Gravity and fictitious forces

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Good evening,

Suppose that there are two bodies, A and B, of masses mA and mB, in vacuum. In an arbitrary inertial reference frame, the gravitational acceleration produced by B in A is [tex]\vec{a}_A[/tex] and the one produced by A in B is [tex]\vec{a}_B[/tex].
Now, I want to calculate the gravitational acceleration of A as seen from B's reference frame, which is non-inertial. It would be [tex]\vec{a}_A-\vec{a}_B[/tex].
Now, suppose I want to calculate the net force in A. It would be [tex]m_A\vec{a}_A-m_A\vec{a}_B = m_A\vec{a}_A+(-m_A\vec{a}_B)[/tex].
My question is: A new force seems to appear here: [tex]-m_A\vec{a}_B[/tex]. Since B is a non-inertial reference frame, should I consider this a fictitious force? If yes, is there a name for this particular fictitious force?

Thank you in advance.
 
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  • #2
K^2
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Yes, it's a fictitious force. I don't think it has a specific name.
 
  • #3
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Good evening,

Suppose that there are two bodies, A and B, of masses mA and mB, in vacuum. In an arbitrary inertial reference frame, the gravitational acceleration produced by B in A is [tex]\vec{a}_A[/tex] and the one produced by A in B is [tex]\vec{a}_B[/tex].
Now, I want to calculate the gravitational acceleration of A as seen from B's reference frame, which is non-inertial. It would be [tex]\vec{a}_A-\vec{a}_B[/tex].
Now, suppose I want to calculate the net force in A. It would be [tex]m_A\vec{a}_A-m_A\vec{a}_B = m_A\vec{a}_A+(-m_A\vec{a}_B)[/tex].
My question is: A new force seems to appear here: [tex]-m_A\vec{a}_B[/tex]. Since B is a non-inertial reference frame, should I consider this a fictitious force? If yes, is there a name for this particular fictitious force?

Thank you in advance.
The extra "force" may be named "inertial force". You need to take it into account if you want to apply Newton;s second law in the non-inertial frame.

I am not sure if it is appropriate to call the difference aA-aB "gravitational acceleration".
 
  • #4
Andrew Mason
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Fictitious force is a term that is normally used to explain inertial effects experienced in a non-inertial (accelerating) frame of reference. Here, there are no inertial effects felt by either A or B.

Gravitational force is fundamentally different than other forces in that the frame of reference of a body accelerating in a gravitational field is locally equivalent to an inertial reference frame. There are no inertial effects, so there are no "fictitious" forces exerienced by either mass.

To B, A would appear to be accelerating at:

[tex]\vec{a}_A = \frac{G(m_A + m_B)}{r^2}\hat r[/tex]

and to A, B would appear to be accelerating at:

[tex]\vec{a}_B = -\frac{G(m_A + m_B)}{r^2}\hat r[/tex]

where r, the separation, has the same magnitude for each mass and [itex]\hat r[/itex] is the unit displacement vector of B from A.

An inertial observer, C (eg. the centre of mass frame) would observe A's acceleration to be:

[tex]\vec{a}_A = F_A/m_A = \frac{Gm_B}{r^2}\hat r[/tex]

and B's acceleration to be:

[tex]\vec{a}_B = -\frac{Gm_A}{r^2}\hat r[/tex]

So the difference between B's and C's measurement of the force on A would be:

[tex]F_{A by B} - F_{A by C} = \frac{Gm_A^2}{r^2}\hat r[/tex]

AM
 
  • #5
K^2
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Andrew, you are contradicting yourself. You are writing equations for accelerations that are different in two frames, and yet you insist that there are no inertial forces.

In frame of A, B accelerates at higher rate than gravity due to A can account for. That's a fictitious force, as OP correctly points out.
 
  • #6
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But isn't the coordinate frame of an object in free fall inertial? And since both A and B are in free fall, does that mean that all three frames are valid inertial frames of reference? And if that is correct, how can there be any fictitious forces?
 
  • #7
D H
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The extra "force" may be named "inertial force". You need to take it into account if you want to apply Newton;s second law in the non-inertial frame.
That's the generic name for any accelerating (as opposed to rotating) reference frame. As far as I know, there is no specific name when the acceleration results from gravitation as opposed to some other force.


