DO objects fall at the same speed? NO!

[johnreed]

I'm not sure if anyone spoke about drag. I just skimmed through the posts. The question did include air resistance though because he stated "know that objects with the same surface area but different densities fall at the same speed in a vacuume" and he did include air resistance in his answer. The paper and cardboard (same shape and different mass) will have different terminal velocities. It is wrong in saying that heavier objects are less affected by air resistance but he has the right idea, kind of.

 

Terminal velocity is where drag is equal to the force of gravity. Fg_y=ma. It requires less drag force for the lighter object to reach terminal velocity.

 

proof: v_paper = sqrt(2fg/CpA) (C is the drag coefficient p is the air density and A is the cross-sectional area of the object) (Drag equation manipulated from D=0.5*CpAv^2 and D-Fg=ma)

= sqrt(2(9.8*1*10^-3)/0.5*1*(20*45)) (not sure what A or what p is and cannot be bothered to measure but is meaningless if keep constant)

=6.6*10^-3 m/s

the only thing in the equation that changes is fg so lets say the cardboard is 10* the mass.

v_cardboard = 0.021 m/s

 

Please note the values I used are not accurate but does not make the calculations wrong. In both equations the only thing that changes is fg so the values are constant in these circumstances.

 

The objects do accelerate at the same speed initially but the velocity at which they hit the ground will be different. It is dependent on the surface area of the object though. If you take the paper and cardboard and drop it where the maximum SA is expossed and then compare it to where the smallest SA is expossed you will see the difference in results.

 

 

 

I disagree to a extent. It is related to the mass in the form of force. Lets say the paper, cardboard and steel were all 100% structually sound. We now drop it off a building where all the dimentions of the maximum SA acts towards the ground. The terminal velocity can be determined of each. Drag - Fg =ma. This is the amount of force needed for the air resistance to counteract the force of gravity. Lighter objects will require much less force by air resistance to reach terminal velocity.

Read my last post.

 

These posts all reflect our ancient beliefs.

 

We used the balance scale to give us weight for thousands of years and for thousands of years we believed that heavy objects fall faster than lighter objects. So when Galileo showed that all objects fall at the same rate when dropped at the same time from the same height we were surprised and we have remained disturbed for 400 years. The problem revolves around what we think is fundamental. A force that we feel and measure and call gravity. We still think we are comparing weight on the balance scale... and we are... but that is superficial. Weight is subject to location.

 

Where we place a balance scale is immaterial to the function of the balance scale (as long as it is placed in a frame within which it can operate). Wherever we place it the magnitude of the acceleration [g] as a factor of the product weight [mg], will be the same on each pan regardless of the mass magnitudes placed on the pans. We calibrate the machine with empty pans or identical weights. So when we define an object in units of weight [mg], the only quantity we are comparing on the balance scale is the quantity of mass [m], since acceleration [g] is a consequence of location, it allows us to calibrate the scale. This uniform action [g] on each pan enables us to compare non-uniform mass resistance on the balance scale. That's pretty simple isn't it? So you might wonder why I bother to point it out.

 

Since what is called gravitational acceleration [g] is a consequence of location and not a consequence of mass [m] (as our tactile sense informs us) all objects MUST fall at the rate of [g]. We should investigate why this is true [1]. We need not waste effort in validating its accuracy. Rather let us determine why it is so.

 

Further, if [g] was not a consequence of location then mass [m] and acceleration [g] could not be combined into the product called weight [mg]. In such a case the balance scale would only give us weight as [w]. However if that were the case we could never have developed.

 

Nonetheless we have engaged in extensive research to verify that all objects fall at the same rate, independent of their mass [m] when dropped at the same time from the same height (in a vacuum). We need not argue that the rate of fall is [g]. Do we need to argue that [g] is a consequence of location? Maybe. If so let us begin. If we agree that [g] is a consequence of location the next question is why.

 

Endnote

[1]

I have made it easier to reference my supporting work by creating a Google Science and Technology Group titled: "The Least Action Consistent Universe and the Mathematics". Currently it contains Sections 1 through 9 for reference. The many sub-sections and work prior to 2007 has not been included. I will develop it further as I have the time and gain familiarity with the venue. Meanwhile my more recent work is available for public review to all, and open to criticism and discussion by any person who joins the group. The latter is a condition established by Google and newsgroups in general. I provide information. I seek no recruits. However, there are no restrictions or requirements to join.

 

Current web address: http://groups.google.com/group/thejohnreed

johnreed

[Axioms]

These posts all reflect our ancient beliefs.

