why are there no tides in a fish tank?

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

1. DrZygote214Registered Member

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If lunar gravity reaches the surface of Earth, namely the oceans, then why are there no tides in a fish tank? I mean, the average difference between low tide and high tide is several feet, so how come my tank doesn't regularly overflow?

Come to think of it, I don't think small ponds experience tides either. I can remember living by several small ponds (100 sq ft or so) and never noticed any difference in water level no matter what time of day it was. But the question remains: why?

3. Arne Saknussemmtrying to figure it all outValued Senior Member

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I will hazard a guess and say that there is a tide, a lunar tug, on aquariums and ponds but the minuscule volume of water, even of a 'large' pond, dictates the rise is too little to observe. However, when the volume of water is the ocean (and there really is just one big ocean, not seven seas, am I right?) oh boy do we notice the difference!

5. billvonValued Senior Member

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There are!
Well, on an ocean 8000 miles across, there are tides of several feet. On a fish tank 3 feet across, you'd get tides of .000002 millimeters.

On a small pond you'd see pretty small tides. On the other hand, on a much larger lake (say one of the Great Lakes) you see tides of almost two inches!

7. RJBeeryNatural PhilosopherValued Senior Member

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This got me thinking. Because the tank water is confined to a container...the lunar pull wouldn't make any difference. The water will remain the same distance from solid Earth whether the tank is on the near or far side of the Earth relative to the moon, and this differential is crucial for what we see as the ocean tides...right?

On the other hand, when the lunar pull is directed at the side of the tank, the water theoretically would be shifted to that side, so there still would be some apparent sloshing around, but it would be incredibly slow and controlled.

8. billvonValued Senior Member

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When the tank is either at far side, the near side, or the 90 degree point, then there will be no net effect. Gravitation will change slightly but since water is nearly incompressible that won't make much difference. But at the 45 degree points you will see some "tilting" if you have instruments accurate enough to detect it.

9. RJBeeryNatural PhilosopherValued Senior Member

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Why do you believe the 90 degree point would not show tilting?

10. billvonValued Senior Member

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In terms of tides there are four points where the surface of the water is essentially flat. The "high" points (at 0 and 180 degrees) and the "low" points (at 90 and 270 degrees.) At other places there is some tilt.

11. RJBeeryNatural PhilosopherValued Senior Member

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This might be true for free flowing water but think in terms of a fish tank; if the force of gravity is not perpendicular to the base there will be tilting.

12. billvonValued Senior Member

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Right. And at the 0, 90, 180 and 270 degree positions, the force of gravity is perpendicular to the base.

13. RJBeeryNatural PhilosopherValued Senior Member

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At 90 and 270 the moon + earth would be putting a net gravity force on the water at an angle

14. billvonValued Senior Member

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Why do you think that? There is a tidal force away from the Earth on the opposite side from the Moon caused by centrifugal force, and there is a tidal force away from the Earth on the same side as the Moon caused by the Moon's gravity. Those two forces cause high tides. Those two forces exactly cancel along a circle between those two points. The lack of tidal force at those points causes low tides. At the high tide and low tide point there is no off-center force acting on water.

15. DrZygote214Registered Member

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Not sure I get this. The acceleration due to gravity does not depend on the mass of the object. So a 30 gallon tank weighing a few hundred kilos should feel the same acceleration from the Moon as does a part of the ocean. Can you explain why your statement is true? Feel free to use as much math/formulas as necessary, I don't shy away from those.

In fact, let me try to write something out for the sake of concreteness.

F = ma
a = F/m
F = (G*m*M)/d^2
a = (G*m*M)/(m*d^2)

so cancel the m's, that's why the acceleration of gravity on an object doesn't depend on that object's mass, although it certainly does depend on the ***Earth's*** mass.

a = G*M/d^2

Now real quick let me see what the a's are for the Moon and Sun acting on the Earth

for the Moon pulling an object on Earth:
a = G*7.347e22 / 378,028,000^2 = 3.431e-5 m/s/s // object on Earth's surface closest to Moon
a = G*7.347e22 / 384,399,000^2 = 3.318e-5 m/s/s // object at center of Earth
a = G*7.347e22 / 390,770,000^2 = 3.211e-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 = 5.932051e-3 m/s/s // object on Earth's surface closest to Sun
a = G*1.989e30 / 149,598,261,000^2 = 5.931546e-3 m/s/s // object at center of Earth
a = G*1.989e30 / 149,604,632,000^2 = 5.931041e-3 m/s/s // object on Earth's surface farthest from Sun

Okay well now I just confused myself a lot more. I read somewhere that the Sun also causes tidal effects, but not as strong as the Moon. Yet here, unless I made a mistake, shows that the Sun's gravity at Earth is about 100 times greater than the Moon's gravity at Earth...?

