Same gravitational acceleration of unequal masses

[steve102]

I understand that objects of different masses will accelerate equally in a vacuum. This is because the force of acceleration from gravity is the same regardless of mass.

I also know I can manipulate the F=ma equation to a=f/m, which also shows the above to the true.

Intuitively, though, I have trouble reconciling this with the notion of gravity "causing" the items to have different masses when sitting still.

I was wondering if anyone had a plain-English explanation. Thank you for considering my question.

 

[Chewy0087]

Masses? It's weights, be careful which you use. Gravity doesn't give anything mass, it gives it weight, which is a force, think about that;

F = ma, as you can clearly see the thing that gravity is GIVING the mass is force.

Incidentally, the reason why acceleration due to gravity is constant for anybody is due to Newton's equation of gravity;

(Mass of Object 1 x Mass of Object 2 x Gravitational Constant) / Distance between them squared.

Or to find the pull of 1 on the other only use the mass of the one pulling.

As you can see, if you take you to be the first object, the centre of the Earth to be the other, now equal this to F = ma, your mass simply cancels out with the mass in Newton's equation.

http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitationEdit : If your question is really related to how things have mass, see Higgs Boson etc, much more abstract & I don't think anyone could give you a real answer right now

 

[TurtleMeister]

Here's my way of thinking of it - in plain English, with no equations. Objects accelerate toward each other (toward the center of mass), not just one toward the other. In a vacuum, the Earth and a massive "ordinary" object WILL have a higher acceleration toward each other than would the Earth and a less massive "ordinary" object. But that difference in acceleration cannot be detected because the difference in the combined mass of the Earth and the two "ordinary" objects is too small to be detected. The rate of acceleration of any two objects toward each other depends on the TOTAL mass of both objects combined. So if you have a 1kg object combined with the mass of the Earth and then you double the objects mass to 2kg, how much have you increased the total mass (as a percentage)? It's a very very small number. So that's why all objects appear to accelerate at the same rate.

Edit:
Because of what I have learned in this thread, I now know that the above statement is incorrect. It is true that two free falling objects of different mass will have an extremely small and undetectable difference in acceleration. But it is not the great mass of the Earth that prevents us from detecting this difference. In fact, the Earths mass has nothing to do with it. Only the difference in mass of the two objects and their distance to the center of the Earth affect their difference in acceleration, expressed as G(M1-M2)/r2. I guess sometimes you just have to use a little math to see things as they really are. :)

 

[Chewy0087]

I'm afraid you're wrong, we don't 'appear' to be attracted more or less, we are all attracted exactly the same, regardless of mass.

You work out the gravitational field (using Newton's law's) to be;

Mass of Object * G / r²

Yes, we also attract the sun as well, but that's entirely separate. As I understand it the force due to gravity is constant (at sea level) for all bodies regardless of mass, another way to look at it is as I said;

\frac{M1 * M2 * G}{r^2} = M1 * A (M1 bieng my mass for example)

The M1's cancel, (our) mass does not affect it.

 

[TurtleMeister]

Yes, my interpretation may indeed be wrong. But I don't think it is. And I welcome anyone who can explain to me why it isn't. The OP asked for a plain English explanation and that was what I gave, stating clearly that it was my way of thinking. I admit that it's not complete because I said that the acceleration of two bodies toward each other depends on the total mass of both bodies. That's true, but I left out that it also depends on the distance between them. Anyway, here's the problem I have with your explanation (which is the one I usually hear):

You're saying that I am wrong because of this:

A = M2 * G / r² where M2 = mass of Earth
(M1 * M2 * G) / r² = M1 * A where M1 = my body mass
The M1's cancel, my body mass does not affect it.

If that's true then why isn't this true:

A = M1 * G / r² where M1 = my body mass
(M1 * M2 * G) / r² = M2 * A where M2 = mass of Earth
The M2's cancel, Earths mass does not affect it.

 

[sylas]

If that's true then why isn't this true:

A = M1 * G / r² where M1 = my body mass
(M1 * M2 * G) / r² = M2 * A where M2 = mass of Earth
The M2's cancel, Earths mass does not affect it.

The "it" here is the acceleration of the Earth towards you, and yes, the Earth's mass doesn't affect that.