But isn't the coordinate frame of an object in free fall inertial? And since both A and B are in free fall, does that mean that all three frames are valid inertial frames of reference? And if that is correct, how can there be any fictitious forces?
This question is being raised in the context of classical physics, and a frame centered on an object in free fall (i.e., subject to gravitation by some other body/bodies) is not an inertial frame in that context.
 
  • #8
Andrew Mason
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Andrew, you are contradicting yourself. You are writing equations for accelerations that are different in two frames, and yet you insist that there are no inertial forces.

In frame of A, B accelerates at higher rate than gravity due to A can account for. That's a fictitious force, as OP correctly points out.
I may be taking a more narrow view of "fictitious force" than others. I am using that term to mean an apparent force causing an apparent acceleration that is independent of mass (ie. the force is proportional to mass) that is observed in the non-inertial frame of reference due to inertial effects. A fictitious force would be the "centrifugal force" felt by a rider on a roller coaster as it turns.

In this case, the observer in B's frame of reference does not observe any inertial effects resulting from the gravitational force between mass A and mass B. Moreover, the observer in the B frame (not taking into account his own accelerating frame) measures mass A's acceleration to be higher than that measured by the inertial observer C by an amount [itex]Gm_A^2/r^2[/itex]. This is not proportional to [itex]m_A[/itex] but to [itex]m_A^2[/itex].

AM
 
  • #9
D H
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I may be taking a more narrow view of "fictitious force" than others.
You are taking too narrow a view.

Suppose we have a system of two massive objects, call the A and B. The forces acting on a test object C of negligible mass as observed from the perspective of an inertial frame is given by the superposition of the forces given by Newton's law of gravitation:

[tex]
\vec F_C =
\vec F_{\text{non-grav}}
- \frac{G m_A m_C}{||\vec r_C - \vec r_A||^3} (\vec r_C - \vec r_A)
- \frac{G m_B m_C}{||\vec r_C - \vec r_B||^3} (\vec r_C - \vec r_B)
[/tex]

where [tex]\vec r_X[/tex] is the position vector of object X with respect to the origin of the inertial frame.

From the perspective of a non-rotating frame with origin at object A's center of mass, the forces on this test object are

[tex]
\vec F_C =
\vec F_{\text{non-grav}}
- \frac{G m_A m_C}{||\vec r_C||^3} \vec r_C
-\frac{G m_B m_C}{||\vec r_C - \vec r_B||^3} (\vec r_C - \vec r_B)
- \left(-\,\frac{G m_B}{||\vec r_B||^3} \vec r_B\right)m_c
[/tex]

where [tex]\vec r_X[/tex] now denotes the position vector of object X with respect to the origin of the (non-inertial) frame with origin at the center of object A. Note that additional final term. This is the "inertial force" that is the subject of this thread.
 
  • #10
Andrew Mason
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You are taking too narrow a view.
I think if you do not take my "narrow" view of fictitious force, you have to consider every "apparent force" observed in a non-inertial frame that is not observed in an inertial frame to be fictitious. I don't think that is how the term "fictitious force" or "pseudo force" is used.

Consider a car moving with a constant forward acceleration, [itex]\vec{a}[/itex], down a straight road (ignore earth curvature). From the reference frame of the car, the earth and everything attached to it is accelerating backward with acceleration [itex]-\vec{a}[/itex]. So the observer in the car might analyse this as being the result of a force being applied to every object of [itex]\vec{F_{obj}}= -m_{obj}\vec{a}[/itex]. Do we call these fictitious forces? How does this force differ from the OP's example?

I think you will find that historically the term "fictitious force" is reserved for forces that appear to affect objects in non-inertial reference frames due to inertial effects experienced in that frame. The only "fictitious force" in the example I have given is the rearward "force" experienced by the car and objects in the car. As in the OP's example, there are no inertial effects felt by any object that is subjected only to a gravitational force.

AM
 
  • #11
D H
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I think if you do not take my "narrow" view of fictitious force, you have to consider every "apparent force" observed in a non-inertial frame that is not observed in an inertial frame to be fictitious. I don't think that is how the term "fictitious force" or "pseudo force" is used.
If by "apparent force" you mean a force that vanishes from the perspective of a (Newtonian) inertial frame, that is exactly how the term "fictitious force" or "pseudo force" is used. An "apparent force", "fictitious force", or "pseudo force" is a force that acts on some object of mass m such that (a) the force is proportional to the mass of the object, and (b) the force vanishes in a (Newtonian) inertial frame.


