How Do Tides Work? | Poal: Say what you want.

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The high tide nearest the Moon is simple enough -- they tell you the Moon pulls the water toward it. OK, fair enough. There's about as much basis to this explanation as there is to the notion of "suction" but at least it seems to make sense. But when it comes to the simultaneous high tide on the opposite side of the Earth, they never explain it. It's counter intuitive that the tide should be high on the side of the planet facing away from the Moon.

How does it not make sense to you? Think of the Earth when that happens. There's a ring around Earth perpendicular to the Moon - Earth axis (imagine the - is a + and the ring around Earth lines up with the vertical line of the plus character). That much water displaced from pulling the water to the side closes with the moon will be insanely enormous, even though it's inches, feet, centimeters whatever, the mass is great. When the ... this is going to be very hard to explain... water pulled by the gravity of the pull on the close half is matched by the force of the water it pulls. The now lower gravity perpendicalr ringe will cause a similar displacement on the opposite half of the Earth that is away from the moon.

' | ' | F-Earth-N ------------------------ Moon ' | ' |

Ignore the ' necessary to preserve placement. So the perpendicular ring I speak of runs through the word "Earth" vertically. The near side of Earth is represented by N. The far side of Earth is represented by F.

So when the Moon pulls water due to it's gravity to the N side it will cause a displacement on the water with subsequently lower pull the farther from the vertex of the N side - the perpendicular ring. With this water "removed" from the ring and now on N a point in which the amount of force to pull more water equals exactly that of the pull back (equal and opposite, it's a law). What this does then is significantly lowers the gravity around the center of the Earth alinged with that ring. This then means the F half of Earth has a higher gravity than the middle section. This will also cause a high tide on the F half, though not as great.

TLDR / don't understand; It comes down to Newton's 3^{^rd} law.

e; Also 2^{^nd} law; https://socratic.org/questions/what-causes-low-tides-and-high-tides-to-occur

>The high tide nearest the Moon is simple enough -- they tell you the Moon pulls the water toward it. OK, fair enough. There's about as much basis to this explanation as there is to the notion of "suction" but at least it seems to make sense. But when it comes to the simultaneous high tide on the opposite side of the Earth, they never explain it. It's counter intuitive that the tide should be high on the side of the planet facing away from the Moon. How does it not make sense to you? Think of the Earth when that happens. There's a ring around Earth perpendicular to the Moon - Earth axis (imagine the - is a + and the ring around Earth lines up with the vertical line of the plus character). That much water displaced from pulling the water to the side closes with the moon will be insanely enormous, even though it's inches, feet, centimeters whatever, the mass is great. When the ... this is going to be very hard to explain... water pulled by the gravity of the pull on the close half is matched by the force of the water it pulls. The now lower gravity perpendicalr ringe will cause a similar displacement on the opposite half of the Earth that is away from the moon. ' | ' | F-Earth-N ------------------------ Moon ' | ' | Ignore the ' necessary to preserve placement. So the perpendicular ring I speak of runs through the word "Earth" vertically. The near side of Earth is represented by N. The far side of Earth is represented by F. So when the Moon pulls water due to it's gravity to the **N** side it will cause a displacement on the water with subsequently lower pull the farther from the vertex of the **N** side - the perpendicular ring. With this water "removed" from the ring and now on N a point in which the amount of force to pull more water equals exactly that of the pull back (equal and opposite, it's a law). What this does then is significantly lowers the gravity around the center of the Earth alinged with that ring. This then means the **F** half of Earth has a higher gravity than the middle section. This will also cause a high tide on the **F** half, though not as great. TLDR / don't understand; It comes down to Newton's 3^^rd law. e; Also 2^^nd law; https://socratic.org/questions/what-causes-low-tides-and-high-tides-to-occur

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