[Halc]

The next question is an interesting one. You introduce the spin of a planet eg Venus that changes the planet’s surface speed. However, I would argue, as I suggested to Le Repteux, that the spin has no effect on the orbit.

I didn’t suggest that the spin has any (short term) effect on the orbit.  Yes, Le Repteux has a thread open about tidal forces slowly pushing orbiting things away, and that is true.  The spin of the sun has a higher angular velocity than any of the planets, so that tide slowly pushes each of them away.  Most planets also spin faster than their orbits, putting thrust on the sun that also contributes to higher orbits.  Venus is an exception there, and its negative spin actually degrades its orbit, but not as much as the tide on the sun expands that orbit.  All this is relevant to the other thread, but your response above seems to concern this point.

The discussion was about tides, with a suggestion that the tides might be partially caused by higher linear speeds at points furthest out, but I pointed out that points on Venus furthest from the sun have the lowest linear speed and should be ‘seeking lower orbit’, and the points closest to the sun have the greatest linear speed and thus should be ‘seeking higher orbit’.  If that were so, the tides on Venus would be to the sides, not towards and away from the sun as all solar tides are.

If you consider the locus of points on the surface the spin does not cause any net increase or decrease in the orbital speed and it is orbital speed which is the prime cause of any centrifugal forces. At any instant in time the forces on any mass on the earth/moon line are the gravitational pull and the centrifugal force and we can treat the spin separately because, it has no effect on the net orbital speed of any part of the mass. Also, as I mentioned to Le Repteux we can treat the spin as a separate component of the overall system and it's only effect is to create centrifugal force causing an equal bulge on the equator of the spin, it does not cause a net motion towards or away from the earth.

All this seems to be about orbital speed, something on which I was not commenting in this topic, until what I wrote above in this post now.

The comment of mine that you quoted in the immediately preceding post concerned centrifugal explanations for the tides.  Instead, I agree with your last line there that we can treat the spin as a separate component of the overall system and yes, all it does is that standing bulge everywhere at the equator.
Tides are partially an effect of spin.  The spin rate coupled with the resonant frequency of oceans and shorelines causes higher tides in some places than others.  Move the continents or alter the spin and the tides will be higher in different places.  But high tide is always towards and away from the gravity gradient and the bulge will form regardless of the axis or intensity of spin.

[rmolnav]

The spin of the sun has a higher angular velocity than any of the planets, so that tide slowly pushes each of them away.  Most planets also spin faster than their orbits, putting thrust on the sun that also contributes to higher orbits

Sorry, but I can´t understand what you mean ...

The discussion was about tides, with a suggestion that the tides might be partially caused by higher linear speeds at points furthest out

Colin2B and me have said several times that idea (about actual linear speeds and orbits), considering as a whole the spinning added to the revolving, is rather a "bad" approach ... Especially when, as in our case of earth tides, linear speeds caused by those different movements are so different ...
Keep in mind that, on the one hand, v (linear speed) = ω r (r: radius of curvature, app. the distance from center of each path to the earth considered point).
The angular speed ω of the spinning is some 28 times bigger than the one of the revolving.
And the radius some 3/2 times bigger too.
And on the other hand, related centrifugal forces are proportional to ω²r ... Enormously higher in the spinning !!
"Normal" bulges are actually quite insignificant compared to the so called equatorial bulge !!
And to tackle earth tides as in what quoted is utterly absurd !!
And regarding Venus special case, I´m afraid you say wrong things, but prefer not to refute them without having quite clear what on my first phrase in bold ... 

[Le Repteux]

I just had a flash. If the earth would stop spinning, the equatorial bulge would get down and the tidal bulges would follow it, but for my proposition to be true, they should also disappear, and eventually reappear if ever the orbital speed was also stopped because then, the two sides of the earth would not be accelerating at the same rate towards the moon. If only the spinning is stopped, then the only fleeing motion left is due to orbital speed, and all the different parts of the earth are fleeing almost the same distance away from the moon in the same time, but there is still a difference in the force from the moon on those different parts since they are not necessarily at the same distance from it: if the cg of the earth is pulled a certain distance towards the moon to absorb the fleeing motion, then the outer part is pulled less and the inner one more. This way, contrary to what I was proposing, the spinning would have effectively no effect on the tidal bulges, and it seems to match David's explanation, but I'm not sure rmolnav's will agree with it since he seems to need a centrifugal force where I only need a tangential motion.

