Equal and opposite force

[johnreed]

Although we can treat any object we lift as a blob and say that the planet acts on the blob with an equal and opposite force to the one we exert, let's define our blob a bit more precisely. Let's say we have 1 gram atom of Gold. This is the relative atomic weight of gold in grams. Let's call that weight x so here so far we have mg gold=x. Let's take the weight of the earth as X. So we have Mg earth as X. So in terms of equal and opposite force we have x=X.

 

But since 1 gram atom of gold equals (N) atoms of gold or Avogadro's number. We have the resistance of (N) atoms of gold equal to the resistance of all the atoms in the Earth. Where it appears to me that the resistance of (N) atoms of gold here is only equal and opposite to the force we exert and has nothing to do with the force the planet exerts other than as a convenience for our successful navigation through the universe in terms that originate with us.

 

How does the planet's cumulative atom Resistance = a cumulative resistance we feel with something we lift, if the planet does not act on all atoms uniformly?

[Klaynos]

Resistance of atoms? What do you mean by that?

[johnreed]

What do I mean by resistance of atoms? On the balance scale we compare one pan of atoms to another pan of atoms. We use units of weight mg. So we are comparing weight. However [g] is acting on the balance scale itself and on each pan so [g] divides out of the equation as far as the balance scale operation is concerned. The balance scale compares the resistance of pans of atoms in units we call mass. This comparison is invariant unlike the comparison of weight which changes with location

[swansont]

One of the parts of relativity is that gravity acts uniformly on all objects, regardless of composition. Your conjecture that "the planet does not act on all atoms uniformly" is dubious. It's hard to tell for sure, since you are using non-standard vocabulary.

[D H]

Let's call that weight x so here so far we have mg gold=x. Let's take the weight of the earth as X. So we have Mg earth as X. So in terms of equal and opposite force we have x=X.

Your x=X means m_goldg = M_Earthg, or m_gold = M_Earth. In other words, 196.97 grams = 5.9736×10²⁴ kg. This is of course nonsense. Start with a nonsensical assumption and everything that follows is nonsensical. Your X≠x. The problem is your expression of the gravitational force of gold ball on the Earth. It is not M_Earthg.

 

How does the planet's cumulative atom Resistance = a cumulative resistance we feel with something we lift, if the planet does not act on all atoms uniformly?

It looks like you are stumbling into yet another line of fallacious reasoning here. Make your case. Do note that the strong equivalence principle strongly argues against whatever line of fallacious reasoning you are trying to follow.

[johnreed]

Your x=X means m_goldg = M_Earthg, or m_gold = M_Earth. In other words, 196.97 grams = 5.9736×10²⁴ kg. This is of course nonsense. Start with a nonsensical assumption and everything that follows is nonsensical. Your X≠x. The problem is your expression of the gravitational force of gold ball on the Earth. It is not M_Earthg.

 

 

It looks like you are stumbling into yet another line of fallacious reasoning here. Make your case. Do note that the strong equivalence principle strongly argues against whatever line of fallacious reasoning you are trying to follow.

 

jr writes> I'm sorry. I don't follow your change. In equal and opposite terms we have mg(gold) = Mg(Earth) where the slack is taken up in the magnitude of [g]. Is that not correct? Since mg(gold) here has been measured as (N) and since [mg] is equivalent to a Force [F] we feel. we say that the force we feel is the force of gravity which acts between all objects as an attraction in an equal and opposite manner. We say (again) that the huge discrepancy is taken up by the miniscule magnitude of [g] in [Mg]. Is this not correct?

