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## Same gravitational acceleration of unequal masses

I guess for now we'll just have to disagree. But - I've changed my mind before. :)

The reason I believe that a more massive object will not accelerate toward the Earth any faster than a less massive object is this: If an object increases it's mass it also increases it's resistance to change in motion (inertia) which offsets it's ability to increase acceleration. At the same time, the gravitational field that it produces will be stronger, which will cause the Earth to increase it's acceleration toward the object.

I still don't get your changing G value. I'll study that later. I have to go right now. Thanks for the thought provoking input.

 Mentor Nobody said G is changing. G is a physical constant. It's numerical value is rather small when expressed in terms of meters, kilograms, and seconds. This is in part a result of our specific choice of units but also in part because gravitation is by far the weakest of the four fundamental interactions.

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 Quote by D_H Nobody said G is changing. G is a physical constant. It's numerical value is rather small when expressed in terms of meters, kilograms, and seconds. This is in part a result of our specific choice of units but also in part because gravitation is by far the weakest of the four fundamental interactions.
Yes, I realize now that he was simply referring to the very small value of G. I was in a hurry when I left and didn't take enough time to read the post thoroughly. Sorry.

 Quote by sganesh88 See the equation i gave in my post. Difference in acceleration of two bodies of masses M1 and M2 falling towards the earth as observed by an observer on earth is G(M1-M2)/r2. This is because earth's mass contributes equally to the accelerations of both the masses towards the earth-again- as observed by someone on earth.
I disagree. I don't understand how you arrived at that. It makes no sense to me. Yes, the mass of the Earth contributes equally to both objects, but it should not be taken out of the equation. It's the ratio of difference between the objects masses and the mass of the Earth that affects the acceleration difference between the two objects in free-fall. I would do it this way: (M1 * (M2 - M3) * G) / r2 where M1 = mass of Earth, M2 and M3 = mass of the two objects. If the Earth were less massive then the difference in mass between M2 and M3 will have more effect and therefore would be more easily detectable.

I would like to add that since all objects accelerate equally toward Earth, regardless of the mass of the objects, what the Earth-bound observer is actually measuring is the combined accelerations of the objects toward the Earth and the Earth toward the objects.

 Quote by sganesh88 But the OP is of the opinion that the great mass of earth forbids us from sensing this difference which is not so. That factor doesn't enter the equation.
Obviously I still have that opinion. Why am I wrong?

 Quote by TurtleMeister I would do it this way: (M1 * (M2 - M3) * G) / r2 where M1 = mass of Earth, M2 and M3 = mass of the two objects.
What does this expression signify? Difference in accelerations between the two objects? If so, you should understand that this is not dimensionally homogenous.

 Quote by TurtleMeister Obviously I still have that opinion. Why am I wrong?
If you have understood the concepts involved, then a small error in the calculations might be the reason.

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 Quote by sganesh88 If you have understood the concepts involved, then a small error in the calculations might be the reason.
Yes, I have realized my error. Actually, I was going to come back and delete the post but since you've already commented, I guess I'll just leave it.

Rather than continuing to go down the road of crackpottery, as D_L mentioned previously, I guess I should just stop posting in this thread and give up on trying to understand the nature of gravity. :) I obviously do not have the math skills required to do such a simple task. I would like to thank you guys for your patients and for working with me in this thread.

 The only reason this cannot be proven experimentally is that any object we create will be minute compared to the Earths mass. The difference in elapsed time would be undetectable.
It can be proven, when an object falls to Earth. The Earth does not move towards the object with the same acceleration as the falling object.

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 Quote by Buckleymanor The Earth does not move towards the object with the same acceleration as the falling object.
No one has made that claim. A falling object would have to have the same mass as the Earth to cause that. Let's hope that never happens. :)

 I would like to add that since all objects accelerate equally toward Earth, regardless of the mass of the objects, what the Earth-bound observer is actually measuring is the combined accelerations of the objects toward the Earth and the Earth toward the objects.
Don't forget the Earth's acceleration around the Sun and the galaxies acceleration due to rotation plus the expansion of the universe.
Why all objects accelerate equally they have a small mass in comparison .
So what you are measuring is in effect, a gravitational force against this background.