 

[TurtleMeister]

The "it" here is the acceleration of the Earth towards you, and yes, the Earth's mass doesn't affect that.

I see your point. have to go right now but will think about this and reply later. Thanks for the input.

 

[D H]

Yes, my interpretation may indeed be wrong. But I don't think it is.

Your explanation is looking at things from the perspective of a non-inertial reference frame. If you look at things from the perspective of an inertial frame, all objects, regardless of mass, will undergo exactly the same acceleration.

 

[Cantab Morgan]

I understand that objects of different masses will accelerate equally in a vacuum. This is because the force of acceleration from gravity is the same regardless of mass. .

I'm sorry but that's where you're going wrong. The force is greater for larger masses. F=GMm/r^2 So, if you increase m (the mass of the object), then the gravitational force on that object increases too.

I think the phrase "force of acceleration" is too confusing and has led you astray. There is force. And there is acceleration. They are different things, measured in different units.

 

[TurtleMeister]

Your explanation is looking at things from the perspective of a non-inertial reference frame. If you look at things from the perspective of an inertial frame, all objects, regardless of mass, will undergo exactly the same acceleration.

What do you mean by inertial frame?

Ok, I checked Wikipedia and I don't get how my view is non-inertial. But I'll admit that I am having doubts about my view. Is it completely wrong? Or is it a matter of perspective?

The "it" here is the acceleration of the Earth towards you, and yes, the Earth's mass doesn't affect that.

Why is it ok to use these equations and say that one object is accelerating toward the other, but it's not ok to say that both objects are accelerating toward each other (toward the center of mass)? It would appear that by setting A = M1 * G / r² or A = M2 * G / r² that you are setting your frame of reference to one or the other object and linking A to only one of the masses. Obviously any changes in the mass not linked to A will have no affect. Isn't it just as true to say that both objects are accelerating toward each other according to (M1 * M2 * G) / r² and that A depends on both M1 and M2?

 

[Chewy0087]

Think of it this way, I'm floating in space, about 3,000 kilometres from the sun, wearing an astronaut suit so I don't die.

Now, me and the sun attract each other due to gravity thanks to the equation

\frac{M1 * M2 * G}{r²} = Vector Force

I accept this, now I think 'how fast am I moving?' I can work out the force on me, so now I decide to use another law of Newtons - F = ma, since the above equation gives me the force, i can equal it to MA

\frac{M1 * M2 *G}{r²} = M1 a (M1 bieng my mass)

I can then divide through by M1, meaning my mass is redundant, independant of my acceleration.

I think maybe some of you are getting caught up on the fact that it's an attraction from the two, and not just one, and you'd be right, but you can see them as independent from each other, the effect of our attraction on the sun, and the sun's to us, when working out the acceleration of one (towards the other), it's independant of (the one that's your trying to work out the acceleration for's) mass

 

[sylas]

Ignore this post. Originally I queried a comment which has now been fixed up in edit, so I am also removing what I quoted. We're all on the same page again.[/size]

 

[Chewy0087]

Yeah ok, didn't really read it my bad, the gravitational pull is greater for objects with larger mass's, yeah. f = ma with a being constant.

 

[D H]

What do you mean by inertial frame?

Ok, I checked Wikipedia and I don't get how my view is non-inertial. But I'll admit that I am having doubts about my view. Is it completely wrong? Or is it a matter of perspective?

Your view is non-inertial because your reference frame is accelerating. Strictly speaking, Newton's laws of motion are not valid in a non-inertial frame. To make them appear to be valid one must add fictitious forces to the mix.

There is nothing wrong per se with working in a non-inertial frame. Suppose you want to explain some phenomenon. Whether you do your derivations from the perspective of an inertial frame or non-inertial frame in a way is irrelevant. Do the math right and both perspectives will yield the same end result. The choice is very relevant if you take into consideration the difficulty of arriving at the end result. Explaining the weather from the perspective of an accelerating and rotating reference frame (i.e., a frame fixed to the rotating Earth) is rather difficult but doable. Explaining the weather from the perspective of an inertial frame (ignoring very tiny effects, a non-rotating frame with origin at the solar system barycenter) is a lot worse than rather difficult. On the other hand, trying to explain the interactions in some remote star system from the perspective of an Earth-fixed frame, while possible, is downright stupid.