I think you will find that historically the term "fictitious force" is reserved for forces that appear to affect objects in non-inertial reference frames due to inertial effects experienced in that frame.
That is exactly what is going on here.

Suppose you have at hand some frame I known to be inertial in the Newtonian sense. Suppose also that you have a test particle of mass m which is being acted upon by a number of external forces and that you have painstakingly enumerated and calculated each of these external forces. Now suppose you want to describe the motion of this test particle in frame I. The answer is given by the superposition principle coupled with Newton's second law:

[tex]m\,\frac{d^2}{dt^2}\vec r_I(t) =\sum_i \vec F_{\text{ext}_i}(t)[/tex]

where the subscript I on the position vector designates that this is the position as observed and expressed in the inertial frame I.

Now suppose that you arbitrarily define another frame A whose origin is accelerating with respect to this inertial frame. I'll denote this acceleration as [tex]\vec a_A(t)[/tex], where the subscript A explicitly designates that this is the acceleration of the frame A's origin wrt inertial. Just as our inertial observer did, you can once again painstakingly enumerate and calculate all of the physical forces acting on the test particle. Simply applying Newton's 2nd law will no longer describe the behavior of the test object. You need to introduce an apparent force in the equations of motion:

[tex]m\,\frac{d^2}{dt^2} \vec r_A(t) =m \vec a_A(t) + \sum_i \vec F_{\text{ext}_i}(t)[/tex]

This additional term [tex]m \vec a_A(t)[/tex] is a "fictitious force". It is (a) proportional to mass and (b) vanishes in the case that frame A is an inertial frame (in which case [tex]\vec a_A(t)=0[/tex]).

As far as the name of this fictitious force: While some use the term "inertial force" or "d'Alembert force" as a pseudonym for any fictitious force, others use limit these terms specifically to mean the apparent force that results from the acceleration of the origin of the reference frame itself. For example, see Lanczos in The variational principles of mechanics, http://books.google.com/books?id=ZW...46Q9NIrYWwn5EmYpPv-LPuZd0#v=onepage&q&f=false.

In aerospace engineering and astrophysics, the apparent force that arises in a frame centered on a planet or star from the gravitational acceleration of that planet/star toward other massive objects is often called the "third-body effect".
 
  • #12
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DH said:
TurtleMeister said:
But isn't the coordinate frame of an object in free fall inertial? And since both A and B are in free fall, does that mean that all three frames are valid inertial frames of reference? And if that is correct, how can there be any fictitious forces?
This question is being raised in the context of classical physics, and a frame centered on an object in free fall (i.e., subject to gravitation by some other body/bodies) is not an inertial frame in that context.
I don't understand. Since all objects are in free fall (the moon is in free fall with the earth, the earth is in free fall with the sun, the sun is in free fall with the galaxy, ect...) does that mean that an inertial frame cannot be centered on any object in classical physics?
 
  • #13
D H
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I don't understand. Since all objects are in free fall (the moon is in free fall with the earth, the earth is in free fall with the sun, the sun is in free fall with the galaxy, ect...) does that mean that an inertial frame cannot be centered on any object in classical physics?
Strictly speaking, yes. Strictly speaking, there is no such thing as an inertial frame, at least insofar as we know (name one if you think otherwise).

Practically speaking, there are plenty of reference frames that are approximately inertial. If you want to describe the flight of an object such as a football, treating an Earth-fixed frame as inertial works just fine. The Coriolis effect on a flying football is very small compared to the known unknowns such as variations in atmospheric drag. The Coriolis effect can become important if you want to accurately model the flight of a long-range missile, but tidal forces are still very small.

On the other hand, if you want to accurately model the trajectory of an artificial satellite you had better incorporate third body effects from the Moon and Sun. In other words, you need to treat an Earth-centered frame as a non-inertial frame.
 
  • #14
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Yes, you are correct. I should have given this more thought before I posted. I think I even remember a similar discussion in another thread some time ago.

