Centrifugal force and Newton 3rd law

[tiny-tim]

reality … ? … don't talk to me about reality … !

FYI Cabraham, tiny tim is a PhD physics candidate.

Who … me??

I'm just a little goldfish who tries to make three-dimensional sense out of the two-dimensional images I see projected onto the boundary of the bowliverse.

oooh … I've just thought of a question …

forces do work …
can centrifugal force do work? ​

 

[rcgldr]

So when are they going to change the name of a centrifuge to a centripuge?

 

[George Jones]

Who … me??

I'm just a little goldfish who tries to make three-dimensional sense out of the two-dimensional images I see projected onto the boundary of the bowliverse.

oooh … I've just thought of a question …

forces do work …
can centrifugal force do work? ​

I can't recall running into this, but it seems to me that work in non-inertial frames can be useful. For example, I think the work-energy theorem can be used to find a particle's change in speed in a non-inertial frame.

Still don't have time to respond like I want; on my out the door again.

I am going to need a vacation to recover from my current vacation at my in-laws!

 

[cmos]

oooh … I've just thought of a question …

forces do work …
can centrifugal force do work? ​

Good question!


It seems counter-intuitive to me, but I want to say yes. Consider a particle placed on a rotating disc. The centrifugal force on the particle is
\vec{F}_{cent}=m\omega^2\vec{r}
where r is directed radially. So in a particle being pushed from a point near the axis towards the edge of the disc, work must be performed:
W=\int \vec{F}_{cent} \bullet d\vec{r}


My problem with this is where does the work go in the inertial frame? Also, clearly the Coriolis force cannot do work:
\vec{F}_{Cor}=-2m \vec{\omega} \times \vec{\dot r}
where dr/dt is the velocity of the particle in the non-inertial frame. So I find it an oddity that one fictious force may do work while another may not.

 

[George Jones]

My problem with this is where does the work go in the inertial frame?

When considering work, usually only one frame is used. Even when restricted to inertial reference frames, a force can do zero work in one frame and non-zero work in another.

Consider the following example.

In an inertial frame, a particle is subject to a (net) force. Suppose that when the force starts acting, the velocity of the the particle is c \hat{e_1}, and that the velocity of the particle when the force stops acting is c \hat{e_2}, where c is a constant. Because the initial and final speeds are the same, there is no change in kinetic energy, and, by the work-energy theorem, no work done on the particle by the force.

Now consider the same situation from the point of view of an inertial reference frame that moves with velocity c \hat{e_1} with respect to the first inertial reference frame. In this frame, the initial velocity of the the particle is \vec{0}, and the final velocity of the particle is c \left( \hat{e_2} - \hat{e_1} \right). In this frame, the change in kinetic energy and work done is m c^2.

Also, clearly the Coriolis force cannot do work:
\vec{F}_{Cor}=-2m \vec{\omega} \times \vec{\dot r}
where dr/dt is the velocity of the particle in the non-inertial frame. So I find it an oddity that one fictious force may do work while another may not.

A "force" of this form can change the direction of a particle's motion, but cannot change a particle's speed. Again, this is true in both inertial and non-inertial frames. For example, a moving charged particle in a "real" magnetic field is subject to a force (proportional to) \vec{v} \times \vec{B}, which does no work.

 

[cmos]

Excellent! Thank you for the very clear explanation. Bravo, Prof. Jones.

 

[rcgldr]

Can centrifugal force do work?

Take the example of the person twirling an object. Change this to a person holding a very low friction pipe with a string going through it. The are objects attached to both ends of the string. The person twirls one of the objects, and the rotating object reacts to the centripetal force from the string by applying a "reactive centrifugal force" to the string, creating a tension, which in turn can lift the object dangling below the pipe at the other end of the string.

Work done is peformed on the hanging object, it's weight times the height the hanging object is raised.

Work is also done on the rotating object by the person, equal to it's change in kinetic energy.

Even when restricted to inertial reference frames, a force can do zero work in one frame and non-zero work in another.

