First, the force itself (being the weight) , we are told by Galileo, does not affect the acceleration of a falling body. — Gampa Dee

The reason for this is that the mass of a falling object is negligible in relation to the mass of the Earth. the mass of the Earth cannot probably still be measured to the ton, and if it could, some of the mass would been the air above and exert a negative force.

In calculating the obit of the planets, it might become significant.I haven't checked your algebra though, to know if your calculations would be correct.

So while m (small mass) can indeed affect the force, what good is it in calculating the acceleration if all different masses fall at the same rate? — Gampa Dee

Different masses accelerate at the same rate towards a reference mass because they also have different inertias, which balances the different forces generated.....

Thank you for your response, unenlightened

The reason for this is that the mass of a falling object is negligible in relation to the mass of the Earth. — unenlightened

Yes, I would agree with that as well; my concern is the Galileo assumption that the mass of a falling body does not affect the acceleration (g) at all. This opposed the Greek philosopher Aristotle, who was said to claim that heavier objects were falling quicker .
I would in fact surely agree to viewing the system as a whole(earth and falling bodies)as being invariant in terms of energy “if” the fallen body was taken from the earth (did not come from outer space), for in this case, the falling body was formally on the ground being part of the cause for the g acceleration, and since while the falling body will attract the earth as well, then,for the whole system, the overall acceleration will not change, in my mind.

the mass of the Earth cannot probably still be measured to the ton, and if it could, some of the mass would been the air above and exert a negative force. — unenlightened

I agree that the air will indeed attract the earth as all bodies will cause gravity......
Since the air has weight, it still will have an effect in being part of the earth’s gravitational acceleration. However you have pointed out something important, as gravity is caused not only by the mass only, but by it’s density as well; for example, if the earth would be squashed to the size of the moon, it would have a much greater surface gravity....

Grampa Dee

Thank you for responding Pantagruel:

Different masses accelerate at the same rate towards a reference mass because they also have different inertias, which balances the different forces generated..... — Pantagruel

I personally see this as being only part of the equation since the falling bodies should attract the earth as well, and the difference of the earth’s acceleration towards the falling body will be different with different masses .

Grampa Dee

Indeed it is, but the same condition applies with respect to the relationship between force and inertia. The force is a composite product of the two masses, as is the acceleration.

Thank you for replaying Pantagruel

ndeed it is, but the same condition applies with respect to the relationship between force and inertia. The force is a composite product of the two masses, as is the acceleration. — Pantagruel

Maybe you could help me in this area. The way I see it would be that Mass (the earth) gravitational energy will be causing an acceleration g at a certain distance ....the whole sphere at that distance will have the same g...
M / 4pi * r ^2 = k g

...in fact, this is all we need in my opinion to calculate the g acceleration.
However, since the same is said for the second mass, the two accelerations will need to be added, and if the second mass is great or small will make a difference in the overall g, it seems.

Grampa Dee

...in fact, this is all we need in my opinion to calculate the g acceleration.
However, since the same is said for the second mass, the two accelerations will need to be added, and if the second mass is great or small will make a difference in the overall g, it seems. — Gampa Dee

But you are conflating a generalized equation for force (which includes two masses) and acceleration (which includes two masses but in which the inertia of the much smaller of the two masses always balances whatever additional force it contributes to the overall system, and so can be factored out). Yes, for any given large body there is an acceleration equation which disregards the mass of the "falling" smaller object. But also yes, the overall force realized does vary if the mass of the smaller object varies.

I'm sorry, Pantagruel, I might very well be misunderstanding what you're saying.

But also yes, the overall force realized does vary if the mass of the smaller object varies — Pantagruel

I do agree that the overall force will vary with different masses.
My post is about reconciling gravity with Galileo’s concept of different masses having exactly the same acceleration in freefall;... this is the problem I’m having.right now.

Grampa Dee

Ok. And they have the same acceleration because they have different inertias. How does this not answer that?