[Colin2B]

All this seems to be about orbital speed, something on which I was not commenting in this topic

In that case we are talking at cross purposes and misunderstanding each other; my reply to @Le Repteux  was only to do with orbital speed.

I just had a flash. If the earth would stop spinning, the equatorial bulge would get down and the tidal bulges would follow it,   ............. contrary to what I was proposing, the spinning would have effectively no effect on the tidal bulges,

The spinning has no effect on tides whatsoever. The centrifugal force causing the equatorial bulge is balanced by earth’s gravity so the ‘new’ level is the geoid ie sea level and any tide raising forces produce an increase relative to this. Not sure if that is what you were trying to say, if so my comments are only for clarification.

[rmolnav]

The sun spins every 25 days, which is faster than the orbit of any planet (88 days of Mercury being the fastest), so the tides on the sun push all of the planets outward, and correspondingly slow the sun’s spin.
Quote
And regarding Venus special case, I´m afraid you say wrong things, but prefer not to refute them without having quite clear what on my first phrase in bold ...
In my quote above your bolded phrase, I was speaking of the tides on the sun caused by any planet (including Venus), not of the tides on any particular planet.  Venus is not a special case in this regard.

But those "tides" on the sun ...
1) ... are almost negligible, even the ones caused by biggest planets ...
2) ... given the distances involved, tangential component of each "bulge" would also be quite negligible ...
2) ... those "individual" bulges would be scattered around sun´s surface kind of at random (and varying with time) ...
3) ... and therefore it is not like the earth-moon scenario whatsoever, where we continuously have a gap between a bulge and actual sublunar meridian (the bulge ALWAYS eastwards from the moon), which is what pulls forward the moon, not just earth eastward spinning ...
So, to extrapolate earth-moon scenario to sun-planets case is quite erroneous ...   

[Le Repteux]

The centrifugal force causing the equatorial bulge is balanced by earth’s gravity so the ‘new’ level is the geoid ie sea level and any tide raising forces produce an increase relative to this. Not sure if that is what you were trying to say, if so my comments are only for clarification.

Yes, that's what I was saying about the spinning, but as I said, I prefer to use the concept of centrifugal motion instead of centrifugal force, just in case @rmolnav would understand what @David Cooper was telling him. If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it, and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation, but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass, as if it was at rest at a certain distance away from the circle. Of course it is different if the trajectory is an ellipse, because then, the force is not often perpendicular to the tangential vector and part of it serves to accelerate or decelerate the body tangentially to its trajectory.

[rmolnav]

If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it, and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation, but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass, as if it was at rest at a certain distance away from the circle.

1) "If we draw a tangential speed vector on a circle, the tip of the arrow is where the body would be if no force was applied on it ...": "Almost" quite right: you should have added "after a second" (or any other used unit of time).
2) "... and the force produces the acceleration needed to bring it back on the circle in the same time it took to get away. That centripetal force can be called inertial if the body is held by a rope, or gravitational if it held by gravitation ...": quite right the first phrase, but not the second.
 The centripetal force that makes an object rotate (instead of following in the direction of a previously acquired "tangential" speed), is always an "active" force (not to be called "inertial"), both if gravitational or the tension of a rope ...
3) "...  but to me, we shouldn't oppose it to a centrifugal one because it is a bit misleading. We could if the body's speed was affected by it, but it is not the case since it is perpendicular to the tangential motion. The only opposition the force faces is from the body's mass..."
It´s not "we" who "oppose it (the centripetal f.) to a centrifugal one" ... Centrifugal f. is the way "inertia" (certainly due to the body´s mass, and its speed) can manifest itself, not always in the same way ...
Even being "perpendicular to the tangential motion", it could for instance cause deformations ...
Rather than "a bit misleading" I would say it is something difficult to grasp, let alone to successfully convey to others ... I´ll keep trying. As I´ve already said, I have some "fresh" ideas for it, but prefer not to post them until having them better elaborated, precisely not to "mislead" anybody (or the fewer the better ...)

[Le Repteux]

Centrifugal f. is the way "inertia" (certainly due to the body´s mass, and its speed) can manifest itself, not always in the same way ...
Even being "perpendicular to the tangential motion", it could for instance cause deformations ...

