You’re making a distinction where there isn’t one; the two situations are analogous. The tidal bulge on the far side of the moon isn’t being pushed away, the Earth is being pulled toward the Moon. The Earth is simply being pulled toward the Moon harder than the water on the far side is because it’s closer. Does that help?
BTW, I neglected the actual tidal forces here (i.e., the difference in force between the two points). But it doesn’t make a difference to the general picture, just the relative size of the bulges. No matter what the tidal configuration, the near side feels less gravity and the far side more.
I’d suppose that if you could subtract out the non-tidal forces, you may well be left with two bulges. But the effect is much less, and dominated by the basic gravitational force.
How, in this thought experiment, are they held into place? If, as Dr. Strangelove has pointed out, you hold just the planet, and not the oceans, into place, as if with a framework of unobtainium, then there’d be no far side bulge. The far side bulge is caused by the Moon’s attraction of the planet being larger than its attraction of the far side water, but if you hold the planet in place somehow, the gravitational pull between the rocky bodies no longer influence their distance.
The part that makes even less sense to me is that the portions of the ocean that are on the sides are actually squeezed in. Is this not right?
I’ve given up trying to understand it. I just accept it and move on. I hope my child never asks me to explain it.
I have been greatly made aware of my ignorance, something I really appreciate. I had no idea the tide worked like this. My mind is blown and I am happy for it.
Thank you SDMB!
It seems I was wrong above - my apologies. Now I am truely perplexed. The patient explanations provided by posters almost penetrate this dim mind but damned if I can really understand it.
The sea has to be rotating with the planet. That is a lot of mass which wants to move in a straight line away from the Earth, but gravity pulls it back. However if I read the above correctly, this outward force is irrelevant?
Here’s a simple way of looking at it. There are two ways of making the water appear higher - you pull the water up or you pull the earth down.
The near tide is caused by the first effect - the water is closer to the moon than the earth is (counting the earth as a solid object with its center of gravity in its midpoint) so the water gets pulled up towards the moon. Meanwhile on the opposite side, the earth is closer to the moon than the water is so it gets pulled down towards the moon. Both effects create a high tide.
They aren’t squeezed in, except in the sense that we’re all squeezed by gravity.
That outward force doesn’t exist. It seems you’re thinking centrifugal force, like the Earth is a bucket, gravity is the handle, and you swing the bucket around. But unlike when twirling a bucket around gravity works on the whole shebang. So yes, the water wants to move in a straight line, but so does Earth, and they’re both affected by the same forces, so inertia doesn’t explain the tides.
I assume you’re talking about Figure 5 in Musicats’s link. That’s due to the Moon as well.
The Moon’s pull is straight to the right at points A and B, but it isn’t straight to the right at the top and bottom of the Moon. It’s pulling towards the Moon, so in Figure 3, if you drew in arrows at the top and bottom of the Moon, they’d be the same length as the one in the center of the Moon, but angled a little bit down and up.
When you move to figure 4 by subtracting the center arrow, they don’t completely cancel, because they aren’t parallel. What’s left, that doesn’t cancel, is the vertical component of the top and bottom arrow, so there’s a tidal pull towards the Earth. The magnitude of that downward pull is about 1/2 the magnitude of the upward pull at the points of the Earth below the Moon and opposite the Moon. So it accounts for 1/3 of the Earth’s tides.