Could it be that tides are not really caused by the strength of gravity, but rather by the differential between gravity's strength from one end of the planet to the other? But I still don't get how a bulge is possible in the ocean's, then, because something is pulling that bulge up and it must be gravity.

16. EnmosValued Senior Member

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I guess that, because of the small volume of your fish tank, the difference between the pull of the moon on the side of the tank nearest to the moon and the pull of the moon on the side of the tank farthest from the moon is negligible.

17. Russ_WattersNot a Trump supporter...Valued Senior Member

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Because that isn't what tides are. Tidal forces don't pull objects to one side or another, the tidal force is a STRETCHING force. A tension.

...and same answer for DrZygote: you are using the wrong equations.

18. billvonValued Senior Member

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In a way. On the Earth there are two bulges - one under the Moon and one on the other side of the Moon. The closer one is due to gravity; the farther one is due to centripedall force, which manifests itself as a force opposing gravity.

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I've been seeing these tide discussions my whole life, and don't think I've ever seen anyone who understood the basics of what tides are. Only a very tiny part of a tide is static. Tides are mostly momentum of the water being shifted a little bit by mostly, solar gravity. Once you get to deep water, you see very little rise and fall of the surface. That's why tides can be two feet or fifty feet. They're mostly from the inertia of the water flowing into bays, straights and shallows.

20. Russ_WattersNot a Trump supporter...Valued Senior Member

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Sorry, but that is wrong too. Tidal force has nothing to do with centrifugal force. It works the same if there is no rotation or revolution:

http://en.m.wikipedia.org/wiki/Tidal_force

Last edited: Jul 19, 2014
21. DrZygote214Registered Member

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Can you provide a citation? That's really fascinating, yet kinda hard to believe; call me skeptical i guess. First of all, tides follow the Moon (not the Sun), offset by a small angle I believe. Second, well, if it's true that in the deep ocean, the water doesn't rise or fall that much, I would be impressed.

The link u posted is partially dead. It just links to a page that says "There is no wiki page with that name". Did you possibly mean "Tidal_force" instead of "Tidal_forge"? Sorry I couldn't quote the link here, but an error says I need 15 posts or more to put in links :/

Here's what i gather so far as an answer to my question: Fish tanks don't experience a noticeable tide because, even if the water is attracted to the Moon when it's overhead, the tank itself, and the table, and your house, and the ground under it, are also attracted. It would take very large distances before the gravitational differential causes some stretching on a noticeable scale.

It would be funny if tides happened the way i thought, tho. Then, even a glass of water would overflow and we'd have to be careful what time of day we pour our drinks.

22. billvonValued Senior Member

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Nope. If you could stop the Earth and Moon from rotating around each other, they would quickly collide. If you put up an ultra-strong scaffold to keep them in their current positions, and just stopped their motion, then all the "tide" would be on the side closest to the Moon.

However since the two bodies orbit each other, there is a tide on both sides. On one side it is due to gravity, on the other side it is due to orbital dynamics (i.e. the water wants to keep going in a straight line, but the moon accelerates the Earth away from the water's path.) That is referred to as centripedal force. From the Wikipedia article you mention:

"The attraction on the far-side oceans could be expected to cause a low tide but since the solid earth is attracted (accelerated) more strongly towards the moon, there is a relative acceleration of those waters in the outwards direction."

23. Russ_WattersNot a Trump supporter...Valued Senior Member

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Corrected, thanks. Odd - it was a copy/paste.
Yes.
Note that since water is a liquid, "stretch" may sound misleading. The volume of water stays consant, it just bulges like a squeezed water balloon. In a rigid container like a glass, it cant really overflow.