A force causes an acceleration in the direction it is applied, whether the body is moving or not. We can even consider that the body is at rest to measure the force. What opposes the force is the mass of the body, and that body has to move for the force to be measured. When the force pulls on the body, the body moves towards the force, not away, and this motion is not clear when studying rotation. We may still imagine that a rotation produces an outward force, thus centrifugal, but it is an outward motion that it produces, and that motion is then pulled back in the direction of the force. If that outward motion is due to the spinning earth, then it produces equatorial bulges, and if it is due to its orbital motion, then it produces the tides, but it is the same kind of motion and the same kind of force, so there is no need to separate them, and it is what you seem to be doing.

[rmolnav]

We may still imagine that a rotation produces an outward force, thus centrifugal, but it is an outward motion that it produces, and that motion is then pulled back in the direction of the force. If that outward motion is due to the spinning earth, then it produces equatorial bulges, and if it is due to its orbital motion, then it produces the tides, but it is the same kind of motion and the same kind of force, so there is no need to separate them, and it is what you seem to be doing.

Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...
But within layman language, I would tell you those water motions (when equatorial bulge and tides) happen in opposition to own water weight (the most important force there), and after the imagined tangential movement and centripetal force pulling back, water remains further from earth CG, actually weighing a little bit less.
I, and eminent physicist, call that diminishing of weight centrifugal force ... If you prefer, I also could call it an "outward" force, certainly one of the ways inertia can manifest itself ...
 

[rmolnav]

Just some additional details about what said on my last post:

...those water motions (when equatorial bulge and tides) happen in opposition to own water weight (the most important force there), and after the imagined tangential movement and centripetal force pulling back, water remains further from earth CG, actually weighing a little bit less

What in bold refers only to antipodal bulge ...
As I´ve said many times, considering only the earth revolving around moon-earth barycenter, all earth particles follow identical circles, and their locations (within own circular path) are ALWAYS the farthest from the moon. That implies centrifugal force is always in the sense opposite to the moon.
Therefore, where sublunar bulge water (and solid elements) weights actually increase a tiny little bit ...(due to that inertial effect, the centrifugal force, in same sense as earth´s own pull).
But as moon´s pull there is bigger than that inertial force, sublunar bulge builds (towards the moon).
But at the antipodes moon´s pull is smaller than centrifugal force, the later prevails, and also a bulge (in the sense opposite to the moon) builds there.
In the case of the so called equatorial bulge all centrifugal forces are radially in the "fugal" sense from earth c.g., and the only existing gravitational pull is own weight of earth particles, uniform around the equator ... That´s why there the weights of all particles diminish, proportionally to ω²r (r the distance to earth axis of spinning, the closer to the equator the bigger ...).
Solid earth get deformed (angular speed ω is some 28 times bigger than in the moon-earth dance), and water from both hemispheres piles up towards the equator.

 
 

[Le Repteux]

Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...

During rotation, the motion is not directly outwards, it is tangential, and the distance being increasing if we cut the force is thus only incidental. That motion is due to inertia, reason why we call it inertial. While saying "an outward motion is not a proper physical variable", what do you mean exactly? Do you mean that inertial motion is not physical?

[rmolnav]

Well, first of all, an "outward motion" is not a proper physical variable, let alone a force ...

During rotation, the motion is not directly outwards, it is tangential, and the distance being increasing if we cut the force is thus only incidental. That motion is due to inertia, reason why we call it inertial. While saying "an outward motion is not a proper physical variable", what do you mean exactly? Do you mean that inertial motion is not physical?

An adjective such as "inertial" can be used scientifically, or in a broader layman sense ...
Therefore, in that last sense, you can use it for any noun with almost any relation to inertia ...
But in physics science, the "variables" can be speed, mass, acceleration, force, momentum ... all with their own units. Never just a "motion" ...
And that "motion" is not actually due to "inertia", it is due to an initial speed, whatever its cause.
Due to "inertia", objects "tend" to keep constant their velocity vector. If no force opposes that inertial "tendency", objects keep moving without any "problem". But if objects are "forced" to change their velocity vector, they somehow react (they are kind of stubborn: "no, no, I don´t want to change my velocity vector"...). And inertia manifests itself, in different ways ...
In our case, inertia manifests itself as a force, equal but opposite to the centripetal force, the "guilty" for the velocity vector change ... An outward force ... centrifugal (in its broad sense) because it is kind of "fugitive" (< > fugal) from the moon (the "center" of gravitational pull, source of acting centripetal forces ...)