Rockets in space and their spent fuel are a good exception. Assume an environment with no external forces (free of gravity). Momentum is preserved. All the work done is internal, and increases the kinetic energy of fuel and/or rocket, depending on the reference frame, but it will turn out that, within reason, the frame of reference won't change the amount of work done, since it's source the the chemical energy of the fuel. Even if the frame of reference is the accelerating rocket itself, all the work is done to the fuel in this frame of reference, but it's the same amount of work done as observed in constant velocity frame of reference.

 

[George Jones]

The person twirls one of the objects, and the rotating object reacts to the centripetal force from the string by applying a "reactive centrifugal force" to the string, creating a tension, which in turn can lift the object dangling below the pipe at the other end of the string.

As`Doc Al already said in https://www.physicsforums.com/showthread.php?p=1785198#post1785198", reactive centrifugal force" is very non-standard terminology in physics.

 

[George Jones]

Excellent! Thank you for the very clear explanation. Bravo, Prof. Jones.

I am just an instructor.

There is no literal Coriolis force actively influencing the missile trajectory, but rather it is a correction term accounting for the deviation AS IF THERE WAS a "real" Coriolis force. The force is not literal, but the missile's path deviation is absolutely real. Coriolis force is a virtual force mathematically defined to account for A VERY REAL NON-FICTITIUOS path deviation.

Yes, As a student, I had to do this type of a problem on my third-year mechanics final exam. We had to determine whether a hunter killed a duck.

I, however, would not call a Coriolis force a correction term, it's a term that appears in rotating reference frames, just as centrifugal force does. The path of a particle is real, but the shape of the path is very dependent on the coordinate system used. Coriolis force and centrifugal account for the shape of the path in a rotating (with respect to an inertial frame) reference frame.

As D H and Doc Al (and possibly others) have said, it's crazy not do some problems in a rotating frame. When doing such problems, Coriolis force and centrifugal force are treated as real forces for which intuition developed doing more elementary Newton's second law problems can be used. But it should be kept in mind that these forces are just artifacts of a non-inertial coordinate system.

Consider accelerometers that consist of two main parts - a hollow sphere like a basketball inside of which is a slightly smaller sphere. Initially, the centres of the spheres coincide, so that there is a small, uniform gap between the spheres. During acceleration, the gap will be closed, and contact between the spheres will be made. An alarm that indicates acceleration motion will sound. For zero acceleration, no alarm will sound, and straight line motion in inertial frames is indicated.

An accelerometer won't measure acceleration due to either Coriolis force or to centrifugal force.

Also, if a freely falling accelerometer is small enough that tidal forces can be neglected, it will measure zero acceleration, since both spheres fall at the same rate. Acceleration due to gravity, just like acceleration due to Coriolis and centrifugal forces, is independent of mass. Can gravity, too, be considered to be an artifact of a coordinate system? No and yes.

Centrifugal and Coriolis forces at all points in space, i.e., globally, can be transformed away by a single transformation from the rotating frame to an inertial frame. For this reason, these forces are called fictitious. No single transformation will transform gravity away globally, but gravity can be transformed away locally by moving to a freely falling frame. If we expand the term "fictitious force" to include to forces that can be transformed away locally, then gravity also is a fictitious force. Pursuing this line of thought leads to the concept of gravity as spacetime geometry.

I repeated the points that D H and tiny-tim have made, using slightly different words.

As Doc Al has said, in mechanics courses, gravity is treated as a real force; it would be crazy to do otherwise. Considering gravity to be due to spacetime geometry, however, expresses something deeper about the physical world.

 

[cabraham]

I am just an instructor.



Yes, As a student, I had to do this type of a problem on my third-year mechanics final exam. We had to determine whether a hunter killed a duck.

I, however, would not call a Coriolis force a correction term, it's a term that appears in rotating reference frames, just as centrifugal force does. The path of a particle is real, but the shape of the path is very dependent on the coordinate system used. Coriolis force and centrifugal account for the shape of the path in a rotating (with respect to an inertial frame) reference frame.