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Buckleymanor, I was referring to objects in free-fall near the Earth. But you are correct that objects such as the Sun and Moon are also in free-fall toward the Earth (and vise versa). But fortunately that's happening in a circle.

 Quote by Buckleymanor Why all objects accelerate equally they have a small mass in comparison . So what you are measuring is in effect, a gravitational force against this background.
Either I don't understand this, or it's just wrong.

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 Quote by sganesh88 Difference in acceleration of two bodies of masses M1 and M2 falling towards the earth as observed by an observer on earth is G(M1-M2)/r2.
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.

 Buckleymanor, I was referring to objects in free-fall near the Earth. But you are correct that objects such as the Sun and Moon are also in free-fall toward the Earth (and vise versa). But fortunately that's happening in a circle. Originally Posted by Buckleymanor Why all objects accelerate equally they have a small mass in comparison . So what you are measuring is in effect, a gravitational force against this background. Either I don't understand this, or it's just wrong.
Yes, the point being that gravity is holding them in these orbits.
The Moon around the Earth.The Earth around the Sun.Sun and solar system around the galaxy.
Therefore all objects have a smaller mass in comparison to the rest of the universe and when they fall they accelerate not just towards the Earth but towards the rest of the universe as well.

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 Quote by Buckleymanor Therefore all objects have a smaller mass in comparison to the rest of the universe and when they fall they accelerate not just towards the Earth but towards the rest of the universe as well.
In what direction do you find the rest of the universe?

The rest of the universe has no effect on accelerations. This is analogous to the well known result that there is no gravitational field inside a massive hollow spherical shell.

 Quote by sylas In what direction do you find the rest of the universe? The rest of the universe has no effect on accelerations. This is analogous to the well known result that there is no gravitational field inside a massive hollow spherical shell.
The rest of the universe is expanding allthough our local group of galaxies are on a collision course.Generaly though the galaxies are moveing away from each other.
How this is analogous to a massive hollow spherical shell I find difficult to imagine.Presumably the shell would have to be perfect with no lumps and there would be nothing else in the universe.
Which clearly is not the case.

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 Quote by Buckleymanor The rest of the universe is expanding allthough our local group of galaxies are on a collision course.Generaly though the galaxies are moveing away from each other.
That is not "acceleration" in the sense you are using here, and there's no preferred direction, which is an essential part of gravitational acceleration. You are comparing apples and oranges.

 How this is analogous to a massive hollow spherical shell I find difficult to imagine.Presumably the shell would have to be perfect with no lumps and there would be nothing else in the universe. Which clearly is not the case.
The basic equations for expansion of the universe involve a uniformly homogenous mix; no lumps at all. All the major parts of the analysis, including expansion, accelerating expansion, and so on, is done with a uniform mix and with no local accelerations whatsoever. The universe is roughly the same in all directions, and that means there's no contribution at all to the accelerations measured for bodies in a gravitational field. There are additional complexities with lumps considered with studies in finer detail; and none of that detracts from the main point.

Look at the units. None of this effect is an "acceleration" measured in ms-2

The rate of expansion of the universe is a rate of change of scale factor, and scale factor is dimensionless. Hence the units are simply inverse time. The accelerating expansion is a second derivative, with unit of inverse square time.

Don't be mislead by the word "accelerating". It's not the same thing as the acceleration of a body in a gravitational field at all.

Cheers -- sylas

 The rate of expansion of the universe is a rate of change of scale factor, and scale factor is dimensionless. Hence the units are simply inverse time. The accelerating expansion is a second derivative, with unit of inverse square time. Don't be mislead by the word "accelerating". It's not the same thing as the acceleration of a body in a gravitational field at all.
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.

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