# 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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45
Uh, so you're saying each point follows a circle around the barycenter, and that circle has constant radius? I don't disagree with that. But sure sounded to me like you said that every point (as in all the different possible points on Earth's surface) have the same radius about the barycenter. That's what I was calling 100% false and I hope you'll understand the misunderstanding of "every point".

If i've interpreted all this correctly then i don't see what i said wrong in post #82 that you pointed out in post #89, when u said i was 90% there but missed something.

Can't completely agree. There are 2 centrifugal forces (Cf) and i don't wanna confuse them, so here they are again: one from rotation and one from revolution about the barycenter. The rotational one doesn't play a role, since the same centrifugal force from it occurs everywhere on the equator. The revolutional one, however, could play a role (but i have to calculate it to be sure, it could be negligible), because different points on Earth's surface have different centrifugal force from revolution. The anti-Moon side of Earth will have the greater Cf since it's farther from the barycenter; the near-Moon side of Earth is closer to the barycenter so it will have a smaller Cf.

Actually we never agreed to that. You're prolly thinking of something agreed between you and billvon.

I suppose i could go find a video that doesn't show any Earth rotation while still orbiting the barycenter, but honestly it will be the same Cf from revolution. Just pause it at any point in the diagram and i hope it clearly shows to you that different surface points on Earth have different radii to the barycenter. Since the angular speed of the revolution is the same, they must have different centrifugal forces. So again, I can't see what you're pointing out in post #89 about what i said wrong is post #82.

One thing I just realized from all this, however, is that the 2 different Cfs can interfere with each other so to hell with it ill just calculate them both right now.

The centripetel acceleration at Earth's equator, from rotation, is constant (thank god). I calculate it at 6378000*(2*pi/86164)^2 = 0.033915057 m/s/s. I used siderial rotation period of Earth of 86,164 s and an equatorial radius of 6,378 km.

Centripetel acceleration, from revolution, at sub-Moon point on Earth's surface = (6,378,000 - 4,600,000)*(2*pi/2,360,585)^2 = 1.259656718 e-5 m/s/s

Centripetel acceleration, from revolution, at anti-Moon point on Earth's surface = (6,378,000 + 4,600,000)*(2*pi/2,360,585)^2 = 7.777565493 e-5 m/s/s

The Ca's from revolution differ from each other by a factor of over 6, but they're still much much smaller than Ca from rotation. Summing or subtracting any of the last 2 values from the 1st one is not gonna make a huge difference.

But now let's get the acceleration from the Moon's gravity. I copy them from post #82 so go there if u wanna see how i derived these:

sub-Moon point on Earth: 3.431 e-5 m/s/s
anti-Moon point on Earth: 3.211 e-5 m/s/s

Low and behold we're in the same order of magnitude (10^-5). I guess I hafta thank you for that misunderstanding or i might never have calculated this. I hereby state, with reasonable certainty based on calculations, that when you google image "tide chart" and find many amplitudes that repeat a second, slightly higher peak, that higher peak represents the anti-moon bulge, and it's higher cuz of the centripetel acceleration due to Earth's revolution around the barycenter, which is over 6 times stronger on the anti-Moon point than the sub-Moon point.

However, this is not the same thing as causing the anti moon bulge. It's just causing it to be higher than the near-Moon bulge.

One more thing. Russ Watters one of your links led me here: http://www.lhup.edu/~dsimanek/scenario/tides101.htm

On that page it says, "The moon's gravitational force on earth acts on all parts of the earth with nearly the same force, differing by only a fraction of a percent on the sides of earth nearest and farthest from the moon."

But that's false. I just calculated it. Difference between them is 3.431 e-5 m/s/s and 3.211 e-5 m/s/s. Divide one by the other and you get 1.0685. This is almost 7%, and I'm afraid that's not "a fraction of a percent" as claimed in the link.

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

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I'm sorry, you're just going to need to try again then. I'll try to find you an animation showing it, but in the meantime you've been shown at least two links that say the same thing.
Huh? You even said it again, just a few lines higher:
"The rotational one doesn't play a role, since the same centrifugal force from it occurs everywhere on the equator."