Thank you for responding Pantagruel

Ok. And they have the same acceleration because they have different inertias. How does this not answer that? — Pantagruel

I don't understand how they can have the same acceleration, Pantagruel.
That's the problem that I have.
Grampa Dee

My post is about reconciling gravity with Galileo’s concept of different masses having exactly the same acceleration in freefall; — Gampa Dee

Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether.

Careful though. This expresses coordinate acceleration relative to some inertial frame. This means that if you drop a 10 cm diameter ball of iron (10 kg say, just guessing) from the leaning tower, it hits the ground in around 3 1/3 seconds. But now if you increase the density of that ball to say the mass of the moon, it will hit the ground in less time, not because its initial coordinate acceleration is any greater, but because it has enough mass to significantly cause the ground to come up and meet it. So R decreases quicker, and as it does so, coordinate acceleration also increases quicker, a secondary effect barely measurable with our dropped moonlet. But the ground coming up is a primary effect and the decreased time to impact would definitely be measurable (not to mention the damage to the planet from dropping something that heavy, I mean the Pisa tower itself would collapse just from having that thing nearby)

Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether. — noAxioms

Yep.

...the mass of a falling object is negligible... — unenlightened

Not quite; as noAxiom shows, the mass of the falling object is irrelevant.

Thank you for responding noAxiom

But now if you increase the density of that ball to say the mass of the moon, it will hit the ground in less time, not because its initial coordinate acceleration is any greater, but because it has enough mass to significantly cause the ground to come up and meet it — noAxioms

I fully agree with your description noanxiom, but to claim that the increased acceleration due to the earth moving upwards is separated from the downward moving mass, I don’t quite get. If someone drives towards you at 20 km/hr and you towards him/her at 20 km/hr, the velocity is indeed 40 km/hr relative to the two cars... we add the two velocities

Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether. — noAxioms

Here is the reason why I have a problem with the equation. As it is written it follows Galileo’s axiom, for it doesn’t matter what you put as the small mass, the acceleration will continue as being (GM/r^2).
But then, what do we do for the mass equal to the mass of the moon? The upper equation will calculate the acceleration still as being (GM / r^2)...
It is here that I personally believe that another acceleration needs to be added onto the first.
being A is also = Gm/r^2... if we add both of them, then we get A = (GM / R^2) + (Gm / r^2)
However , this equation does not agree with Galileo since a change in mass for the small m will indeed change the total acceleration of the system.

but to claim that the increased acceleration due to the earth moving upwards is separated from the downward moving mass, — Gampa Dee

No it isn't, since Earth accelerating upward will decrease r more quickly, and since the coordinate acceleration of the dropped mass is a function of that r, it affects the acceleration of the dropped thing. That's the secondary effect I was talking about.

Keep in mind that all this is Newtonian physics, which doesn't describe reality. I squashed the mass of the moon into a grapefruit, and that pushes the limits of that simple formula. For instance, one has to start asking about the location of the clock used to measure this acceleration, because unlike in Newtonian physics, it makes a difference.

I don’t quite get. If someone drives towards you at 20 km/hr and you towards him/her at 20 km/hr, the velocity is indeed 40 km/hr relative to the two cars... we add the two velocities

Under Newtonian physics, yes. But we're not adding speeds here, we're computing coordinate acceleration.

Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether.
— noAxioms

As it is written it follows Galileo’s axiom, for it doesn’t matter what you put as the small mass, the acceleration will continue as being (GM/r^2)

Until r starts changing...

But then, what do we do for the mass equal to the mass of the moon?

Still works, at least under Newtonian physics. Same coordinate acceleration.

It is here that I personally believe that another acceleration needs to be added onto the first.
being A is also = Gm/r^2

That would be wrong. The acceleration of the moon would not have that component. Earth does. Remember, I was stating that the formula gives acceleration relative to some inertial frame. I think you are trying to use the accelerating frame of Earth when adding them like that. But the force is given by F=GMm/r², and since acceleration is A = F/m, the moon accelerates by the simple formula, not adding the Earth part to it. Either that or F=ma is wrong, which is a denial of some pretty basic laws.