What you have done wrong is to think that because the acceleration of an object toward the Earth does depend on the object's mass somehow falsifies the equivalence principle. It doesn't. The equivalence principle now stands as one of the most precisely verified axioms in all of physics. It has been verified to about 1 part in 10¹³ (see the Physics World article cited below). Three future (in development or proposed) satellite missions, MICROSCOPE (CNES), Galileo Galilei (ASI) and STEP (NASA), intend to increase this accuracy by two, four, and five orders of magnitude respectively.


References:
Relativity at the centenary, Physics World (2005), http://physicsworld.com/cws/article/print/21148

Microscope: Exploring the limits of the equivalence principle, CNES, (2008), http://www.cnes.fr/web/CNES-en/2847-microscope.php

Galileo Galilei, GG project home page, http://eotvos.dm.unipi.it

STEP, STEP project home page, http://einstein.stanford.edu/STEP

 

[TurtleMeister]

Thanks for all the info. It has provided me with a lot to think about.

Your view is non-inertial because your reference frame is accelerating. Strictly speaking, Newton's laws of motion are not valid in a non-inertial frame. To make them appear to be valid one must add fictitious forces to the mix.

Yes, I can see now that it is a matter of perspective and that my view would indeed have to be accelerating (unless both objects were the same mass).

What you have done wrong is to think that because the acceleration of an object toward the Earth does depend on the object's mass somehow falsifies the equivalence principle.

No. I do not think that. Equivalence experiments is one of the things I have read quite a lot about. I am especially interested in equivalence experiments involving active and passive gravitational mass. I've only found one (Lloyd B. Kreuzer 1966). See my first PF post: https://www.physicsforums.com/showthread.php?t=305189

 

[TurtleMeister]

Never mind, I found my error. :)

 

[D H]

It just seems unintuitive to have to set my frame of reference to one or the other objects.

Choose a different reference frame and you can get a vastly different view of things. What's so unintuitive about that?

But it is unintuitive, and in my opinion even deceiving, to say that any two objects will fall at the same rate regardless of their mass. ... Now, will both the 1kg object and the 1/4 mass of the Earth object take the same amount of time to hit the surface? I don't think so.

You're comparing apples to oranges.

The equivalence principle is about the gravitational acceleration as observed by an inertial observer. You are talking about relative acceleration. It's well known that the masses of both objects contribute to relative acceleration. For example, this is one reason why Kepler's laws are only approximately correct. (Kepler's laws implicitly assume planets have negligible mass compared to that of the Sun, and that is not a valid assumption given with the precision with which we can measure orbits.)

 

[TurtleMeister]

D_H, since you quoted something that I deleted I will re-post it here so that other readers can see it in it's entirety.

Yes, I can see now that it is a matter of perspective and that my view would indeed have to be accelerating (unless both objects were the same mass). It just seems unintuitive to have to set my frame of reference to one or the other objects. Let's take the following example: An object in Earth free-fall will accelerate at the same rate regardless of it's mass. That is true. The key word being accelerate. But it is unintuitive, and in my opinion even deceiving, to say that any two objects will fall at the same rate regardless of their mass. And here is the reason I think so: Suppose we have an object (lets say 1kg) which is positioned at a certain distance from the Earths surface. We let the object free-fall and measure the time it takes for it to hit the surface (assuming there is no atmosphere). Now suppose we make the mass of the object 1/4 the mass of the Earth, set it at the same distance and let it free-fall. We once again measure the time it takes it to hit the surface. And keep in mind that regardless of what frame of reference we use to make the measurement (disregarding any minute differences due to relativity), the time lapse will not be affected. Now, will both the 1kg object and the 1/4 mass of the Earth object take the same amount of time to hit the surface? I don't think so. But of course you can get around this by saying that by increasing the mass of the object we increased the Earths acceleration toward the object. But I think you can see by this example how it is deceiving, and unintuitive to say the least, to say that any object, regardless of it's mass will fall at the same rate. It may be technically true to say that increasing an objects mass will not change it's acceleration, but it will indeed change the time it takes for it to reach the surface. 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.

