Heliocentrism from a relativistic standpoint

[deuced]

This is one of those questions that I've always wanted to ask, but felt stupid for it.

If we look at celestial mechanics from a relativistic standpoint, considering that all inertial reference frames are equivalent and we may choose any arbitrary reference frame, how is it that we can say that the Earth truly revolves around the sun?

Of course, it does if we choose the sun as a stationary frame of reference. But what if we choose the Earth as our frame of reference (which, being our home, seems like a natural one)? Isn't it then perfectly legitimate to say that the sun goes around the earth? It seems to me that the heliocentrism/geocentrism question simply depends on the reference frame from which the system is viewed.

 

[mathman]

For practical purposes the center of mass of the system is the most convenient origin for analysis. For the solar system, it is essentially the center of the sun, since there is relatively very little mass (in the solar system) outside the sun.

 

[JesseM]

This is one of those questions that I've always wanted to ask, but felt stupid for it.

If we look at celestial mechanics from a relativistic standpoint, considering that all inertial reference frames are equivalent and we may choose any arbitrary reference frame, how is it that we can say that the Earth truly revolves around the sun?

The sun can be considered to be moving in an almost inertial manner over normal timescales (although over large enough timescales it's orbiting the center of the galaxy), but over the course of a year the Earth definitely does not move inertially. Inertial movement means not just constant speed, but also movement in a straight line--any change in direction is a type of acceleration, even if the speed is constant. An observer will know he's accelerating because he feels G-forces--he'll feel them if he changes speed or if he changes direction (in the case of moving in a circle at constant speed, he'll feel a 'centrifugal force').

 

[pervect]

This is one of those questions that I've always wanted to ask, but felt stupid for it.

If we look at celestial mechanics from a relativistic standpoint, considering that all inertial reference frames are equivalent and we may choose any arbitrary reference frame, how is it that we can say that the Earth truly revolves around the sun?

Of course, it does if we choose the sun as a stationary frame of reference. But what if we choose the Earth as our frame of reference (which, being our home, seems like a natural one)? Isn't it then perfectly legitimate to say that the sun goes around the earth? It seems to me that the heliocentrism/geocentrism question simply depends on the reference frame from which the system is viewed.

As various other people have said, the center of mass of the solar system is a true inertial frame (ignoring pertubations from the rest of the galaxy, at least, which are small).

Because the sun is so much heavier than any of its planets, it's _almost_ true that the sun is in an inertial frame of reference. But actually the position of the sun is perturbed (very slightly) by the motions of the planets.

The notion of what can be considered to be an inertial frame really depends on how big an area of space one wants to model.

In GR, it's convenient to model gravity by its tidal forces. When the tidal force is so small as to be negligible even when multipled by the distance scale of interest, one can consider that region of space to be "inertial", though one usually says instead "flat space-time", or one of the more technical equivalents.

When the tidal force (acceleration/meter) multipled by the distance scale (meters) is not negligible, the region of space can't really be considered to be "inertial" or "flat".

So we are left with the fact that to an astronaut in the international space-station, space-time will be reasonably "flat" in a region of a few tens of meters.

But when you go to a broader distance scale, of millions of kilometers, the gravity of the Earth and sun become very important - the Earth, for instance, doesn't follow a straight line path as it would in an inertial frame, but moves instead in an elliptical orbit.

 

[Creator]

But what if we choose the Earth as our frame of reference ... Isn't it then perfectly legitimate to say that the sun goes around the earth? .

You've forgotten one thing. We are not alone. There are thousands of other stars which bear witness to our orbital motion - Its called parallax.
Even if you were unable to detect the Earth's acceleration (about the sun), the semi-annual shift in our position relative to the stellar background makes Earth's orbital motion about the sun indisputable.

Creator

 

[JesseM]

You've forgotten one thing. We are not alone. There are thousands of other stars which bear witness to our orbital motion - Its called parallax.
Even if you were unable to detect the Earth's acceleration (about the sun), the semi-annual shift in our position relative to the stellar background makes Earth's orbital motion about the sun indisputable.

But if the Earth was moving inertially rather than accelerating, it would be just as valid to look at things from the perspective of a frame where the Earth was at rest and all the stars were moving.

 

[Saw]

I also wanted to ask this question. The answers here are very helpful. I just wanted to ask a little further.

This is my understanding (considering a simplified system where there is only the Sun and the Earth and none of them rotates):

- It is clear that the Sun, being more massive, accelerates (changes the direction) of the Earth more than viceversa. Because of that, centrifugal force due to interaction between the Sun and the Earth is much stronger on the Earth than on the Sun. Nevertheless, centrifugal force on the Earth exclusively due to gravitational attraction of the Sun (leaving rotation aside) is also very tiny. Is it discernible at all?

- In any case you can draw a picture of the solar system where the Earth's frame is the coordinate system, the Earth is deemed to be motionless and the Sun revolves around.

- This picture, however, would be more complicated to draw than a Sun-centered picture. Why? Because in the Sun-centered picture you can only care about gravitational interaction (considering that the effect due to the Earth's pull is negligible), while in the Earth-centered system most of your work, if you want the picture to be faithful, would be based on fictitious forces moving the Sun? For example, if a car (the Earth) stops and I want to describe the situation from the standpoint of the car's (accelerated) frame, I have to imagine that the driver is pushed forward by a fictitious force. Is this analogy correct?

- These comments are based on purely Newtonian mechanics. Does GR alter anyhow the conclusions?

Thanks.

 

[JesseM]

It is clear that the Sun, being more massive, accelerates (changes the direction) of the Earth more than viceversa. Because of that, centrifugal force due to interaction between the Sun and the Earth is much stronger on the Earth than on the Sun.

The centrifugal force is a fictitious force in Newtonian mechanics that only appears in rotating frames--which rotating frames are you talking about in order to compare centrifugal forces? Perhaps a frame where both the center of the Sun and the center of the Earth are at rest vs. a frame where both the center of the Sun and the surface of the Sun are at rest? Of course in the latter case the centrifugal force experienced on the surface of the Sun would just depend on the rotation rate of the Sun, not on "interaction between the Sun and the Earth". If that's not what you meant, can you specify what two centrifugal forces you're talking about when you say the force is "much stronger on the Earth than on the Sun"?

 

[PeterDonis]

But if the Earth was moving inertially rather than accelerating, it would be just as valid to look at things from the perspective of a frame where the Earth was at rest and all the stars were moving.

The Earth *is* moving inertially, at least as far as its orbit about the Sun goes. It is in free fall, following a geodesic worldline. The presence of the Sun means that that worldline is "curved", but it's still inertial.

It is true that, in a frame of reference in which the line connecting the Earth and the Sun is considered "at rest"--i.e., a frame which is "revolving about the Sun along with the Earth"--the Earth would be considered to be "held up against the Sun's gravity by centrifugal force". But the Earth doesn't *feel* this force.

 

[JesseM]

The Earth *is* moving inertially, at least as far as its orbit about the Sun goes. It is in free fall, following a geodesic worldline. The presence of the Sun means that that worldline is "curved", but it's still inertial.

