Is my weight during the night a little bit more than during the day?

[A.T.]

I mentioned near the top that considering it as a system in free fall, the two bulges will separate out and remain separated out because of the orbit.

Not because of the orbit, but because of the gravity gradient. The two bulges would exist in linear free fall as well.

With no orbit, you would just get a linear, one - off free fall then a crash.

Who cares for how long you would have two bulges? The point is that would still have two bulges, without the orbit around the barycentre. And of course you could keep them longer, if you move the Moon away by some other force, as @DrStupid points out.

The period depends on all the masses involved and the separation.

The tidal period is 12h because of 24h Earth's spin period, and has nothing to do with the 27d orbit period.

 

[sophiecentaur]

There’s no question that the Moon’s field varies with distance. Also the spin of an object is not too much affected. There are two ‘periods’ involved with our tides. The lunar orbital period determines the tidal height range. (I can’t believe we are revisiting this yet again). The range of the Moon’s gravity on either side of the Earth is not arbitrary. It’s due to the monthly period which depends on the two masses. A different orbital period / separation would mean different tidal max min.

 

[A.T.]

The lunar orbital period determines the tidal height range.

The gravitational gradient determines the tidal height range.

The range of the Moon’s gravity on either side of the Earth is not arbitrary. It’s due to the monthly period which depends on the two masses.

This is confusing relation and causation.

The combination of masses and distance determines both:
- The gravitational gradient which determines the tides
- The orbit period for a nearly circular orbit

That doesn't imply that the tides are caused by the orbiting, or depend on the orbital period. In particular the tides don't require a circular orbit or any orbiting at all.

 

[sophiecentaur]

The gravitational gradient determines the tidal height range.

And what determines the gradient? Is it independent of distance?

The combination of masses and distance determines both:
- The gravitational gradient which determines the tides
- The orbit period for a nearly circular orbit

You can rearrange those to taste. What determines the distance?

I have to disagree that you would have time for any tides if there was not rotation of the Earth Moon system. You would just get 'stretch'.

This is confusing relation and causation.

?

 

[sophiecentaur]

Assuming that the lump of goo rotates like the Earth, yes.

If the goo was spinning then the more distant parts would lag and the nearer parts would lead. It would form a spiral, I think. An added complication.

 

[Orodruin]

The lunar orbital period determines the tidal height range. (I can’t believe we are revisiting this yet again).

It does not. It is determined by the tidal force strengths. The one-bulge of your stationary Earth is conpletely canceled out by the free fall mechanics and therefore has absolutely nothing to do with the amplitude of the tidal bulge. This has been pointed out to you repeatedly but you simply won't accept it. You enter the free fall and the same tidal forces remain regardless of whether you are in circular orbit or in linear free fall. Indeed, I really cannot believe that we are having this conversation.

 

[sophiecentaur]

You imply that the distance is not relevant? Of course the orbit affects the Moon’s field or you would get the same (quantitative) effect anywhere.
if you have decided that my way of looking at this is wrong then try re-reading what I have written.

 

[Orodruin]

You imply that the distance is not relevant?

I did not and I do not see how you could possibly read that into my post. Tidal forces depend on distance as 1/r^3, which is well known.

if you have decided that my way of looking at this is wrong then try re-reading what I have written.

I have, several times, and honestly - I believe your posts in this thread are bordering on misinformation.

 

[DrStupid]

The lunar orbital period determines the tidal height range.

As it seems we are running in circles, let me try it with an analogy:

Storks Deliver Babies

Urbanisation determines both, the birth rate and the stork population. Thus, there is a strong correlation between them. You can conclude from one to the other. But that does not mean that the stork population determines the birth rate.

Now let's come back to the tides: As @A.T. told you in #38, the combination of masses and distance determines both, the tidel hight range and the orbit period. Thus there is a strong correlation between them. You can conclude from one to the other. But that does not mean that the orbital period determines the tidal height range.

 

[A.T.]

I have to disagree that you would have time for any tides if there was not rotation of the Earth Moon system. You would just get 'stretch'.

Why do you keep ignoring the Earth's spin around it's own axis in your strawmans. If the Earth is being stretched along a line, while spinning around it's own axis, how can you not have tidal periods on the Earth's surface?

 

[sophiecentaur]

Why do you keep ignoring the Earth's spin around it's own axis in your strawmans.

Way back, I agreed / stated that the daily tide rate relates to the spin. It's pretty much independent of what actually causes the tide; the same effect would be there with a solar day of 30 hours and the bulges would have the same heights. Would you not agree? So I considered that taken care of in this context. No need to mention the resulting change to Spring and Neap tides.

In an orbital system, there are a number of variables but, somehow, people are claiming that some variables are more important than others.

 

[Orodruin]

The lunar orbital period determines the tidal height range.

Ok, so let us take this statement at face value and compare it to the Sun-Earth system. By the same reasoning you would arrive at the orbital period would determine the height of the corresponding tide. This is about a factor 12 different from the lunar orbital period. Yet, the effect is about half the size of the lunar effect. Similarly, the orbital period depends on the reduced mass of the system in relation to the system’s total mass, whereas the actual tidal effect only depends on the mass of the other body. Your assertions here simply do not compute.

 

[A.T.]

Way back, I agreed / stated that the daily tide rate relates to the spin.

Your claim was:

The two cycles a day is due to the Earth moving around the Barycentre too.

Is this dispelled now?

 

[sophiecentaur]

Your claim was:

Is this dispelled now?