You are talking about relative acceleration. It's well known that the masses of both objects contribute to relative acceleration.

Choose a different reference frame and you can get a vastly different view of things. What's so unintuitive about that?

The unit of measure used in my thought experiment is time, not acceleration. And it works the same regardless of whether it's in an inertial or non-inertial frame of reference. Why isn't that more intuitive?

D_H, I am not trying to dispute the equivalence principle. And I do not believe that my thought experiment shown above contradicts it. I have realized my error as far as the inertial frame is concerned. However, I still do not think my thought experiment is incorrect. I am willing to change my mind but I haven't been convinced yet. I have learned something from this thread and feel that I understand gravity better than I did before.

 

[D H]

It just seems unintuitive to have to set my frame of reference to one or the other objects. Let's take the following example: An object in Earth free-fall will accelerate at the same rate regardless of it's mass. That is true.

As perceived by whom? Let's ignore the Earth's rotation. Suppose we have two observers with infinitely accurate means of assessing acceleration. One is fixed with respect to this non-rotating Earth and the other is fixed with respect to some inertial frame. Both measure the acceleration of an object falling toward their Earth at the same time with their infinitely accurate sensors. The two will measure different accelerations.

Now suppose we use a different test object whose mass is orders of magnitude greater than that of the first test object. Both observers measure the acceleration of this new test object after placing it at exactly the same position with respect to the Earth as the first test object. While the Earth-based observer will measure a different acceleration than measured for the first test object, the inertial observer will see this new object as undergoing exactly the same acceleration as the first test object.

Another way to look at this: Imagine two cars, a blue one and a red one, in a drag race. From the perspective of someone in the grandstands, the blue car's acceleration is amazingly high. From the perspective of the driver of the red car, the blue car's acceleration is fairly small. In fact, the driver of the red car most fervent wish is that the blue car appears to be accelerating backwards.

 

[TurtleMeister]

It just seems unintuitive to have to set my frame of reference to one or the other objects. Let's take the following example: An object in Earth free-fall will accelerate at the same rate regardless of it's mass. That is true.

As perceived by whom?

A = M2 * G / r² where M2 = mass of Earth
The objects mass has no affect.

Let's ignore the Earth's rotation. Suppose we have two observers with infinitely accurate means of assessing acceleration. One is fixed with respect to this non-rotating Earth and the other is fixed with respect to some inertial frame. Both measure the acceleration of an object falling toward their Earth at the same time with their infinitely accurate sensors. The two will measure different accelerations.

Now suppose we use a different test object whose mass is orders of magnitude greater than that of the first test object. Both observers measure the acceleration of this new test object after placing it at exactly the same position with respect to the Earth as the first test object. While the Earth-based observer will measure a different acceleration than measured for the first test object, the inertial observer will see this new object as undergoing exactly the same acceleration as the first test object.

I'm not sure how to comment on this because I don't know what "some inertial frame" is. If this observer who is at "some inertial frame" measures the acceleration relative to himself then what he measures will depend on what his velocity and acceleration is relative to the objects being measured. However, I can use this to illustrate my point where I say that using A = M2 * G / r² is unintuitive. If your observer at "some inertial frame" and the Earth observer measured the time lapse of the objects free-falls, wouldn't their results be the same - regardless of the frame of reference of the observer at "some inertial frame"? Your post seems bolster my argument!

 

[TurtleMeister]

Ok, I have a 2D gravity simulator. No, this is not a thought experiment. I actually have a 2D gravity simulator program that I will use to illustrate my point. I start out with the Earth and the Moon at a stand-still in space. In other words their velocity and acceleration relative to each other are zero. We as the observers are also at a stand-still. I start the simulation. The Earth and Moon crash into each other. I record the time lapse and note the acceleration of each body. Now I do another simulation, except this time I increase the mass of the moon by a factor of 20. The Moon is now more massive than the Earth (I had to use an extreme change in order to see the effects in real time). I start the simulation. The Earth and Moon crash into each other. The time lapse is much shorter as would be expected. But the acceleration of the Moon does not change (A = M * G / r²). The Earth however did increase acceleration, thus the shorter time lapse. Just as the equation states, the mass of the moon did not affect it's acceleration. But an observer on Earth would not be able to tell whether the moons acceleration increased or his own acceleration increased. However, all observers would be able to compare the time lapse and be in total agreement.

