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Consider a flat body of water on the Earth. On all tiny elements of water in that body, the buoyant force is equal to the force of gravity due to Earth. Now if I introduce the moon to this body, (or any other force which would act to oppose the earths gravitational pull), and attract the body of water upwards, how would the forces balance out to prevent the water from constantly accelerating upwards. As now it appears as though there is a net force upwards, any by NL2 a constant acceleration in that direction.

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closed as unclear what you're asking by John Rennie, Jon Custer, GiorgioP, ZeroTheHero, M. Enns Mar 24 at 1:34

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  • $\begingroup$ Possible duplicate of How does the Moon cause the tides? $\endgroup$ – John Rennie Mar 22 at 16:45
  • $\begingroup$ Sorry I need edit this question considerably as my issue is more about equilibrium of forces if including the tidal force, rather than the origin of the tidal force. $\endgroup$ – Vishal Jain Mar 22 at 17:03
  • $\begingroup$ Related (and required reading): Does Earth really have two high-tide bulges on opposite sides?. $\endgroup$ – dmckee Mar 22 at 17:06
  • $\begingroup$ I have changed the phrasing of my question to better represent what I originally meant as I mislead people into thinking I was asking for an explanation of the tidal force. $\endgroup$ – Vishal Jain Mar 22 at 17:12
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    $\begingroup$ @VishalJain, the moon pulls on the ground as well. So if both the water and the ground accelerate upward, there is no change in their relative locations. $\endgroup$ – BowlOfRed Mar 22 at 18:42
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The upwards force from the Moon isn't even close to the downward force from the Earth. The only way the Moon can cause water to move upwards is if the downward force from the Earth is counteracted by other forces, such as buoyancy from surrounding water molecules.

If a molecule of water is surrounded by other water molecules, those other water molecules will be exerting a buoyancy force that counteracts the Earth's gravity, leaving the Moon's force free to pull it up. But as the molecule rises above the surrounding water molecules, the buoyancy force decreases, and eventually there's a point at which the Moon's gravity isn't able to counteract the Earth's. In addition, as a water molecules rises above the surface, the surface tension increases, further creating a restoring force.

For a similar effect, you can take a lightweight bowl and push it into a sink of water. As you push it further down, the force needed to push it down increases. If you let it go, the buoyant force will push it up, and as it goes up, the buoyancy force decreases, until eventually the buoyancy force is equal to its weight. That is what floating is: it's a state of equilibrium between weight and buoyancy. If you decrease the net weight of the bowl, say by tying a helium balloon to it, that will pull it up until the buoyancy decreases to its new, lower net weight.

Of course, all of the water molecules are exerting forces on, and experiencing forces from, other molecules simultaneously, so there's a complex process for which the end result is that the surface of the water is locally "flat", for a particular definition of "flat".

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On all tiny elements of water in that body, the buoyant force is equal to the force of gravity due to Earth.

This is true only if the Earth is not rotating and there are no other forces with components parallel to the gravitational field. The buoyant force is actually a net composite force which can change based on other forces. As soon as you introduce the moon's gravitational field you have an additional vertical component. The buoyant force changes in response to that. Or you can consider that the weight of the top layer of water decreases because the magnitude of the total gravitational field has decreased.

If the body of water is at some latitude less than $90^{\circ}$, there will be a Coriolis component which further reduces the buoyant forced needed for the water to be in equilibrium.

Bottom Line: the buoyant force is not a fixed magnitude force. It is similar to what we call the "normal" force. Its magnitude changes based on other forces and how the object accelerates. Ten newton (weight) box which is not accelerating resting on a horizontal floor. The normal component of the total force from the floor is 10 N upward. How do I know? Because I know the acceleration is zero. Now, attach a string to the box with a 1 newton mass at the other end, draped over a pulley. The box is still resting on the floor, but the normal force is now 9 N. How do I know? Because the acceleration is still zero.

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Is this an "in theory" question, or an "in practice" question?

There is no such thing and as flat body of water. In practice the Mean Sea Surface (MSS) is the time averaged surface of the ocean. It includes currents, bathymetry, tides, wind, whatever. In theory, their is the Mean Sea Level (MSL). This is the imaginary surface of the ocean in a tide free, current free state.

MSL is the nearest "in theory" thing to a flat ocean. MSL is an equipotential surface, and when it is extended through land masses, it is called The Geoid--of which there are various models. The most popular unclassified ones are EGM96 and EGM08.

So now you introduce the Moon. If you just do that, the Earth falls into it along with the ocean, so, you put the Moon in a orbit. Now there's centrifugal force preventing the fall.

Since the Moon's gravity is stronger (weaker) on the near (far) side, and centrifugal force is stronger (weaker) on the far (near) side--the ocean and the solid Earth deform to a new equipotential shape that has a big quadrupole term accounting for the tidal effects (this hypothetical is for a tidally locked earth).

In theory, maybe you don't want the Earth to deform--so what happens if an orbiting moon is added to a tidally locked stiff-Earth? Then the geoid can be supplemented with a big quadrupole lunar term that causes the near and far sides to bulge, and the ocean follows that bulge.

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As the sea bulges up there is more material (water) below the water surface and its extra gravity balances the attraction of the earth.

See problem 3 on the following homework set:

https://courses.physics.illinois.edu/phys508/fa2018/508hw10.pdf

for some details

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See https://en.wikipedia.org/wiki/Tidal_force

"The Moon produces a greater tidal force on the Earth, than the tidal force of the Earth on the Moon. The distance is the same, but the diameter of the Earth is greater than the diameter of the Moon, resulting in a greater tidal force.

What matters is not the total gravitational attraction on a body, but the difference from one side to the other. The greater the diameter of the body, the more difference there will be from one side to the other."

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  • $\begingroup$ I understand that there will be a difference in force, but what i don't understand is how this difference reduces to 0 as the sea level increases, which it must do for it to stop stretching. $\endgroup$ – Vishal Jain Mar 22 at 17:01
  • $\begingroup$ The quoted materials is simply wrong. (Nor do I find it in the linked page as of this hour or the last few revisions.) It neglects the differing effect of the masses. Tidal force is proportional to separation from the free-fall center, yes, but it is also proportional to the mass of the object causing the tide. $\endgroup$ – dmckee Mar 23 at 0:29

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