So which is it?

5. ### DrZygote214Registered Member

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45
Found one: https://en.wikipedia.org/wiki/File:Orbit3.gif

There are different points on Earth that do not have the same distance to the barycenter. Therefore they have different centrifugal forces (from revolution, not from rotation). The points im interested in are the sub-Moon point on Earth's surface and the anti-Moon point on Earth's surface. Their respective differences from the barycenter are 6,378 - 4,600 km and 6,378 + 4,600 km.

Also, i looked at those links and read them, but will need to re-read them to be sure they sunk in and prevent as far as possible any more misunderstandings. In the meantime, i quoted one of them with a factual error about "a fraction of a percent", which is not true according to my calculations, and about which i have not heard back from you. This makes me wary of everything else the article says so i'm gonna take my time scrutinizing it.

I see what you're saying now. You said we "agreed to remove rotation," but i never agreed to "remove" anything. I claimed that centrifugal force doesn't play a role, but please don't assume that means i'm joining the same hypothetical scenario in which the Earth is not rotating.

You made a point that the video i quoted involved the Earth rotating, which you seemed to not like. Me, i have no qualms about an animation that involves rotation (since it doesn't matter one way or the other in regards to Cf from revolution), so i couldn't understand what you were talking about when you said we agreed to "remove" rotation. If it doesn't matter, then a video may have it or it may not, shouldn't matter when only talking about revolution about the barycenter. At least that's they way i see it.

[EDIT]. I watched the youtube vid again and realized that it's rotation is not right. It's rotating at the same rate that the Moon and Earth orbit each other, which is a very wrong rate of rotation. If that's what you were pointing out, then I agree completely it's a poor example video.

I'll try to use the word "orbit" instead of "revolution" from now on. Earth orbits about the barycenter and rotates around its center. The barycenter and Earth's center are not the same points. If we agree so far then i'm not sure how else to explain that one side of Earth and the other have different distances to the barycenter and therefore experience a different Cf in regards to orbit. Of course, one side and the other have the same distance to the normal center of Earth (at the equator anyway), and so have the same Cf in regards to rotation.

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

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5,051
That doesn't help because it doesn't show any detail of the planet. So your imagination is free to picture if it is rotating or not. It doesn't show the effect we are discussing (revolution without rotation).
I haven't had time to check into that yet. Frankly, it isn't a high priority for me because it doesn't have anything to do with what we are discussing and given that you have errors in your understanding, I'm not inclined to be a believer that you found an error. Better to become thoroughly versed in the subject first, otherwise you will create a mental block against learning.
"Doesn't play a role" means it has no impact so it can safely be ignored. Sure, you can leave it in, but all it can do is create complexity and confusion, which it very clearly is. It is causing you to misunderstand the revolution motion.
And indeed, that is the error you are making. The only way I can see to fix this is for you to view the revolution by itself. Perhaps you could try it with a roll of tape on your desk?

8. ### DaveC426913Valued Senior Member

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OK, but that's not what you were saying on p1 of this thread:

(Note: I think you meant centrifugal force. Centripetal force points inward, same as gravity. Centrifugal force points outward.)

The centrifugal force you are talking about really does manifest in unrelated scenarios. It is what keeps the water in a bucket on a string, when flung around your head. Note that there is no gravitational gradient in this scenario.

So, if it were a factor in the Earth-Moon scenario - where we have shown that gravitational potential nicely accounts for all forces and all tides - then the centrifugal acceleration would - as Russ says - be added on top, and the farside bulge would be twice as high.

I know, I know. You're arguing that they are one-and-the-same force, so you're claiming there would be no doubling.

But my pail-on-a-string scenario shows that they are indeed two distinct forces - I can create a measurable centrifugal effect - without any gravitational potential. So if both were a factor in the Earth-Moon scenario, they would indeed be additive.

Conclusion: centrifugal acceleration is a distinct force in a revolving system (isolatable and measurable, independently of other forces such as gravitational gradient) but it does not play a part in tides.