There would be one tide but for the orbital motion. The Moon on a long pole, with no rotation, would cause a single bulge - towards it. I'm not sure what I should retract. Just not using your language, perhaps?

 

[Orodruin]

There would be one tide but for the orbital motion. The Moon on a long pole, with no rotation, would cause a single bulge - towards it. I'm not sure what I should retract. Just not using your language, perhaps?

Again, missing the point that the Moon does not need to be in orbit to produce the tide. The case with the Moon on a pole is completely irrelevant as the Moon is not on a pole, but the Earth is in free fall in an inhomogeneous gravitational field, which is the important thing. If you could somehow produce a time dependent, but homogeneous gravitational field producing the same kind of orbit for the Earth, then there would be no tides.

Your argumentation produces faulty results as has already been pointed out. That is the time when you should start retracting statements instead of doubling down.

 

[A.T.]

There would be one tide but for the orbital motion.

This is wrong, as the example of linear fall shows. It has no orbiting but still two tides for each spin around the Earth's own axis:

Not really. Even if Moon and Earth were falling straight towards each other along a line (no motion around the common Barycentre), the Earth would still be stretched along that line (due to the Moon's gravity gradient), and have two bulges, and two tidal circles per one rotation around its own axis.

In order to avoid the collision you could attach thrusters on the Moon and use it as gravity tractor for Earth. There would still be two tidal bulges.

In my example above (Moon powered with rocket engines to the same acceleration as Earth) there is no crash and no orbit but the same stretching as in orbit. All that matters is the gravitational field of the Moon and the relative position of Earth in this field. Orbit or not makes no difference.

Do you really claim there would be just one tide for each spin around the Earth's own axis in the above scenario?

 

[sophiecentaur]

Do you really claim there would be just one tide for each spin around the Earth's own axis in the above scenario?

In the scenario where the Moon is not accelerating relative to the Earth, the Field of the Moon is always towards the Moon. So I can't understand how there can be any resultant force on the water at the opposite side of the Earth, to pull / push water 'upwards'. I appreciate that, when in free fall, the Earth will be pulled to the Moon with greater acceleration than the water on the far side.

I can't argue against the criticism that it's unrealistic but, if the 'rocket' scenario is allowable , what can cause the outward bulge? Whatever the gravity gradient, it's still in the same direction. The direction of gravitational force from the Moon would always be in the same direction but the gravitational force from the Earth would be in different directions on different sides.

I appreciate that a direct hit is just a special case of an orbit so you could call the result a 'tide' but it would change height as the distance reduces and it would be temporary.

 

[Orodruin]

Whatever the gravity gradient, it's still in the same direction.

Again you fail to see the point that the Earth is being accelerated too by the gravitational field. The tide is a result of the outer parts being accelerated less and therefore, relative to the frame of the Earth, there is a force outwards on the far side. This is explained by any of the many videos you will find if you just search for "Tides" on YouTube. Taking one of the first hits (and skipping the first hit by Niel de Grasse Tyson because its Niel de Grasse Tyson and the video is a bit ... well NGT - not because it says anything else):

 

[A.T.]

So I can't understand how there can be any resultant force on the water at the opposite side of the Earth, to pull / push water 'upwards'.

what can cause the outward bulge?

The gravitational gradient stretches things. If you stretch a sphere you get a sphere with two bulges.

In the inertial frame there is no resultant force in the opposite direction on the far side. The far side still accelerates towards the Moon. But since the Moon pulls it less, it must be pulled more by the rest of the Earth, hence deformation.

 

[Ibix]

There would be one tide but for the orbital motion.

There would be one tide if the Earth's center of mass were not in freefall, which was the point I thought you were making back on page 1. (I thought that was wrong, but I was wrong myself.) The orbital motion of the moon affects the timing of the tides slightly, but a vertically falling moon (assuming it took more than a day to fall) would produce two tides too.

 

[sophiecentaur]

The gravitational gradient stretches things. If you stretch a sphere you get a sphere with two bulges.

We already went into this and I suggested, instead of a sphere (rigid implied) with water round it, we used goo there in free fall and that the goo would stretch. That's from why back in the thread and I thought that was all sorted. What exactly are the ground rules about this? You are quoting disembodied sentences of mine with no context.

If you won't allow a rigid coupling between Earth and Moon (or the equivalent rocket motor) then the question about upward and downward forces can't apply - of course. But if you take a rigid dumbbell (make it simple as possible) with water round one of the spheres, are you still telling me there will be an outward bulge on the water? How can this be? Surely such a system would have a flattened ocean on the other side. Sorry if you think it's a non sequitur but an isolated statement about stretching things in a gravitational gradient needs to be put in context. Why can't I suggest this rigid model - with just one flexible element? And there is no point in saying that it's not the way things are - simplified models are allowed in Physics. You don't seem to realize that I am not arguing against your correct model because the goo argument and free fall are fine by me.
Perhaps we should just stop now.

 

[Orodruin]

Why can't I suggest this rigid model - with just one flexible element?

Because it has very little to do with how tides actually work.

simplified models are allowed in Physics

Simplified models are useful when the demonstrate a particular phenomenon and make qualitatively correct predictions. This is not the case here. It is predicting a single bulge instead of two and does not at all demonstrate the basic physics behind the actual phenomenon. It has furthermore led you to make incorrect statements about the lunar orbital period determining the tidal amplitude, etc. I would say those are pretty good reasons not to use this particular simple model.

But either way, this thread has long since ran its course.

 

[berkeman]

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