What happens to the water on the surface of the Earth if the Earth is not rotating about its axis in the Earth-Moon system?

Question

Suppose the Earth is not rotating. As usual, the Moon follows its normal path around the Earth. Let's assume it's a circular motion and that there are no other gravitational influences.

A test particle on the surface of the Earth directed to the Moon on the line that connects the middle points of both will feel the Earth's gravity directed to its center and the Moon's gravity directed to the Moon.

A test particle on the opposite side of the Earth will also feel the Earth's gravity directed to its center (equal but opposite to the gravity experienced by the above-mentioned test particle) and the (smaller because of the greater distance to the Moon) gravity caused by the Moon.

So the surface water on Earth is attracted on both sides where the test particles reside with equal strength to the Earth. But the gravitational influence of the Moon on the water is bigger on the side directed to the Moon than it is on the opposite side. The difference is the greatest for the two test particles.

You would expect that because the water is pulled to the Moon by a little bigger gravitational force on the side of the Earth facing the Moon than on the opposite side (not facing the Moon), a bulge of water, directed to the Moon, will emerge that rotates around the Earth in sync with the Moon's motion.

But we must of course not forget that in this case there are also centrifugal forces that pull on the water due to the rotation of the Earth and the Moon around their CM (this is of course not the rotation of the Earth around its axis which I put zero). The CM between the mass and the Moon lies at 4600(km) from the center of the Earth in the direction of the Moon. The centrifugal force is bigger on the part of the Earth the furthest from the CM, where the gravitational effect of the Moon is the smallest (when the Earth is rotating also the centrifugal forces due to the rotation of the Earth itself come into play, which makes the situation more complicated).

So the question reduces to: What is the ratio between the centrifugal force plus the gravitational force caused by the Moon on the furthest part on Earth to the Moon and the centrifugal force plus the gravitational force caused by the Moon on the closest part of the Earth.

On the far side (from the Moon) of the Earth the Moon's gravitation is smaller but the centrifugal force bigger, while on the close side the Moon's gravitation is bigger and the centrifugal force smaller. Anybody who can do this quite straightforward calculation (for finding the ratio)? One thing is sure: two opposite bulges will develop which each go around the Earth in the same time as the Moon makes one complete cycle around the Earth. The ratio gives us information about the height of the bulbs.

EDIT
I made an obvious error (to make my error clear I let the question as it is). The Earth isn't rotating (as can be read in my question), so no centrifugal forces are present. It's true that, in this case, the CM (4200(km) from the center of the Earth) rotates around the center of the Earth in sync with the rotation of the Moon, but the Earth doesn't rotate around the CM. So there are only tidal forces at work which causes the two bulges rotating around the Earth in one Moon cycle.

Answer 1

The ratio is 1. It is possible, and often done, to analyze the situation as you have described. But it is not necessary to do it that way. You can just think of the Earth as being in free fall toward the Moon. The forces on it are just the same as if it were positioned at the same distance, but with no transverse velocity, so that the two bodies would soon violently slam together. They don't, because the transverse velocity causes them to orbit about each other endlessly, but that transverse velocity does not affect the variation of the gravitational fields, or the acceleration of the Earth, which is directly toward the Moon in either case.

The only thing the enters into the calculation when considered this way is the difference between the Moon's gravitational field on either side of the Earth, and that at the center of the Earth (these are known as tidal forces). When the distance from the Earth to the Moon is much greater than the radius of the Earth (as it is) these differences are very nearly identical.

You will get the same answer using centrifugal force, but it is unnecessarily complicated.

Answer 2

The center of gravity of the Earth orbits around the common center of gravity of the Earth-Moon system. (Of course, the CG of the Moon also orbits around the the common CG of the Earth-Moon system too.) At the center of the Earth, the centrifugal force balances the gravitational force of the moon. Away from the center of the Earth toward the moon, the gravitational force toward the Moon is stronger than the centrifugal force away from the Moon. In the opposite direction, the centrifugal force away from the Moon is stronger than the gravitational force toward the Moon. The radius of the Earth is much smaller than the distance to the Moon, so the net force (away from the center of the earth and along the line connecting the Moon, Earth, and common CG) changes approximately linearly in both directions. So you are right: there will be two bulges: one on each side of the Earth, approximately along that line. The straightforward calculation is up to you. But the answer will be "zero": the bulge is just enough to increase the gravitational attraction of the Earth at the bulge to compensate for the centrifugal force. Of course tidal friction causes the bulge to be slightly delayed as the Earth-Moon system rotates, so "zero" is just a close approximation.