... if we add both of them, then we get A = (GM / R^2) + (Gm / r^2)
However , this equation does not agree with Galileo since a change in mass for the small m will indeed change the total acceleration of the system.

Total acceleration of the system is zero by conservation of momentum. So don't add them like that. It would be wrong to do so.


In reality, it would take not 3.3 seconds to hit the ground, but probably just a fraction of a second in the dense moon case. You're not going to compute that by just adding the accelerations, which treat both objects as point masses or rigid spheres, which they are not.

In reality, it would take not 3.3 seconds to hit the ground, but probably just a fraction of a second in the dense moon case. You're not going to compute that by just adding the accelerations, which treat both objects as point masses or rigid spheres, which they are not. — noAxioms

I understand the case for the dense moon would be extreme....would you have a problem with the equation I have written to Pantagruel?

M / 4pi * r ^2 = k g

Here, if r is small then g can become extremely large.

However, I would also claim that this is only half of the equation.

Any other mass g acceleration would need to be added at every point in space...

...just a thought.


Grampa Dee

I understand the case for the dense moon would be extreme....would you have a problem with the equation I have written to Pantagruel?

M / 4pi * r ^2 = k g — Gampa Dee

Context is needed for that. This seems to come from here:

The way I see it would be that Mass (the earth) gravitational energy will be causing an acceleration g at a certain distance ....the whole sphere at that distance will have the same g...
M / 4pi * r ^2 = k g — Gampa Dee

First of all, gravity isn't energy. I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 π r² gives you, well, I don't know what. It seems vaguely related to area of a circle. You equate this to 'k g', but no idea what 'k' is (kilo?). 'g' is the constant 9.8 m/sec², so you're seemingly equating this function of mass and radius to the constant 9800 m/sec²

Earth does have a sort of potential energy (negative) which represents the energy required to separate all the mass to infinity. I don't think you're talking about that.

Here, if r is small then g can become extremely large.

You seem to be equating g with A. 'g' is a constant magnitude of acceleration (a scalar), so it cannot be smaller or larger. A is a variable, and a vector, not a scalar. A = GM/r², so yes, 'A' becomes quite large if r is small enough. Saying 'g' can be quite large is like saying a meter can be larger if my table is wide enough.

Any other mass g acceleration would need to be added at every point in space...

'mass g' doesn't parse. I don't know what you mean by this. Are you now adding random objects here and there? Then you need to separately compute the acceleration of each and add those accelerations. Newton showed (via shell theorem) that any spherical distribution of mass of radius r can be treated as a point mass by objects outside of r.


Sorry, but I kind of came in late to this discussion and I don't know what you're trying to do beyond what is described by these very simple equations. For small objects, they all fall at the same rate, and the only reason the feather falls slower is due to air friction. They brought a feather to the moon to show it falling at the same rate as a rock. PR stunt..,. tax dollars at work.

I'm not sure where your formula comes from, but let me see if I can understand and answer your question.

Take a rock of mass m, on earth of mass M, a distance r from the center of mass of the earth, and let us ignore air resistance.

The force on the rock:
F=G*Mm/r^2 (1)
Also
F=ma (2)

substituting 2 into 1 gives

a=G*M/r^2

So the acceleration of the rock in earth's gravitational field is not dependent on the mass of the rock. It is dependent on the distance of the rock from the center of mass of the earth, and on the mass of the earth.

You seem to be asking, what if M were different? If M were different, then the acceleration due to gravity will also be different. This can obviously be seen in that the same rock will have a lower acceleration dropped on the moon than dropped on earth. Galileo was talking about objects of different masses being dropped in the same gravitational field and at the same distance.

-1kg and 100kg dropped from the same height on earth in a vacuum, will have the same acceleration. -1kg and 100kg dropped from the same height on the moon in a vacuum, will have the same acceleration.
-1kg dropped from the same height on earth and on the moon will NOT have the same acceleration.

Is this what you are getting at?

