Same gravitational acceleration of unequal masses

[TurtleMeister]

Thanks for your input sganesh88. I am in total agreement with your second post. The only reason I used the simulator as my example was convenience.

Yes you are right. W.r.t an observer on earth, heavier objects will have a higher acceleration and hence fall down faster than the lighter ones.

No, it may appear that way to an Observer on Earth (in his non-inertial frame), but in reality a falling object will always accelerate toward the Earth at the same rate regardless of it's mass (A = M * G / r² where M = mass of Earth). The reason a more massive object appears to accelerate faster is because the Earth (which the observer is attached to) accelerates faster toward the object (A = M * G / r² where M = mass of the object). It would be accurate to say that the time of free-fall decreases for a more massive object. That can be agreed upon by any observer in any inertial frame.

In the vacuum experiments we don't observe this difference in accelerations of the bodies(w.r.t Earth mind you!) because of the vanishingly small value of G and a considerably high value of r.

I don't understand. G is a constant. The reason we don't observe this difference is because the mass of the Earth is much much greater than any ordinary objects we can use for our observation. The difference in their "fall times" would be far to small to be measured with any known instrument.

 

[sganesh88]

No, it may appear that way to an Observer on Earth (in his non-inertial frame), but in reality a falling object will always accelerate toward the Earth at the same rate regardless of it's mass

Very tempting to believe this right? Even i thought so initially. But nature has no reason to favor the inertial frames. There is no absolute motion that you can assign to an object universally. So you can't say that moon's absolute acceleration is G*M/r². Laws take on the simplest form in inertial frames. Not the truest. Not that there is something called "truest".

I don't understand. G is a constant. The reason we don't observe this difference is because the mass of the Earth is much much greater than any ordinary objects we can use for our observation. The difference in their "fall times" would be far to small to be measured with any known instrument.

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.

 

[D H]

I don't understand. G is a constant. The reason we don't observe this difference is because the mass of the Earth is much much greater than any ordinary objects we can use for our observation. The difference in their "fall times" would be far to small to be measured with any known instrument.

What sganesh88 (correct me if I'm wrong, sganesh88) meant is that G is pretty dang small: 6.6730×10^-11 m³kg^-1s^-2. Couple that with a largish distance, say 6738 km, and you get exceedingly small differences in acceleration for a pair of normal-sized test objects.

 

[sganesh88]

What sganesh88 (correct me if I'm wrong, sganesh88) meant is that G is pretty dang small: 6.6730×10^-11 m³kg^-1s^-2. Couple that with a largish distance, say 6738 km, and you get exceedingly small differences in acceleration for a pair of normal-sized test objects.

Ya. That's precisely what i meant. 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.

 

[TurtleMeister]

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.

 

[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.

 

[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.