So, if we do a Galileo drop tower experiment using infinitely accurate instruments (in a vacuum), the more massive object will strike the Earth quicker than the less massive object. Not because the acceleration of the object increased, but because the acceleration of the Earth toward the object increased. And that's the reason I say that A = M * G / r², even though accurate and true, is unintuitive.

Edit:
Just wanted to add that the reason an objects acceleration does not increase when it's mass is increased is because of inertia. Increasing an objects mass also increases it's resistance to change in motion. But by increasing the objects mass we have also increased the gravitational field that it produces. So any other objects nearby will experience an increase of acceleration toward it.

 

[D H]

A = M2 * G / r² where M2 = mass of Earth
The objects mass has no affect.

One last time, that equation is strictly valid in an inertial frame only. Your lack of understanding of the concept of an inertial reference frame is what is getting you in trouble.

I'm going to be blunt: You're posts in this thread make you appear to be a crackpot. That is not a good road to go down.

 

[TurtleMeister]

D_H, thank you for being blunt with me. I appreciate it when someone is honest with me about their oppinions. But I would like to say that I am only trying to improve my understanding of gravity. If my notions are lunatic I would like to know why. You say that I do not understand the concept of an inertial reference frame. It would help me a great deal if you, or anyone else, would read over my previous post #21 and explain to me where I am going wrong in reference to inertial frames. Or, if there happens to be anything that is correct about my post then I certainly wouldn't mind hearing that also :). I currently feel like my understanding of gravity has improved since I've been posting in this thread. But now you are bringing more doubts to my mind. Any input from other members would be great also.

 

[TurtleMeister]

Here is another thread which is basically saying the same thing I am:
https://www.physicsforums.com/showthread.php?t=95366

 

[nealh149]

TurtleMeister, if you're in a car and you accelerate you feel it. If you accelerate faster you feel it more. Why wouldn't the people on Earth be able to feel the increased acceleration in your experiment of increasing the mass of the moon by a factor of 20?

 

[TurtleMeister]

Why wouldn't the people on Earth be able to feel the increased acceleration in your experiment of increasing the mass of the moon by a factor of 20?

Because they would be in free-fall toward the moon. Objects in free-fall do not feel the force, even though they are accelerating.

But that's not the point of my argument. I only used the large masses so that I could get an easy real time animation on the simulator. Read my first post in this thread.

 

[D H]

This is neither that hard nor that counterintuitive a concept.

It's all about reference frames.Let's work our way up the derivative hierarchy. Suppose we have two reference frames with collinear axes but different origins. The position of some object will have different representations in the two frames because the origins of the two frames are distinct. Next suppose the distance between the origins is changing with respect to time. Now the velocity of some object will have different representations in the two frames because of the two frames are moving with respect to one another. Finally, suppose the distance between the origins is changing non-linearly with respect to time. Now the acceleration of some object will have different representations in the two frames because of the two frames themselves are accelerating with respect to one another.

Next, suppose we know that no net external force act on this object. Now suppose an observer fixed with respect to one of these frames sees the object as moving at a constant velocity. The object is obeying Newton's first law. This might well be an inertial reference frame! If some other object, also known not to be subject to a null net force and not collinear with the observer first object is also observed to move at a constant velocity, the frame is an inertial frame. Newton's second law also applies, as does Newton's law of gravitation.

What about that other frame? That frame is accelerating with respect to the frame now known to be an inertial frame. Objects with no net force acting on them will not move with a constant velocity. F≠ma, and F≠Gm₁m₂/r². Newton's laws are not valid in non-inertial frames.

Two objects with non-zero masses m₁ and m₂ separated by some distance r will accelerate toward one another because of gravity. Sans any other forces acting on the two objects, a reference frame based on either object is, strictly speaking, a non-inertial frame. An observer fixed with respect to an inertial frame will see object 1 accelerate toward object 2 with acceleration a₁=Gm₂/r² and will see object 2 accelerate toward object 1 with acceleration a₂=Gm₁/r². An observer fixed with respect to object 1 will see object 1 as non-moving and will see object 2 accelerating toward object 1 with acceleration a_rel=G(m₁+m₂)/r². An observer fixed with respect to object 2 will see object 2 as non-moving and will see object 1 accelerating toward object 2 with acceleration a_rel=G(m₁+m₂)/r².

