Same gravitational acceleration of unequal masses

[sylas]

Ok. so expansion of the universe is not the same thing as the acceleration of a body in a gravity field.
Should we expect this expansion to have no effect upon our local gravitational field in the future.

Correct. Expansion has no effect upon the local gravitational field.

Expansion alters mean density on large non-local scales, but it does not tear apart gravitationally bound objects. It is not a force. The Earth does not expand. The solar system remains the same size. The distance between sufficiently remote galaxies increases, but the size of an individual galaxy remains defined by its own mass and local evolution.

Cheers -- sylas

 

[sganesh88]

sganesh88, I just wanted to add that your post has added greatly to my understanding of gravity. It's the most significant change of mind I've had about gravity in years. Yeah, I tried very hard to hold on to my misconception, but the simple logic of G(M1-M2)/r² prevailed. But at least I can take comfort in knowing that I'm not the only one who has made this mistake. Do you realize how widespread the misconception is? Anyway, thanks for opening my eyes to this.

Yes deriving that expression is just as simple as computing relative velocity between two trains running in opposite directions on parallel rails. Anyway delighted to be of help.

 

[TurtleMeister]

Yes deriving that expression is just as simple as computing relative velocity between two trains running in opposite directions on parallel rails. Anyway delighted to be of help.

Yes, I think it's pretty simple when looking at it mathematically. But when thinking visually, as with thought experiments, it's not as simple as the train example. It's easy to assume that the great mass of the Earth is the reason that it's difficult to detect the difference in acceleration of two falling objects when you know that the acceleration depends on the total mass of all bodies, and that the ratio of the masses of the falling bodies and the mass of the Earth is astronomically large. There are others who have the same misconception that I did. One example is the author of this article:
http://www.misunderstooduniverse.com/Where_Aristotle_Galileo_and_NASA_went_wrong.htm

 

[Buckleymanor]

Correct. Expansion has no effect upon the local gravitational field.

Expansion alters mean density on large non-local scales, but it does not tear apart gravitationally bound objects. It is not a force. The Earth does not expand. The solar system remains the same size. The distance between sufficiently remote galaxies increases, but the size of an individual galaxy remains defined by its own mass and local evolution.

Cheers -- sylas

You have probably seen this http://http://www.newuniverse.co.uk/Big_Rip.html"
I was wondering if this hypothesis has been refuted.

 

[sganesh88]

But when thinking visually, as with thought experiments, it's not as simple as the train example. It's easy to assume that the great mass of the Earth is the reason that it's difficult to detect the difference in acceleration of two falling objects when you know that the acceleration depends on the total mass of all bodies, and that the ratio of the masses of the falling bodies and the mass of the Earth is astronomically large. There are others who have the same misconception that I did. One example is the author of this article:
http://www.misunderstooduniverse.com/Where_Aristotle_Galileo_and_NASA_went_wrong.htm

Ya. You can't believe in intuition all the time, though it does give some great beautiful insights occasionally. Regarding that site, i think the author is primarily concerned with finding fault with others' arguments. Understanding takes a back seat then.

P.S: I still don't understand how he declares both Galileo and Aristotle wrong at the same time.

 

[TurtleMeister]

I still don't understand how he declares both Galileo and Aristotle wrong at the same time.

He thinks they both made a mistake by not considering the mass of the Earth when they were contemplating the difference (or lack thereof) in the free-fall rates of objects of different mass. Of course it was not a mistake, and he is the one making the mistake by thinking this.

Where they both made their mistake was in comparing the insignificant mass of the objects to each other and excluding the primary mass of the Earth, which is 5.985 trillion trillion kilograms.
http://www.misunderstooduniverse.com/Where_Aristotle_Galileo_and_NASA_went_wrong.htm

 

[Chewy0087]

Kinda ridiculous that he whoever made the article doesn't know what they're on about, unless I'm sorely mistaken.

I still don't understand the failure to see the independence of acceleration due to gravity when describing our motion;

\frac{GmM}{r^2} = ma

\frac{GM}{r^2} = a

Of course we attract the Earth a bit aswell, but there's no way you can quantify that pull without taking into account all of the other people & things on the earth, never mind mountains & cars and stuff.