As D H and Doc Al (and possibly others) have said, it's crazy not do some problems in a rotating frame. When doing such problems, Coriolis force and centrifugal force are treated as real forces for which intuition developed doing more elementary Newton's second law problems can be used. But it should be kept in mind that these forces are just artifacts of a non-inertial coordinate system.

Consider accelerometers that consist of two main parts - a hollow sphere like a basketball inside of which is a slightly smaller sphere. Initially, the centres of the spheres coincide, so that there is a small, uniform gap between the spheres. During acceleration, the gap will be closed, and contact between the spheres will be made. An alarm that indicates acceleration motion will sound. For zero acceleration, no alarm will sound, and straight line motion in inertial frames is indicated.

An accelerometer won't measure acceleration due to either Coriolis force or to centrifugal force.

Also, if a freely falling accelerometer is small enough that tidal forces can be neglected, it will measure zero acceleration, since both spheres fall at the same rate. Acceleration due to gravity, just like acceleration due to Coriolis and centrifugal forces, is independent of mass. Can gravity, too, be considered to be an artifact of a coordinate system? No and yes.

Centrifugal and Coriolis forces at all points in space, i.e., globally, can be transformed away by a single transformation from the rotating frame to an inertial frame. For this reason, these forces are called fictitious. No single transformation will transform gravity away globally, but gravity can be transformed away locally by moving to a freely falling frame. If we expand the term "fictitious force" to include to forces that can be transformed away locally, then gravity also is a fictitious force. Pursuing this line of thought leads to the concept of gravity as spacetime geometry.

I repeated the points that D H and tiny-tim have made, using slightly different words.

As Doc Al has said, in mechanics courses, gravity is treated as a real force; it would be crazy to do otherwise. Considering gravity to be due to spacetime geometry, however, expresses something deeper about the physical world.

Begging to differ, I don't quite follow you when you say Coriolis is not a correction term, but a term which "appears" in rotating ref frames. I guess different physicists, including the ones that taught me at my university, can view things differently. Call me a Newtonian curmudgeon if you please, but I don't see how an *accelerated" ref frame can be 1 to 1 "mapped" or "transformed" into a stationary one.

Let's take Coriolis. If we were to assume that the Earth is NOT rotating, i.e. a stationary ref frame, while it actually IS rotating, what are the consequences? We err by not accounting for the rotation. So, we introduce a correction term in the kinetics/kinematics equations to account for this. But by doing so we are acknowledging the existence of rotation and an accelerated frame of ref. We say that we refer to Earth as a frame of ref, but we do so with a priori knowledge that it is rotating. Thus we include the Coriolis term, twice the cross product of omega and u, to account for the rotation.

I guess one could look at it your way, that the Coriolis force is "there" in the rotating ref frame. When a projectile is launched near the equator, in a direction away from the equator, the rotation of the Earth gives the missile a large eastward component of velocity. The eastward component of points on the Earth further from the equator is less. I'm speaking of linear velocity, not rotational, i.e. u = r* omega. In flight, the missile moves north (or south) away from the equator, while maintaining its eastward velocity imparted via Earth rotation. When the missile lands, it has outdistanced its *intended* landing spot wrt eastward velocity. Hence it lands east of the spot expected if Earth rotation was neglected. I guess you could view it that way, but it can also be viewed as a correction term. With respect to the Earth ref frame, a "virtual" force knocked the missile of course in the eastward direction.

The Coriolis term is intuitive and logical. There seems to be unanimous agreement that there is no actual "Coriolis force" knocking the missile eastward, i.e. Coriolis is "virtual". But in the course of the missile flight, where does "centrifugal" come into play? No one seems to produce the origin of this force, but are too quick to defend its significance. Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one? BR.

Claude

 

[tiny-tim]

The Coriolis term is intuitive and logical. There seems to be unanimous agreement that there is no actual "Coriolis force" knocking the missile eastward, i.e. Coriolis is "virtual". But in the course of the missile flight, where does "centrifugal" come into play? No one seems to produce the origin of this force, but are too quick to defend its significance.

Hi Claude!

Centrifugal force comes in as part of g.