[Le Repteux]

In our case, inertia manifests itself as a force, equal but opposite to the centripetal force, the "guilty" for the velocity vector change ... An outward force ... centrifugal (in its broad sense) because it is kind of "fugitive" (< > fugal) from the moon (the "center" of gravitational pull, source of acting centripetal forces ...)

To me, the definition of inertia is a bit weird: a body goes straight ahead because it has inertia, and it resists to change direction or speed also because it has inertia. A force is not a motion, and inertia is about both, so it means that it is probably still misunderstood. When nothing forces a body to change direction or speed, it goes straight ahead, as if something inside it would know how to move this way, and whenever something is on the way, it resists to change direction or speed, but it still changes them after a while if it is free to move, and it is probably that time that we call inertia. If it would change instantly, there would be no resistance and no inertia, and there would be no motion either. We can't measure it at our scale, but if I'm right about inertia, the atoms are probably going straight ahead for a while during rotation for us to be able to measure a force, and they probably feel no force during that time otherwise they would not be free to move. That's what makes me think that the equatorial bulge and the tidal ones are driven by the same phenomenon, and I think that you could admit it on that basis, but you probably have another viewpoint on inertia that makes you think your way. We behave like atoms as far as inertia is concerned, we all resist to change, and we all finally change since we are free to move, but it takes time. The problem with us is that even if we cannot know if the direction we took is the right one, we still think it is if nothing forces us to change it, but why would an atom change its direction or speed when it is not forced to? :0)

[rmolnav]

... a body goes straight ahead because it has inertia, and it resists to change direction or speed also because it has inertia

To me those two phrases of you are kind of the two sides of a coin: the body, either "feels" a null total force exerted on it, or it feels a not null force ... Inertia phenomenon makes it maintain its velocity vector (including the case of a null vector) in the first case (without any other "inertial" effect), or, in the second case, it experiences an acceleration vector (a=f/m) ... It is in this last case when other inertial effects may occur, not always the same way, depending on the individual forces acting on the object (the breakdown of mentioned total force vector).
 

When nothing forces a body to change direction or speed, it goes straight ahead, as if something inside it would know how to move this way, and whenever something is on the way, it resists to change direction or speed, but it still changes them after a while if it is free to move, and it is probably that time that we call inertia. If it would change instantly, there would be no resistance and no inertia, and there would be no motion either .

Sorry, but that´s utterly erroneous !!
Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ... (dv=a*dt: infinitesimal speed vector change equals to acceleration vector multiplied by the infinitesimal fraction of time the total force has been acting).
And please, try not to make more difficult for people to understand the already "tricky" concept of "inertia" saying things such as "it is probably that time that we call inertia" ...!!

[Le Repteux]

Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ...

If nothing is faster than light, then nothing changes instantly, thus when we apply a force on a body, it necessarily takes a while before it changes speed or direction. If bodies could move instantly, we wouldn't have the time to feel any force, and they would go from one point to the other without going through the transitional ones. Now, if bodies can absolutely not change speed or direction instantly, it is straightforward to deduce that this may be the cause for their resistance to change, and I don't feel I'm misleading people while suggesting that. You may not be tempted to discuss that point, but others might be interested. On the other hand, if you could just admit that possibility for a while, I think you may understand more easily what I'm saying about rotational motion.

[rmolnav]

Since the very first instant (an infinitesimal fraction of time), mentioned acceleration vector changes the object speed, though logically only an infinitesimal amount of change ...

If nothing is faster than light, then nothing changes instantly, thus when we apply a force on a body, it necessarily takes a while before it changes speed or direction. If bodies could move instantly, we wouldn't have the time to feel any force, and they would go from one point to the other without going through the transitional ones. Now, if bodies can absolutely not change speed or direction instantly, it is straightforward to deduce that this may be the cause for their resistance to change, and I don't feel I'm misleading people while suggesting that. You may not be tempted to discuss that point, but others might be interested. On the other hand, if you could just admit that possibility for a while, I think you may understand more easily what I'm saying about rotational motion.

Sorry, but if you have NO IDEA of what the maths of infinitesimal values of variable and functions are, as clearly your comments show, this is not the place for me to try and explain that too ...

