# why are there no tides in a fish tank?

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

1. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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Still wrong, billvon. Do some math! Since the tidal force equation accounts for the far side bulge, adding a second force (centrifugal) would yield a far side bulge twice as big as the nearside bulge!

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3. ### billvonValued Senior Member

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No, there's no adding a second force, nor is there any separate centrifugal force. That second tidal bulge comes about from momentum, and is the same whether the two bodies orbit each other or are falling towards each other. When it involves an orbit you can call it centrifugal force, but it's all the same force.

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5. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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The second bulge comes from the same cause as and is calculated in the same way as the first bulge: the tidal force. If the tidal force is not calculated with a centrifugal force equation, it cannot be said to be a centrifugal force.

Also, you're piling more errors on: neither tidal force nor centrifugal force have anything to do with momentum. They are forces due to accelerations (or attempted accelerations).

Look, you've been saying many of the same wrong things over and over. By this point, you really should either be reading the citations people have posted or start looking for your own. Etiquette, if not forum rules (not certain) requires substantiating claims with cited sources. At least by failing to find sources that support your claim, you might recognize on your own that it is wrong -- I certainly wouldn't expect you to just believe me.

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7. ### billvonValued Senior Member

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There is no such thing as "tidal force", no magic forces created by objects close to each other. There is just gravity and momentum, which gives rise to all the behavior you call "tides."

Centrifugal force has nothing to do with momentum? What do you think centrifugal force is?

Russ, you seem like a smart guy, but you're saying stuff that even a high school physics student should know. There is no such thing as a separate "tidal force" - just gravity and the free motion of bodies affected by it. There is no such thing as a centrifugal force separate from momentum; indeed, centrifugal force is simply the momentum of objects constrained to follow a circular path.

8. ### billvonValued Senior Member

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Here's a good graphical depiction of what you asked:
http://en.wikipedia.org/wiki/File:Field_tidal.svg
At 12, 3, 6 and 9 o'clock there are no left or right forces; all forces are perpendicular to the center of the earth. Away from those four positions there is a net gravitational force to the left or right of center.

9. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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You should start a letter-writing campaign to get it removed from textbooks - and think of a new name to call the equation.

Oh, and you should definitely get a Wikipedia account so you can delete the article.

Anyway - again - I request that you cite a reputable source for your claims.

10. ### DaveC426913Valued Senior Member

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You're both right. You're just in different frames of reference.

The tidal force is as real (or as fictional) as centrifugal force and Coriolis Force. It is dependent on the frame of reference you want to observe from. Just as the centrifugal force is real to someone on a merry-go-round, and Coriolis Force is real when looking at weather patterns, so is tidal force real when examining bodies of water from the vantage point of the Earth's surface. (And, as useful phenomena, they are naturally used in textbooks and cited in Wiki).

However, examining these forces within the accelerating/rotating frame of reference can get complicated, even misleading. Sometimes, for the purpose of elucidation, one must observe them from an external, inertial frame of reference. When this is done, these forces disappear, or more accurately, are accounted for by simpler forces, such as gravitational attraction and inertia. If we try to examine a system too closely without considering its non-inertial frame of reference, we're going to run into problems - which is what's happening in this thread.

Russ, the tidal equation works, but it works within the accelerating frame of reference of the Earth. billvon is looking at it from the inertial frame of reference, where - as Wikipedia states it - "The tidal force is a secondary effect of the force of gravity...".

Last edited: Sep 2, 2014
11. ### KittamaruAshes to ashes, dust to dust. Adieu, Sciforums.Valued Senior Member

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If I'm not mistaken... part of the reason for tidal forces, especially with regards to the oceans, is that the water is, quite literally, pulled from one side of the planet towards the other, around/between/through all the "solid bits".

You would not see this in a fish tank as there is no solid middle "bit" for the water to go around, thus no noticeable effect.

12. ### DaveC426913Valued Senior Member

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Not quite. The tides would still exist even on an entirely ocean planet. The mean sea level (above some arbitrary depth such as the ocean-bottom) would still rise and fall with the passage of the Moon, albeit they would be milder without landmasses impeding their movement.

But you are right in the sense that uncharacteristically high tides are caused by the interference of landmasses. where the flow of water into narrowing channels - such as the Bay of Fundy - literally piles up.

13. ### billvonValued Senior Member

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You would still see the water pulled to maxima (high tides) and minima (low tides) - but without interactions with land, you would not see the much higher and lower tides we see around the world.

As a simple example, I grew up on Long Island. The presence of Long Island Sound caused a funnelling effect which gave us tides of ~8 feet; only 20 miles south, on the side facing the ocean, tides were ~3 feet. On the open ocean you would see tides of only 2 feet.

14. ### DaveC426913Valued Senior Member

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What he said.

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15. ### KittamaruAshes to ashes, dust to dust. Adieu, Sciforums.Valued Senior Member

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*nods* Of course, of course -factor in landscape effects (funneling) and such of course - but I was always under the impression the bulk of the reason we see the tides as pronounced as they are is because of the shifting of water from around the planet?

16. ### Captain KremmenAll aboard, me Hearties!Valued Senior Member

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You could look at the earth as a very big Fish Tank.
So, in that case, it does have tides.

17. ### KittamaruAshes to ashes, dust to dust. Adieu, Sciforums.Valued Senior Member

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18. ### Russ_WattersNot a Trump supporter...Valued Senior Member

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You are wrong about all of that: My position, billvon's and our relationship to the wiki. Again: the wiki says nothing about rotation. And the equation doesn't include motion so it doesn't say anything about inertial vs non-inertial reference frames. It works the same for an object in orbit, free fall, on a pole, etc.

19. ### billvonValued Senior Member

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Well, there isn't much shifting going on in total. Water isn't racing around the planet to follow the moon or the sun; locally water just redistributes itself so that the surface of the ocean is 'flat' with respect to local gravity.

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

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Average is average: if they are higher (more pronounced) somewhere, they must be lower somewhere else by an equal amount. If anything, more obstructions means the energy dissipates faster.

21. ### KittamaruAshes to ashes, dust to dust. Adieu, Sciforums.Valued Senior Member

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Oh? I was looking at things such as this (yes, I know, high school textbook stuff)

http://en.wikibooks.org/wiki/High_School_Earth_Science/Ocean_Movements

Exactly - hence why what is nothing more than a small ripple far at sea can become a huge wave as it closes in on land

22. ### billvonValued Senior Member

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They are basically saying the same thing. Water seeks to distribute itself evenly with respect to local gravity, such that its surface is "flat" with respect to the gravity vector. As gravity changes, water redistributes. In general you get "mounds" of water where the net effect of gravity is lessened by outside influences (acceleration, gravity from other bodies.)

23. ### DaveC426913Valued Senior Member

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It was not my intention to suggest that the tidal force is caused by rotation, simply that the tidal force is derived from more fundamental forces.

I can't help that think we're all talking about the same elephant, but describing different ends of it.

Would you acknowledge that the tidal force is the net difference between the gravitational potential at two points? That, if the net difference in gravitational potential between those two points happened to be zero, then the tidal force would also be zero?