Thank you for responding noAxiom

First of all, gravity isn't energy. I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 π r² gives you, well, I don't know what. It seems vaguely related to area of a circle. You equate this to 'k g', but no idea what 'k' is (kilo?). 'g' is the constant 9.8 m/sec², so you're seemingly equating this function of mass and radius to the constant 9800 m/sec² — noAxioms

I view gravity as a potential for movement observed as the weight of a body, if the mass is stationary, located at a certain height above the ground, or the increase of k.e. when it accelerates towards the ground.

Here, if r is small then g can become extremely large.

You seem to be equating g with A. 'g' is a constant magnitude of acceleration (a scalar), so it cannot be smaller or larger. A is a variable, and a vector, not a scalar. A = GM/r², so yes, 'A' becomes quite large if r is small enough. Saying 'g' can be quite large is like saying a meter can be larger if my table is wide enough. — noAxioms

For me, g is the potential gravitational acceleration given to a “test mass” caused by some other mass (usually a large one), not necessarily the earth; it could be the moon, mars or Jupiter. I was not using the letter “a” because we cannot speak of a force as being the cause. However, if you want me to write down “A”, instead of “g”, then, no problem, noAxiom,, I will use the letter “A”.

So , you wrote: . “A = GM/r², so yes, 'A' becomes quite large if r is small enough.”

This is all that I was saying.

Now, concerning the example of the high density ball pulling the earth towards itself ; How would you write the equation, if A = GM/r² is the acceleration caused by the earth?

Any other mass g acceleration would need to be added at every point in space...

'mass g' doesn't parse. I don't know what you mean by this. Are you now adding random objects here and there? Then you need to separately compute the acceleration of each and add those accelerations. Newton showed (via shell theorem) that any spherical distribution of mass of radius r can be treated as a point mass by objects outside of r. — noAxioms

...I am sorry for having used the letter “g”, noAxiom...just replace it by “A” instead, from now on. Now you mentioned that when adding masses we also need to “add” accelerations... this is all that I am saying in this post.
The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.

Sorry, but I kind of came in late to this discussion and I don't know what you're trying to do beyond what is described by these very simple equations. For small objects, they all fall at the same rate, and the only reason the feather falls slower is due to air friction. They brought a feather to the moon to show it falling at the same rate as a rock. PR stunt..,. tax dollars at work. — noAxioms

I am privileged to have any response at all, noAxiom....I have merely a layman viewpoint on this issue, and am simply trying to explain the problems that I have...
Thank you for your time
Grampa Dee.

Thank you for replying,PhilosophyRunner

So the acceleration of the rock in earth's gravitational field is not dependent on the mass of the rock. It is dependent on the distance of the rock from the center of mass of the earth, and on the mass of the earth.

You seem to be asking, what if M were different? — PhilosophyRunner

No, I am asking how is it that we do not include the acceleration of the earth caused by the second mass?

Grampa Dee

I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 π r² gives you, well, I don't know what. It seems vaguely related to area of a circle. — noAxioms

4 π r², being the surface area of a sphere, would be dependent to the surface Acceleration

M / 4 π r² = A * constant....It was meant to what you have written concerning a high density mass.
While, this is not directly the density of mass, it seems to identify a mass having a certain spherical area as having a certain gravitational surface acceleration... if the radius is small the acceleration will be greater...

Of course, the equation could be erroneous, but I can't see why....that is why I asked you what you thought about the equation

Grampa Dee.

First of all, for Newton's equations as you and I have written, they only apply when used from an inertial frame of reference. The derived acceleration equation is from the frame of reference of the larger object and assumes this object (the Earth or Moon, etc) is not accelerating, which simplifies the math.

If you want to model a system where both bodies are accelerating you will have to use the two-body equations of motion, and set up a inertial frame of reference as the origin (eg: the barycenter). In the simplified calculations, the larger body can be taken as the inertial frame of reference as it is not accelerating, this is not true if it is accelerating.

