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High tides and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. ... When you're in one of the bulges, you experience a high tide.

If ocean water rises on full moon. And gravitational acceleration is not dependent on the mass of the attracted body. Just as a metal ball and feather falls at the same speed, why doesn't both bottle water and ocean water rise by same levels on a full moon?

If air is the reason, then on an atmosphere less planet does bottle water and ocean water rise by same levels?

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    $\begingroup$ The tidal bulge theory is an over-simplification. Please see physics.stackexchange.com/q/121830/123208 $\endgroup$ – PM 2Ring Mar 26 at 18:01
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    $\begingroup$ see also "earth tides" en.wikipedia.org/wiki/Earth_tide $\endgroup$ – anna v Mar 26 at 18:16
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    $\begingroup$ @annav thanks for the link, I can understand why ocean tides bulge but I cant understand why water in the bottle doesn't rise. $\endgroup$ – VARUN.N RAO Mar 26 at 18:30
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    $\begingroup$ I gave the link for ou to see that also the ground changes radius with the pull of the moon (and sun), but we all go up and down with it folowing the bulk.The individual small volumes are is too small to notice it, as the answer by tom10 says. $\endgroup$ – anna v Mar 27 at 4:56
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    $\begingroup$ Where would the water come from to make the water rise in a bottle? Waters in the oceans rise and fall because tidal forces make the oceans waters flow from one place to another. $\endgroup$ – David Hammen Mar 28 at 18:16
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Does bottle water rise a little bit on full moon days?

No. Tidal forces are about the difference in gravitational pull at different points in the same body. For oceans and other very large bodies of water, this difference causes water to flow from one region to another, which causes the rise in tides.

For example, this is why, even though the sun's gravitational pull is much larger on the earth than the moon's, the moon dominates the tides because it is closer to the earth and therefore the difference in gravitational pull is larger.

So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed.

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    $\begingroup$ Even with a very long bottle that was as tall as the earth's diameter, you still wouldn't have an observable tide inside the bottle. Water is incompressible, and there's nowhere else on "the sides" to draw water from - as an ocean tide rises, it has to draw water from somewhere else. There'd be some tidal force, but you wouldn't see the height of the water in the bottle change at all, since you can't stretch a column of water. $\endgroup$ – Nuclear Hoagie Mar 26 at 19:27
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    $\begingroup$ @NuclearHoagie I'm not sure that's right: you are assuming that tidal compression (see e.g.) would not squeeze the water bottle. $\endgroup$ – rob Mar 26 at 19:32
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    $\begingroup$ @NuclearHoagie it can still form bulges at the end and become thin in the middle like a dumble bell. $\endgroup$ – VARUN.N RAO Mar 26 at 20:19
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    $\begingroup$ Although, technically everything on the ground rises due to the moon. en.wikipedia.org/wiki/Earth_tide $\endgroup$ – Shufflepants Mar 27 at 3:27
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    $\begingroup$ @ReversedEngineer The side of the earth near the moon gets pulled harder than the center of the earth, so that's the tidal bulge on the near side that most people understand. The far bulge is due to the fact that the center of the earth gets pulled harder than the far side, so that water gets "left behind" as the earth accelerates away from it. Both bulges are due to the gradient in the moon's gravitational force as you move further away. $\endgroup$ – Nuclear Hoagie Mar 29 at 13:21
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Even the most rewarded answer here has missed out on the fact that the entire bottle of water will rise by up to a meter due to the phenomenon of earth tides. The body of earth is deformed by the same lunar and solar tidal forces that cause the sea tides. See https://en.wikipedia.org/wiki/Earth_tide.

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  • $\begingroup$ I was going to say that but You beat me to it $\endgroup$ – Bill Alsept Apr 6 at 21:03
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So for the bottle, the difference in gravitational pull from one side of the bottle to the other side of the bottle is extremely small because the distance is extremely small relative to the distance to the moon, and the tidal forces can not be observed.