However, we can still speak of approximate inertial frames, such as the OP's scenario. I tend to agree with AM's point of view on this. But I'm having trouble following your response in post #11. A dumbed down version would be helpful. But if that's too much trouble then don't worry about it.
 
  • #15
D H
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However, we can still speak of approximate inertial frames, such as the OP's scenario. I tend to agree with AM's point of view on this. But I'm having trouble following your response in post #11. A dumbed down version would be helpful. But if that's too much trouble then don't worry about it.
A dumbed-down version: The origins of all inertial reference frames are related to one another via a galilean transform. If the origin of a reference frame is accelerating with respect to some inertial frame the frame in question is not an inertial frame. Without the invention of an appropriate fictitious force, Newton's laws do not apply in such frames.

It doesn't matter whether the acceleration of the frame results from an arbitrary designating some point as the origin or from choosing a point that is known to be accelerating (such as from gravitation) as the origin. The frame is accelerating, period. Fictitious forces are needed to give the appearance that Newton's laws still apply.
 
  • #16
Andrew Mason
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If by "apparent force" you mean a force that vanishes from the perspective of a (Newtonian) inertial frame, that is exactly how the term "fictitious force" or "pseudo force" is used. An "apparent force", "fictitious force", or "pseudo force" is a force that acts on some object of mass m such that (a) the force is proportional to the mass of the object, and (b) the force vanishes in a (Newtonian) inertial frame.
I was using "fictitious force" in the sense that it is normally used to refer to inertial effects experienced in a non-inertial frame of reference.

In the case posed by the OP of two bodies accelerating towards each other due to gravitational attraction the question was whether the force needed explain the relative acceleration (ie the relative acceleration of the other body that an observer on one of the bodies measures) from the frame of one of those bodies included a component that could be termed "fictitious". I agree that this component is not a real force. But the term "fictitious force" has a particular meaning - it refers to certain inertial effects experienced in the non-inertial frame of reference - that did not seem to fit this situation.

Suppose I jump out of an airplane and I fall to the earth in freefall. If I treat my frame of reference as an inertial frame, I could only explain the acceleration of the earth toward me by the presence of an enormous force acting on the earth that is accelerating it toward me at 9.8 m/sec^2. That would be:

[tex]F_{apparent} = M_{earth}g = GM_{earth}^2/r^2[/tex]

This was essentially the situation posed by the OP, with a large difference in mass between the two gravitating objects. All I am saying is that we do not include such an apparent force in the term "fictitious force" as that term has been come to be used.

One of the characteristics of a fictitious force is that it is proportional to mass. In this case, the force is proportional to mass^2. There are also no inertial effects experienced by an object in free fall. So I suggested that the "apparent force" that would be required to accelerate the earth toward me at the rate of g does not fit the criteria normally used for fictitious forces.

AM
 
  • #17
D H
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I was using "fictitious force" in the sense that it is normally used to refer to inertial effects experienced in a non-inertial frame of reference.
So am I, Andrew. I gave you one reference already, See post #11.
Here's another:http://physics-help.info/physicsguide/mechanics/noninertial_mechanics.shtml.
Here's yet another (warning: powerpoint): http://web.njit.edu/~gary/430/assets/physics430_lecture20.ppt.
Here is a Zoho viewer link to this last reference: http://viewer.zoho.com/docs/urlview.do?url=http%3A%2F%2Fweb.njit.edu%2F%7Egary%2F430%2Fassets%2Fphysics430_lecture20.ppt [Broken].
Note: Zoho's viewer isn't perfect, but it is much better than google docs on embedded math.


Suppose I jump out of an airplane and I fall to the earth in freefall. If I treat my frame of reference as an inertial frame ...
In your free-fall, where you are falling toward the Earth with an acceleration aA, you can describe the flight of a football f as observed in your frame so long as you subtract aA*mf from all of the calculated real forces on the football. You can describe the orbit of a satellite s as observed in your frame so long as you subtract aA*ms from all of the calculated real forces on the satellite. For any object x the fictitious force -aA*mx appears. When x is the Earth itself, the fictitious force is -aA*mE. The fact that aA is also proportional to the mass of the Earth is a bit irrelevant, and you are hung up on this irrelevancy.