Also, in my mind the 'acceleration due to gravity' has always meant 'due to the earth', which is constant. Galileo was a genius, and his deduction is imo the best ever in science :D, seeing someone mereley dismiss it is =((((

 

[TurtleMeister]

I still don't understand the failure to see the independence of acceleration due to gravity when describing our motion;

\frac{GmM}{r^2} = ma

\frac{GM}{r^2} = a

I don't think anyone is failing to see that. I know I'm not. Or were you referring to the author of the article?

Of course we attract the Earth a bit aswell, but there's no way you can quantify that pull without taking into account all of the other people & things on the earth, never mind mountains & cars and stuff.

That's right, the amount that a non-astronomical sized object attracts the Earth is trivial and unmeasurable. That's the reason it's left out of the equation a = (G * M) / r². For all practical purposes it would be pointless to include it, because the amount of uncertainty of the mass of the Earth would far exceed the mass of the object. But that doesn't mean the attraction is not there. If you wanted to include it you could say a = (G * (M1 + M2)) / r².

Also, in my mind the 'acceleration due to gravity' has always meant 'due to the earth', which is constant.

Again, I agree as long as the free-falling object is non-astronomical. If the free-falling object were massive enough to make a detectable difference then it should be included in the equation.

Galileo was a genius, and his deduction is imo the best ever in science :D, seeing someone mereley dismiss it is =((((

Yes, I agree his contributions were enormous. A lot of people think he proved Aristotle wrong by performing drop tower experiments (leaning tower of Pizza) and rolling objects down an inclined plane. But actually he used logic and reason to prove him wrong.

 

[sylas]

That's right, the amount that a non-astronomical sized object attracts the Earth is trivial and unmeasurable. That's the reason it's left out of the equation a = (G * M) / r². For all practical purposes it would be pointless to include it, because the amount of uncertainty of the mass of the Earth would far exceed the mass of the object. But that doesn't mean the attraction is not there. If you wanted to include it you could say a = (G * (M1 + M2)) / r².

It depends what you are measuring. If you are measuring an acceleration of an object, then no, you would not say that.

If you are measuring the rate of change in the separation distance (second derivative), then yes, you would say that.

This difference has been explained in the thread already. It's not hard. Whether you leave M1 in, or out, depends on what you are measuring. If you are measuring the acceleration of an object with mass M1 due to gravity, then you leave out M1, always. No matter how large it is.

Cheers -- sylas

 

[D H]

You are still having difficulties with reference frames, Turtle.

 

[TurtleMeister]

You are still having difficulties with reference frames, Turtle.

No, I did not state my frame of reference. But since Chewy used A = GM/r² I just assumed the reader would know that my reference frame was the Earth.

Here's what I'm thinking:

F = (G * M1 * M2) / r²
where F = force between M1 and M2.
Reference frame doesn't matter since F applies to M1 and M2 equally.

A1 = (G * M2) / r²
where A1 = acceleration of M1 toward the center of mass of M1 and M2.
Reference frame is center of mass of M1 and M2.

A2 = (G * M1) / r²
where A2 = acceleration of M2 toward the center of mass of M1 and M2.
Reference frame is center of mass of M1 and M2.

A = A1 + A2 = (G * (M1 + M2)) / r²
where A = acceleration of M1 and M2 toward each other.
Reference frame is either M1 or M2.

Where am I wrong?

 

[D H]

Here's what I'm thinking:

F = (G * M1 * M2) / r²
where F = force between M1 and M2.
Reference frame doesn't matter since F applies to M1 and M2 equally.

A1 = (G * M2) / r²
where A1 = acceleration of M1 toward the center of mass of M1 and M2.
Reference frame is center of mass of M1 and M2.

A2 = (G * M1) / r²
where A2 = acceleration of M2 toward the center of mass of M1 and M2.
Reference frame is center of mass of M1 and M2.

A = A1 + A2 = (G * (M1 + M2)) / r²
where A = acceleration of M1 and M2 toward each other.
Reference frame is either M1 or M2.

Where am I wrong?

The first and last. (You got the last one right, but did so by compounding errors).

1. First off, it is the magnitude of the gravitational force is F=GM_1M_2/r^2. That's only half the picture. Forces are vectors; they have magnitude and direction.