How small? Let's work it out. The acceleration on a bottle of water due to the Moon is...

$$\text{Gravitational constant} \times \text{mass of the Moon} \times \frac{\text{diameter of the bottle}}{\text{distance to the Moon}^3}$$

Let's assume our bottle has a diameter of about 0.06m. The distance to the Moon varies, I'll use the semi-major axis. It won't make a difference.

$$\text{mass of the Moon} = 7.34 \times 10^{22} kg$$ $$\text{distance to the Moon} = 3.84 \times 10^8 m$$ $$\text{Gravitational constant} = 6.674 \times 10^{−11} m^3⋅kg^{−1}⋅s^{−2}$$

Which is about $6 \times 10^{-15} m⋅s^{−2}$. This is an order of magnitude smaller than the smallest acceleration measured. Not just extremely small, unmeasurably small.

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As correctly stated in the deleted answer, the tides have nothing to do with the fact that the moon is full. Tides occur always, regardless of the moon being full or not.

Imagine your bottle filled with water is free-falling towards the moon, the bottle parallel to the moon's gravity field. What will happen to the water inside? On one side of the bottle, the gravity field will be higher than on the other end. Though very, very little (this isn't the case for a large body of water like that of the oceans). This difference in the gravitational force is called the tidal force. I think you can see that this tidal force is easily overcome by the force which holds the water level inside your bottle fixed. The water level increases a tiny, tiny bit, up to the point where the molecules in the water are separated enough to counteract the differences in the gravitational field. Only when the gravitational field is high enough, then the level will visibly change. The water will be stretched into droplets, and maybe into individual molecules. And the molecules? Think about it...

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A less known fact about the tides is that the lithosphere has few centimeters of tides, too.

So yes, your bottle as a whole will rise that high two times a day.

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    $\begingroup$ Not sure why this was down-voted. It's accurate and the effect it's referencing is more relevant than the effect on the water inside the bottle which is the other aspect the OP was asking about. $\endgroup$ – Brondahl Mar 29 at 11:13
  • $\begingroup$ @Brondahl: I see that this answer has 1 downvote and zero upvotes. If it is correct, then why do you not upvote it? I do not know if it is or not, so I'm not voting. $\endgroup$ – dotancohen Mar 29 at 13:29
  • $\begingroup$ I did upvote it :( $\endgroup$ – Brondahl Mar 29 at 13:54
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    $\begingroup$ @dotancohen Have re-applied my upvote. Don't know what's happened there. $\endgroup$ – Brondahl Mar 29 at 13:55
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Since water is compressible it will expand when there is less gravitational pull towards the center of the Earth, which is the case on the Moon-facing side of the Earth. It's just not very much because water is not particularly well compressible.

The water surface in the bottle will also, like the oceans in the simplified tide model, have a bulge — a minuscule deviation from the curvature it has anyway because of the inhomogeneous gravitational field of Earth.

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    $\begingroup$ Come on! You cannot just unleash whatever effect crosses your mind. Make a few computations! I wouldn't be surprised to find out that what you are talking about is order of magnitudes smaller than the noise due to Brownian motion. $\endgroup$ – DarioP Mar 28 at 20:11
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    $\begingroup$ Water is, effectively, incompressible. This is why hydraulics work. To say the effect across a bottle of water would be "not very much" at the scale of a bottle of water is grossly misleading understatement. $\endgroup$ – Schwern Mar 28 at 21:11
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    $\begingroup$ @Schwern "Anything I can't see is irrelevant" is an engineer's attitude. This is SE Physics. The OP asked; it was a naive question. I gave the only correct answer. $\endgroup$ – Peter - Reinstate Monica Mar 28 at 23:59
  • $\begingroup$ @DarioP The tidal bulge per horizontal meter should be in the order of magnitude of, say 1m/1E7m, i.e. 1E-7 (height of the tidal bulge assumed as 1m for convenience, distributed over a quarter Earth circumference). For a bottle with a diameter of 0.1m (again, for convenience) we arrive at a height difference of 1E-8m, which corresponds, if I didn't make a mistake, to a layer of 40 water molecules, each having a a diameter of 0.27E-9m. There is probably some boundary layer chaos making measurements difficult but that's a different (and more complicated) discussion. $\endgroup$ – Peter - Reinstate Monica Mar 29 at 11:01
  • $\begingroup$ @DarioP The height difference due to compression is probably smaller but I'm not entirely sure how to come up with an order of magnitude. Perhaps you can provide one? The compressibility of 5E−10/Pa is indeed small. $\endgroup$ – Peter - Reinstate Monica Mar 29 at 11:02

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