 

So I want to know what [mg] represents. Well it represents weight and we have used the balance scale for 6000 years to compare weight. So as far as we are concerned the balance scale compares weight. We use it to compare the weights of atoms. One group of atoms in one pan another group in another pan. They can all be uniform or they can all be different. In either case, on the balance scale the quantity [g] divides out at any location the balance scale is functional and we are left with one pan of atoms balancing another pan of atoms. We standardize one of the pans into mass units and we are measuring the comparative resistance of two pans of atoms which we call mass. WE are comparing weight. The balance scale is comparing the resistance of atoms. This resistance is conserved in the classical frame. Weight changes with location in the classical frame. Weight changes according to the magnitude of [g]. Weight happens to coincide with what we feel. So what we feel changes where mass remains the same. But what we feel in all cases is the resistance of atoms to our effort which depends on the pull on those atoms and not on their comparative resistance as measured on the balance scale.

 

For sure I have been stumbling. That seems to be all I ever do. Ever now and then I stumble into something I have to examine. The poe. What does it mean? The force we feel is equivalent to the force we feel ma=mg and so the force we feel controls the least action consistent universe? Just because we feel the units we measure it by?

 

Thanks for your sincere answer. I hope I don't discourage you.

johnreed

 

 

 

[swansont]

|F|=GMm/r^2

That holds for the mass M and the mass m. The acceleration will be different for each, in both magnitude and direction. The object feels an acceleration of g = GM/r^2 The earth feels an acceleration of Gm/r^2

[johnreed]

One of the parts of relativity is that gravity acts uniformly on all objects, regardless of composition. Your conjecture that "the planet does not act on all atoms uniformly" is dubious. It's hard to tell for sure, since you are using non-standard vocabulary.

 

 

jr writes>

Again I apologize for my lack of clarity. I am saying that the planet attractor acts on all atoms uniformly. We measure the resistance of atoms on the balance scale and we call this mass. The planet attractor acts on all objects uniformly and on all atoms uniformly. When we lift an object we are acting on the cumulative resistance of the atoms that the planet acts on uniformly. Gravity is the force we feel and it is equivalent to the resistance of the atoms we lift at location. That resistance is caused by the planet's uniform action on atoms. Which action then becomes electromagnetic of a type we feel as gravity.

 

To carry it further here. In sum gravity is a form of electromagnetism that acts on all atoms comparatively weakly (wrt our experience) but originates from the planet core electromagnetic nature. This last paragraph is not part of this post as it is too speculative so far. I'm trying to work out the ground rules only here. So if you shoot me out of the water let your aim destroy this paragraph. Thanks.

johnreed

[D H]

Word salad. What bits I could make of it is wrong.

 

You don't feel gravity. Suppose you take one of those amusement park rides where you are very briefly in free fall. Or you could take a ride on the Vomit Comet, where you are in free fall for a slightly longer duration. Gravitational acceleration changes by only tiny bit during the ride. What you feel in your guts and in your inner ear changes by a whole lot. You don't feel gravity. You feel everything but gravity.

 

Gravity is not a form of electromagnetism.

[swansont]

We can measure the electric and magnetic properties of atoms and of conglomerations of atoms. They do not give the result that is anywhere close to being consistent with gravity.

[johnreed]

Word salad. What bits I could make of it is wrong.

 

You don't feel gravity. Suppose you take one of those amusement park rides where you are very briefly in free fall. Or you could take a ride on the Vomit Comet, where you are in free fall for a slightly longer duration. Gravitational acceleration changes by only tiny bit during the ride. What you feel in your guts and in your inner ear changes by a whole lot. You don't feel gravity. You feel everything but gravity.

 

Gravity is not a form of electromagnetism.

 

 

jr writes> Talk about word salad. Who said we don't feel gravity? That is precisely what we feel gravity and we define it in terms of what we feel [mg] and call it force. It is force to us and it is equal and opposite to the resistance we encounter. That resistance is caused by the planet's attraction on atoms which is uniform for all atoms. That is what I say. Please quote me otherwise.