Secondly, and more importantly. the frame most certainly does matter because in an non-inertial frame you need to add fictitious forces into the picture to make F=ma appear to be valid. If you do not take those fictitious forces into account, F is not equal to mass times acceleration in a non-inertial frame.

4. You added accelerations. You should have subtracted them, vectorially. The end result is correct. The means of arriving at it is anything but correct.You have been harping about the equivalence principle not being true for several pages now. The whole source of your misunderstanding results from either intentionally ignoring or not understanding the difference between inertial and non-inertial frames.

 

[TurtleMeister]

1. The magnitude of the gravitational force is LaTeX Code: F=\\frac {GM_1M_2}{r^2} . The frame most certainly does matter because in an non-inertial frame you need to add fictitious forces into the picture to make F=ma appear to be valid. If you do not take those fictitious forces into account, F is not equal to mass times acceleration in a non-inertial frame.

What is the, or a, correct reference frame?

4. You added accelerations. You should have subtracted them, vectorially. The end result is correct. The means of arriving at it is anything but correct.

I don't understand. Seems like subtracting will give me the difference of acceleration. Can you give me an example.

You have been harping about the equivalence principle not being true for several pages now. The whole source of your misunderstanding results from either intentionally ignoring or not understanding the difference between inertial and non-inertial frames.

Which equivalence principle? Do you mean the equivalence of gravitational and inertial mass?

 

[sylas]

What is the, or a, correct reference frame?


I don't understand. Seems like subtracting will give me the difference of acceleration. Can you give me an example.

Acceleration is a vector. By considering the line of sight between the two objects, you can let two of the co-ordinates be zero.

In this case A1 and A2 are the co-ordinate along line of sight, and one is positive, the other negative. The correct way to get a relative value is subtraction, and the magnitude neatly turns out to be the magnitude of your addition, because actually A1 and A2 have opposite direction and hence opposite sign.

Cheers -- sylas

 

[TurtleMeister]

Acceleration is a vector. By considering the line of sight between the two objects, you can let two of the co-ordinates be zero.

In this case A1 and A2 are the co-ordinate along line of sight, and one is positive, the other negative. The correct way to get a relative value is subtraction, and the magnitude neatly turns out to be the magnitude of your addition, because actually A1 and A2 have opposite direction and hence opposite sign.

Ok, I think I understand now. I was not thinking in terms of a co-ordinate system. Only the absolute values of the acceleration of the bodies toward each other.

 

[D H]

What is the, or a, correct reference frame?

All reference frames are equally valid. You just need to make sure you are doing a proper accounting in developing your equations of motion. You are not doing that.

I don't understand. Seems like subtracting will give me the difference of acceleration. Can you give me an example.

Start with position. Bob is 2 meters to the north west of you. Jim is 2 meters to the north east of you. Where is Jim with respect to Bob? One way to solve this is to formulate the problem in terms of vectors. Let's call north the xhat direction and east, yhat. Bob's location with respect to you is \surd 2 \hat x - \surd 2 \hat y. Jim's location: \surd 2 \hat x + \surd 2 \hat y. There relative position is found by subtracting these two vectors: (\surd 2 \hat x + \surd 2 \hat y) - (\surd 2 \hat x - \surd 2 \hat y) = 2\surd 2 \hat y. Jim is about 2.8 meters to the east of Bob.

Now suppose Bob is 2 meters to the west of you and Jim is 2 meters to the right of you. You still subtract vectorially, in this case, 2 \hat y - (-2 \hat y) = 4 \hat y. In this example Jim is 4 meters to the east of Bob.

The same thing is going on here with accelerations. Object 1 is accelerating toward object 2, and object 2 is accelerating toward object 1. These accelerations point in the opposite directions. To arrive at the relative acceleration you need to subtract, not add, but you need to do the subtraction vectorially.

Which equivalence principle? Do you mean the equivalence of gravitational and inertial mass?

If all bodies fall with the same acceleration in a gravity field then gravitational and inertial mass are the same. Conversely, if gravitational and inertial mass are the same then any two test bodies will fall with the same acceleration. Saying one is the same as saying the other. Disputing one is the same as disputing the other. You said early on in this thread that you were not trying to dispute the equivalence principle. Yet you have been disputing for several pages.

