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

[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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