Discussion in 'Physics & Math' started by DrZygote214, Jul 18, 2014.
Right: billvon is. That's what you misunderstood.
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No, I didn't misunderstand, but I think you're both talking about the same thing, just different viewpoints, and for the general edumacation of the thread, I hope to reconcile them.
Can you to elaborate on this? Can you explain what force "from the Moon" would pull water away from the far side of the Earth i.e. away from the Moon?
I'm going to speed this up by anticipating your response. Please do not hesitate to correct me if I presume too much.
I presume you're going to say that "the Moon does not impart a force on the farside water, pulling it away from the Moon - rather - the Moon imparts a force on the Earth, pulling it away from the farside water (this is because the Earth is in a higher gravitational potential than the farside water is)".
Did I anticipate correctly? If not, I'll redact the whole thing and let you speak for yourself.
Ok...well, if you say so. So far I haven't seen anything to suggest that you understood what billvon and I were arguing about.
Your understanding is correct.
What does the Moon do to the Earth, to cause it to pull away from the farside water? It accelerates it, yes?
I contend that the farside water lags behind the Earth's acceleration toward the Moon. From a stationary observer's POV, they would see Earth AND farside water fall toward the Moon, the Earth falling faster.
It is only from the accelerated reference frame of the Earth that a force appears to be acting upon the farside water to lift it into a bulge. In fact, it is simply the acceleration of the Earth (under higher g) minus the acceleration of the farside water (under lower g).
So now we set it up so that the Earth does not fall into the Moon. They are set rotating. Now the Earth remains a constant distance from the Moon. But if the Earth is no longer falling toward the Moon, how can it pull away from the farside water? It still needs to accelerate to do so.
Rotation is acceleration. (I know this is child's-play for you, forgive the pedantism.)
The Earth is accelerating around the Moon. The farside bulge is also accelerating around the Moon, but lags behind.
There is no actual tidal force - it is simply the difference between two gravitational forces, that when seen from the Earth, appears to be its own force.
You made a bunch of errors there...
1. The earth is not "accelerating around the moon", it is accelerating toward the moon. You described it better in previous posts.
2. The Earth and moon are revolving, not "rotating" around their common center of gravity.
3. "Rotation is acceleration" is too non-descriptive to be useful.
...but generally your logic is correct, despite your tripping over a bunch of the terminology. Unfortunately, then, you wrap up with a bizarre error:
If I can measure it on a spring-scale, it is an "actual" force. It is real. Furthermore, billvon knows it is real. So none of this has anything to do with billvon's error!
This new error you're introducing appears to be related to a misunderstanding of peoples' complaints about the centrifugal force. The centrifugal force is not said to be "fictitious" because there is no "actual" force at play, it is said to be "fictitious" because it is pointing in the wrong direction when viewed from an inertial frame.
I've got one piece to go yet. I haven't yet reconciled it with billvon's contention that it is merely inertia (in a rotating frame) that lifts the farside water. That gravitational gradient hasn't actually gone away yet.
Well, the sum of the inertial component with the overall gravitational field of both the Earth and the Moon.
Accelerating toward it but missing it because of its proper motion. Better?
The system is set rotating. The individual bodies are revolving around each other.
I took it for granted that you knew that one.
Now I think you're being pedantic, but i'll let it pass.
Why would "pointing in the wrong direction" result in it being labelled "fictitious"? Those two concepts are worlds apart.
It is fictitious because it raises more complications than it solves. Like" where does the force come from that reconciles the difference between pointing where we think it should be pointing, and pointing where it is pointing.
In fact, you are moving in the right direction but the wrong cause is applied to it (your inertia, driving you in a straight line).
EDIT: Russ, we can drop this, or sidebar it. It's a side discussion about the labels of forces in rotating frames of reference (which I acknowledge, are perfectly valid way to analyze some problems. (Some problems.)). It actually has no impact on the main discussion. We don't disagree on the outcome, simply on the semantics.