Before delving into more advanced topics such as the equivalence principle, I suggest you first study on more basic things. Specifically, vectors and reference frames.

 

[Canticle]

Sorry to come in late, but I hope I can offer a better 'plain english' understanding;

Consider how much energy it takes to push a car from rest. Then how much energy to push a go-cart.

The difference is the car has far more inertia. Yes ..it does 'weigh' much more, which is actually what gives it that additional inertia.

The really interesting thing is that gravity just happens to exactly equal that inertia - that's equivalence - which is where Einstein started. Gravity = Acceleration.

Everything's actually rushing through space, i.e. inertia = momentum. Imagine the car and go-cart side by side rushing past a planet, maybe earth. The planet's gravity starts affecting them. The car has more momentum as it has more mass so it's more difficult to deflect. But the go-cart is lighter so it's less affected by the gravity. The differences balance each other out.

Hope that sorts it?

If you want to go further; This gives us a good hint about what gravity is, that Dirac understood best, unfortunately better than many physicists today! Simply again; it may be seen as the 'slope' of the sides of the depression the large sphere in the relativity demonstration makes in the gridded sheet it sits on.
But best you stop thinking about it there or you'll be put under house arrest!

 

[Buckleymanor]

Everything's actually rushing through space, i.e. inertia = momentum. Imagine the car and go-cart side by side rushing past a planet, maybe earth. The planet's gravity starts affecting them. The car has more momentum as it has more mass so it's more difficult to deflect. But the go-cart is lighter so it's less affected by the gravity. The differences balance each other out.

So does the same apply to all objects in a gravitational field.
Light passing a massive object like our Sun gets deflected or bent but a large object side by side with the light beam would probably fall into the Sun.
Same would happen to a car or go-kart rushing past a planet though the light would probably not be so nowticebly bent and would not fall towards it.
If the differences balanced out why does it not apply to very light or massless objects.

 

[TurtleMeister]

All reference frames are equally valid. You just need to make sure you are doing a proper accounting in developing your equations of motion. You are not doing that.

I may have just worded that poorly, because to me "reference frame doesn't matter" means the same thing as "all reference frames are equally valid". What do you mean by "doing proper accounting"?

Thanks for the detailed description of vectors and the co-ordinate system. I am familiar with using this type of system. I used an xy co-ordinate system when I wrote the gravity simulator program. I think I understand your argument but what I don't understand is why it necessary for me to use this system in #4 when I clearly state that the reference frame is M1 or M2? To an observer on M1 measuring the acceleration of M2 towards him (lets say with a radar gun), he will measure A1+A2 (the acceleration of M1 toward M2 plus the acceleration of M2 toward M1) The whole point is to show that an observer on M1 or M2 will measure both accelerations and not just one.

edit1:
Ok, I think I understand better now as to what your argument is about and what you mean by proper accounting. Since #2 and #3 are in opposite directions A1 and -A2, I should have written it that way in #4 (A1 - A2). If #4 were by itself it would be ok, but since it's referencing #2 and #3 it should take the direction into account.

If all bodies fall with the same acceleration in a gravity field then gravitational and inertial mass are the same. Conversely, if gravitational and inertial mass are the same then any two test bodies will fall with the same acceleration. Saying one is the same as saying the other. Disputing one is the same as disputing the other. You said early on in this thread that you were not trying to dispute the equivalence principle. Yet you have been disputing for several pages.

I do not remember disputing the fact that all bodies fall at the same acceleration in a gravitational field regardless of their mass. If I did then it was a mistake or a misunderstanding and it was not what I meant to say. Let's say two objects M1 and M2 are moving toward each other under their mutual gravitation. M1 will accelerate toward M2 at the same rate regardless of M1s mass. M2 will accelerate toward M1 at the same rate regardless of M2s mass. Increasing an objects mass also increases it's inertia, so it's acceleration will not change. However, increasing an objects mass will also increase the gravitational field that it produces, so any other objects will increase their acceleration toward it. I have a firm understanding of this. I am NOT disputing the equivalence principle.