[rmolnav]

MY ULTIMATE GO ?
Once again I´m going to try and convey a core idea of my stand, now in a “fresh” way …
It will have three parts. I´ll post them neither on same post, nor on three different ones, but on two posts … (that way readers will have more time to “ruminate” what proposed).

A) Not to need to include a figure, please kindly IMAGINE one with two celestial objects on same “vertical” line (on your screen).
Let us imagine the lower one (M) on a fix location, and the upper one (E) moving “horizontally” from right to left, and let us consider M´s gravity starts to pull downwards E just when reaching M´s vertical (remember, M is considered somehow on a fix location).
If within certain game of relation between masses, distance and speed of E, E will start “orbiting” around M.
M´s gravity, acting as centripetal force, changes E´s speed vector direction, but it is not “able” to change its size. A “free fall” to many, but rather a “partially free fall” to me, because E´s inertia avoids a direct straight line fall.

B) Let us now imagine M, at the very moment E gets at M´s vertical, starts moving also from right to left, with same velocity as E.
Being now M´s pull on E always vertical, E will fall towards M, following a parabolic line.
Now M´s pull is able not only to bend E´s speed vector, but also to increase its vertical speed with an acceleration proportional to M´s pull … same way as if neither of them had an initial horizontal speed …
That´s what I consider is a “fully free” fall ...
And now we should not say E is orbiting “around” M, because it will fall onto it !!
 

[Le Repteux]

Sorry, but if you have NO IDEA of what the maths of infinitesimal values of variable and functions are, as clearly your comments show, this is not the place for me to try and explain that too ...

To study a relativistic phenomenon, we need Relativity math, but even then, math won't help us if our theory is wrong, so since I find my motion simulations interesting, for the moment, I prefer to think that our theory on motion is wrong. By the way, in my simulations, the particles do move while making very smal steps. They progress by steps whose length is .01 times the steps of the photons, which are only 1 pixel long on the screen. I could have used a lot smaller steps, but it would have slowed down the simulation too much. There is no way to move anything on a screen without using steps, and I can't see how nature could avoid that.

[rmolnav]

MY ULTIMATE GO ? (2nd part)

C) Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right … And not just as big as E´s (as in case B) but much, much bigger (see below how much).
If with the correct values for masses, M-E distance, and initial speeds, we would have our real case, E being the Earth and M the Moon …
We know the result is that E, instead of orbiting around M as in case A, and somehow opposite to what in case B (when E got “fully free” to fall directly onto M), now E is “forced” to bend its trajectory much more, because M´s location is changing in a sense opposite to E´s speed … E and M revolves/rotates around their common center of mass. E´s trajectory curvature is much, much bigger than when orbiting around M (case A).
For similar (but kind of opposite) reasons to case B, it is erroneous to consider E “orbits” around M, and to say that E is in a “free fall” … In cased A we had a “partially free" fall, and in B a “fully free” fall … But now I consider we have no free fall at all, because initial speeds and mutual pulls are causing a kind of dance of the couple, keeping their separation actually constant …
With this and last post I´ve already delivered what “promised”: a fresh explanation of one of the main roots of my stand …
But I´ll leave for another post the elaboration of main consequences of that, as far as I can understand ...   

[Le Repteux]

Let us now suppose that, when E reaches M´s vertical with its initial “horizontal" speed, apart from starting M´s gravitational pull, M somehow is given an initial speed parallel to E´s, not to the left as in case B but to the right …

Notice that when the pulling begins, the motion is straight. If it was already curved, no force would be needed to curve it. A force that curves a trajectory needs to be applied to a straight trajectory, so since we don't observe such a motion at our scale, atoms are probably going straight line for a while when we apply a force on them. They probably change their direction, and then go on straight line, and change their direction again, and go straight line, indefinitely. That's probably the main reason for their quantified energy. Their quantified light is probably linked to their quantified motion. They probably move by steps to accommodate individually each photon that strikes them, and it is probably while they are executing a step that they emit a photon. That would produce straight steps, and each of them would have a specific speed and a specific direction. In the case of a curved motion, the steps would have a constant length in the direction which is perpendicular to the force, and their length in the direction of the force would depend on its intensity. This way, there can be no outward force during an orbital motion, only inward. Nothing forces a body to go straight line, it moves all by itself.