9. ### Neddy BateValued Senior Member

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2,221
Very interesting discussion. The only part that I had trouble following was the idea that every point of the earth moves in the same way relative to the Earth-Moon barycenter. For some reason I could not picture that in my mind, but I did find this:

And that cleared the matter up quite nicely.

Here is a link to the entire document:

http://www.vialattea.net/maree/eng/index.htm

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

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Thank you, that's exactly what I was looking for!

11. ### Neddy BateValued Senior Member

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You're welcome! And thank you for the information on how tides really work. I had some of the common misconceptions before reading this thread.

12. ### billvonValued Senior Member

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There are two separate causes. They sum everywhere. One causes the effect on the far side, one causes the effect on the near side. Neither is sufficient by itself, and neither solely causes either bulge. (Indeed, they sum such that each resultant bulge is considerably _smaller._)

Local gravity determines the height of our tides, via the ratio to the net tidal force. Thus it certainly influences the tides. It does not change the tidal FORCE, which is why I mentioned "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." "

. . .

Uh, OK. If you think that, then I am good with that.

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

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Please provide an equation describing that and a source. If it makes it easier for you, you can use words:
(Tidal acceleration) = (farside lunar gravitational acceleration) - ......

14. ### DrZygote214Registered Member

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45
No need to believe. Only need is to calculate.

Frankly in reply, for someone who multiple times pleaded to do some calculations (okay it was at bivillon, not me), that's a pretty harsh response to someone who actually took up your advice.

Thank you for that animation and article. It explained my misconception about Cf from the barycenter very well. Is there a secret to finding these brilliant things? vialattea.net sounds pretty obscure, pretty deep in the deep web, but if there's more to it than that, please let me know.

For those of you in the US and UK, be warned: The numerical values in that article use a comma as the radix point (decimal point). This threw me off since I skipped right to Case 2 (inertial reference frame which makes most sense to me) where right below Figure 8 there is a 4-digit number with a comma right where i thought it should be, but it's a radix point.

Only one question I have for that article right now. It says below Figure 9: "However, it is correct to say that all the points of the Earth describe a circle whose radius is c and thus undergo an acceleration 3.322 e-5 m/s² pointing towards the circle's center, i.e. towards the Moon."

However, Earth and Moon are orbiting each other, so they're in free fall right? So actually, there should be no centrifugal force from the orbital motion...it's just free fall, no acceleration felt. Shouldn't matter if that free fall is in a straight line, like with 2 planets gonna collide, or if the free fall is orbital. If this is true then I was wrong before when I was trying to clarify 2 distinct Cfs. There's only one, from rotation, which plays no role in the tides.

This and my other calculation (nearly 7% difference between Moon's gravity from one end of Earth to the other, as opposed to a fraction of a percent), are my only claims of error. I'm not claiming with absolute certitude they're wrong, I'm just claiming that my calculations differed from their claims and i hope one of you can corroborate me or else point out my mathematical or conceptual error.

15. ### DaveC426913Valued Senior Member

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In post 94, I showed that this cannot be true.

In a nutshell, the gravitational gradient accounts for the tidal bulge on both the nearside AND the farside - without any requirement for another force - and it does so symmetrically.

So, if there were another force in play (such as centrifugal), it would have to add, necessarily causing the farside bulge to be higher than nearside by that much.

16. ### billvonValued Senior Member

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20,216
The two forces are:
Gravity from the Moon. This falls off as 1/R^2. Thus the Moon exerts more force on the side nearest the Moon than the side farther away from the Moon. To put numbers to this, the attraction on the near side is about 3.5 microgees towards the Moon, and on the far side about 3.27 microgees towards the Moon. Note that this force alone generates a "nearside bulge" but not a farside one. Again, this is only one of the two forces, and is modeled by the "moon-on-a-pole" thought experiment.

The second force is caused by the Earth being accelerated towards the Moon. This thought experiment is to accelerate the Earth towards the Moon without the Moon's gravity being there. In that case, the force is symmetrical across the entire planet since acceleration is the same everywhere. In that case you will see 3.38 microgees away from the Moon. This would form a farside bulge but not a nearside one.