Yes, that's true from a GR standpoint, but since the original poster was talking about inertial frames I thought we were dealing with a simplified case where we ignore spacetime curvature and just treat the Earth as moving in a circle in flat SR spacetime, where only objects moving at constant speed and direction (relative to inertial frames) would qualify as inertial...I probably should have spelled this out though.

 

[PeterDonis]

An observer will know he's accelerating because he feels G-forces--he'll feel them if he changes speed or if he changes direction (in the case of moving in a circle at constant speed, he'll feel a 'centrifugal force').

Per my previous post, the Earth doesn't *feel* this force. There is nothing wrong with setting up an "inertial" coordinate system centered on the Earth. In fact, there is such a system, called the "Earth Centered Inertial" system, as described http://celestrak.com/columns/v02n01/" . It's used for calculations of the orbital mechanics of satellites. (I put "inertial" in quotes because, since spacetime is curved, there are still issues that need to be dealt with that don't arise with inertial frames in flat spacetime.)

The problem with such a frame if you're looking at the dynamics of the whole solar system, as mathman pointed out, is that the Earth is nowhere near the center of mass of the solar system, so everything will look a lot more complicated in such a frame.

 

[PeterDonis]

Yes, that's true from a GR standpoint, but since the original poster was talking about inertial frames I thought we were dealing with a simplified case where we ignore spacetime curvature and just treat the Earth as moving in a circle in flat SR spacetime, where only objects moving at constant speed and direction (relative to inertial frames) would qualify as inertial...I probably should have spelled this out though.

I would agree, since you mentioned *feeling* G-forces as a distinguishing factor, which *would* be true for an object moving in a circle in flat spacetime, but isn't for the Earth moving in a slightly elliptical orbit in curved spacetime.

 

[Saw]

JesseM:

It is clear that the Sun, being more massive, accelerates (changes the direction) of the Earth more than viceversa. Because of that, centrifugal force due to interaction between the Sun and the Earth is much stronger on the Earth than on the Sun. Nevertheless, centrifugal force on the Earth exclusively due to gravitational attraction of the Sun (leaving rotation aside) is also very tiny. Is it discernible at all?

The centrifugal force is a fictitious force in Newtonian mechanics that only appears in rotating frames--which rotating frames are you talking about in order to compare centrifugal forces? Perhaps a frame where both the center of the Sun and the center of the Earth are at rest vs. a frame where both the center of the Sun and the surface of the Sun are at rest? Of course in the latter case the centrifugal force experienced on the surface of the Sun would just depend on the rotation rate of the Sun, not on "interaction between the Sun and the Earth". If that's not what you meant, can you specify what two centrifugal forces you're talking about when you say the force is "much stronger on the Earth than on the Sun"?

I’ll tell you what was in the back of my mind and how I have thought of re-formulating it (though I am not sure at all it’s correct):

It is my understanding, in terms of Newtonian mechanics, that whenever the acceleration of an object is measured from a given frame, it’s because a real force has acted. A different thing is that if we measure from a certain reference frame, assuming it is motionless, we do not measure our own acceleration and thus attribute it to the object being observed, as if it were accelerated by a “fictitious” force. The term is a little of a misnomer (strictly speaking what is fictitious is not the force, but the assumption that the event causing the acceleration has acted on the observed object), but let’s maintain it for discussion purposes.

For example, talking about the gravitational interaction between the Earth and the Sun, leaving aside effects due to rotation of the Sun or the Earth about their respective axes…

If we choose the Earth as reference frame (as if it were non-accelerated), then it can be that:

(i) The observed object is the Sun. We infer that there is a force acting because the Sun is observed to orbit the Earth. If we explain this by assuming that all the force is suffered by the Sun, that force is mainly fictitious and only to a very small extent real.

(ii) The observed object is the distant stars (if we assume that they are so far away that, virtually, they do not interact with the Sun or the Earth). We infer again that there is a force acting because the stars are observed to orbit the Earth. If we explain this by assuming that all such force is suffered by the stars, all that force is fictitious.

(iii) The observed object is, for example, the waters of the ocean, that is to say, “part” of the Earth. (We leave aside the influence of the Moon, more relevant in practice). We infer again that there is a force acting because the waters tend to build up a bulge. At the surface of the Earth, closer to the Sun, that force is real (because the gravitation is stronger at that point). At the other side, that force is fictitious (actually, what happens is that the gravitation is stronger for the rest of the Earth).

If we chose the Sun as reference frame, the principles would not change, although the proportion between fictitious and real forces would be much in favour of the latter.

If we chose as reference frame the centre of mass of the Sun-Earth system, then all acceleration observed would be explained by real forces, except for the orbital movement of the distant stars.

Above I talked about measurement or observation, as reflected in a space-time system of coordinates. I suppose the expression “feeling” is reserved for noticing the acceleration in constituents of the frame from which you measure. That would be point (iii), i.e., tidal effects, wouldn’t it? These effects are not noticeable in small regions of space or time.

Then I get lost about the implications on this picture of GR.

PeterDonis: You mention that the orbit is a circle in flat ST, but elliptical accounting for the the curvature of ST. But isn't it also en ellipses in Newtonian mechanics? What does GR add? Is it that also time is curved?

 

[DrGreg]

PeterDonis: You mention that the orbit is a circle in flat ST, but elliptical accounting for the the curvature of ST. But isn't it also en ellipses in Newtonian mechanics? What does GR add? Is it that also time is curved?

Circles are possible in both GR and Newtonian theory. The difference arises when you have fixed ellipse in Newtonian theory which becomes in GR a "precessing ellipse", an ellipse whose axis is slowly rotating. (As famously measured in the orbit of Mercury.)

 

[PeterDonis]

PeterDonis: You mention that the orbit is a circle in flat ST, but elliptical accounting for the the curvature of ST. But isn't it also en ellipses in Newtonian mechanics? What does GR add? Is it that also time is curved?

That isn't quite what I said. What I said was: in SR, since spacetime is flat, an inertial worldline *has* to be a *straight* line--straight in the Euclidean sense of "straight". In GR, spacetime can be curved, and in a curved spacetime, inertial worldlines can be curves other than Euclidean straight lines.

So in SR, any object traveling in a circle *must* be accelerated (i.e., must feel a G-force), because a circle isn't a Euclidean straight line. (Actually the object's worldline would be a helix, like a coiled spring, because of the time dimension being included--but its projection into a suitably chosen spacelike slice would be a circle.) This would also be true of an object moving in an ellipse in flat spacetime--I'm sorry if my previous post didn't make this clear. There's nothing special about circles; what I've said applies to any worldline in flat spacetime that isn't a Euclidean straight line.

But in GR, in a curved spacetime (for example, spacetime near the Sun), an object (like the Earth) can be moving on a worldline that isn't a Euclidean straight line (for example, a circular or elliptical orbit), but can still be moving inertially, because of the spacetime curvature. That's why the Earth doesn't feel any G-force from the Sun--it's moving on an inertial worldline, which is curved because of the curvature of spacetime.