Let's get down to Earth and look at the tides. I'll designate the positions with respect to the center of the Earth of the Moon as [tex]\vec r_m[/tex] and that of a drop of water on the surface of the Earth as [tex]\vec r_d[/tex]. From the perspective of an inertial frame, the gravitational force on this drop of water by the Moon is

[tex]\vec F_{\text{inertial}} =
- \frac{Gm_mm_d}{||\vec r_m - \vec r_d||^3}(\vec r_m - \vec r_d)[/tex]

The Earth is accelerating toward the Moon as well as the droplet. To obtain the net force on the droplet from the perspective of an Earth-centered frame we need to subtract the fictitious force resulting from the Earth's acceleration toward the Moon:

[tex]\aligned
\vec F_{\text{Earth-centered}} &=
- \frac{Gm_mm_d}{||\vec r_m - \vec r_d||^3}(\vec r_m - \vec r_d)
- \left(-\,\frac{Gm_m}{||\vec r_m||^3}\vec r_m\right)m_d
\\[6pt]
&=
- Gm_m
\left(
\frac{\vec r_m - \vec r_d}{||\vec r_m - \vec r_d||^3}
- \frac{\vec r_m}{||\vec r_m||^3}
\right)m_d
\endaligned[/tex]

Here is a graphical depiction of these tidal forces:

stress.gif
 
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  • #18
Andrew Mason
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All of what you say is quite correct. All I was saying is that the term "fictitious force" is usually used in a narrower sense.

For example, if we considered the earth to be an inertial frame, the planets would require strange forces to explain their strange Ptolemaic motions. Would you refer to these as "fictitious forces"? I would not.

There are forces that we call "fictitious forces" that result from inertial effects in a rotating frame - the coriolis force, centrifugal force, for example. But forces imputed to external objects to explain their peculiar motions relative to the rotating observer are not usually referred to as "fictitious forces".

AM
 
  • #19
D H
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Andrew: Please find ONE reference that says that the apparent force that arises from the acceleration of the origin of the frame of reference is not a fictitious force. I have found many that say that it is. Here is yet another, good old wikipedia, http://en.wikipedia.org/wiki/Fictitious_force
Four fictitious forces are defined in accelerated frames: one caused by any relative acceleration of the origin in a straight line (rectilinear acceleration), two caused by any rotation (centrifugal force and Coriolis force) and a fourth, called the Euler force, caused by a variable rate of rotation, should that occur.​
(The term "straight line" is not needed here, but then again it is wikipedia.)

What, exactly, would you call the apparent force that arises from the acceleration of the origin of the frame of reference if not a fictitious force? It has both of the distinguishing characteristics of a fictitious force: The force vanishes in an inertial frame, and the force is proportional to mass (specifically, it is -a*m).
 
  • #20
Andrew Mason
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Andrew: Please find ONE reference that says that the apparent force that arises from the acceleration of the origin of the frame of reference is not a fictitious force.
That is not what I am saying. I am just saying that not all the forces that the non-inertial observer imputes to external objects (ie objects not in the observer's non-inertial frame) are "fictitious forces" as that term has come to be used. The term "fictitious force" normally refers to the imputed forces listed in the Wikipedia article to which you referred and all are entirely created by inertial effects of the object to which the imputed force applies. This means that the apparent force is proportional to mass and disappears completely if you eliminate the observer's acceleration.

The force that one imputes to the other gravitating mass to explain the relative acceleration is proportional to the square of that mass. So that part doesn't fit. Besides, the force that you are wanting to call "fictitious" is simply an augmentation of a real force, so it is not created by the acceleration of the non-inertial observer. The magnitude changes but it does not disappear in the inertial (eg. centre of mass) frame.

What, exactly, would you call the apparent force that arises from the acceleration of the origin of the frame of reference if not a fictitious force? It has both of the distinguishing characteristics of a fictitious force: The force vanishes in an inertial frame, and the force is proportional to mass (specifically, it is -a*m).
My point is one of terminology. Since not all forces that are imputed due to the acceleration of the observer have the same fundamental characteristics, they should not be given the same name. The additional gravitational acceleration that the OP refers to has neither of the distinguishing characteristics of what we call a "fictitious force". We could call them Ptolemaic forces, but that might be confused with the military forces of King Ptolemy.