 

The above is the subject of my most recent posts. I can add direction to that substance in a very general way using few words. Please do not consider this paragraph a statement I am arguing at thr present time. But let conjecture that what we call gravity does not collapse into a black hole but instead the electromagnetic properties of atoms constructively combine to commandeer the core of the collapsed star to create an electromagnetic dynamo that attracts all form of matter at a certain distance and leaves shells of matter that coalesce into planets. The collapse is accompanied by an explosion that sends star parts everywhere. These star parts eventually return to their origin but their origin location is gon and so they pass closely around the star and are thrown back into their distant orbit. This is just conjecture for now and won't be entertained seriously until all my preliminary arguments are worked out so thy are clear to all. But I have always sought assistance.

johnreed

 

|F|=GMm/r^2

That holds for the mass M and the mass m. The acceleration will be different for each, in both magnitude and direction. The object feels an acceleration of g = GM/r^2 The earth feels an acceleration of Gm/r^2

 

jr writes: I have no argument with this. We can also say that mg=GMm/r^2 where [m] divides out leaving g=GM/r^2. That's fine. But where do you get it that [M] divides out? And once you divide [m] out all you are left with is least action consistent motion. But I am not arguing this right now. In fact I've got to revisit my original few paragraphs to remind myself what I am arguing for. Thank you for not shooting me out of the water.

 

One of the parts of relativity is that gravity acts uniformly on all objects, regardless of composition. Your conjecture that "the planet does not act on all atoms uniformly" is dubious. It's hard to tell for sure, since you are using non-standard vocabulary.

 

Jr writes>

I might be getting my sea legs here a bit better. No what I say is that the planet acts on all atoms uniformly, That allows us to measure the resistance of each pan of atoms in a conserved unit we call mass. In fact if the planet did not act on all atoms uniformly and instead acted on object mass we could never have evolved. I apologize for my non-standard terminology. Resistance has a specific meaning in physics and force does as well. We act on resistance and we feel a force that is equal and opposite to the resistance we act on anywhere in the universe we can occupy. But the force is initiated by us and is not a property of inanimate objects.

johnreed

 

Word salad. What bits I could make of it is wrong.

 

You don't feel gravity. Suppose you take one of those amusement park rides where you are very briefly in free fall. Or you could take a ride on the Vomit Comet, where you are in free fall for a slightly longer duration. Gravitational acceleration changes by only tiny bit during the ride. What you feel in your guts and in your inner ear changes by a whole lot. You don't feel gravity. You feel everything but gravity.

 

Gravity is not a form of electromagnetism.

 

jr writes> Rereading this comment you are taking freefall as an example where we don't feel gravity. I agree. In freefall we are traveling in the direction our atoms are being pulled. However when we accelerate against the pull on our atoms we feel this pull and we call it gravity. When we are in contact with a non-freefalling object we also feel the pull on our atoms. Sorry I think I mis read your original paragraph. People do it all the time.

[D H]

jr writes> Talk about word salad. Who said we don't feel gravity?

Physicists. Sensor designers. Albert Einstein.

 

Did you read my post at all? I gave a specific example about amusement park rides that illustrates that we don't feel gravity. We feel everything but gravity.

 

Here's another example. You've probably heard that the astronauts aboard the Space Station feel weightlessness, or that the Space Station is a zero-g (technically, micro-g) environment. The gravitational force on the astronauts is reduced by only 8% or so. So why are they floating around, and why are they called weightless? The answer: The astronauts don't feel gravity. You can't feel gravity. Nothing does. An accelerometer onboard the Space Station will register a near zero acceleration. That same accelerometer at rest on the surface of the Earth will indicate that it is accelerating 32 feet/second²upward. The accelerometer is sensing everything but gravity. It can't sense gravity.