Secondly why would we include the acceleration of the earth caused by the second mass? That is not what we were calculating - we were calculating the acceleration of the rock. If you wanted to calculate the acceleration of the Earth caused by the rock, that is a separate calculation

Thank you for replaying, PhilosophyRunner

Secondly why would we include the acceleration of the earth caused by the second mass? That is not what we were calculating - we were calculating the acceleration of the rock. If you wanted to calculate the acceleration of the Earth caused by the rock, that is a separate calculation
an hour ago — PhilosophyRunner

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference?

If you want to model a system where both bodies are accelerating you will have to use the two-body equations of motion, — PhilosophyRunner

ok; so, what would be this equation?

Grampa Dee

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference? — Gampa Dee

No, because you have to measure the acceleration from an inertial frame of reference, which is neither the earth nor the rock in our example if we are including the earth's acceleration.

"Acceleration of the rock towards the earth" implies measuring the acceleration of the rock from earth's frame of reference. If the earth is accelerating towards the rock, this is not a valid frame of reference.

The same goes for " the acceleration of the earth towards the rock" as the rock is not a valid frame of reference.

Instead, what we should be doing is measuring the acceleration of the earth in 3d coordinates from a point in space that is not accelerating (any non accelerating point), and measuring the acceleration of the rock from this same point. And from the point, the rock will be seen to accelerate quickly towards the earth, while the earth will barely accelerate towards the rock.

This way of analysis is not just correct for gravity, but for anytime you are using classical mechanics to analysis motion. Think of a stationary passerby watching a car hitting a tennis ball. The ball will have a high acceleration as a result of the collision, while the car will hardly accelerate at all. From an inertial frame of reference, the car and tennis ball do NOT both have the same acceleration from the collision.

ok; so, what would be this equation? — Gampa Dee

a1 = Gm2r/r3

Where a1 is the acceleration of body 1 from an inertial frame of reference. Note this is a vector in 3d.

For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html

I view gravity as a potential for movement observed as the weight of a body, if the mass is stationary, located at a certain height above the ground — Gampa Dee

Very little of Earth is 'at a height above the ground', so by this definition, Earth has negligible gravitational energy. What you are describing is the positive potential energy of a small amount of mass relative to nearby places of lower gravitational potential. It has no requirement that the material be stationary relative to any particular thing.

For me, g is the potential gravitational acceleration given to a “test mass” caused by some other mass (usually a large one), not necessarily the earth; it could be the moon, mars or Jupiter. I was not using the letter “a” because we cannot speak of a force as being the cause.

The symbol for that is 'a', not 'g'. 'a' is a vector variable acceleration. g is a scalar constant acceleration. Neither are a force. Force is measured in Newtons and uses the symbol F. Try to use standard symbols when discussing such things, as personal preferences only lead to confusion.

However, if you want me to write down “A”, instead of “g”, then, no problem, noAxiom,, I will use the letter “A”.

We've been doing that, but it's actually lowercase. 'A' is used for Area (mathematics) and electrical current (physics). So I'm committing the same offense; :sad:

Now, concerning the example of the high density ball pulling the earth towards itself ; How would you write the equation, if A = GM/r² is the acceleration caused by the earth?

Well, I would make A lowercase to fix that problem, and the rest is correct. The acceleration of the moon when it is at radius r is exactly that in Newtonian physics. That does not mean it will hit the ground in 3 1/3 seconds like the 10 kg ball. As I said, a small fraction of a second is more likely.

Any other mass g acceleration would need to be added at every point in space

I'm assuming that the dense super-mass is rigid, so yes, the acceleration applies to every point in the moon-ball. I am not assuming Earth is sufficiently rigid to not deform under the ungodly tidal stress the ball would apply to it. The sidewalk slab 60 meters below will be yanked up without waiting for Earth to catch up with it.

Now you mentioned that when adding masses we also need to “add” accelerations... this is all that I am saying in this post.