AM
 
  • #21
D H
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That is not what I am saying. I am just saying that not all the forces that the non-inertial observer imputes to external objects (ie objects not in the observer's non-inertial frame) are "fictitious forces" as that term has come to be used.
Where did you get this idea? The meaning of the term "fictitious force" has come to mean any and all of the forces that arise solely from the reference frame being non-inertial.

This means that the apparent force is proportional to mass and disappears completely if you eliminate the observer's acceleration.
Exactly. Andrew, you are appear to have an undue obsession here with this m2 term. I have shown you that the force is proportional to mass. Specifically, the apparent force acting on some object is the acceleration of the origin of the reference frame times the mass of the object. That the acceleration of the origin of the reference frame is also proportional to the mass of the object in question in the special case of the two body problem with the origin selected to be the center of mass of one of the two masses is irrelevant. The frame is accelerating with respect to inertial, so fictitious forces arise, and this is one of them.

I have given many references that say exactly this. I have asked you to give me just ONE reference that says otherwise.
 
  • #22
Andrew Mason
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Where did you get this idea? The meaning of the term "fictitious force" has come to mean any and all of the forces that arise solely from the reference frame being non-inertial.
So you would say that the "forces" which cause planets to make epicycles about the earth would be "fictitious forces"? Show me one reference that refers to these as "fictitious forces".

Exactly. Andrew, you are appear to have an undue obsession here with this m2 term. I have shown you that the force is proportional to mass. Specifically, the apparent force acting on some object is the acceleration of the origin of the reference frame times the mass of the object.
If you change the mass you also change the acceleration. That is not an example of an inertial effect. See: the Politzer reference below.

The imputed force is Gm_A^2/r^2. The acceleration is proportional to mass. The force is proportional to m^2.

That the acceleration of the origin of the reference frame is also proportional to the mass of the object in question in the special case of the two body problem with the origin selected to be the center of mass of one of the two masses is irrelevant. The frame is accelerating with respect to inertial, so fictitious forces arise, and this is one of them.

I have given many references that say exactly this. I have asked you to give me just ONE reference that says otherwise.
I gave you the Wikipedia source that you provided. http://www.scientificamerican.com/article.cfm?id=what-is-a-fictitious-force".

Gravity is a special kind of force. It behaves very differently than other forces. In the OP's example, you cannot separate the mass from the acceleration. The two are inextricably tied together. Gravity itself is indistinguishable from an inertial force (principle of equivalence). I think it is a mistake to say, as in the OP's example, that part of the "force" that is observed is an inertial force and another part isn't.

AM
 
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  • #23
D H
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So you would say that the "forces" which cause planets to make epicycles about the earth would be "fictitious forces"? Show me one reference that refers to these as "fictitious forces".
Please. Now you are resorting to argument from ridicule.

If you wanted to model the behavior of the solar system from the perspective of an Earth-centered frame you would of course need to subtract the acceleration of the Earth toward all the bodies in the solar system from the computed accelerations in an inertial frame. Nobody does that because accuracy will suffer compared to using a solar system barycentric model. Accuracy suffers because the planets are well outside the Earth's Hill sphere.

What you will find in literature is that in precision orbit determination of an artificial Earth satellite from the perspective of an Earth-centered frame one must model the gravitational acceleration of the satellite toward the Sun, Moon, Jupiter, and Venus as third body effects -- i.e, you need to subtract the arth's acceleration toward those bodies from the acceleration calculated by Newton's law of gravitation. In other words, fictitious forces arise.

The imputed force is Gm_A^2/r^2. The acceleration is proportional to mass. The force is proportional to m^2.
This is the heart of your problem. You are missing the big picture, which is that the apparent force on any object in this non-inertial frame is the mass of the object times the acceleration of the origin of the frame.

I gave you the Wikipedia source that you provided.
Did you read that article? It clearly says that the acceleration of the origin of a reference frame results in a fictitious force.

Did you read that article? It opens with an example of a linearly-accelerating frame of reference rather than a rotating one.


I have given multiple reference that explicitly identify the acceleration of the Earth toward the Moon results in a fictitious force in an Earth-centered frame. You have yet to give one that says this is something other than a fictitious force.


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