[johnreed]

jr writes>

I apologize for the non standard vocabulary. I took a different path from day one and have had to learn the vocabulary of mainstream science as I go. But my initial arguments here are simple and involve the meaning of words like mass and the function of simple machines like a balance scale. Take the idea of a constant of proportionality. Where K is a constant of proportionality for T^2/r^3. I grasp that perfectly clear. G on the other hand is a constant of proportionality that purports to assign masses to planets based on our measure of mass here and Newton's basic assumption that since its true here its true there. Paraphrased. I think that our acceptance of G as a constant magnitude is shaky in this day and time. Einstein just in this case set what has been known and perplexing to us a a principle. It doesn't make G any the more correct. But I am getting way ahead of my little patch of ground I'm trying to carve out here. Yes all I am trying to say is apparently already accepted in part. That the planet attractor axts on atoms.

One of the parts of relativity is that gravity acts uniformly on all objects, regardless of composition. Your conjecture that "the planet does not act on all atoms uniformly" is dubious. It's hard to tell for sure, since you are using non-standard vocabulary.

 

I hope I've done this correctly. Have a good time everyone and thanks for not shooting me out of water yet.

johnreed

 

Hit wrong button

[swansont]

jr writes>

I apologize for the non standard vocabulary. I took a different path from day one and have had to learn the vocabulary of mainstream science as I go. But my initial arguments here are simple and involve the meaning of words like mass and the function of simple machines like a balance scale. Take the idea of a constant of proportionality. Where K is a constant of proportionality for T^2/r^3. I grasp that perfectly clear. G on the other hand is a constant of proportionality that purports to assign masses to planets based on our measure of mass here and Newton's basic assumption that since its true here its true there. Paraphrased. I think that our acceptance of G as a constant magnitude is shaky in this day and time. Einstein just in this case set what has been known and perplexing to us a a principle. It doesn't make G any the more correct. But I am getting way ahead of my little patch of ground I'm trying to carve out here. Yes all I am trying to say is apparently already accepted in part. That the planet attractor axts on atoms.

 

 

That applies to all formulas, though. We can't be sure of electrostatics, because k could be different. The fine structure constant could be different. But then one would have a hard time explaining many, many observations, because we see a consistency in atomic spectra from distant sources.

 

Add to this that a change in G over space implies a loss of translation symmetry, which implies that momentum is not a conserved quantity.

[johnreed]

That applies to all formulas, though. We can't be sure of electrostatics, because k could be different. The fine structure constant could be different. But then one would have a hard time explaining many, many observations, because we see a consistency in atomic spectra from distant sources.

 

Add to this that a change in G over space implies a loss of translation symmetry, which implies that momentum is not a conserved quantity.

 

jr writes> I have copied and pasted the above because it deserves A clear and comprehensive reply. I draw a distinction between linear and angular momentum that is critical here. Thanks for holding that rifle at bay. I will return soon.

johnreed

 

 

[johnreed]

jr writes> I have copied and pasted the above because it deserves A clear and comprehensive reply. I draw a distinction between linear and angular momentum that is critical here. Thanks for holding that rifle at bay. I will return soon.

johnreed

 

 

 

 

jr writes>

I apologize for the non standard vocabulary. I took a different path from day one and have had to learn the vocabulary of mainstream science as I go. But my initial arguments here are simple and involve the meaning of words like mass and the function of simple machines like a balance scale. Take the idea of a constant of proportionality. Where K is a constant of proportionality for T^2/r^3. I grasp that perfectly clear. G on the other hand is a constant of proportionality that purports to assign masses to planets based on our measure of mass here and Newton's basic assumption that since its true here its true there. Paraphrased. I think that our acceptance of G as a constant magnitude is shaky in this day and time. Einstein just in this case set what has been known and perplexing to us as a principle. It doesn't make G any the more correct. But I am getting way ahead of my little patch of ground I'm trying to carve out here. Yes all I am trying to say is apparently already accepted in part. That the planet attractor acts on atoms.

 

Swanson

That applies to all formulas, though. We can't be sure of electrostatics, because k could be different. The fine structure constant could be different. But then one would have a hard time explaining many, many observations, because we see a consistency in atomic spectra from distant sources.

