almost exists outside its radius. Make a fist and shake it in a tight circle. That's what the Earth is doing.

    So there is centrifugal force from rotation and revolution. I don't believe the back-facing bulge is caused by centrifugal force FROM ROTATION...because that force exists constantly everywhere at the equator, yet the tidal bulges are on opposite sides and do not encircle all around the equator.

    But maybe the centrifugal force FROM REVOLUTION plays a role? Well, in this case, the centrifugal force would NOT have the same magnitude at opposite sides of Earth, because its not revolving around earth's geometric center. The moon-facing side would have a weak centrifugal force accelerating things away from Earth (and towards the Moon), while the moon-opposite side would have a stronger centrifugal force accelerating things away from Earth (and away from the Moon).

    If this were true, then the cumulative effects of this would make the near-Moon bulge SMALLER than the far-Moon bulge. Can anyone find charts to support this? If i recall correctly, the charts showing water level, which peaks 2x daily, have the same amplitude, which would mean this centrifugal force FROM REVOLUTION is too small to be noticeable.

    I'm still leaning to russ-water. The differential of the Moon's gravity causes nearer water to be pulled closer, the center of Earth (solid mass) to be pulled a little less closer, and the farther water to be pulled a lot less closer, hence the far-Moon bulge.

    Daring to get into the "what if" no rotation, no revolution, scaffold between Earth and Moon... There would still be a differential of the Moon's gravity across Earth, so I think the two bulges would still exist as far as the LAND masses of Earth are concerned. As for the water, maybe it would flow into one big bulge near the Moon, idk.



    Now that I've dipped my foot into the water (OIY!), here's something I just asked myself. Does the atmosphere also have tidal bulges?
     
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  5. DaveC426913 Valued Senior Member

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    Yes.
     
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  7. Russ_Watters Not a Trump supporter... Valued Senior Member

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    [Deleted]
     
  8. DaveC426913 Valued Senior Member

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    I demonstrate that the tidal force is a derived force, comprised of
    - the gravitational force from the other body and
    - the centrifugal force of revolution (i.e. inertia, if viewed from an inertial FoR) about the other.


    We separate out the forces acting upon the oceans of the Earth and show that, independently, they do not produce expected results, but that together, they behave exactly as we see.

    We can substitute other forms of acceleration for gravity, so that we can replace the system while removing only the gravitational gradient. There is nothing weird or physics-bending about any of this.

    If you balk at the preposterousness of the scale here (a rigid structure that can spin the Earth?) feel free to scale the masses down by 9 or 10 zeroes, and pretend M and E are asteroids, within the limit of conceivable technology. The principles will still apply.

    Finally, this is a qualitative diagram only. It demonstrates direction and RELATIVE magnitude. The next step would be to put numbers to those mags.

    Scenario 1: We set up the E-M system with no proper motion of E wrt to M (i.e. the system is not rotating, E is not revolving around M, nor M around E). To keep this a static situation, we use our rockets to ensure E and M keep a constant distance.

    View attachment 7311
    Only the gravitational force from M acts on E's oceans. Gravitational force is always positive so, magnitude aside, it will always point left. There is no force here to create a farside bulge. The oceans will (to a greater or lesser extent) flow toward M.

    Scenario 2: We re-establish the rotation but replace the force keeping E "in orbit" of M. We set M's mass to neglible, so no gravity well, then use rockets at the point of rotation to keep M stationary while E rotates around it.
    View attachment 7312
    Only the centrifugal force acts upon E's oceans. It acts asymmetrically, with more force the farther form the centre of rotation. There is no force the to create a nearside bulge. The oceans will (to a greater or lesser extent) flow away from the CoR (i.e. M).


    Scenario 3: We sum the two sets of forces. The net force acting on each individual point on E is the sum of the gravtational pull of M and the centrifugal pull of rotation.
    View attachment 7313
     
  9. billvon Valued Senior Member

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    Thanks for the images; they explain it quite well. They will likely be more clear to readers than my textual explanations.
     