(And actually, as Dr. Greg pointed out, the Earth's orbit--more precisely, its projection into a suitable spacelike slice--isn't actually a closed ellipse, because it precesses. The precession of Earth's orbit is too small for us to measure with our current tools, but the precession of Mercury's orbit has been measured and agrees with the GR prediction.)

 

[PeterDonis]

But isn't it also en ellipses in Newtonian mechanics? What does GR add? Is it that also time is curved?

I should also have mentioned that, as far as actual predictions of observations go, the difference between Newtonian gravity and GR, as far as the orbits of objects are concerned, is the precession that Dr. Greg mentioned, which is not predicted by Newtonian gravity.

Conceptually, however, there is a huge difference between the two theories. According to GR, there is no such thing as a "force of gravity"--what we normally think of as "gravity" is just spacetime curvature. So the Earth, for example, doesn't move in an elliptical orbit because of a "force" from the Sun; it does so because of the spacetime geometry in its immediate vicinity, which "tells" it that the orbit it is following is the best geodesic path it can follow. The only influence the Sun has, in GR, is in determining the spacetime curvature in the Earth's immediate vicinity.

In GR, the way you tell whether there is a force acting is by asking whether objects *feel* a force acting. And GR's prediction is that only objects moving on non-geodesic worldlines will feel a force acting. The Earth moves on a geodesic, so it feels no force; but you and I, standing on the Earth's surface, move on non-geodesic worldlines, so we feel a force. In GR, this force is not "gravity"--it's the force of the ground pushing up on us, preventing us from traveling along the nearest accessible geodesics. Similarly, if you look at any other such case, where an object feels a force that in Newtonian language we would be inclined to call "fictitious", you will find that in GR, the actual force--what the object *feels*--is due to whatever it is that is pushing the object on a non-geodesic path. And those forces are always "real" forces--the Earth pushing up on us, or a rocket engine pushing on a rocket, or the walls and floor of a subway car pushing on us as we lurch about.

 

[JesseM]

It is my understanding, in terms of Newtonian mechanics, that whenever the acceleration of an object is measured from a given frame, it’s because a real force has acted.

Assuming you are talking about inertial frames, yes.

A different thing is that if we measure from a certain reference frame, assuming it is motionless, we do not measure our own acceleration and thus attribute it to the object being observed, as if it were accelerated by a “fictitious” force.

I would say a simpler description is that fictitious forces are used in accelerating frames, but I think that's basically what you're getting at here--that if you're an accelerating observer, if you use an accelerating frame where you are at rest, then objects may appear to accelerate in this frame even though they are really moving inertially, and their apparent acceleration can be accurately calculated using the same laws that apply to inertial frames as long as you add in a fictitious force. Note that the fictitious force trick works for calculating the acceleration of any object in an accelerating frame, though, not just objects which are really moving inertially--i.e. an object may be measured as accelerating in both inertial frames and a certain accelerating frame, and in that accelerating frame you can accurately predict the object's motion using a combination of real forces and fictitious ones.

The term is a little of a misnomer (strictly speaking what is fictitious is not the force, but the assumption that the event causing the acceleration has acted on the observed object), but let’s maintain it for discussion purposes.

I don't really understand what you mean here. Perhaps you are saying that even if there is no actual force acting on the object that appears to be accelerating in the accelerating frame, there still must be a force acting on the accelerating observer. I suppose this is true, but the magnitude of the fictitious force seen in the observer's accelerated frame doesn't have to match the magnitude of the real force on the observer seen in inertial frames, so it's not like the fictitious force is just a coded way of talking about the real force (for example, consider an observer being moved in a circle of fixed radius by some force, who is calculating the motion of an object at a different radius from himself...in this case the fictitious centrifugal force on the object in the observer's rotating frame is different from the actual centripetal force on the observer himself). Also, in principle you are free to calculate how things behave in an accelerated frame even if there are no actual observers or objects at rest in that frame--a frame is just a coordinate system after all--although in practice this usually isn't done.

For example, talking about the gravitational interaction between the Earth and the Sun, leaving aside effects due to rotation of the Sun or the Earth about their respective axes…

If we choose the Earth as reference frame (as if it were non-accelerated), then it can be that:

(i) The observed object is the Sun. We infer that there is a force acting because the Sun is observed to orbit the Earth.

I'm not sure what kind of frame you're talking about. Ordinarily if an object is rotating in a circle around some central point, then to consider a rotating frame where the object is at rest usually means considering a frame where both the object and the central point (in this case the Sun) are at rest (this is the type of frame where the fictitious forces can be broken down into a centrifugal force and a coriolis force, for example). Think of looking at a spinning record, and then placing a camera above it looking down, with the center of the lens directly above the center of the record, and the camera spinning around the center of its lens at the same rate that the record is spinning, so that the camera will show a motionless record with the entire room appearing to spin around it.

Perhaps you mean a frame where not only is the center of the Earth at rest, but points on its surface are at rest too, so the Sun appears to orbit around it once per day? This would be a much more complicated accelerating frame to analyze in terms of fictitious forces, but it could be done. However, even in this case I would still have trouble making sense of your comment "It is clear that the Sun, being more massive, accelerates (changes the direction) of the Earth more than viceversa. Because of that, centrifugal force due to interaction between the Sun and the Earth is much stronger on the Earth than on the Sun." The rotation of the Earth about its own axis has nothing at all to do with any real force from the Sun, and thus nothing to do with the Sun's own mass.

 

[Saw]

Well, there are many things in your post and Dr. Greg's and PeterDonis' to learn from and comment, but let's clarify something:

The rotation of the Earth about its own axis has nothing at all to do with any real force from the Sun, and thus nothing to do with the Sun's own mass.

Of course! In all my comments (I stated it but it may be somehow masked afterwards) I'm assuming that there's NO

rotation of the Sun or the Earth about their respective axes

as if simply that didn't exist, precisely because that spinning or rotation about the axis exists, let us say, for historical reasons, but is not connected with the gravitational interaction between Earth and Sun. I was just talking about orbital motion, revolution of the Earth and Sun about the common centre of mass, if you take that RF, or of the Earth about the Sun or vice versa, if you take as RF any of them.

 

[Saw]

Perhaps you mean a frame where not only is the center of the Earth at rest, but points on its surface are at rest too, so the Sun appears to orbit around it once per day?

Linking with the attempt at clarification above... As noted, I am considering a non-real situation where the Earth does not rotate about its own axis. Given this, I assume that if I start on the side of the Earth that is facing the Sun at a given moment, during the course *of the year* there will be times when Sun-light comes laterally to me, times when it will not reach me at all (because it is shadowed by the other side of the Earth), times when again the Sun-light will come laterally from the other side and finally a time when I will be facing the Sun once again. Thus the overall picture, from the Earth's standpoint, would be one where the Sun completes a revolution around the Earth once a year. Is that what would be observed from the Earth if the latter did not spin around its own axis? I supposed so, on the grounds that such would be the reason why we experience the seasons every quarter of a year. Is it right? Can we consider that scenario?

