why are there no tides in a fish tank?

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

  1. Russ_Watters Not a Trump supporter... Valued Senior Member

    "Quickly" is relative: for the first day you wouldnt notice any change - except that the tides didn't advance as much as usual and the moon is in the same place in the sky as yesterday.
    I'm sorry, but repeating it won't make it true. This is a common misconception.

    Right: no mention of rotation or centripedal acceleration, just normal acceleration. The differential acceleration is caused by distance and the inverse square law of gravity.
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  3. sideshowbob Sorry, wrong number. Valued Senior Member

    It seems to me that the rotation of the earth would cause the water to bulge out slightly at the equator but that would be constant.
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  5. Russ_Watters Not a Trump supporter... Valued Senior Member

    It does - and not just the water, the land too.
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  7. billvon Valued Senior Member

    Right. It would take several days for the collision to occur.

    Do you deny that there is a tidal bulge on the _opposite_ side of the Earth, where the Moon's gravity is actually helping to flatten, rather than raise, the oceans?

    No, it's caused by the fact that the Earth/Moon system rotates around a point that is not in the center of the Earth.
  8. river

    The fish tank is simply to small
  9. Janus58 Valued Senior Member

    Go back the the Moon falling directly towards the Earth. What would happen to the Earth tidal bulges? (other than their not advancing the same as before). Would the Earth still have two bulges? Yes.

    The nearside water is closer to the the Moon, so it will want to accelerate towards the Moon at a rate greater than the center of the Earth and thus the Earth's main body. The far side water, being further away will want to accelerate at a rate less than the center of the Earth. The Earth's gravity will hold things together, but there will still be an effective weakening on the Near and far sides, producing two tidal bulges.

    Now as the Earth and moon get closer, the tidal bulges will grow. (but never to the point where they separate from the Earth, as the Earth's Roche limit distance from the Moon is less than its own radius. )
  10. Russ_Watters Not a Trump supporter... Valued Senior Member

    You dropped your argument. The point is that for the first day there would be virtually no change in the far side tidal bulge because it is caused by the moon's tidal force, not it's orbit around earth.

    Look: it is called a "tide" because it is caused by the "tidal force". It almost seems like you've never heard of the tidal force. You should read the wiki on what it is and how it works.

    The way you worded that it is wrong: There is a tidal bulge on the far side of the earth, which is caused by the tidal force from the moon, just like the bulge on the near side.
    The quote you posted didn't say that. It was clearly talking about normal acceleration due to gravity.

    This link specifically discusses your common misconception:
  11. river

    The friggen fish tank is to small , to notice the effect by the moon
  12. billvon Valued Senior Member

    Yes - since, again, the water on the far side of the Earth wants to remain at rest. The Earth is now accelerating towards the Moon at some rate. This acceleration results in the bulge on the far side, since the inertia of the water acts to effectively decrease the net force on the water.

    Now put up that scaffold to keep them from accelerating towards each other. Now the bulge is only on the side facing the Moon.

    During a normal orbit, there is little relative change in the distance between the Earth and the Moon. Thus one might expect the far side bulge, caused by acceleration towards the Moon, to disappear. However, the water on the far side of the Earth is moving about the Earth-Moon center of rotation - and is farther from that center than the water on the near side. Thus there is "centrifugal force" - a force caused by all that water wanting to continue in a straight line, instead of being constrained to circle around the Earth-Moon rotational center.

    Note that this is no different from the case where they are falling freely towards each other, at least for the instant at which they start falling towards each other. In both cases the gravity of the Moon is effectively nullified at the center of mass of the Earth since they are falling freely towards each other (or orbiting each other, which can be described as falling towards each other and always missing.) Thus another, equally valid, way to describe it is to say that the Moon's gravity is "positive" on the near side and "negative" on the far side. (Gravity is never, of course, negative, but the differential makes it appear that way.)

    (* - Centrifugal force is in quotes because that term gives some people heartburn; it is simply momentum in a rotating system.)
    Last edited: Jul 20, 2014
  13. TBodillia Registered Senior Member

    Moon Myths: The Truth About Lunar Effects on You

    "On the other side of the Earth, another high tide occurs, because the center of Earth is being pulled toward the moon more than is the ocean on the far side. The result essentially pulls the planet away from the ocean (a negative force that effectively lifts the ocean away from the planet).

    However, there's no measurable difference in the moon's gravitational effect to one side of your body vs. the other. Even in a large lake, tides are extremely minor."

    Water levels in the Great Lakes change primarily because of meteorological effects

    "Studies indicate that the Great Lakes spring tide, the largest tides caused by the combined forces of the sun and moon, is less than five centimeters (two inches) in height. These minor variations are masked by the greater fluctuations in lake levels produced by wind and barometric pressure changes.