Before delving into more advanced topics such as the equivalence principle, I suggest you first study on more basic things. Specifically, vectors and reference frames.

I can agree that I have a lot to learn. But is my understanding of these things really so poor? If so, how was I able to write a 2d gravity simulation program (nbodies) in less than 2 weeks?

 

[atyy]

Again, I agree as long as the free-falling object is non-astronomical. If the free-falling object were massive enough to make a detectable difference then it should be included in the equation.

Yes, if we use a reference frame attached to the surface of the earth: http://en.wikipedia.org/wiki/Reduced_mass

 

[TurtleMeister]

Yes, if we use a reference frame attached to the surface of the earth: http://en.wikipedia.org/wiki/Reduced_mass

Yes, that's what I meant. Sorry, I should have included "as viewed by an Earth bound observer".

 

[atyy]

Yes, that's what I meant. Sorry, I should have included "as viewed by an Earth bound observer".

BTW, the wikipedia reduced mass article has the two minus signs you missed in the post which DH said the errors canceled out.

 

[TurtleMeister]

BTW, the wikipedia reduced mass article has the two minus signs you missed in the post which DH said the errors canceled out.

Yes, I noticed that. a - (-a) = a + a.

 

[Canticle]

So does the same apply to all objects in a gravitational field.
Light passing a massive object like our Sun gets deflected or bent but a large object side by side with the light beam would probably fall into the Sun.
Same would happen to a car or go-kart rushing past a planet though the light would probably not be so nowticebly bent and would not fall towards it.
If the differences balanced out why does it not apply to very light or massless objects.

Yes, the same applies to everything. But you don't quite have it yet- Still in plain english;

If the 'large object' was going the same speed as the light beam it would certainly NOT fall into the sun! (momentum / intertia) It's track would bend the same.

Gravity is curved space time, so curves everything, massive or not, including all zero mass disturbances from gamma rays to radio waves.

The mass of a body defines the 'slope inclination' of space time. And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more. Too little to notice of course, but make it the size of the moon then you'd notice!

Can you get your head round that lot?

 

[Buckleymanor]

If the 'large object' was going the same speed as the light beam it would certainly NOT fall into the sun! (momentum / intertia) It's track would bend the same.

Would not that require rather a large amount of energy.

And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more.

So the metal ball would hit the Earth first.

 

[TurtleMeister]

And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more.

So the metal ball would hit the Earth first.

Yeah, that's pretty much been the theme of this thread. The metal ball would strike first but only by an extremely small undetectable amount. The difference in the accelerations would be a = G(M1-M2)/r² where a = difference in acceleration, G = gravitational constant, M1 = metal ball, M2 = wood ball, and r = the distance between M1, M2 and the center Earth. Notice that the mass of the Earth has noting to do with it, which was my earlier misconception.

 

[sylas]

You guys don't actually mean it would hit "first". That would imply being dropped at the same time, in which case they'd hit at the same time (ignoring friction etc)

What you really mean is that if the experiment was repeated on two occasions, then the metal ball would hit the ground in slightly less time.

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

 

[TurtleMeister]

You guys don't actually mean it would hit "first". That would imply being dropped at the same time, in which case they'd hit at the same time (ignoring friction etc)

What you really mean is that if the experiment was repeated on two occasions, then the metal ball would hit the ground in slightly less time.

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

I've been thinking about that myself. The only reason I haven't brought it up is that I wanted to come to a conclusion first. But since you've already brought it up I'll say that my intuition tells me that the metal ball would still strike first even if they were dropped at the same time. But hey, my intuition has been known to be wrong. :) Dropping both at the same time complicates things a bit.

 

[Buckleymanor]

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

Why not,
Both balls "perfectly" round Earth "perfectly" round.
Earth's round edge would touch metal ball first because the edge would be pulled to it via the centre of gravity.

 

[Buckleymanor]

Dropping both at the same time complicates things a bit.

Certainly does.
The metal ball would allso accelerate towards the wood ball and vice versa but the Earth would accelerate more towards the metal ball.