It depends only on position, just like gravitational force, and so it's just part of what we measure as g (which isn't exactly "vertical" anyway, because of mountains etc).

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one?

Which post was that?

 

[Doc Al]

Begging to differ, I don't quite follow you when you say Coriolis is not a correction term, but a term which "appears" in rotating ref frames. I guess different physicists, including the ones that taught me at my university, can view things differently. Call me a Newtonian curmudgeon if you please, but I don't see how an *accelerated" ref frame can be 1 to 1 "mapped" or "transformed" into a stationary one.

Let's take Coriolis. If we were to assume that the Earth is NOT rotating, i.e. a stationary ref frame, while it actually IS rotating, what are the consequences? We err by not accounting for the rotation. So, we introduce a correction term in the kinetics/kinematics equations to account for this. But by doing so we are acknowledging the existence of rotation and an accelerated frame of ref. We say that we refer to Earth as a frame of ref, but we do so with a priori knowledge that it is rotating. Thus we include the Coriolis term, twice the cross product of omega and u, to account for the rotation.

I suppose you can consider analyzing motion that takes into account the Earth's rotation as a "correction" to an analysis that ignores the Earth's rotation. But that's not what is being discussed here. We are talking about analyzing things from an inertial, non-rotating frame (in which the Earth rotates, of course!) versus analyzing things from a non-inertial, rotating frame (in which the Earth is at rest). We are not talking about ignoring the Earth's rotation versus taking it into account. (Both frames take rotation into account!)

When we say that coriolis and centrifugal forces are artifacts of using a non-inertial, rotating frame that does not mean that new physical effects magically appear when using a rotating frame of reference. The physical effects exist even in an inertial frame--they are just much harder to analyze! (In an inertial frame, the effects are "simply" due to the fact that while a projectile goes "straight", the Earth rotates.) If you do your analysis from an inertial frame, there is no need to introduce "forces" such as centrifugal or coriolis. (But good luck carrying out that analysis!)

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one?

I thought I answered that in https://www.physicsforums.com/showpost.php?p=1785157&postcount=51".

 

[George Jones]

I thought I answered that in https://www.physicsforums.com/showpost.php?p=1785157&postcount=51".

I, too, thought that you did.

 

[cmos]

The Coriolis term is intuitive and logical. There seems to be unanimous agreement that there is no actual "Coriolis force" knocking the missile eastward, i.e. Coriolis is "virtual". But in the course of the missile flight, where does "centrifugal" come into play? No one seems to produce the origin of this force, but are too quick to defend its significance.

For this specific example, no one has defended its significance. The reason is that, relative to gravity, Coriolis force, and force due to burning of fuel; the centrifugal force is usually negligible.

Please don't use this as an argument to refute its existence. There are many times when something exist but is negligible. In analyzing the hydrogen atom, you care only about the Coulomb potential. Gravity is negligible in this case, but it is still there physically.

Also, no one has yet explained the origin of cf in the moon orbit question I raised earlier. Is anybody going to attampt to tackle that one? BR.

Yes, Doc Al already explained this. But also, this is the reason I presented the geosynchronous satellite example. It is a simpler example but generalizes to the moon.

 

[George Jones]

Yes, Doc Al already explained this. But also, this is the reason I presented the geosynchronous satellite example. It is a simpler example but generalizes to the moon.

https://www.physicsforums.com/showthread.php?p=1783976#post1783976".

 

[tiny-tim]

For this specific example {a missile}, no one has defended its significance. The reason is that, relative to gravity, Coriolis force, and force due to burning of fuel; the centrifugal force is usually negligible.

Hi cmos!

I don't think that's right:

Coriolis force = -2m\,\mathbf{\Omega} \times \mathbf{v}_{rel}

Centrifugal force = m\,\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})\,=\,m\,\mathbf{\Omega} \times \mathbf{v}_{rot} where \mathbf{v}_{rot} is the velocity of rotation.

So centrifugal force is usually much larger than Coriolis force:

\frac{|centrifugal|}{|Coriolis|}\,=\,\frac{v_{rot}}{2v_{rel}}[/itex] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

 

[cmos]

Hi cmos!