 

[Saw]

I am still curious about the answer to the previous post... If neither the Earth nor the Sun spinned around their respective axes, would an Earth-observer feel

(i) like a bucket in a "noria" (you can see a picture of a "noria" in http://asterion.almadark.com/wp-content/uploads/2008/12/noria1.gif), so that from the Earth he would see the Sun orbiting around (and completing one orbit per year), or

(ii) like someone in a merry-go-round, so that he would see the Sun spinning around its own axis (and completing one spin per year)?

 

[JesseM]

Linking with the attempt at clarification above... As noted, I am considering a non-real situation where the Earth does not rotate about its own axis. Given this, I assume that if I start on the side of the Earth that is facing the Sun at a given moment, during the course *of the year* there will be times when Sun-light comes laterally to me, times when it will not reach me at all (because it is shadowed by the other side of the Earth), times when again the Sun-light will come laterally from the other side and finally a time when I will be facing the Sun once again. Thus the overall picture, from the Earth's standpoint, would be one where the Sun completes a revolution around the Earth once a year. Is that what would be observed from the Earth if the latter did not spin around its own axis? I supposed so, on the grounds that such would be the reason why we experience the seasons every quarter of a year. Is it right? Can we consider that scenario?

OK, I understand what you mean now, but this would be a rather complicated non-inertial frame to consider in detail--it wouldn't be a typical sort of rotating frame where the fictitious forces can be broken down into the centrifugal force and the coriolis force. Again, if the Earth is moving in a circle around the Sun, the standard type of rotating frame would be one where both the center of the Sun and the center of the Earth are at fixed position coordinates. If you're set on thinking about the type of coordinate system you describe, what specific questions do you have about how things would work in this system? I agree that the Sun would go around the Earth in a circle once a year (and it would also be rotating in this system), if that's all you're asking about. The motion of other fixed stars at rest relative to the Sun would be more complicated though.

I am still curious about the answer to the previous post... If neither the Earth nor the Sun spinned around their respective axes, would an Earth-observer feel

(i) like a bucket in a "noria" (you can see a picture of a "noria" in http://asterion.almadark.com/wp-content/uploads/2008/12/noria1.gif), so that from the Earth he would see the Sun orbiting around (and completing one orbit per year), or

(ii) like someone in a merry-go-round, so that he would see the Sun spinning around its own axis (and completing one spin per year)?

It'd be #1--if the Earth had no angular momentum then a person at a point on the Earth directly facing the Sun at one time of year would not be directly facing the Sun at all times of year, that would only happen if the Earth's rotation period about its own axis was the same as its orbital period (i.e. 1 day = 1 year). Note that this question doesn't have anything to do with inertial vs. non-inertial frames though, it's a frame-independent question about what would actually be seen by an observer on Earth.

 

[Saw]

OK, I understand what you mean now, but this would be a rather complicated non-inertial frame to consider in detail

Well, the idea was not to create complication, but to reduce it. The original question was about the choice between heliocentrism and geocentrism in the solar system. Considering at the same time the effects of two phenomena (orbital motion due to gravity + rotation around the axis) is difficult. So the point was focusing on one instead of two at the same time. The choice for orbital motion seemed obvious, since spinning around the axis is not due to the interaction between Earth and Sun and hence seems out of place in a comparison between geocentrism and heliocentrism.

Again, if the Earth is moving in a circle around the Sun, the standard type of rotating frame would be one where both the center of the Sun and the center of the Earth are at fixed position coordinates.

Sorry, I am not familiar with that. How can it be that, as measured from a given reference frame, either the Earth or the Sun, the centers of both are “at fixed position coordinates”? If you mean by “fixed” a point at rest (the origin of the corresponding coordinate system), it’d seem that, if the center of the Earth is the origin of the RF, then the Sun is orbiting and vice versa. Presumably you meant something else, but I didn’t catch it.

If you're set on thinking about the type of coordinate system you describe, what specific questions do you have about how things would work in this system? I agree that the Sun would go around the Earth in a circle once a year (and it would also be rotating in this system), if that's all you're asking about.

My question would simply be whether there are effects that betray the fact that the Earth and the Sun are subject to gravitational interaction. If so, since the attraction is reciprocal, those effects should be more apparent in the Earth RF (which is less inertial and more accelerated) than in the Sun RF (which is more inertial and less accelerated), although strictly speaking they would be present in both frames, the difference just being a question of degree.

it wouldn't be a typical sort of rotating frame where the fictitious forces can be broken down into the centrifugal force and the coriolis force.

Text-books usually say that, in the RF of the orbiting body, the centripetal force and a centrifugal force are balanced and cancel out. This centrifugal thing is actually just the inertia of the orbiting body and it is called a “force”, as if it were due to an external agent, in a fictitious manner, just as it is fictitious that the orbiting body in question is inertial.

Certainly, this “force” is not “felt” precisely because of that, because the orbiting bodies do not crash against the attracting planet and do not lose harmony among themselves, if they all have the same speed and are close enough to each other. But if you consider longer distances, the tidal effects show up and they are usually explained as the result of fictitious centrifugal forces, aren’t they? For example, at the surface of the Earth that, at a given time of the year (assuming, as always, that there's no spinning around the axis), were farther from the Sun, a bulge of water will appear and in the RF of the Earth it would be said that this is due to a stronger fictitious centrifugal force, wouldn’t it?, although in fact it’d be due to the gravitational attraction of the Sun being weaker there. (Similarly, yes, another bulge would appear at the surface closer to the Sun, but it'd still be a queer thing that would have to be explained by resorting to a fictitious force, which in this case would be deemed to be... weaker?)

And if you sent a rocket from the Earth to the Sun, wouldn’t the Earth have to account for something similar as the Coriolis force?

 

[ryuunoseika]

This is all true, even without relativity. Heliocentric models are simply preferred because they are mathematically simpler. I dare you to try at calculating the orbit of jupiters moons in a geocentric model.

 

[JesseM]

Sorry, I am not familiar with that. How can it be that, as measured from a given reference frame, either the Earth or the Sun, the centers of both are “at fixed position coordinates”? If you mean by “fixed” a point at rest (the origin of the corresponding coordinate system), it’d seem that, if the center of the Earth is the origin of the RF, then the Sun is orbiting and vice versa. Presumably you meant something else, but I didn’t catch it.

No, I did mean that both the center of the Earth and the center of the Sun are at rest. Consider the Sun as a spot of paint at the dead center of a record player, and the Earth as a spot of paint on the rim of the record, and then think about the example I posted earlier:

Think of looking at a spinning record, and then placing a camera above it looking down, with the center of the lens directly above the center of the record, and the camera spinning around the center of its lens at the same rate that the record is spinning, so that the camera will show a motionless record with the entire room appearing to spin around it.

If you rotate the camera so that the record appears motionless while the room is spinning around it, then of course any spot on the record will be motionless too, whether at the center or at the rim.

My question would simply be whether there are effects that betray the fact that the Earth and the Sun are subject to gravitational interaction. If so, since the attraction is reciprocal, those effects should be more apparent in the Earth RF (which is less inertial and more accelerated) than in the Sun RF (which is more inertial and less accelerated), although strictly speaking they would be present in both frames, the difference just being a question of degree.