  10. Russ_Watters Not a Trump supporter... Valued Senior Member

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    No: the Roche limit describes the impact of tidal stretching, so it has nothing to do with what you are describing. What you are describing is a much, much bigger force than the tidal force. Do the math!
    No. The tidal force has a definition and that isn't it. And it has that definition for a reason: it is describing a specific phenomena.
    Right: so after some period of time, all of the water would be on the side of the bulge (except water captured in basins, of course). Again: this bears no resemblance to the tidal bulge. There is no stretching here, it is just one force pulling all of the water to one side of the earth.

    You keep ignoring this, so I'll keep posting it until you address it:
    You omitted the next sentence of my post from your quote of me: All you've done is create a new force that is vastly stronger to mask it (to mask the tidal force). The Earth isn't free to move because you applied a new force to it.

    You also ignored this:
    Then, calculate the magnitude of the force on a kg of water in the "nearside bulge" and compare it to the magnitude of the tidal force.

    At this point, I think you ignored that for a reason: I think you know that what you are suggesting is much, much larger than the tidal force, so you are avoiding doing the math to compare it to the tidal force.

    In any case, none of this helps any of your previous argument. It is all an unrelated aside.
     
  11. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Still wrong.
    That's billvon's scenario, but using rockets instead of a large column. You describe it fine. Still has nothing to do with the tides though. As with billvon: calculate the force and find out!
    Like water in a bucket on a rope. You describe it fine. But again: still has nothing to do with the tides. Calculate the force and find out!

    And while it is fine as its own irrelevant scenario, when comparing it to the tides, it contains an error: you are (as I said before) confusing rotation and revolution. The earth's rotation makes a doughnut shaped bulge around the center of the earth. The earth's revolution creates no bulge anywhere because the entire non-rotating earth is tracing-out the same circle.

    In any case:
    Where did you do that? All I'm seeing is a picture. If you say you need to sum the forces (and I agree), then do it!

    Hint: The problem is, you may actually be able to get the right answer here, but I don't think even then you'll know what it means. If you get the right answer, then what you've proven is that you did too much work, not that the work you did was needed (because if done correctly, the extra stuff you throw in simply sums to zero and leaves the tidal force intact). The tidal force equation is what it is and it is simpler than what you describe.

    I previously posted this link which describes your error in great detail. Please read it!
    http://www.lhup.edu/~dsimanek/scenario/tides.htm
     
  12. Russ_Watters Not a Trump supporter... Valued Senior Member

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    You're 90% there with the post, but that's your error: Every point on earth traces a circular path of the same radius around the barycenter.

    From the previously posted link:
     
  13. billvon Valued Senior Member

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    I did. Do you disagree?
    Yes, and that phenomena is the superposition of the three forces I listed. DaveC has a good graphic describing that superposition. If that's not clear to you try the University of Washington's Applied Physics laboratory:

    ===============
    The figure demonstrates the tidal bulge of the Earth due to the combination of the Moon's direct gravitational pull and the centrifugal force in the Earth's and Moon's revolution about their center of mass (which happens to be inside the Earth, but not at its center). Earth's rotation on its axis is ignored here - when the Earth rotates under this bulge we see the twice-daily tidal maxima as points on the Earth surface rotate under the bulges, but here we will only calculate and compare the tidal acceleration causing the bulge itself. If we neglect the Earth's rotation, then as the Earth revolves about the Earth/Moon center of mass, all points in the Earth circumscribe the same size circle. This means that all points in the Earth experience the same centrifugal acceleration from this revolution. In this scenario, at any given time in the revolution the centrifugal acceleration leads away from the Moon while the Moon's gravitational attraction pulls toward the Moon, as we see in the figure. Since the strength of the gravitational acceleration from the Moon's attraction is distance dependent (falls off as 1/r2), it doesn't balance equally with the centrifugal acceleration everywhere; the difference between the two gives the tidal acceleration. At the center of the Earth (point C) the two are equal and there is no residual tidal acceleration. On the Earth's surface closer to the Moon (point B) the gravitational attraction is stronger so the tidal acceleration points toward the Moon. On the far surface (point A), the gravitational acceleration is much less so the tidal acceleration points away from the Moon. If one considers the Earth's rotation on its axis in this scenario after all, one finds subtleties like the bulge not pointing exactly toward and away from the Moon, leading to higher order (smaller) effects, but again, let us ignore that and just compare the main accelerations here.
    http://staff.washington.edu/aganse/europa/tides/tides.html
    =========================

