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I'm wondering exactly in which situations forces between bodies are, and are not consider to be tidal forces in the context of planetary and astronomical science.

If two rigid, non-deformable spherical masses with shperically symmetric mass distributions orbit each other, can we say there are no tidal forces or tidal effects there?

If one of them has a permanent, static deformation (e.g. quadrupole moments and higher) are there tidal forces then? If both do?

Or are dynamic, induced deformations required before we invoke the "T-word"?

Can tidal forces be radial, such that they do not tend to raise or lower an orbit over time, or must they have tangential components, like the one that is slowly raising the Moon's orbit?

Is it possible to draw a clear line here, at least within the scope of this SE site and This month's focus tag which is ?

According to that tag's Tag Info:

There is no usage guidance for this tag … yet!


Slightly related: Constraints on the mass distribution within each body such that their mutual orbits are Keplerian?

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    $\begingroup$ "non-deformable" Saying 'these bodies are not going to be distorted by the applied forces' does not mean that there are no forces, just that you've decided to ignore the effects there-of or that those effects are negligible in comparison with some other feature of the situation. $\endgroup$ – dmckee --- ex-moderator kitten Jan 13 '19 at 6:20
  • $\begingroup$ @dmckee I haven't meant to decided anything yet. These are only intended as example demarcations, each adding some additional aspect(s) to the problem. I'm just looking for constraints/guidelines/definitions on effects that should or shouldn't be labeled as "tidal". $\endgroup$ – uhoh Jan 13 '19 at 7:49
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The "tidal force" is simply a manifestation of good old inverse-square-law gravity. It doesn't need a rigid body to be present to manifest.

Consider a giant circular ring of dozens of space stations established in interstellar space. They're far away from any star and just sit there happily doing whatever it was they were set up to do. Suddenly, a star sweeps in from stage left (perhaps a fast-moving star created by a supernova-disrupted close binary system). What happens to the ring of stations?

They are attracted towards the rogue star, of course and start moving noticeably towards it when it gets close enough. They accelerate towards it pulled by the force of the star's gravity, which follows the inverse square law. But they don't all accelerate at the same rate. The station on the edge of the circle closest to the star accelerates fastest, stations half-way around the circle accelerate, but not as much, and the station on the far side of the circle accelerates least.

Soon the circle becomes an oval with its long axis pointing towards the star. This is the tidal effect. Let's say that the rogue star was dark -- a neutron star, perhaps, which had been stripped of its accretion disk by the explosion. What would the people on the stations observe? They wouldn't notice the acceleration per se because they're in free fall. Unless they were doing doppler spectroscopy of distant stars, they wouldn't detect their velocity (at least not at first), but they would see their neat ring of stations turning into an oval. From their point of view, some force was tugging on them and dragging them out of alignment. That's the tidal force.

From the point of view of the omniscient observer (us!) there are no tides, there's just simple gravity. But from the point of vie of the freely falling observers in the gravity field, most of the effects of gravity disappears and what's left looks like a force.

Wikipedia has an article on the tidal force which gives a mathematical treatment. (Personally, I don't think the article is terribly clear, though it appears to be correct.)

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  • $\begingroup$ I think it does, actually. Note that the tidal force exists entirely independently of any bodies, rigid or otherwise. (Ultimately, if I understand what you're getting at, I think your question is completely ill-formed. I was trying to say answer by saying what tides are rather than attacking your question directly.) $\endgroup$ – Mark Olson Jan 13 '19 at 14:46
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I'm wondering exactly in which situations forces between bodies are, and are not consider to be tidal forces in the context of planetary and astronomical science.

Don't overthink it. All a tidal force is, is the net force you get when two gravitational forces aren't quite the same. The force of gravity is the first derivative of potential (the "slope"), whilst the tidal force is the second derivative of potential (the change in "slope"). The classic example is spaghettification. You're falling into a black hole, and the force of gravity is greater at your feet than at your head. So you get stretched by a tidal force.

If two rigid, non-deformable spherical masses with spherically symmetric mass distributions orbit each other, can we say there are no tidal forces or tidal effects there?

No. Like Mark said, the tidal force is independent of your non-deformable spherical masses.

If one of them has a permanent, static deformation (e.g. quadrupole moments and higher) are there tidal forces then? If both do?

There are tidal forces because one side of each object is closer to the other object than the other side is.

Or are dynamic, induced deformations required before we invoke the "T-word"?

No deformations are required!

Can tidal forces be radial, such that they do not tend to raise or lower an orbit over time, or must they have tangential components, like the one that is slowly raising the Moon's orbit?

They're usually radial. The simplest case is where one object is falling straight towards another much larger object. Things get more complicated when you've got two large orbiting objects like the Earth and the Moon.

Is it possible to draw a clear line here, at least within the scope of this SE site and This month's focus tag which is tidal-forces?

Probably not, because to really understand the tidal force you have to understand the force of gravity. A lot of people don't, even though Einstein made it clear. I said something about that yesterday in this answer, but I don't think it was appreciated.

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