If we're talking GR, then "inertial frames" aren't possible in curved space time except in an infinitesimal local neighborhood, but if we're talking Newtonian gravity, then you will need to account for the gravitational force to explain the motion of the Sun and Earth regardless of what frame you choose (if you choose an inertial frame, that's the only force you need to explain the Earth's motion in this frame, while if you choose a non-inertial one, you'll need to consider a combination of fictitious forces and the real gravitational force). I'm not sure if this addresses your question though--are you talking about "effects" other than just the paths of objects in a given frame, and what do you mean by "those effects should be more apparent"?

Text-books usually say that, in the RF of the orbiting body, the centripetal force and a centrifugal force are balanced and cancel out.

In a rotating coordinate system constructed in a way similar to my example with the record, this is true. In other types of non-inertial coordinate systems the fictitious forces would be more complicated. Also, the centrifugal force only cancels out with the gravitational force for objects that are at rest in the rotating frame (i.e. they are in perfect circular orbits), for other objects (like an object with an elliptical orbit) they don't.

But if you consider longer distances, the tidal effects show up and they are usually explained as the result of fictitious centrifugal forces, aren’t they?

Only if you choose to analyze them in a rotating frame--in Newtonian gravity you can of course understand tidal forces just fine from the perspective of an inertial frame, in which case no fictitious forces will be involved in the explanation (instead the explanation will have to do with the way the strength and direction of the gravitational force varies across different points on the body).

For example, at the surface of the Earth that, at a given time of the year (assuming, as always, that there's no spinning around the axis), were farther from the Sun, a bulge of water will appear and in the RF of the Earth it would be said that this is due to a stronger fictitious centrifugal force, wouldn’t it?

I've never heard it explained that way. The simplest way of explaining tidal bulges I've seen is to consider the gravitational force vector at each point on a body's surface, then from each of these vectors subtract whatever the gravitational force vector is at the center, giving the "gravity differential field" shown in Fig. 2 of this wikipedia page (also explained in a little more detail here). Since the force at the part of the Earth nearer to the Sun is slightly stronger than the force at the center, the gravity differential you get when you subtract the center vector points slightly inwards towards the Sun, causing a bulge in that direction; but then since the force at the part of the Earth farther from the Sun is slightly weaker than the force at the center, the gravity differential there will point slightly outwards away from the Sun, so you'll have an outward bulge on the far side.

And if you sent a rocket from the Earth to the Sun, wouldn’t the Earth have to account for something similar as the Coriolis force?

If you wanted to use a rotating coordinate system where both the center of the Earth and the center of the Sun were at rest, then there'd be a Coriolis force. But of course, nothing stops scientists on Earth from just calculating everything from the perspective of an inertial frame where the center of the Earth isn't at rest.

 

[D H]

Linking with the attempt at clarification above... As noted, I am considering a non-real situation where the Earth does not rotate about its own axis. Given this, I assume that if I start on the side of the Earth that is facing the Sun at a given moment, during the course *of the year* there will be times when Sun-light comes laterally to me, times when it will not reach me at all (because it is shadowed by the other side of the Earth), times when again the Sun-light will come laterally from the other side and finally a time when I will be facing the Sun once again. Thus the overall picture, from the Earth's standpoint, would be one where the Sun completes a revolution around the Earth once a year. Is that what would be observed from the Earth if the latter did not spin around its own axis? I supposed so, on the grounds that such would be the reason why we experience the seasons every quarter of a year. Is it right? Can we consider that scenario?

OK, I understand what you mean now, but this would be a rather complicated non-inertial frame to consider in detail--it wouldn't be a typical sort of rotating frame where the fictitious forces can be broken down into the centrifugal force and the coriolis force.

That is neither an inertial frame nor a rotating frame. It is an accelerating frame with non-rotating axes. This is in fact a widely-used reference frame in the aerospace community that is (somewhat incorrectly) called Earth-Centered Inertial. NASA sent people to the Moon using this frame (specifically, the M50 ECI frame) and tracks the orbits of Earth-orbiting satellites in this frame (specifically, the EMEJ2000 and more lately, the GCRF). About all those acronyms: They refer to refinements of where scientists think the center of the Earth lies and refinements of what constitutes a non-rotating reference frame.

As this is an accelerating frame, fictitious forces arise due to the Earth's acceleration toward the Sun, Moon, and to a lesser extent, the planets. The solution is simple: Model those fictitious forces. They are essentially tidal gravity forces. People use this frame because its much more accurate for modeling the behavior of things orbiting the Earth compared to using a heliocentric frame.

 

[D H]

Again, if the Earth is moving in a circle around the Sun, the standard type of rotating frame would be one where both the center of the Sun and the center of the Earth are at fixed position coordinates.

Sorry, I am not familiar with that. How can it be that, as measured from a given reference frame, either the Earth or the Sun, the centers of both are “at fixed position coordinates”?

Simple: The reference frame rotates at one revolution per year. JesseM is talking about rotating reference frames. Imagine you are on a merry-go-round. You define a reference frame with origin at the center of and rotating with the merry-go-round. You are sitting on a plastic horse fixed with respect to the platform; your position in this frame is fixed.

Now imagine that a sibling is standing on the ground outside the merry-go-round. From your perspective, your sibling is rotating around you. According to Newton's laws of motion, some force must be acting on your sibling. This apparent force is called the centrifugal force. Note well: From the perspective of an inertial frame, you are the one rotating, not your sibling. Your sibling does not feel this centrifugal force. It is a “fictitious” force.

Text-books usually say that, in the RF of the orbiting body, the centripetal force and a centrifugal force are balanced and cancel out. This centrifugal thing is actually just the inertia of the orbiting body and it is called a “force”, as if it were due to an external agent, in a fictitious manner, just as it is fictitious that the orbiting body in question is inertial.

Rant mode onI really hate that explanation. You don't even know what a rotating frame is! How can a textbook say such things without first describing what centrifugal force is? Textbooks that explain orbits as a balance between centrifugal force and gravitational force are doing an immense disservice. There is no centrifugal force in an inertial frame. None. Rant mode offOK, back to the frame JesseM introduced. One reason for using this rotating frame is because there are five special points away from the Earth at which centrifugal and gravitational forces balance. Two of these, the Sun-Earth L1 and L2 points, are currently occupied by a quite a few satellites (actually, they are in pseudo orbits about these points). The L1 point lies between the Sun and the Earth. By all rights, an object orbiting the Sun at less than 1 AU should orbit the Sun in less than a year. Yet something at the L1 point orbits the Sun in sync with the Earth. Understanding why these five points are special is something every physics major has to go through, typically in their sophomore or junior year in college. The explanation is much easier from the perspective of a rotating frame.

Another place where rotating frames are very handy is in displaying the trajectory of a vehicle going from the Earth to the Moon. The Earth and Moon will have fixed positions in a frame that rotates once per month.

 

[D H]

Heliocentric models are simply preferred because they are mathematically simpler. I dare you to try at calculating the orbit of jupiters moons in a geocentric model.