    Correct. You would have to add a second force - the centrifugal/inertial force - to produce the two-lobed tide we see. As I mentioned above, the tides we see are a result of THREE forces - gravity of Earth, gravity of the Moon and inertia. Or, if you like, the tidal force as distinct from the Earth's gravity (i.e. the perturbing force) is the product of two forces, the gravity of the Moon and inertia, which is often referred to as "centrifugal force."

    ?? No, it is the gravity of the Moon. Surely you are not suggesting that the Moon's gravity is somehow reduced or eliminated when the Earth is orbiting the Moon normally? It is always there, and is part of the equation whether the two bodies are orbiting each other or are fixed in place. The only difference is that when the bodies are fixed in place it is the ONLY effect - and would give us a one-lobed tide.

    Refer to DaveC's scenario 1.
     
  14. DrZygote214 Registered Member

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    I'm afraid that's 100% false. Every point on the surface of Earth does not have the same distance to the barycenter, because the barycenter is not the same as Earth's geometric center. They are not the same thing.

    [EDIT] that youtube video is a poor example because it shows Earth rotating at the same rate as it orbits the barycenter. A much better animation is here: https://en.wikipedia.org/wiki/File:Orbit3.gif

    Have a look at this video: https://www.youtube.com/watch?v=CnLwLF4e6uQ
    From 0:05 to 0:08, it shows the Earth and Moon going around the barycenter. As you can see, the barycenter is not the center of Earth, and therefore, each point on the surface of Earth has a different distance/radius to the barycenter. So the centrifugal force from revolution has different magnitudes at different places.

    I did a google image search of "tide chart" and found a bunch of amplitude waves. On many of them, the second peak is a little higher than the previous. This was a fascinating discovery for me, so im starting to suspect that centrifugal force from revolution is important (but still not centrifugal force from rotation). However, I can't find one that labels each peak as the near-Moon bulge and far-Moon bulge. I might have to end up going down to the beach and measuring for myself.
     
    Last edited: Sep 4, 2014
  15. DaveC426913 Valued Senior Member

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    Of course it does. Tides are a direct result of a gravitational gradient. Exactly what we have here.

    If you don't see that we have a gravitational gradient across points on E, and that this causes the oceans to bulge and flatten in the form of tides, then we have a serious disconnect in understanding here.




    Doing calculations, if they're the wrong calculations, only serves to further convince one that they're right, even if they're not.

    This is why I am vehemently objecting to your insistence that calculations are the be-all. We must first establish and agree that they are appropriate.

    The mechanic applies the equation for max height of a ballistic object: h = vy2/ 2g and gets the correct answer.
    Until someone says: 'er, we're on an asteroid with a small radius. That equation is not appropriate in this context. You must view this as an elliptical path not a parabolic path'
    Mechanic says: 'That is the formula for the height of a ballistic object. Plug the numbers in yourself.'



    I am not confusing the two. As I have said: the E-M system is rotating. It is essentially a single fixed unit (in my case, by rockets, in billvon's case, with a spar) that is rotating about a point. if it pleases you, I will say from now on that 'the Earth is revolving around the Moon', rather than 'the Earth-Moon system is rotating'. But please acknowledge that, as I have described them, they both describe the same thing.

    That being said, we agree that the farside is by definition the point on the Earth opposite the Moon. Whether the Earth rotates on its own axis or not is irrelevant.


    So: the farside point traces out a longer path than the nearside. (the nearside point traces out a path whose radius is approx. 8,000 miles - one Earth diameter - less than farside.) Farside is under greater centrifugal force.


    I am pretty sure you know perfectly well that you can sum forces, both in terms of magnitude and direction, on a graph, without numbers.

    http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

    In scenario 3, the purple force vectors are the sum of the red and blue vectors. i.e. they are the net result.