I don't think that's right:

Coriolis force = -2m\,\mathbf{\Omega} \times \mathbf{v}_{rel}

Centrifugal force = m\,\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})\,=\,m\,\mathbf{\Omega} \times \mathbf{v}_{rot} where \mathbf{v}_{rot} is the velocity of rotation.

So centrifugal force is usually much larger than Coriolis force:

\frac{|centrifugal|}{|Coriolis|}\,=\,\frac{v_{rot}}{2v_{rel}}[/itex] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

<br /> <br /> tiny-tim,<br /> <br /> I think you are missing a term in your centrifugal force. Regardless, my counter-argument:<br /> <br /> I was thinking more in terms of the centrifugal force and the force due to gravity. At the equator, the ratio of the magnitudes of the centrifugal to gravitational force is 0.0034413. This ratio tends to zero as we approach either the north or south poles. So for these two forces, I would say the centrifugal is negligible.<br /> <br /> Regarding your assertion, in your final equation, you are missing at least one sine term (there may be more). So in some cases, the centrifugal force will still be negligible. But I do have to say, I could imagine some trajectory in which this would not be the case.<br /> <br /> Well played.

 

[cmos]

I was thinking more in terms of the centrifugal force and the force due to gravity. At the equator, the ratio of the magnitudes of the centrifugal to gravitational force is 0.0034413. This ratio tends to zero as we approach either the north or south poles. So for these two forces, I would say the centrifugal is negligible.

I just want to add, we are talking here about the specific example of an intercontinental ballistic missile traversing the surface of the Earth.

The ratio I mentioned, which monotonically decreases as we approach the poles, helps to explain the phenomenon known as equatorial budge. Clearly not necessarily negligible (depending on the problem).

 

[cabraham]

So what is the consensus? I think we can say that those who legitimize cf do so as follows. In a rotating ref frame, cf is necessary to cancel another force, say gravity, in order to satisfy the sum of forces going to zero.

With the moon orbiting the earth, if we adopt the moon as a ref frame, then from that perspective the moon is attracted to the Earth via gravity. But it doesn't accelerate towards the earth, leaving one to conclude that there must be an equal and opposite force to that of gravity to maintain static equilibrium, i.e. forces summing to zero. Hence, cf is conceptualized.

But who says that the gravitational force is "real" in the moon's ref frame? Here in NE Ohio, there is a city called Sandusky, with an amusement park named "Cedar Point". Maybe you know of it. A ride known as the "Demon Drop" features a platform that the passengers are loaded on to. Then the platform literally drops in free fall, the platform and its passengers accelerate downward at near 1 g.

Let us adopt the free falling platform as our ref frame, an accelerated one at that. If I hold a ball in my hand and release it when the platform is at rest at the top, the ball falls to the floor. Gravity is real and present in this static ref frame. But if the platform commences free fall, and I release the ball in descent, in my accelerated frame of ref, the ball floats in mid-air and does not fall to the floor. Of course to a stationary observer on the ground, the platform, passengers, and ball are all falling downward at 1 g and gravity is active. But in the accelerated ref frame the ball hangs in mid-air. In order to cancel gravity, there must be a "counter gravity force" present to account for the non-motion of the ball. Thus can we conceptualize a "counter gravity" force present in the accel ref frame of the platform?

Or, is it that in the accel ref frame, there is no gravity? The passengers are weightless, and the ball hangs in mid-air. In this frame, gravity is NOT present. Hence there is no counter gravity present either. There is no need to balance the force of gravity as it does not exist in the platform frame of ref.

With the moon or geosat orbiting the earth, many presume that since the geosat does not accelerate towards the Earth and that gravity is present which would force it to do so, that there MUST BE another force opposing gravity, hence centrifugal. But, if I'm on the geosat, and let go of a ball while orbiting the earth, it does not "fall" in my ref frame. Hence there is no gravity in my accel frame of ref. Since there is no gravity, there is no counter force to gravity, the net sum of the forces on the ball in this ref frame is already zero. No gravity, no centrifugal force.