Acceleration is absolute, not relative, so adding them seems to result in a fairly meaningless value. I suppose you can use it to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration. For instance, if an apple detaches from a tree and the distance between me and it decreases at 9.8 m/sec², that in no way suggests that I am accelerating at that rate, and in fact I'm accelerating (coordinate acceleration, not proper acceleration) away from it a little bit.
If we switch to proper acceleration, both the apple and I have a proper acceleration of about 1g until the apple detaches, at which point I have 1g proper acceleration upward and the apple has none. So it is me that hits the apple, not the other way around. But that isn't really the Newtonian way of looking at it.

The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.

But there is an equation for each object, each dependent on only that mass, and not on the mass of the thing accelerating towards it.

4 π r², being the surface area of a sphere, would be dependent to the surface Acceleration — Gampa Dee

OK, but the surface area seems to be a needless complication. We know the acceleration as a=GM/r². Where the surface is is irrelevant so long as it is below r.

While, this is not directly the density of mass, it seems to identify a mass having a certain spherical area as having a certain gravitational surface acceleration..

That would not work. A grapefruit has a similar area, but far less gravitational acceleration at its surface. So acceleration is not a function of just area.

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference? — Gampa Dee

As P-R points out, the coordinate acceleration described by Newton's equations is relative to any inertial coordinate system, and the equations don't work when used with a non-inertial coordinate system such as an accelerating or rotating one (Earth is both).
Mind you, plenty of books treat things like the Earth-moon system in isolation, which is to reference an accelerating coordinate system. It works to an extent, but adds confusion. For instance, few are aware that the moon always accelerates towards the sun since the sun exerts more force on it that does Earth. That means that when the moon is between the two (solar eclipse), the moon is accelerating directly away from Earth.

From the accelerating frame of Earth, the moon seems to maintain a fairly constant distance that isn't much a function of which direction the sun is. Yes the orbit of the moon (like any orbit) is eccentric to a degree, but that again isn't due to the sun.

I suppose you can use it to compute the rate of change in distance between the two objects, but that rate isn't acceleration. — noAxioms

Great posts all around. :up:

Thank you PhilosophyRunner

For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html — PhilosophyRunner

Yes, I will....
Thank you again.

Grampa Dee

Thank you for replying, noAxioms

The symbol for that is 'a', not 'g'. 'a' is a vector variable acceleration. g is a scalar constant acceleration. Neither are a force. Force is measured in Newtons and uses the symbol F. Try to use standard symbols when discussing such things, as personal preferences only lead to confusion. — noAxioms

I will try to do that noAxiom...I’m not well versed in the scientific terminology .

:(
— noAxioms

The acceleration of the moon when it is at radius r is exactly that in Newtonian physics. That does not mean it will hit the ground in 3 1/3 seconds like the 10 kg ball. As I said, a small fraction of a second is more likely. — noAxioms

I agree with you, but don’t understand how Newton’s equation can show this particular case.

I'm assuming that the dense super-mass is rigid, so yes, the acceleration applies to every point in the moon-ball. I am not assuming Earth is sufficiently rigid to not deform under the ungodly tidal stress the ball would apply to it. The sidewalk slab 60 meters below will be yanked up without waiting for Earth to catch up with it. — noAxioms

Sure....but we’re only using thought experiments ...gedank...or whatever

:)
— noAxioms

Acceleration is absolute, not relative, so adding them seems to result in a fairly meaningless value — noAxioms

I agree that there is only one “observed” acceleration.

I suppose you can use it to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration. For instance, if an apple detaches from a tree and the distance between me and it decreases at 9.8 m/sec², that in no way suggests that I am accelerating at that rate, and in fact I'm accelerating (coordinate acceleration, not proper acceleration) away from it a little bit. — noAxioms

I do agree that the 1 dimensional is much simpler. As for the acceleration of the apple at that rate relative to you, if we identify acceleration as simply a change in the rate of velocity, then, I would say that it’s only a relative situation to claim yourself as the one who is accelerating... and you feel the force to say so :)

If we switch to proper acceleration, both the apple and I have a proper acceleration of about 1g until the apple detaches, at which point I have 1g proper acceleration upward and the apple has none. So it is me that hits the apple, not the other way around. But that isn't really the Newtonian way of looking at it.