    An enviable type of 'problem' to have.

    Please Register or Log in to view the hidden image!



    Russ, the tidal force is not a magical force above and beyond gravity. They do not sum to zero. I've accounted for gravity in my diagrams (think of the vectors as numerical, but accurate to only one sig dig). Is there some extra invisible gravity that you think I'm not accounting for?



    Of course it is. This extra work was for your benefit, because you're not seeing that these two viewpoints are the same. I'm pulling them apart and showing you that the tidal force is comprised of these basic forces. Pull em apart, examine em one-by-one, and assure that - when you put them aback together - you've got the tidal force.

    And this is what I've been trying to get at. You have an equation, and you apply it in a "tidal" circumstance. That gets you the result.

    But it's like you thinking like a mechanic. What a mechanic needs is the practical application. He doesn't care hoe it was derived, but that also means he can't examine it. He's stuck rigidly. Worse, if he applies the formula blithely, without ensuring he's looking a the big picture, or without considering whether his formula is appropriate in this circumstance, he finds himself at odds with others, gets answers that conflict with his colleagues.

    There's is more than One Way to arrive at a correct answer in physics. Do you acknowledge that the tidal equation is derived from underlying principles of gravity and inertia? Because that's what billvon is looking at.


    I will.
     
    Last edited: Sep 3, 2014
  16. DaveC426913 Valued Senior Member

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    I have read it.

    [Lots of stuff redacted after lying awake for an hour]
     
    Last edited: Sep 4, 2014
  17. DaveC426913 Valued Senior Member

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    Dagnabit. Russ, you're right. billvon, he's right.

    Reading Russ' article did not, in itself, cause me to see the misconception, but caused me to re-examine the vector of rotational acceleration.

    Look at this diagram:
    View attachment 7315
    Look at the direction of the accelerative force. It is always towards the Moon.

    Let's back up for a second.
    If we were to examine the Moon pulling on the Earth, and there were no revolution of the Earth about the Moon, we would see two bulges. Nearside is in the highest g-potential, therefore is pulled strongest. Earth is in a middling g-pot, and is pulled less. Farside is in lowest g-pot, therefore pulled least.

    So far, we should all agree on the above scenario.

    Of course, the Earth and Moon are plunging headlong toward each other. They are accelerating and the direction of that acceleration is toward the Moon.
    But we could take a photo and we would see two bulges at just the right height.
    So, we have an accelerative force pulling the Earth directly toward the Moon and we have two bulges.

    We should still all agree.

    Now, we don't want the Earth to crash into the Moon, so we give it a sideways nudge - a big nudge - of proper motion, perpendicular to the Earth-Moon axis.

    We are now using rotational acceleration to keep them from colliding. Notice I am not applying any constant force; all I did was give it some proper motion.

    But - and this is the key - what direction does the accelerative force have? Look at the diagram. It is the same as in scenario 1. The direction of that acceleration is toward the Moon.

    This is identical to scenario 1.

    When only gravitational potential is considered, and a compensating accelerative force balances the fall of the Earth into the Moon (a force that always points directly toward the Moon) - the tidal forces balance out exactly. The rotation itself has provided no input to this. All it has done is provide a way to suspend the Earth at a constant distance from the Moon so that we have scenario 1 but it is static.

    Russ, you're right.
     
  18. billvon Valued Senior Member

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    Right. And while this changes the orbits of the moon and earth, it does not change the instantaneous acceleration both of them see - and hence does not change the tides. In other words, it doesn't matter if they are orbiting each other or falling freely towards each other; in both cases, you have two tidal bulges.

    Right again. Both cases are the same with respect to tides.

    Right. The varying gravitational potential, summed with the inertial forces caused by the acceleration of the Earth towards the Moon, produce the tides. The fact that it's orbiting is coincidental; they could be falling towards each other and you'd see the same thing. The only potential point of confusion here is that in such an orbit, people often refer to those inertial forces as "centrifugal" and thus people think they are somehow different forces - but they're really not.
     