Centrifugal force does not exist.

 

[cmos]

With the moon or geosat orbiting the earth, many presume that since the geosat does not accelerate towards the Earth and that gravity is present which would force it to do so, that there MUST BE another force opposing gravity, hence centrifugal. But, if I'm on the geosat, and let go of a ball while orbiting the earth, it does not "fall" in my ref frame. Hence there is no gravity in my accel frame of ref. Since there is no gravity, there is no counter force to gravity, the net sum of the forces on the ball in this ref frame is already zero. No gravity, no centrifugal force.

Centrifugal force does not exist.

Wrong again. If you are standing on the geosynchronous satellite, then we can say that it is the gravitational force between you and the satellite that keeps you on the satellite (obviously this will be balanced by the normal force since you are not accelerating with respect to the satellite).

So now you take a ball out of your pocket and let it go. I will thus fall to the satellite due to gravity.

Do you still seriously believe that there is no centrifugal force at all or was your last sentence relative to the specific example you erroneously were talking about?

 

[D H]

So what is the consensus? I think we can say that those who legitimize cf do so as follows. In a rotating ref frame, cf is necessary to cancel another force, say gravity, in order to satisfy the sum of forces going to zero.

No. Nobody is saying the centrifugal force is a "real" force. It is a pseudo-force that appears when doing physics from the perspective of a rotating frame and trying to ascribe Newton's second law to describe motion as observed from the perspective of that frame.

From the very end of your post,

Centrifugal force does not exist.

It is a fictitious force. That does not mean it does not exist. It is a force that depends only on the observer, and fictitious forces can appear to be very real to those observing things from the perspective of a rotating frame. This is, I think, why you are stuck on the Coriolis force being a "real" force. It certainly appears to be real! It is not a real force. It, like the centrifugal force, arises solely from the perspective of the observer rather than from the physical envifornment.

On to gravity ...

Let us adopt the free falling platform as our ref frame, an accelerated one at that.

This is your key mistake. The free-falling platform is an inertial frame from the perspective of general relativity. You may not like that definition, but using it gives more accurately describes orbits (particularly Mercury's orbit) than does Newtonian physics.

Gravity is real and present in this static ref frame.

Your "static ref frame" is not an inertial frame.

 

[cabraham]

The gravity you refer to is that between the ball and the geosat. The mass of the geosat is tiny vs. that of the earth. The ball will be attracted to the geosat, but the force is very very small. The ball does not feel the Earth's gravity. Of course the gravity between me and the geosat is there, as well as the normal force. The normal force balances the local geosat gravity. The forces sum to zero WITHOUT introducing centrifugal force.

I believe this sums it up. The Earth's gravity is not felt on the geosat ref frame. There is a local gravity field, but not the earth's. The downward force of Earth gravity in the geosat frame of ref is ZERO. The upward centrifugal force relative to the geosat frame of ref is ZERO.

There is no centrifugal force to an observer on the geosat moving with it.

There is no centrifugal force to a stationary observer.

There is no centrifugal force.

 

[rcgldr]

Getting back to the OP:

Lets say I have an object who moves in a circular path, we know the object is accelerating because the velocity is constantly changing direction, the acceleration is towards the center.

Now, where does the centrifugal force comes to play?

Instead of centrifugal force, call it "reaction force to centripetal force", since the meaning of "centrifugal force" was changed sometime in the past.

Is it equal to f = m*w^2*R = m*v^2/r ?
Does it comes from Newton 3rd law to the centripetal force?

Yes, it's simply Newtons 3rd law about action + reaction forces. Something applies a force to an object, and the object responds with a reaction force applied to that "something". Doesn't matter what the direction of the force is (parallel to path, perpendicular to path or some combination).

 

[jostpuur]

Now, where does the centrifugal force comes to play?

Instead of centrifugal force, call it "reaction force to centripetal force", since the meaning of "centrifugal force" was changed sometime in the past.