Summing both forces you get both bulges - each of which is significantly lower. To sum both cases above, the total tidal force on the near side would be .115 microgees acting TOWARDS the Moon, and on the far side would be .109 microgees acting AWAY from the Moon, giving us our two lobed tide.

(There is a slight error in the above, because the above assumes that the Moon's gravity attracts the Earth as a point source. Since the Earth is not a point source, and gravity falls off as 1/R^2, the Moon's "center of attraction" is moved closer to the Moon, away from the center of the Earth. The result is that the acceleration of the Earth towards the Moon is slightly higher, and the resultant tidal forces are .112 microgees on _both_ sides.)

They do add - but they add to result in the two-lobed tides we see.

17. ### Neddy BateValued Senior Member

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1. It seems you are saying the first force (simple garvity) works like this:
_a. The attraction on the NEAR side is +3.500 microgees TOWARDS the Moon
_b. The attraction on the FAR side is +3.270 microgees TOWARDS the Moon

2. And you are saying the second force (acceleration of earth towards moon) works like this:
_a. The attraction on the NEAR side is -3.388 microgees TOWARDS the Moon
_b. The attraction on the FAR side is -3.382 microgees TOWARDS the Moon

3. Such that the end result is like this:
_a. The attraction on the NEAR side is +0.112 microgees TOWARDS the Moon
_b. The attraction on the FAR side is -0.112 microgees TOWARDS the Moon

Near Side:
+3.500
-3.388
+0.112

Far Side:
+3.270
-3.382
-0.112

18. ### billvonValued Senior Member

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20,216
Yes

There's no attraction in that case. That case is simply the result of an object being on a planet that is under acceleration. Inertia will want to keep the object stationary, which means the force acts away from the moon. (And it is the same everywhere in the simple case since the entire earth is under acceleration.)

"Attraction" (force) on the far side is -.112 microgees towards the Moon, or .112 microgees away from the Moon - giving us the two lobed tide.

19. ### danshawenValued Senior Member

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3,950
A marvelous analysis of the Earth-moon system. I'm a little surprised that no one threw the Sun into the mix. For the benefit of those of you who missed the very short debate topic which bears on this thread also:

Orbiting bodies and the things they orbit are as ignorant as any other stones, and there is no reason to expect that they could find the direction that is the center, or barycenter, or otherwise, 1) without instruments of any kind, or 2) in the dark (with no means of determining either density, mass concentrations, or contours), without some sort of interaction with the "stuff" that lies between them. This is true whether you are using Newton's "divine hand" mechanics, or Einstein's "curved space" formulation of General Relativity, or anything else that does not take vacuum energy interacting with matter into account.

These are holes in your science (as well as math) that need amending in order for any theory of gravity to be complete. What are you waiting for? Divine intervention?

Really cool cartoon, but this is only the situation if density is uniform within shells.

20. ### Neddy BateValued Senior Member

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2,221
Yes, I purposefully used the convention "TOWARD the moon" so that vectors which were pointed away from the moon would have negative signs. That way they could be added algebraically.

21. ### Neddy BateValued Senior Member

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2,221
Let's take the simple case, where the force is the same everywhere, since the entire earth is under acceleration.

In that case, your numbers might look something more like this:

1. The first acceleration (simple gravity):
_a. The acceleration on the NEAR side is +3.500 microgees TOWARDS the Moon
_b. The acceleration on the FAR side is +3.270 microgees TOWARDS the Moon

2. The second acceleration (acceleration of earth towards moon):
_a. The acceleration on the NEAR side is -3.385 microgees TOWARDS the Moon
_b. The acceleration on the FAR side is -3.385 microgees TOWARDS the Moon

3. Such that the end result is like this:
_a. The acceleration on the NEAR side is +0.115 microgees TOWARDS the Moon
_b. The acceleration on the FAR side is -0.115 microgees TOWARDS the Moon

Near Side:
+3.500
-3.385
+0.115

Far Side:
+3.270
-3.385
-0.115

In which case, I believe Russ Watter's argument is that the tidal force can be calculated simply as ±(3.500-3.270)/2 or ±0.115.

22. ### DaveC426913Valued Senior Member

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16,061
nm.

Last edited: Sep 9, 2014