Actually, I do something very much like that as a test of my orbital dynamics models: I calculate Clementine's orbit about the Earth's Moon from the perspective of a Neptune-centered inertial frame. This approach does lose accuracy rather quickly, but if it doesn't capture the basic motion of Clementine there's something drastically wrong with my models.

 

[Saw]

JesseM is talking about rotating reference frames. Imagine you are on a merry-go-round. You define a reference frame with origin at the center of and rotating with the merry-go-round. You are sitting on a plastic horse fixed with respect to the platform; your position in this frame is fixed.

Now imagine that a sibling is standing on the ground outside the merry-go-round. From your perspective, your sibling is rotating around you. According to Newton's laws of motion, some force must be acting on your sibling. This apparent force is called the centrifugal force. Note well: From the perspective of an inertial frame, you are the one rotating, not your sibling. Your sibling does not feel this centrifugal force. It is a “fictitious” force.

No, I did mean that both the center of the Earth and the center of the Sun are at rest. Consider the Sun as a spot of paint at the dead center of a record player, and the Earth as a spot of paint on the rim of the record, and then think about the example I posted earlier:

Think of looking at a spinning record, and then placing a camera above it looking down, with the center of the lens directly above the center of the record, and the camera spinning around the center of its lens at the same rate that the record is spinning, so that the camera will show a motionless record with the entire room appearing to spin around it.

If you rotate the camera so that the record appears motionless while the room is spinning around it, then of course any spot on the record will be motionless too, whether at the center or at the rim.

I understand the merry-go-round and record examples of rotating frames, but they differ with the frames I was referring to in that:

(i) The orbit of the Earth around the Sun is not a circle but an ellipsis, even if a little eccentric one.
(ii) If the Earth were a spot on the rim of the record or a plastic horse close to the edge of the platform, someone looking from the spot or the horse to the center of the rotating thing (the Sun) would always see that center and be seen from that center, whereas we had agreed in post #21 that such would not be the case in reality for the Earth and the Sun.

Maybe what you are telling me is that all that does not prevent us from modelling the Sun-Earth system from ideal dimensionless points at the center of the Earth and at the center of the Sun, which would be at rest with each other in a common rotating frame? Is it more or less that?

 

[JesseM]

(i) The orbit of the Earth around the Sun is not a circle but an ellipsis, even if a little eccentric one.

Well, we were already considering a bunch of idealizations, like ignoring curved spacetime in GR and your suggestion about changing the rotation rate of the Earth, so I guess I figured the simplification of a circular orbit was part of that (also, you had talked about an orbit representing a balance of the gravitational and centrifugal force in a rotating coordinate system, but this would only be precisely true for a perfectly circular orbit). In any case, my main point was just about how "rotating coordinate systems" work in classical physics, again making the point that these are the only types of coordinate systems where we can say that the fictitious forces are just the centrifugal and coriolis force. You could use a rotating coordinate system to analyze an object in an elliptical orbit, it just wouldn't be perfectly at rest in this system; likewise, you could come up with a coordinate system where an object in elliptical orbit was at rest at all times, but it wouldn't be what is meant by a "rotating coordinate system", and the fictitious forces would be more complicated than just the centrifugal and coriolis forces.

(ii) If the Earth were a spot on the rim of the record or a plastic horse close to the edge of the platform, someone looking from the spot or the horse to the center of the rotating thing (the Sun) would always see that center and be seen from that center, whereas we had agreed in post #21 that such would not be the case in reality for the Earth and the Sun.

Sure, but no coordinate system has any preferred role in telling you what someone sees, what you see can be broken down into purely local questions of what light hits your eyes and at what time according to your clock, so any possible coordinate system would give the same answer. If your point has something to do with a connection between what is seen and what coordinate system is used, you'll have to explain more, because I don't understand why what you see visually should have any implications for what coordinate system you use.

Maybe what you are telling me is that all that does not prevent us from modelling the Sun-Earth system from ideal dimensionless points at the center of the Earth and at the center of the Sun, which would be at rest with each other in a common rotating frame? Is it more or less that?

That's what I had been suggesting, although again it would depend on idealizing the Earth's center as moving in a perfect circle. In any case I was just trying to illustrate what is meant by a rotating frame, previously it seemed you hadn't understood the idea that idealized points at the centers could both be at rest when you said:

How can it be that, as measured from a given reference frame, either the Earth or the Sun, the centers of both are “at fixed position coordinates”?

As long as you understand how rotating frames work, and that this term is reserved only for these types of coordinate systems where an object in a perfectly circular orbit would be at rest along with the center, and that only in rotating frames do the fictitious forces consist solely of the centrifugal and coriolis forces, then I think we can move on from this point.

 

[Saw]

Ok, I understand the rotating frame better now…

The other alternative is what I was calling the Earth or the Sun frame. These two frames, as viewed from an inertial frame, are accelerating (just like the rotating frame) but (unlike the rotating frame) they do not have points at rest with each other: in the Sun frame, the center of the Sun is always at rest with all points of its sphere, but it’s never at rest with the center of the Earth, which is orbiting around; and vice versa. Let us suppose that this orbit is a perfect circle (another idealization), just to make things closer to the rotating frame model. Can’t we say that at the center of any of these frames gravity force (exerted by the other body) cancels out centrifugal force? Well, I suppose this can never be exactly true, precisely because in my definition I have ruled out ideal dimensionless points: the “centers” are here spots with some size and so the two forces would not exactly cancel out in any part those spots, except in their ideal infinitesimally small centers… But couldn’t we say that that is true to a good approximation?

I say so because then, on that basis, we could still talk about gravity and centrifugal force as a way of describing tidal effects. We could thus explain, for example, the bulge of water on the surface of the Earth farther from the Sun as follows:

- From the reference frame of the Sun, as arising from the non-uniformity of the gravity field across distance: at that location the force is slightly weaker than the force at the center of the Earth (because of longer distance from the center of the Sun) and that is why the water is centripetally accelerated less.
- From the reference frame of the Earth, as arising from the fact that at that location centrifugal force overcomes gravity force.

See these texts:

Britannica:

At the surface of the Earth closest to the Moon, the Moon's gravity is stronger than the centrifugal force. The ocean's waters, which are free to move in response to this unbalanced force, tend to build up a small bulge at that point. On the surface of the Earth exactly opposite the Moon, the centrifugal force is stronger than the Moon's gravity, and a small bulge of water tends to build up there as well.

http://scijinks.jpl.nasa.gov/weather/howwhy/tides/

The Moon's gravitational pull on the side of Earth nearest the Moon is strong enough to overcome the centrifugal force and pull the oceans toward the Moon. The Moon's gravity is tugging on the far side of Earth too, but because that side is farther away, the Moon's gravity is too weak to overcome the centrifugal force. Thus the oceans bulge on that side as well.

Aren’t these statements made as if the authors were speaking from the (accelerating and not part of a rotating) reference frame of the Earth?