 

[worldrimroamr]

Why is everyone making such a complicate deal out of something that can be expressed so straightforwardly? Gravitational force is exactly proportional to mass. But inertial resistance to motion is also exactly proportional to mass. (That's Einstein's equivalence principle -- the founding insight of general relativity.) So for twice the mass, you have twice the force and twice the inertia. They exactly cancel each other out. Thus, all objects fall at equal rates in vacuum.

 

[TurtleMeister]

Yes, worldrimroamr, you are correct. But what we are talking about is the way it would be viewed by an Earth bound observer, who is in a non-inertial reference frame. I hope I get it right this time. :)

A1 = G * M2 / r²
acceleration of M1 toward M2

A2 = G * M1 / r²
acceleration of M2 toward M1

A = A1 - A2 = G * (M1 + M2) / r²
relative acceleration between M1 and M2

If you were located on M1 and you measured the acceleration of M2 toward you, then A is what you would measure, not A2. And if you were located on M2 and you measured the acceleration of M1 toward you, then A is also what you would measure, not A1. So the "relative" acceleration of an object in Earth free-fall is not independent of the objects mass "as measured by an Earth bound observer".

 

[D H]

An Earth bound observer is not in an inertial frame, no matter how you cut it.

• Newtonian mechanics. The Earth is accelerating toward the object. An accelerating frame is not inertial in Newtonian mechanics.
• General relativity. A stream of falling apples accelerates with respect to the observer. A frame with a stream of falling apples is not inertial in general relativity.

 

[TurtleMeister]

Thanks DH, I changed it.

 

[Canticle]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

No matter what the 'share' of mass between them the total acceleration is the same, (only the distance changes that - 2nd law).

So both balls would hit at the same time whatever. That's equivalence!

Except for Godels incompleteness theorem etc. There are loads of other tiny factors, there IS an ether so relative atmospheric drag comes into it, your spaceship taking off and landing, the relative pull of the sun whichever its closest to, etc. Thats why Mr Planck invented Plancks, and we have Quantas, supposedly if it's smaller than that it's undetectable (thanks Turtle) and we can ignore it. (though one day we'll find we need it!) Not using Quantas is fine, BA fly to Aus anyway.

 

[sylas]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

Good point. If you repeat the experiment with different balls, you are effectively using the other ball as part of the mass of the Earth!

 

[Buckleymanor]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

No matter what the 'share' of mass between them the total acceleration is the same,

The total acceleration is the same but how can you recognise if the big one or the small one falls or moves towards you at all.

 

[TurtleMeister]

Canticle, your thought experiment is correct but incomplete in so far as what we're talking about. In order to make it complete you should identify your reference frame as non-inertial and then repeat the experiment using an inertial reference frame.

Edit1
I noticed something a little different in your thought experiment. You said: "No matter what the 'share' of mass between them the total acceleration is the same". Thus far we have not talked about redistribution of mass between the two objects. In this case the relative acceleration A = G * (M1 + M2) / r² in the non-inertial frame will always be the same, as you stated. But the individual accelerations of the two objects in the inertial frame will change.

If you repeat the experiment with different balls, you are effectively using the other ball as part of the mass of the Earth!

That is correct. It is the combined mass that determines the relative acceleration of the two bodies toward each other. And that's the reason for (M1+M2) in the equation A = G(M1+M2)/r² for the non-inertial frame. So regardless of which object (M1 or M2) you are observing from, the mass of the other object WILL affect the observed acceleration. However, if you are an observer in an inertial reference frame you will see M1 and M2 accelerating at different rates. M1 will accelerate at A1 = G * M2 / r² and M2 will accelerate at A2 = G * M1 / r². It's really a very simple thing. It's just that we're describing the same thing from two different points of view. If you are located on M1 or M2 then you are accelerating, and so your reference frame is non-inertial.

In short, for the acceleration of an object(M2) who's mass is variable, in Earth(M1) free-fall:

A = G * M1 / r²
This is the inertial reference frame. The mass of the falling object(M2) does NOT affect the acceleration because of the equivalence principle.

A = G * (M1 + M2) / r²
This is the non-inertial reference frame. The mass of the falling object(M2) DOES affect the relative acceleration because we are located on Earth(M1) which would include our own acceleration toward the object. This is not to say that the equivalence principle is being violated because the equivalence principle is only valid in an inertial reference frame.