 

jr writes> Atomic spectra is electromagnetic. The displacement of distant stars along our line of vision near the Sun was and as far as I know, still is, attributed to gravity (lensing). It is a possibility that if classical conserved object mass cannot be accurately proportioned dynamically to least action consistent star and planet masses niether can it be proportioned to the least action consistent sub atomic regions below protons, neutrons and electrons outside of a fill in for least action consistent equations proportioned to classical magnitudes of least action consistent mass.

 

Swanson

Add to this that a change in G over space implies a loss of translation symmetry, which implies that momentum is not a conserved quantity.

 

jr writes Where mass is the conserved cumulative resistance of planet and moon surface object atoms and is conserved independently of the celestial least action motion. Recall that we have spin angular momentum and linear momentum from Newton's first law. We don't have orbital angular momentum from that law. We acquire orbital angular

momentum from Newton's mathematical derivation for centripetal force where he used a perfect circle and uniform motion to argue for centripetal acceleration.

 

The spinning perfect circle angular velocity is an artifact of the uniformly spinning circle itself. The angular velocity of a spinning disk, sphere, or solid object, is an artifact of the

uniformly spinning disk, sphere, or solid. So we have least action consistent single object spin angular momentum in fact, and as an artifact of the spinning perfect circle angular velocity.

 

Newton then used the least action consistent angular velocity of Kepler's empirical time controlled law of areas for 2 body planet orbital motion, to mathematically carry his perfectly circular 2 body uniform motion, spin angular momentum analog, to the planet's

non-uniform but least action consistent 2 body orbital motion.

 

It's consistent with least action time-space parameters where the emergent conserved cumulative resistance of planet and moon surface atoms is either proportioned to (as the cause of) the least action consistent celestial motion (Newton's gravity), or as the consequence of the least action consistent motion, as space-time curvature (Albert Einstein and peers). This where planet surface object mass is independent of the celestial frame.

 

So the ideas for locally measured linear momentum magnitudes remain viable but our starting points to proportion local mass magnitudes from (planet surface object mass and G) do not necessarily apply beyond the frame of their origin.

 

It may be just an advanced form of "cargo cult" science, where we duplicate the least action consistent orbital motion (Kepler) using the force we apply and then assign that force to the entire universe because we can take it with us wherever we go.

Thanks for your time and effort and any corrections you can apply.

Have a good time.

johnreed

[johnreed]

jr writes> Talk about word salad. Who said we don't feel gravity? That is precisely what we feel gravity and we define it in terms of what we feel [mg] and call it force. It is force to us and it is equal and opposite to the resistance we encounter. That resistance is caused by the planet's attraction on atoms which is uniform for all atoms. That is what I say. Please quote me otherwise.

 

The above is the subject of my most recent posts. I can add direction to that substance in a very general way using few words. Please do not consider this paragraph a statement I am arguing at thr present time. But let conjecture that what we call gravity does not collapse into a black hole but instead the electromagnetic properties of atoms constructively combine to commandeer the core of the collapsed star to create an electromagnetic dynamo that attracts all form of matter at a certain distance and leaves shells of matter that coalesce into planets. The collapse is accompanied by an explosion that sends star parts everywhere. These star parts eventually return to their origin but their origin location is gon and so they pass closely around the star and are thrown back into their distant orbit. This is just conjecture for now and won't be entertained seriously until all my preliminary arguments are worked out so thy are clear to all. But I have always sought assistance.

johnreed

 

jrwrites> I am only guessing as to who this response is directed to. But here it is:"I have defined mass as the conserved comparative resistance of non-uniform (and uniform) atoms in response to and as a consequence of a uniform attraction on all atoms"

So, electrons don't have mass the?