 

[Austin0]

Britannica:


At the surface of the Earth closest to the Moon, the Moon's gravity is stronger than the centrifugal force. The ocean's waters, which are free to move in response to this unbalanced force, tend to build up a small bulge at that point. On the surface of the Earth exactly opposite the Moon, the centrifugal force is stronger than the Moon's gravity, and a small bulge of water tends to build up there as well.

http://scijinks.jpl.nasa.gov/weather/howwhy/tides/


The Moon's gravitational pull on the side of Earth nearest the Moon is strong enough to overcome the centrifugal force and pull the oceans toward the Moon. The Moon's gravity is tugging on the far side of Earth too, but because that side is farther away, the Moon's gravity is too weak to overcome the centrifugal force. Thus the oceans bulge on that side as well.

Aren’t these statements made as if the authors were speaking from the (accelerating and not part of a rotating) reference frame of the Earth?

Hi Saw

I have a question: Why would the gravitational pull on the moon side be acting in opposition to the centrifugal force?
Rather than acting in conjunction with the centrifugal force in opposition to the centripetal force of the Earth's gravity?
It does seem to be considered from a frame where the Earth is not rotating , otherwise the concept of centrifugal force would not make sense.

 

[D H]

The other alternative is what I was calling the Earth or the Sun frame. These two frames, as viewed from an inertial frame, are accelerating (just like the rotating frame) but (unlike the rotating frame) they do not have points at rest with each other ...

They're accelerating frames. Accelerating frames and rotating frames are different beasts. Did you read post #25? There is no centrifugal force in these frames because they are not rotating frames.

I say so because then, on that basis, we could still talk about gravity and centrifugal force as a way of describing tidal effects. We could thus explain, for example, the bulge of water on the surface of the Earth farther from the Sun as follows:

- From the reference frame of the Sun, as arising from the non-uniformity of the gravity field across distance: at that location the force is slightly weaker than the force at the center of the Earth (because of longer distance from the center of the Sun) and that is why the water is centripetally accelerated less.
- From the reference frame of the Earth, as arising from the fact that at that location centrifugal force overcomes gravity force.

The first explanation works even more clearly from an Earth centered frame. The latter explanation happens to work, but it is wrong. Suppose some alien spacecraft with the ability to violate Newton's first law gets stranded in the vicinity of the Earth. This spacecraft has the ability to steal linear momentum from a nearby object orbiting a star. The vehicle is stranded; what can it do? Simple: It can steal all of the Earth's linear momentum, making it plunge sunward. The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun. The centrifugal force argument completely fails in this case. The tidal gravity explanation still works in this example.

 

[Saw]

I have a question: Why would the gravitational pull on the moon side be acting in opposition to the centrifugal force? Rather than acting in conjunction with the centrifugal force in opposition to the centripetal force of the Earth's gravity?

Hi Austin0, I don’t know how I have ended up here defending fictitious forces, when really I hate them. Everything is easier if you analyze things from an inertial frame or at least an almost inertial one for our purposes. Let me consider as such for this comment the Sun.

I visualize it as follows. The Earth has a tangential velocity in a straight line (inertial motion). If left alone, it would fly away. Now we introduce gravity. Think of one single pull from the Sun, as if that were possible. The composition of the vector tangential velocity and the vector centripetal velocity imparted by such pull would be a diagonal path. If the ocean water is closer to the Sun, then gravity force is stronger there and the second vector is of higher magnitude. This should result in a compounded vector with a steeper slope, closer to the center of the Sun. Now we allow for the fact that gravity is constantly there (BTW, with or without time gaps between its pulls?). The composition of all the little imaginary diagonal paths is the orbital motion, but in the case of the water closer to the Sun. But that’s the explanation I gave myself. Experts may correct if I went wrong.

Did you read post #25?

Yes, I read all your posts. They were very helpful, thank you.

Apparently, you are telling me that in, for example, the Earth frame I’m talking about, which

is neither an inertial frame nor a rotating frame. It is an accelerating frame with non-rotating axes.

,

fictitious forces arise due to the Earth's acceleration toward the Sun

It’s only that you do not reserve any role in that fictitious force for the concept of “centrifugal force” but note that

They are essentially tidal gravity forces.

Well, if you promise not to go into “Rant mode”, I’ll tell you frankly what your view is suggesting me.

I interpret that, contrary to the authors I have quoted, you are taking the easy-going way of explaining things in an accelerated frame.

For example, you are driving a truck with a tank half-full of water. You suddenly step on the brake and friction of the tyres with the ground makes the car decelerate. The water keeps going forward by inertia and spills over the cabin. However, if you still want to keep your accelerated truck as a reference, as if it were an inertial frame, even if it is not, in order to preserve the validity of Newton’s laws, you could say that a fictitious force equal to the mass of the water times its acceleration has pushed it forward. Here there’s no more complicated way of explaining things.

But what if, instead of stopping suddenly, you do it progressively, thus acquiring uniformly accelerated motion? The water, in my opinion, would form a constant bulge on the front part of the tank. Why? Because it would collide with the front wall of the tank and tend to recede, but then the truck’s speed would diminish more, thus reproducing the effect of collision with the front wall, and so on. The easy-going way of explaining this in the frame of the truck would be to say that there is a fictitious force making the water form precisely that bulge. Since here the attraction of the water towards the front wall of the tank is caused by the friction of the tyres with the ground, which has decelerated the tyres and communicated this new state of motion to the rest of the truck through its components and ultimately reached the front wall of the tank, we could say, mimicking your nomenclature, that in the truck frame the water has been subjected to a “contact tidal force”. And why does this force have this magnitude, which causes this effect and not another? Why does the water not spill over the cabin…? I suppose you know, I don’t, on the basis of that explanation. I would understand things better with the explanation of the quoted authors, who would take an apparently more complicated approach, but more explanatory in my view. They would say that in the truck frame the bulge of water is due to the net result of offsetting the contact decelerating force against a truck-fugal fictitious force.

I think this answers the example you have proposed:

The first explanation works even more clearly from an Earth centered frame. The latter explanation happens to work, but it is wrong. Suppose some alien spacecraft with the ability to violate Newton's first law gets stranded in the vicinity of the Earth. This spacecraft has the ability to steal linear momentum from a nearby object orbiting a star. The vehicle is stranded; what can it do? Simple: It can steal all of the Earth's linear momentum, making it plunge sunward. The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun. The centrifugal force argument completely fails in this case. The tidal gravity explanation still works in this example.

Think of it in terms of my example: in yours gravity plays the part of the contact force in mine (contact with the ground and with the front wall of the tank) and a fictitious inertial force pulling away from the Sun plays the part of the truck-fugal force.

You say, “The Sun will still cause tides. In fact, the tides will get higher and higher as the Earth gets closer and closer to the Sun.” Of course, nobody discusses what happens, it must be the same for all frames.

“The centrifugal force argument completely fails in this case”. Just call it a Sun-fugal force and it should also work here.

