Same gravitational acceleration of unequal masses

[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.

 

[TurtleMeister]

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.

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.

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?

 

[sganesh88]

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.

 

[sganesh88]

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.

 

[TurtleMeister]

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.

 

[Buckleymanor]

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.

 

[TurtleMeister]

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. :)

 

[Buckleymanor]

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.

 

[TurtleMeister]

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.

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.

 

[TurtleMeister]

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]

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.

 

[sylas]

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.

 

[Buckleymanor]

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.

 

[sylas]

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

 

[Buckleymanor]

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.

 

[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. :grumpy:

 

[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