 

jr writes> Electrons have not been shown to exist as objects inside atoms. Electrons do not remain free outside the atom except when contained by powerful electromagnetic fields. Since (g) in (mg) is a consequence of location (g) divides out of the balance scale comparison. This leaves a comparative resistance of atoms at any (g). We call this comparative resistance of atoms Mass. And we think it is the cause of celestial motion ie gravity. We think that what we feel (mg) is the cause of gravity where gravity is defined as what we feel (mg). Where (g) is a consequence of location and (m) is the conserved comparative resistance of atoms as measured on the balance scale. I'm just making sense out of non sense.

 

If this "The relative atomic weight of an atom expressed in grams as weight [mg] using the periodic chart represents one mole of that element." is the "plain English" version, can I try the ornate English version instead please?

 

"That number represents the number of atoms in a gram atom, or the gram atomic number of an element."

Nope, its a gram atomic weight.

 

jr writes> It may be called weight. But (g) divides out on balance. You measure weight because you cannot count the number of atoms. Since (g) divides out please tell me how it is that weight is the fundamental here.

 

jr writes> You can call it a gram atomic weight. This works because (g) as I noted above divides out of the equation. While you measure weight, the balance scale compares mass resistance. This relative resistance is quantified on the periodic table.

 

Here is a plainer English explanation of Avagadro's number.

 

Sometimes we find it easier to work with groups of items rather than individual ones.

For example we commonly buy eggs in dozens.

The same thing happens with atoms but, because they are very small, we use a very large number.

We could choose any big enough number , but the one we use is called Avagadro's number.

It's usually written as

6.0221415 x 10^23

That's about 6 followed by 23 zeros.

 

jr writes> You could at the very least spell Avogadro correctly.

 

The reason we chose that number is that it's how many hydrogen atoms there are in a gram of hydrogen.

 

jr writes> How many atoms are in a gram of Oxygen? Of helium? Of anyium? It appears that we could measure the number of atoms per gram in any weight system we choose. Whatever that system is will provide a common number of atoms to serve as a mole measure in chemical combination notation.

The measure of gram atoms does not change with location. The measure of weight does change with location. You can call what you do a measure of weight but what the balance scale is doing is comparing the relative resistance of atoms which incidentally does not vary even when you use your subjective notion of weight.

 

 

 

 

 

jr writes> Since my support of machines that produce more energy than our present conservation laws provide as possible has deemed me a speculator after I provide a rational argument on apparently unrelated topics... I will continue in the speculation department

 

Plain English Physics 101-4

Excerpts from

Published under new title

Modified Monday, May 21, 2012

johnlawrencereed jr

 

Excerpt on Newton:

 

In any event, our problem did not begin with J.J. Thompson. Some 2000 years after the Ancient Greeks, Tycho Brahe's careful observations on the behavior of celestial planetary bodies and Kepler's subsequent careful analysis of those observations revealed that the symmetry is in time and space. The predictable solar time-space least action consistent symmetry was subsequently co-opted by Isaac Newton, and used as the carrier for our tactile sense of attraction to the planet, quantified in terms of our least action consistent locally isolated (surface planet) "inertial mass" and regarded as the controlling cause of the order we observe in the celestial, least action consistent universe. This was heralded as Newton's great synthesis [*] and is so considered even today.

 

Isaac Newton defined centripetal force in terms of his second law to act at a distance by setting his first law planet surface object on an imaginary circular path of motion at a uniform orbital speed. Newton allowed his moving (planet surface like object) to impact the internal side of the circle circumference at equidistant points to inscribe a regular polygon. He dropped a radius to the center of the circle from each vertex (B) of the polygon to describe any number of equal area triangles.

 

"...but when the body is arrived at (B), suppose that a centripetal force acts at once with a great impulse".

 

Taking the length of each triangle base to the limit (approaching zero) the force vector [ma, mv/t, or dp/dt] at the vertex (B) is by definition directed along the radius toward the center of the circle as [mv^2/r][*].

 

Again, as with Ptolemy we have a perfect circle and perfect motion where here the law of areas clearly falls out as an artifact of the circle itself.