 

[D H]

I visualize it as follows. The Earth has a tangential velocity in a straight line (inertial motion). If left alone, it would fly away. Now we introduce gravity. Think of one single pull from the Sun, as if that were possible. The composition of the vector tangential velocity and the vector centripetal velocity imparted by such pull would be a diagonal path. If the ocean water is closer to the Sun, then gravity force is stronger there and the second vector is of higher magnitude. This should result in a compounded vector with a steeper slope, closer to the center of the Sun. Now we allow for the fact that gravity is constantly there (BTW, with or without time gaps between its pulls?). The composition of all the little imaginary diagonal paths is the orbital motion, but in the case of the water closer to the Sun. But that’s the explanation I gave myself. Experts may correct if I went wrong.

That's just wrong. Note well: You did not invoke the centrifugal force in this explanation. More on centrifugal force below.

Here is the tidal gravity explanation. The gravitational acceleration toward the Sun experienced by some object located at a point \mathbf r' from the center of the Sun is, from the perspective of an inertial frame,

\mathbf a_{\text{inertial}} = - \,\frac{G M_{\text{sun}}}{r'^3}\mathbf r'

The Earth itself is accelerating toward the Sun. Because the oceans move more or less with the Earth, it is the relative acceleration with respect to the Earth that drives the tides. Denoting the position of the Earth with respect to the Sun as \mathbf R,

\mathbf a_{\text{relative}} =<br /> - GM_{\text{sun}}\left(\frac{\mathbf r}{r^3} - \frac{\mathbf R}{R^3}\right)

Finally, express the position of the object relative to the center of the Earth rather than the center of the Sun. Defining \mathbf r = \mathbf r&#039;-\mathbf R,

\mathbf a =<br /> -GM_{\text{sun}}\left(<br /> \frac{\mathbf R + \mathbf r}{||\mathbf R + \mathbf r||^3} -<br /> \frac{\mathbf R}{||\mathbf R||^3}<br /> \right)

Note that the position \mathbf r[/itex] and acceleration \mathbf a[/itex] are relative to the center of the Earth. The term GM_{\text{sun}}/R^3\,\mathbf R is a fictitious force.&lt;br /&gt; &lt;br /&gt; Because r&amp;lt;&amp;lt;R, to first order this becomes&lt;br /&gt; &lt;br /&gt; \mathbf a \approx -\frac {GM_{\text{sun}}}{R^3}(\hat r - 3 \cos\theta \hat R)r&lt;br /&gt; &lt;br /&gt; Here \hat r[/hat] and \hat R are unit vectors from the center of the Earth toward the object in question and toward the Sun. r and R are the magnitudes of the vectors \mathbf r and \mathbf R.&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; It’s only that you do not reserve any role in that fictitious force for the concept of “centrifugal force” &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;Physicists are very careful with nomenclature, Saw. Centrifugal force is a very specific thing: It is the fictitious force&lt;br /&gt; &lt;br /&gt; -m\boldsymbol \Omega \times \mathbf r&lt;br /&gt; &lt;br /&gt; where \boldsymbol \Omega is the rate at which the reference frame is rotating with respect to the remote stars. The centrifugal force is obviously not constant with respect to position. Assuming a constant rotation rate, the centrifugal force at a fixed point is constant over time. Compare this to the fictitious force due to the Earth as whole accelerating toward the Sun. This force is constant with respect to position but varies with respect to time. It doesn&amp;#039;t look anything like the centrifugal force.

 

[Saw]

Ok, it seems I've been misled by some texts. I'll study that. Regards.

 

[Saw]

Physicists are very careful with nomenclature, Saw. Centrifugal force is a very specific thing: It is the fictitious force

-m\boldsymbol \Omega \times \mathbf r

where \boldsymbol \Omega is the rate at which the reference frame is rotating with respect to the remote stars.

I'm studying your text. Shouldn't \boldsymbol \Omega be squared?

 

[D H]

Good catch. However, it is not omega squared. It is

-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)

which reduces to

-m \Omega^2 r

in the special case of \mathbf r being normal to \boldsymbol \Omega.

 

[Saw]

it is not omega squared. It is

-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)

which reduces to

-m \Omega^2 r

in the special case of \mathbf r being normal to \boldsymbol \Omega.

Ok, but according to the sources I am reading, if we make the analysis of the Earth-Sun system from a rotational frame (I talk now of acceleration instead of force):

- The acceleration due to the fictitious centrifugal force can be decomposed into two components: ω^2 * c (c being the distance between the center of mass of the Sun-Earth system and the Earth center) and ω^2 * r (r being the distance between the center of the Earth and the relevant point of the Earth where we're analysing the tidal effect).

- The component ω^2 * r cancels out somehow.

- So we are left exclusively with a "uniform" (the same size in all points of the Earth) centrifugal acceleration of intensity ω^2 * c.

Is that correct, in your opinion? Thanks.

 

[Saw]

Well, after deleting a last post with some blunder, I come back with a better understanding now.

Your analysis, DH, is clear:

Let us say M = mass of the Sun.
N and D are two points on the surface of the Earth separated from its center by the radius (r) of the Earth, N (from near) being the one facing the Sun and D (from distant) the one facing away from the Sun, at a given moment.
R = the distance from the center of the Sun to the center of the Earth.
CM = center of mass of the Sun-Earth system.
c = distance from the CM to the center of the Earth.

- In an inertial frame with origin at the CM (disregarding its acceleration due to other objects in the universe), objects at N and D would tend to be accelerated as follows:

* For N, a = - GM/(R^2 - r)
* For D, a = - GM/(R^2 + r)

Since the frame is inertial, these accelerations are due to "real" forces.

- In an accelerated frame with origin at the Earth center, we have to add to the real force a "fictitious" force pointing away from the Sun = + GM/R^2. Thus the accelerations would tend to be:

* For N, a' = - GM/(R^2 - r) + GM/R^2 = - net result, away from Earth center and toward the Sun
* For D, a' = - GM/(R^2 + r) + GM/R^2 = + net result, away from the Earth center and away from the Sun

However, all this is the analysis at a frozen moment, an ideal instant. It's not strange, hence, that the resulting net force is the same as if the Earth were falling in a straight line toward the Sun.

But here it is not that simple. To describe the kinematics from the simplest perspective, that of the inertial frame, the Earth is orbiting the CM, always with a fixed orientaton of its axes wrt the fixed stars: the center of the Earth is translating in a circle about the CM, whereas Points N and D are also translating in circles with the same radius (c), albeit different centers.

One solution could be what I tried to do in post #33. Why is that wrong? There I was taking the frame of the Sun, assuming that its motion around the CM is negligible and hence inexistent. But to be more precise, let's take the frame with origin at the CM (also disregarding its own acceleration due to gravitational interaction with the rest of the universe). Isn't the Earth's acceleration, wrt the CM and after a time lapse, the result of compounding the gravity force toward the Sun with the Earth's inertial tangential motion?

Another solution could be shifting to the perspective of a rotating frame and then transforming again back into the accelerated Earth frame or the inertial frame. I've got lost both in the rotating frame and in the way back. Does anyone know how to do it?

In any case, is the final result, wrt the Earth center, the following?

* For N, a' = - GM/(R^2 - r) + ω^2c
* For D, a' = - GM/(R^2 + r) + ω^2c