    Consequently, the Great Lakes are considered to be non-tidal."

    So, if a body of water with 5,349 cubic MILES (5,988,976,164,800,000 US gallons) of water is considered to be non-tidal, do you really think a 1, 5, 25, or even 200 gallon fish tank would show a difference in tides?
  14. Layman Totally Internally Reflected Valued Senior Member

    That would make it seem like the tides would be caused more by the rotation of the Earth, rather than the moon. I would guess that it is due to the butterfly effect, or the whole theory is just bologna. Come to think of it, I never heard about what scientist ever discovered this theory. If no one actually discovered this, it would really hint to it just being an urban legend.
  15. Captain Kremmen All aboard, me Hearties! Valued Senior Member

    What do fish do when the tide is out in a fish tank?
  16. James R Just this guy, you know? Staff Member

    Also worth mentioning: for the water in the fish tank to rise, it has to flow from somewhere else. Even when the Moon is directly above the tank, there isn't much difference between the force the Moon is exerting on the water at the sides of the tank and the force it is exerting on the water at the centre. So there's very little flow from the sides to the centre, and not much of a tidal bulge.

    When you consider the oceans of the Earth, on the other hand, a lot of water can flow from the sides towards the points directly below the Moon.
  17. Stryder Keeper of "good" ideas. Valued Senior Member

    Offtopic: but I'd say "Flap".
  18. Arne Saknussemm trying to figure it all out Valued Senior Member

  19. RJBeery Natural Philosopher Valued Senior Member

    Picture the Earth as a clock face. The fish tank is at 12, and the Moon is off in the distance at 3. Are you telling me that the water in this cubical tank would not have a net gravitational force to the right of center between the combined Earth + Moon effects? The centrifugal forces from the Earth's rotation don't "magically" counterbalance the water; the Moon would remain in orbit even if the Earth was not rotating at all.
  20. matterdoc Registered Senior Member

    Terrestrial tides are mere distortions of earth, caused by action of linear 'forces' on a spinning body. They neither pull nor push at parts of earth. That is, actions of linear 'forces' do not cause but they help to produce distortions of earth. This is why two linear 'forces' (central forces towards sun and moon) produce two sets of tides instead of a single set in resultant direction (as required by dynamics). According to dynamics, action of a 'force' should be in the direction of 'force'. However, directions of tides are rarely so.
    Last edited: Jul 26, 2014
  21. tedrowilson Registered Member


    This is exactly right...you guys are looking too far into it...

    The water has to come from somewhere, and flow to someplace. Only when there is a changing difference in gravity between 2 points in the water, will a tidal change occur. In a fish tank, any 2 points have virtually the same gravitational pull. even when the moon is at 3 o'clock to the fish tank, the right side of the tank and the left side of the tank are only a few feet apart....not enough to cause a change(a wave)...also in a walled container there is a meniscus, which is a result of the water's surface tension, causing the outer edges of the container to have higher water levels than the middle. if tides affected this, at 12'oclock the meniscus would flatten, and at 6 would be greater. this might happen at a very small level, but nothing significant.
  22. billvon Valued Senior Member

    If the Earth and the Moon were being held in a rigid frame, so that they were stationary with respect to each other - then yes, there would be an off-center force on the water; indeed the only tidal bulge would be towards the Moon.

    However that's not the case. Since the Earth and the Moon are moving around a common center of gravity, other forces act on the water. The force-often-called-centrifugal-force dominates on the far side of the system, the gravity of the Moon dominates on the near side of the system. (And would be the same if the two bodies were falling freely towards each other, although you wouldn't call it centrifugal force then.) That's why there are two tidal bulges. So in your example, there is a tidal bulge at 3 and 9 (where forces combine to act against gravity) and a tidal trough at 12 and 6 (where forces do not act against gravity). At 3, 6, 9 and 12, the net force on the water is perpendicular to the center of the Earth. At other places there is an off-center component to the force.

    Think about it a different way. Let's suddenly remove the gravity of the Earth but keep the system orbiting itself in the same fashion. Now consider three bodies - one on the side of the Earth closest to the Moon, one on the farthest side, one midway. We'll assume they have zero speed relative to the center of the Earth. The near and far objects, no longer under the influence of the Earth's gravity, will now want to find new orbits. The midway object will tend to want to stay in about the same orbit, since it is close to the center of gravity of the Earth, which itself was in a stable orbit.
  23. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

    Welcome to Sciforums. A rather strange thread in which to make your entrance!

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