 

Note that Newton arbitrarily inserted inertial mass [m] into the least action consistent equation for circular planetary motion.

 

Newton generalized the equal areas in equal times artifact of the perfect circle uniform motion to any curved path directed radially around a point.

 

"Every body that moves in any curve line... described by a radius drawn to a point... and describes about that point areas proportional to the times is urged by a centripetal force... to that point"

 

Newton extended the property of his planet surface like orbiting object to all orbiting celestial bodies.

 

"Every body that by a radius drawn to the center of another body.. and describes about that center areas proportional to the times, is urged by a force.."

 

Newton then tied the force directly to the a priori force he felt and called gravity as [mg].

 

... "For if a body by means of its gravity revolves in a circle concentric to the earth, this gravity is the centripetal force of that body." Newton brought his notion of gravity to the mathematics as an a priori fundamental given. And as the cause of his planetary centripetal force

 

In short the force acted on any orbiting object as though that object is identical to Newton's first law planet surface object where the 2nd law force [ma] could then be proportioned to the areas and times of orbiting celestial bodies. Where here Newton arbitrarily inserted the locally independently derived quantity of inertial mass as [ma] into the equation proportioning the least action consistent celestial universe to the independent locally derived planet surface least action consistent quantity inertial mass [m] as [ma].

 

It has heretofore been a mystery to many as to why so called gravitational mass as [mg] is equal to inertial mass as [ma]. That is how Newton defined it.

 

The result is that the least action consistent universe is defined in terms of what we feel as [mg] and [ma] as least action consistent objects. How's that for a centrist view?

 

All that was supposedly left to do to calculate the mass of planets was to acquire a constant of proportionality [*] based on interactions between planet surface inertial mass objects and the planet using the mathematics supplied by Newton.

 

This was eventually determined by Henry Cavendish by measuring minute magnitudes of torque from hanging balls that twisted a wire that could just as easily and better be explained by unseen electromagnetic causes.

 

Rather than delete it let me provide the above to further embarass my kids and their kids etc. If I am responding out of accepted protocol please advise. Thanks.

johnreed

 

 

 

Ophiolite, on 27 June 2012 - 11:16 PM, said:

Good advice. You could always ask the admins to delete your post. Only a few of us saw it.

In Topic: DO objects fall at the same speed? NO!

1 November 2011 - 11:11 AM

 

jr wrote "If [g] in the expession [mg] for weight is a consequence of location then all atoms must fall at the rate of [g] at that location. Forget pressing air. Think vacuum."

 

 

Moneypoo questioned the meaning and DrRocket questioned the effectiveness of my reply. So I returned to my quote above. Altho it is based on the balance scale... it also requires a vacuum. My response was off point. Thank you.

In Topic: DO objects fall at the same speed? NO!

31 October 2011 - 01:18 PM

Axioms, on 31 October 2011 - 04:56 AM, said:

I think you missed the point of what I said.

_____________________________________________________________________

You say that mass is not responsible for gravity. You also say that mass does not generate the force of attraction.

 

jr writes> I do say that mass does not generate the force of attraction. Gravity is a force that we feel which force is defined as F=mg. I agree that we feel (mg). And that we feel the force we call gravity.

 

If we say that Fg=mg what exactly is generating the force of gravity?

 

jr writes> Earlier I did not follow Fg=mg What exactly is generating the force of gravity? We generate the force we call gravity when we act on the cumulative resistance of atoms

 

Why must there be a field of force equal and opposite to a force we generate in response to a resistance?

 

There must be a gravitational field but what generates this?

 

jr writes> I appreciate your "There must be a ..................... field.

But its not a field we generate.

 

We know from Newtons laws that Fg= G*m1m1/d^2.

 

jr writes> Which I would write as mg=GMm/r^2. Where when we divide m out we have g=GM/r^2. And from this we think we can compute the mass of plaets and stars.