That's where I've come up short. No matter how you cut it, rotation or not, the gravitational potential at the location the farside water is indeed less than that of the Earth. Without considering anything else, we can state that this should result in a tide. Since the Earth and the bulges are symmetrical, so is the potential difference, so will the tides be.
That's fine. And if you think I'm being pedantic there, this distinction is very important here and will come back up when (if) you get to billvon's error.
Agreed -- that's why I don't like it when people call it "fictitious". It often leads to confusion*. There has been much debate on the issue and I'm not completely clear about the history of where the label came from. But in either case, there is no disagreement here relevant to the thread, so I don't see any value in discussing the history of the term "fictitious force".
No, it is called fictitious because it is an inertial force instead of an applied force. In an accelerated frame, you replace the applied force with something akin to gravity to transform it away. Whether that makes analysis more or less complicated depends on what you are using it for.
*Student: But if I can feel it pressing me against the side of the car, then how can it be called "fictitious"?
Teacher: that isn't why it is called "fictitious".
Agreed - unless the two planets are fixed in place _without_ rotation or revolution. Then there is no inertial force (or rotational force, or centrifugal force, or whatever you prefer to call it) acting on the water on the far side of the planet. In that case the highest tide will be solely on the side of the planet facing the Moon, with the minimum tide on the far side of the planet.
This situation is almost too nonsensical to even be wrong (google: "not even wrong"). If the planets are fixed in place, all of the water - and anything else that will move - will slide over to the side facing the other planet. This isn't and has nothing to do with a tidal force and the tidal force does indeed still exist even in this situation. All you've done is create a new force that is vastly stronger to mask it.
You also appear to have forgotten to exclude non-orbit freefall from your options...
Well, no, no more so than normal tides make everything "slide over to the side facing the moon." The total gravity caused by the moon would be a tiny fraction of the gravity caused by the earth, when measured at the surface of the earth. It would cause a "tide" on the near side that was much larger than a normal tide, but you would need good instruments to measure the change in apparent gravity.
No, you've just decomposed the forces. Now there is _only_ the moon's gravitational pull. You still have the gravitational gradient acting across the Earth, but since the Earth is not free to move, there is no inertial force on the opposite side acting against gravity and causing the local apparent gravity to decrease.
Yes...and depending on the specifics of the scenario, the water could all slosh all the way off of one body and onto the other. Remember: the earth isn't the only body in the solar system that experiences tides, so you should be able to apply your changed scenario to any of them.
Either way, the size of the "bulge" you are describing bears no relation to the size of a "tidal bulge" because..... it isn't and has nothing to do with a "tidal bulge".
Really? Prove it: calculate the forces we're talking about. For example, what is the net force (magnitude and direction) on a kg of water perpendicular to the moon?
Hint: if the direction isn't straight down, it will move...
Then, calculate the magnitude of the force on a kg of water in the "nearside bulge" and compare it to the magnitude of the tidal force.
You omitted the next sentence of my post from your quote of me: All you've done is create a new force that is vastly stronger to mask it. The Earth isn't free to move because you applied a new force to it.
Anyway, creating a new scenario that is also wrong but has nothing to do with the tides doesn't help you here. It's just more errors on top of your previous errors.
Yes, you could. However, unless the planet is close to its Roche limit, no water will be "sloshing off one body and onto the other."
It is one of the three forces which, summed together, produces the force you call "tidal force." In terms of the Earth/Moon system they are:
1) Gravity of the Earth. This is obvious, and is by far the dominant force on any water on the Earth.
2) Gravity of the Moon. This causes the bulge towards the Moon.
3) Inertial/centrifugal effect on the water on the opposite side of the Earth. This causes the bulge away from the Moon.
You mean the force on a kilogram of water 90 degrees from the moon's position? The total force is about .00003 newtons. Thus if you normally stood at a 90 degree angle to the surface of the Earth at that location, you would have to stand at an 89.9998 degree angle to counteract the force. Put another way, things near you would see a .0002 degree slope. Hardly enough to make everything "slide over to the side facing the other planet." (Although plenty to cause a slow movement of water to the location of the "bulge.")
Exactly. And when this is caused by the gravity of the moon, we call this a "tide."
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