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## Same gravitational acceleration of unequal masses

 Quote by D_H 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.

 Quote by D_L 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.
 Quote by D_L 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?

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 Quote by TurtleMeister 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

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 Quote by atyy 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".

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 Quote by TurtleMeister 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 cancelled out.

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 Quote by atyy BTW, the wikipedia reduced mass article has the two minus signs you missed in the post which DH said the errors cancelled out.
Yes, I noticed that. a - (-a) = a + a.

 Quote by Buckleymanor 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!

 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.

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Quote by Buckleymanor
 Quote by Canticle 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.
 Blog Entries: 9 Recognitions: Science Advisor 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".

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 Quote by 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".
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.

 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.

 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.
 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.
 Recognitions: Gold Member 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".
 Mentor 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.
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 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.

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