 

The propotional constant G= 6.678*10-11 is known as the universal gravitational constant. This force tends to pull objects towards each other.

 

jr writes> I can show that the planet attractor uniformly acts on uniform and non-uniform atoms. I can also show that mass is the name we give to the conserved comparative resistance of uniform and non-uniform atoms.

And I can show that F=mg and F=ma and that they are quivalent because we have defined both in units we happen to feel. And I can show that Newton defined gravitational force consistent with the units we feel

 

If we manipulate these equations we find that g= fg/m = GMm/md^2 = GM/d^2. Let us add values. g = (6.67*10^-11)(5.98*10^24)/(6370*10^3)^2 = 9.83m/s^2. These values are the mass of the Earth and the distance of the center of mass to the surface of Earth. If you change the mass of the object you see that the strength of the gravitational field changes. This illustates that the force of gravity is dependent on the mass of the object.

 

jr writes> If the force of gravity is equivalent to the force we feel which is how Newton defined it. Then we can define the least action consistent motion in the universe proportioned to the force we generate and this will work for us anywhere in the universe we can occupy.

 

It is easy to think of examples. When we landed on the moon, which has less mass than the Earth, its gravitational field is much weaker than here on Earth. If we go to visit the sun the gravity will crush us. If mass did not aid in determining the strength of gravity then theoretically everything would have the same gravitational pull.

 

jr writes> I did not say that mass is not significant. Nor that it did and does not aid us immensely. In fact it aided us so well that we never sought a precise definition for mass. I think the accepted definition is an "amount of matter". Atoms are also amounts of matter.

 

If we think in terms of Einstein he illustrated that mass bends space.

 

jr writes>

If you mean that the displacement of stars occurs along our line of vision when that line is near the eclipsed Sun, calling the displaced line of vision a consequence of gravity just carries forward our primitive belief that we can define the universe in units that we feel.

 

If you want to go into an argument about what gravity really is we can. If we deny that mass does not play a fundamental role in determining the force of attraction between objects we are blind.

 

Mass is a convenient term that represents the comparative resistance of atoms on the balance scale (the simplest example), It is an independent measure of force (mg). At any place we can use a balance scale the quantity g divides out, and on balance this clearly leaves a comparative resistance of atoms on each scale.

 

You say we aquired mass initially from the definition of [mg] and [ma]? How exactly if we already have m in the equation? You cannot have a force if there is no mass? Mass is determined atomically and we can relate a large mass to have a center of mass but this is a different topic.

 

jr writes> I will continue asap with the rest of this. Have a good time.

johnreed

 

 

jr writes: I have no argument with this. We can also say that mg=GMm/r^2 where [m] divides out leaving g=GM/r^2. That's fine. But where do you get it that [M] divides out? And once you divide [m] out all you are left with is least action consistent motion. But I am not arguing this right now. In fact I've got to revisit my original few paragraphs to remind myself what I am arguing for. Thank you for not shooting me out of the water.

 

 

 

Jr writes>

I might be getting my sea legs here a bit better. No what I say is that the planet acts on all atoms uniformly, That allows us to measure the resistance of each pan of atoms in a conserved unit we call mass. In fact if the planet did not act on all atoms uniformly and instead acted on object mass we could never have evolved. I apologize for my non-standard terminology. Resistance has a specific meaning in physics and force does as well. We act on resistance and we feel a force that is equal and opposite to the resistance we act on anywhere in the universe we can occupy. But the force is initiated by us and is not a property of inanimate objects.

johnreed

 

 

 

jr writes> Rereading this comment you are taking freefall as an example where we don't feel gravity. I agree. In freefall we are traveling in the direction our atoms are being pulled. However when we accelerate against the pull on our atoms we feel this pull and we call it gravity. When we are in contact with a non-freefalling object we also feel the pull on our atoms. Sorry I think I mis read your original paragraph. People do it all the time.

 

 

 

[ACG52]

Very large word salad.