# I Why exactly does the ocean bulge on both sides of the Earth?

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1. Nov 15, 2016

### vector

Here's what my prof says:

"Define F_{mean} to be the mean force, F_close to be the force on the side of the Earth closer to the moon, and F_far to be the force on the side of the Earth furthest away from the moon.

On the closer side the net force is F_close - F_mean > 0
On the further side the net force is F_far - F_mean < 0

So on the far side, the pull is weaker than it is on the rest of the Earth, hence the second bulge."

But I don't understand this explanation, because if the forces on the closest side of the earth and the farthest side of the earth are both pulling forces, then it doesn't make sense that the bulge should occur on both sides, but, rather, intuitively, it would seem that there should only be one bulge, that is on the closest side of the earth to the moon, not two sides.
When I reviewed the process as described mathematically, it made sense, but I would like to understand this on the level of physical intuition. Would appreciate your input.

2. Nov 15, 2016

### Simon Bridge

It's all about POV. If you just have the Moon - and you put two pebbles in space at different radii from the Moon, on the same radial line, then the close pebble will accelerate towards the Moon faster than the far pebble. An observer floating half way between them will accelerate slower than the near pebble and faster than the far pebble. To that observer, then, it will look like the two pebbles are moving away from him.

For more details in a pop-sci format:

3. Nov 15, 2016

### Drakkith

Staff Emeritus
Gravity pulls on the near side harder than it does on the Earth and both the near side and the Earth are pulled harder than the far side. This leads to a difference in acceleration, and, thus, two bulges. If it weren't for the fact that Earth's own gravity pulls on the water on the far side, it would get whipped off into space. Or rather it and the rest of the Earth would wind up in different orbits owing to their different orbital radii. For small objects like asteroids, or large tidal forces like what are generated from being close to stars or other massive objects, this difference in acceleration between the near and far sides can easily pull them apart.

4. Nov 15, 2016

### Simon Bridge

Isn't that an oversimplification?... if that was all there was to it then wouldn't you get noticeable tides in a glass of water (near the equator)? Isn't the acceleration difference across the Earth really small - too small to account for the size of the tide? (The first question would occur to my Y7-8 students and Y10-11s would come up with the second one.)

I think the original question was more about how the Moon's gravity doesn't just pull all the water more to one side than the other.
I was going to go into more detail but found the video so posted than instead. It's worth the watch.

5. Nov 15, 2016

### Staff: Mentor

Simon, I don't see how your explanation and Drakkith's differ: you said acceleration and he said force, but both were pointing out the same phenomena; the difference in force or acceleration between the three points (near, far and center). Afterall, for a unit mass, f=a

6. Nov 15, 2016

### Simon Bridge

My explanation included a video which elaborated on the description I gave in text.
The text is not the whole story. Please see the vid.
(I'd rather not have to type it out...)

7. Nov 15, 2016

### Drakkith

Staff Emeritus
No, I don't think it is.

8. Nov 16, 2016

### A.T.

Depends on how you do the "accounting for" bit. Real world tides water are complicated by many factors, but the OP question is quite basic about the tidal deformation of an idealized object.