  19. Russ_Watters Not a Trump supporter... Valued Senior Member

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    I certainly don't: you did not compare the size of the "billvon force" to the size of the tidal force. The calculation I had you do was to correct one of your added-wrongs. It is tangential to the real issue of the thread. This one:
    [quote order changed for flow]
    I guess I need to remind you of how you got into this mess. Last night, in post #78, you said this:
    Now you agree that in your moon-on-a-pole scenario, all of the water would move over to the nearside. And yes, that is indeed a big difference between the "billvon force" and the tidal force.
    Source, please. I provided the actual definition and sources. You need to do the same.
    It is a cumbersome and unnecessary complication to mention the centrifugal acceleration, but it doesn't have anything to do with your claim. Your claim - and now Dave's - is that each bulge has a different cause. Your source doesn't say that and even though he cites centrifugal acceleration in the explanation, it doesn't appear in the equations, while it does appear on both sides of the moon in the free body diagram. Because obviously both sides experience the same centrifugal acceleration.
    Still no. Still no math from you to show it. Still no references. You keep making the claim, but it is wrong and empty/unsupported. The tidal force equation works fine as it is and does not include a centrifugal force component. Your own source confirms that you are wrong!
    [tap, tap, tap] Please calculate that force and compare it to the tidal force. You will see that it is much, much larger than the tidal force: it isn't a tide, even though you keep calling it that.
     
  20. Russ_Watters Not a Trump supporter... Valued Senior Member

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    That isn't what I said. I said every point traces a circular path of the same radius around the barycenter. It's a wobble without a rotation. And billvon's source agrees (and my previous source said the same thing):
    Which means that it doesn't do anything because all points on earth are experiencing the same centrifugal acceleration all the time. With no difference, there is nothing to cause any stretching. Hence, this force plays no role in the tides.
    You aren't understanding what I said and the video is likely adding to your confusion: In the video, the earth is rotating and we agreed we'd remove rotation because it doesn't do anything but create the equatorial bulge. This is one of the main sources of confusion in the thread.
     
  21. billvon Valued Senior Member

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    No. BOTH bulges are the superposition of three forces. Take any of the three away and the result is very different.
    I just provided you a reference; you chose to ignore it. That's fine - but don't expect further links when you ignore the ones I give you.
    Correct. It is not a tide. It is ONE COMPONENT of a tide. And unless you can provide a means whereby your tidal force magically cancels the gravity of the Moon entirely, it acts on the Earth. There is no way you can get around that fact.
     
  22. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Wait, what? You now seem to have taken-back everything you've been arguing for the entire thread and now are agreeing that you have been wrong all along. Please confirm that's where we are -- and if so, no need to go through my post #96.
     
  23. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Or maybe not...
    That isn't what you said in post #11:
    So, previously you said that there were two separate causes for the two bulges and now you are saying that they both have the same cause. You're moving in the right direction, at least.

    You also said 2 forces and have now changed it to 3. Neither are correct, but at least 2 was closer. On that issue you are getting further away, but I think that'll go away once you realize (and you seem to be starting to) that your other forces cancel each other out, which is why they aren't part of the definition or equation.
    I did no such thing: I directly addressed it in pointing out to you that it disagrees with you in several key areas! The equation in your link provides the definition: it doesn't include centrifugal force/acceleration. The tidal force is only the difference between the gravitational forces from one body on opposite sides of a second body.
    What I am asking for that you continue to refuse to provide is a source with a definition that includes centrifugal force/acceleration and an equation to match (an equation that includes the centrifugal acceleration).

    Here's the correct definition again, from the wiki:
    "The tidal force is a secondary effect of the force of gravity and is responsible for the tides. It arises because the gravitational force exerted by one body on another is not constant across it; the nearest side is attracted more strongly than the farthest side."

    Or:
    "A differential gravitational force acting along an extended body as a result of the varying distance from a gravitational source to the different parts of the body, such as the force of the moon on the earth's oceans closest to and farthest from the moon."
    http://www.yourdictionary.com/tidal-force

    That's not what you said in post #76:
    I've never made such a claim. You're getting lost in your own over complications.
     

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