This was not a good reply. What is the "it" supposed to mean? If it meant the centrifugal force is modern sense in the original post, then you shouldn't call it a reaction force. The truth probably is, that asi123 didn't know what he was meaning with the centrifugal force, and that's why he was asking. But we should respond by explaining what the term "centrifugal force" means (with modern conventions), and not by what something else means.

 

[cmos]

The gravity you refer to is that between the ball and the geosat. The mass of the geosat is tiny vs. that of the earth. The ball will be attracted to the geosat, but the force is very very small. The ball does not feel the Earth's gravity. Of course the gravity between me and the geosat is there, as well as the normal force. The normal force balances the local geosat gravity. The forces sum to zero WITHOUT introducing centrifugal force.

This is more or less what I said, ok. But the relative masses can be whatever you want them to be, within reason I suppose. Also, the ball not feeling the Earth's gravity, I just want to point out that this would be an approximation. Valid in the same sense that I can approximate that I moon's gravity does not directly affect me.

I believe this sums it up. The Earth's gravity is not felt on the geosat ref frame. There is a local gravity field, but not the earth's. The downward force of Earth gravity in the geosat frame of ref is ZERO. The upward centrifugal force relative to the geosat frame of ref is ZERO.

This is completely false. Go study Newton's law of gravitation.

There is no centrifugal force to an observer on the geosat moving with it.
There is no centrifugal force to a stationary observer.
There is no centrifugal force.

At this point, are you just doing this to waste people's time?

 

[jostpuur]

Let us adopt the free falling platform as our ref frame, an accelerated one at that. If I hold a ball in my hand and release it when the platform is at rest at the top, the ball falls to the floor. Gravity is real and present in this static ref frame. But if the platform commences free fall, and I release the ball in descent, in my accelerated frame of ref, the ball floats in mid-air and does not fall to the floor. Of course to a stationary observer on the ground, the platform, passengers, and ball are all falling downward at 1 g and gravity is active. But in the accelerated ref frame the ball hangs in mid-air. In order to cancel gravity, there must be a "counter gravity force" present to account for the non-motion of the ball. Thus can we conceptualize a "counter gravity" force present in the accel ref frame of the platform?

The formula for the effective force in a non-inertial frame is this:

<br /> \vec{F}_{\textrm{eff}}\; =\; \vec{F} \;-\; m\ddot{R} \;-\; m\dot{\vec{\omega}}\times\vec{r} \;-\; m\vec{\omega}\times(\vec{\omega}\times\vec{r}) \;-\; 2m\vec{\omega}\times\vec{v}<br />

(copied from the page 392 of the Classical Dynamics of Particles and Systems, fifth edition, by Thorton and Marion.)

Now the term -m\ddot{R} is the relevant one for your question. It means that if the coordinate set itself is accelerating, there is a corresponding pseudo force in the opposite direction. So if you choose to think the gravity as a force (so that inertial frames are the ones which are in rest with the surface of the earth), and then choose to use an accelerating frame that falls in a free fall, then yes, there is a pseudo force that you chose to call "counter gravity".

Or, is it that in the accel ref frame, there is no gravity? The passengers are weightless, and the ball hangs in mid-air. In this frame, gravity is NOT present. Hence there is no counter gravity present either. There is no need to balance the force of gravity as it does not exist in the platform frame of ref.

This is the case if you choose to think about the gravity as not a force, but as an alternative definition of the inertial frames.

 

[cabraham]

This is more or less what I said, ok. But the relative masses can be whatever you want them to be, within reason I suppose. Also, the ball not feeling the Earth's gravity, I just want to point out that this would be an approximation. Valid in the same sense that I can approximate that I moon's gravity does not directly affect me.



This is completely false. Go study Newton's law of gravitation.



At this point, are you just doing this to waste people's time?

Not at all, I assure you. Here is a summary.