I agree, it’s Einstein who recognized that a freefall object didn’t experience any force.

The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.
But there is an equation for each object, each dependent on only that mass, and not on the mass of the thing accelerating towards it. — noAxioms

Ok; I’m not sure what you’re saying but this might be where I’m not thinking it through.

ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference?
— Gampa Dee
As P-R points out, the coordinate acceleration described by Newton's equations is relative to any inertial coordinate system, and the equations don't work when used with a non-inertial coordinate system such as an accelerating or rotating one (Earth is both).
Mind you, plenty of books treat things like the Earth-moon system in isolation, which is to reference an accelerating coordinate system. It works to an extent, but adds confusion. For instance, few are aware that the moon always accelerates towards the sun since the sun exerts more force on it that does Earth. That means that when the moon is between the two (solar eclipse), the moon is accelerating directly away from Earth.

From the accelerating frame of Earth, the moon seems to maintain a fairly constant distance that isn't much a function of which direction the sun is. Yes the orbit of the moon (like any orbit) is eccentric to a degree, but that again isn't due to the sun. — noAxioms

I’ll ponder on this, noAxiom...thanks again for your time
Grampa Dee

If we identify acceleration as simply a change in the rate of velocity, — Gampa Dee

Yes, that's the physics definition. Never confuse it with the common language definition which is the 'rate of increase in speed'.

I suppose you can use [the two added equations] to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration.
— noAxioms

I do agree that the 1 dimensional is much simpler.

It's not that it's simpler. Adding the equations only produces a useful result if there is no motion except along one axis. So for instance, under Newtonian mechanics, the ISS is continuously accelerating (coordinate acceleration) towards Earth at about 8.7 m/s² and yet its distance from Earth is roughly fixed, and its speed relative to Earth is also roughly fixed. This is because it is not a 1d case. The ISS has motion in a direction other than just the axis between it and Earth.
The ISS example also serves as a wonderful example of the difference between the physics definition of acceleration (rate of change in velocity=8.7 m/s² down) vs street definition (zero change in speed).

As for the acceleration of the apple at that rate relative to you, if we identify acceleration as simply a change in the rate of velocity, then, I would say that it’s only a relative situation to claim yourself as the one who is accelerating.

No. Under Newtonian mechanics, relative to any inertial coordinate system, it is the apple and only the apple that is accelerating.

Only under Einstein does it become you that experiences acceleration, as measured by an accelerometer that you carry with you. The accelerometer on the apple would read zero. But this is now proper acceleration, not coordinate acceleration. Coordinate acceleration becomes dependent on the arbitrary coordinate system of choice and it even varies from one inertial coordinate system to the next. The equations become quite different.

and you feel the force to say so

That's what the accelerometer does. Multiply what it says by your mass and you get your weight, which is why one is weightless on the ISS.

Thank you PhilosophyRunner

For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html
— PhilosophyRunner — Gampa Dee

I went through the example, and I will try to explain how I have understood to the best of my ability.what was written.....I'm not claiming to be correct, only that this is how I have interpreted the text.
Also beware of my use of symbols...they're awful as I don't have the use of the correct symbols.
For example R(double dot) simply means acceleration, however, use the text beside you and you will be able to follow......Here goes nothing :)

Two-Body Equations of Motion in an Inertial Frame
The position vectors R1 and R2 of two point masses in this reference frame are:
R1 = X1Ihat + Y1Jhat + Z1Khat
R2 = X2Ihat + Y2Jhat + Z2khat

Here, we are given a 3 dimensional reference frame X,Y,Z where two bodies are positioned at (x1,y1 z1) and (x2,y2,z2)...I don’t understand how they could be identified as vectors though, since position is not a vector.

Let r be the vector pointing from m1 to m2, which we also phrase as m2 relative to m1. Then:
r = r2 – r1
r = (x2 – x1)Ihat + (y2 – y1) Jhat + (Z2 – Z1)Khat

Since the mass is said to be pointing towards the other mass, then ok, r is then a vector.