In short, from the perspective of an inertial observer, the gravitational acceleration of some object toward another depends only on the distance between the objects and the mass of the other object. From the perspective of an observer fixed with respect to one of the objects, the gravitational acceleration of the other object depends on the distance between the objects and the mass of both objects.

 

[TurtleMeister]

D_H, thank you for that detailed description of inertial and non-inertial frames.

Two objects with non-zero masses m1 and m2 separated by some distance r will accelerate toward one another because of gravity. Sans any other forces acting on the two objects, a reference frame based on either object is, strictly speaking, a non-inertial frame. An observer fixed with respect to an inertial frame will see object 1 accelerate toward object 2 with acceleration a1=Gm2/r2 and will see object 2 accelerate toward object 1 with acceleration a2=Gm1/r2. An observer fixed with respect to object 1 will see object 1 as non-moving and will see object 2 accelerating toward object 1 with acceleration arel=G(m1+m2)/r2. An observer fixed with respect to object 2 will see object 2 as non-moving and will see object 1 accelerating toward object 2 with acceleration arel=G(m1+m2)/r2.

I understand, and I am in complete agreement. I actually breezed through this part. Something I probably could not have done at the start of this thread.

In short, from the perspective of an inertial observer, the gravitational acceleration of some object toward another depends only on the distance between the objects and the mass of the other object.

I had to read this a couple of times but I think it was because you did not name the objects, but rather called them "some object" and "the other object". But I get it, and I am in complete agreement.

From the perspective of an observer fixed with respect to one of the objects, the gravitational acceleration of the other object depends on the distance between the objects and the mass of both objects.

I am in total agreement with this also. And very good that you put "both" in italics. :)

The first part of your post is a little difficult for me to follow. Unless there is something there that I should know in addition to what I have quoted, I will assume that I have the correct concept of inertial and non-inertial frames. To verify, I will use an example from my simulator. If I have my simulator programmed to show the Sun and the four inner planets, and I set the frame of reference to the center of mass (which would be inside the Sun), then I would be at an inertial frame. If however, I set my frame of reference to the Earth (all other bodies revolving around the Earth) then I would be at a non-inertial frame.

I noticed that you did not comment on my post# 21. Does this mean there are no inertial reference frame errors?

 

[sganesh88]

Addressing your post #21:
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. 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. If the difference is really huge(like the second input of the simulation), then the change in acceleration will definitely be felt. Enough of words,
Difference in acceleration of two bodies at a distance , r from the center of Earth is given by
G(M1-M2)/r²
M1 and M2 refers to the masses of two objects. (for instance the moon with its original mass and the moon that suddenly inflated to 20 times its original mass). Note that Earth's mass does not enter the picture.

 

[sganesh88]

D_HIf I have my simulator programmed to show the Sun and the four inner planets, and I set the frame of reference to the center of mass (which would be inside the Sun), then I would be at an inertial frame. If however, I set my frame of reference to the Earth (all other bodies revolving around the Earth) then I would be at a non-inertial frame.

Forget the electronic simulator. You've a much sophisticated biological one. Let's start.
Imagine you're in a room, such that no part of it has any relative motion w.r.t you. You've all the facilities to observe and record the position, velocity, acceleration and whatnot of other objects. Now you observe an object that is sufficiently isolated from other objects. So we come to a conclusion that it is not subjected to any external forces. How does it move? observe properly. If it has a constant velocity- meaning its speed and direction doesn't change w.r.t time- then you classify YOUR room as an inertial frame. Any other frame that has such a uniform motion w.r.t your room also becomes an inertial frame. If the object doesn't exhibit a uniform motion-it might be forming a circle or accelerating linearly- then you classify YOUR room as a non-inertial one. This frame is not favored by Newton's laws.
P.S: The motion of distant stars in a circle around the north-south axis of the Earth makes us realize that Earth is non-inertial.

 

[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.