For a geosat orbiting earth, let's use 2 frames of ref. A stationary observer in ref frame S, and an observer on the geosat revolving around the earth, ref frame R. In the early part of this thread, I was rebuked because I persistently used the stationary S frame for my reference. I was told that there are instances when it is more appropriate to consider ucm from the moving frame R. Also it was emphasized that in the R frame, the geosat feels the force of the Earth's gravity but does not accelerate towards the earth. Therefore it was reasoned that an equal and opposite counter force to gravity must account for the forces on the geosat summing to zero in its own ref frame R. Hence centrifugal force is conceptualized. It actually sounds logical. If Earth gravity pulls the geosat down, yet it stays put in the R frame of ref, centrifugal force must be countering gravity so that the net forces acting on the geosat are zero in ref frame R.

But in an accelerated ref frame, R in this case, the force of Earth gravity is not present. With the "demon drop" amusement park ride, a platform full of passengers is in free fall. They are weightless and objects released from their grip float weightless in space along with them in ref frame R, the free falling platform. In the R frame of ref, no gravity is present, hence the forces on the ball released by a passenger are zero without any force to counter gravity since it is zero.

In the geosat frame of ref R, the ball, the observer, and the geosat have the same speed of zero in frame R. If the observer releases the ball it will float weightless with him and the geosat. In frame R, the forces sum to zero as the Earth gravity force is zero downward, and there is zero upward force as well.

In frame S, the geosat does feel the gravity force. It has an acceleration towards the earth. The net force on the geosat is equal to the centripetal acceleration times the mass in frame S.

Back to frame R, yes cmos the geosat has ITS OWN gravity. It pullsdown on the observer, and a normal force counters this local gravity. The forces sum to zero, local gravity and normal. All is balanced without centrifugal.

There is no centrifugal force in frame R.

There is no centrifugal force in frame S.

There is no centrifugal force period.

Regarding Newton's law of gravitation, of course the Earth attracts the geosat **in frame S**. Newton's law is not being challenged. Just like the free falling platform, there is no gravitation **in frame R**. Hence no counter force. That is the crux.

If there was a centrifugal force, even if only in one of the ref frames, my physics prof Dr. M would have taught me so. Ditto for the other physics profs and ME profs as well. I trust them. They know their stuff. I trust them more than myself. The reason I have such confidence in "my position" is because it is NOT my position, but that of others more learned than me regarding these issues.

Thanks to every one ESPECIALLY THOSE who opposed me. These debates keep all of us sharp and I've been well educated. I know much more now than a few days ago. To those who disagree with me, I extend my hand in peace and friendship. You've helped me immensely. Have a great day!

Claude

 

[D H]

The formula for the effective force in a non-inertial frame is this:

<br /> \vec{F}_{\textrm{eff}}\; =\; \vec{F} \;-\; m\ddot{R} \;-\; m\dot{\vec{\omega}}\times\vec{r} \;-\; m\vec{\omega}\times(\vec{\omega}\times\vec{r}) \;-\; 2m\vec{\omega}\times\vec{v}<br />

(copied from the page 392 of the Classical Dynamics of Particles and Systems, fifth edition, by Thorton and Marion.)

I already said this in https://www.physicsforums.com/showthread.php?p=1783888#post1783888".

 

[rcgldr]

What is the "it" supposed to mean? If it meant the centrifugal force is modern sense in the original post, then you shouldn't call it a reaction force. The truth probably is, that asi123 didn't know what he was meaning with the centrifugal force, and that's why he was asking. But we should respond by explaining what the term "centrifugal force" means (with modern conventions), and not by what something else means.

Since the OP specifically mentioned Newtons 3rd law, it should be clear that he's interested in reaction force, not the more complicated modern meaning of centrifugal force. Plenty of othres have already given the new modern definition.

I would have preferred that centripetal and centrifugal were used to describe action / reaction pair forces, but I'm not in charge of physics terminology, and there's already a thread about this.

When did centrifugal force die?

 

[jostpuur]

cabraham, frames that move along orbiting planets are either examples of accelerating frames, with the direction of acceleration changing, or examples of frames that are both accelerating and rotating. Using them requires good understanding of pseudo forces. You need to learn pseudo forces with simpler examples first, instead of trying to jump into complicated examples in attempt to find confusion so that you could prove pseudo forces inconsistent.