We also define a unit vector pointing from m1 toward m2:
Uhat*r = r / r

This, being a unit vector, will not do anything to the equation as such except maybe to identify r^2(denominator), which is not a vector,....as being identified as a vector....in my opinion.

where r (denominator) is the magnitude of r ( nominator), or the distance between the two masses.

Forces in the Two-Body System#
The two masses are acted upon only by their mutual gravitational pull. F12 is the force exerted on m1 by m2 and F21 is the force exerted on m2 by m1. By Newton’s third law:
F12 = -F21

We have two forces coming from the two bodies...1 being + and the other -.


Newton’s second law says that the force is equal to the mass times the acceleration:
(17)#
F12 = m1 R(double dot)1
F21 = m2 R(double dot)2
where R¨ is the absolute acceleration of the subscripted mass. Absolute means that the acceleration is taken relative to an inertial reference frame. This is important because Newton’s second law only applies for absolute accelerations.

This is simply, in my opinion,

F1 = m1*a1
F2 = m2*a2
F1 =- F2

Since the only force in this system is the gravitational attraction, the force is also equal to Newton’s law of gravitation, (1). The force of m2 on m1, F12, points in the positive direction of u^r. Because of Newton’s third law, as represented by (16), the force of m1 on m2, F21, points in the negative direction of u^r. This is shown in (18):

F12 = Gm1m2 / r^2 (uhatr)
F21 = - Gm1m2 / r^2 (uhatr)

Here we have two of the Newtonian gravitational equations... one of them is negative identifying the reverse direction of the second force.




Finding the Equations of Motion *********

Here’s where our focus of discussion lies.



Combining (17) and (18), we find:
m1R1(double dot) = Gm1m2 / r^2 (uhatr)
m2R2(double dot) = - Gm1m2 / r^2 (uhatr)

This would mean, in my opinion, m1*a1 = Gm1m2 / r^2 ... u(hat)r is simply a unit vector...I think we can ignore it.The same goes for m2R2(double dot)

Finally, we divide through by the mass on the left side of each equation and replace u^r with its definition, (15) to arrive at the two-body inertial equations of motion:
R1(double dot) = Gm2 r / r^3
R2(double dot) = -Gm1r /r^3

Notice here that there are two accelerations, which are not going to have the same magnitude.
a1 = Gm2 / r^2 ( one of the r in r^3 cancels out with the r in the numerator)
a2 = - Gm1 / r^2

So, now we have 2 accelerations; one is negative, the other positive. What do we do at this point?
If we have two cars having velocities of v1 and –v2(since it is going towards v1), relative to the road, what will the observers within the car measure the relative velocity between themselves?
It’s not going to be v1 – v2, but v1 + v2....I see the same problem for the two accelerations.

I hope this post wasn't too long.

Grampa Dee

So, now we have 2 accelerations; one is negative, the other positive. What do we do at this point?
If we have two cars having velocities of v1 and –v2(since it is going towards v1), relative to the road, what will the observers within the car measure the relative velocity between themselves?
It’s not going to be v1 – v2, but v1 + v2....I see the same problem for the two accelerations. — Gampa Dee

There you have done an excellent job, and now have acceleration from an inertial frame of reference. Great. You wanted the acceleration, now you have them.

Then you try to move that to the frame of reference of the earth and of the rock, at which point Newton's laws are no longer valid. This step is wrong. You can change the coordinate system, that is fine, but when you change it so that the origin is an accelerating body, Newton's laws no longer work correctly.

To use your car example, I am standing on the pavement not accelerating. I see one car (A) moving accelerating at 3m/s2 towards my left and another car (B) accelerating at 5m/s2 towards my right. That is simply the acceleration of the cars. It is meaningless (from a classical mechanics sense) to then add these up and say car A is accelerating at 8m/s2 from the point of view of the car B. Neither car is accelerating at 8m/s2. Car A is accelerating at 3m/s2 and car B at 5m/s2.