 

We used the balance scale to give us weight for thousands of years and for thousands of years we believed that heavy objects fall faster than lighter objects. So when Galileo showed that all objects fall at the same rate when dropped at the same time from the same height we were surprised and we have remained disturbed for 400 years. The problem revolves around what we think is fundamental. A force that we feel and measure and call gravity. We still think we are comparing weight on the balance scale... and we are... but that is superficial. Weight is subject to location.

 

Where we place a balance scale is immaterial to the function of the balance scale (as long as it is placed in a frame within which it can operate). Wherever we place it the magnitude of the acceleration [g] as a factor of the product weight [mg], will be the same on each pan regardless of the mass magnitudes placed on the pans. We calibrate the machine with empty pans or identical weights. So when we define an object in units of weight [mg], the only quantity we are comparing on the balance scale is the quantity of mass [m], since acceleration [g] is a consequence of location, it allows us to calibrate the scale. This uniform action [g] on each pan enables us to compare non-uniform mass resistance on the balance scale. That's pretty simple isn't it? So you might wonder why I bother to point it out.

 

Since what is called gravitational acceleration [g] is a consequence of location and not a consequence of mass [m] (as our tactile sense informs us) all objects MUST fall at the rate of [g]. We should investigate why this is true [1]. We need not waste effort in validating its accuracy. Rather let us determine why it is so.

 

Further, if [g] was not a consequence of location then mass [m] and acceleration [g] could not be combined into the product called weight [mg]. In such a case the balance scale would only give us weight as [w]. However if that were the case we could never have developed.

 

Nonetheless we have engaged in extensive research to verify that all objects fall at the same rate, independent of their mass [m] when dropped at the same time from the same height (in a vacuum). We need not argue that the rate of fall is [g]. Do we need to argue that [g] is a consequence of location? Maybe. If so let us begin. If we agree that [g] is a consequence of location the next question is why.

 

Endnote

[1]

I have made it easier to reference my supporting work by creating a Google Science and Technology Group titled: "The Least Action Consistent Universe and the Mathematics". Currently it contains Sections 1 through 9 for reference. The many sub-sections and work prior to 2007 has not been included. I will develop it further as I have the time and gain familiarity with the venue. Meanwhile my more recent work is available for public review to all, and open to criticism and discussion by any person who joins the group. The latter is a condition established by Google and newsgroups in general. I provide information. I seek no recruits. However, there are no restrictions or requirements to join.

 

Current web address: http://groups.google...oup/thejohnreed

johnreed

 

I did not deny that gravity effects all objects on earth at the same rate. You used the term in a vacuum where there is no drag. I did specify that I am working outside a vacuum. If I were working in a vacuum then they would all have exactly the same rate of decent of around 9.8m/s^2. There would be no force to oppose the force of gravity.

 

The main question asked if objects fell at the same speed when outside a vacuum. [g] is the gravitational acceleration that an object experiences. Speed is related to velocity, although [g] still acts the same on both objects, the objects speeds will differ. If this were not true then somebody jumping out of a plane with a parachute would meet a terrible fate. A lot of research has gone into the study of drag which does rely on [g].

 

I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics.

 

A vacuum scenario only illustrates a part of physics which proves that objects experience the same gravitational acceleration. It has been further developed to explain why we can jump out of a plane with a parachute and reach the ground with a constant terminal velocity that we can survive. If you disagree with this then take a peice of cardboard and paper with the same dimentions and drop them. I did say that SA of the objects is a main component when determining drag. When determining how much force needs to be applied by air resistance to let the objects reach terminal velocity is dependent on its mass.

 

We deal with more than the force of gravity on Earth. There are many forces that we interact with and we need to take this into consideration. If we look at a specific case of falling in a vacuum it is impractical. If you can prove to me that mass does not play a role in determining the speed of an object that is dropped outside of a vacuum I will be impressed.

 

Please note that I did not deny [g] but am mearly stating that Fg=mg is the amount of force needed to act against the falling object upwards to cause it to reach terminal velocity. I stand by my comment that heavier objects will fall faster than lighter objects when comparing their terminal velocities. If you compare the amount of force needed to make a heavier object reach terminal velocity to that of a lighter object there will be a big difference. The force, SA, air density and drag coefficients will determine how long it would take an object to reach its terminal velocity.

 

I did specify that in a vacuum they will have the same acceleration and velocity. I did not write all of this for a vacuum, I include air resistance. The study of aerodynamics would be meaninless if we only considered objects falling in a vacuum.

[johnreed]

I did not deny that gravity effects all objects on earth at the same rate. You used the term in a vacuum where there is no drag. I did specify that I am working outside a vacuum. If I were working in a vacuum then they would all have exactly the same rate of decent of around 9.8m/s^2. There would be no force to oppose the force of gravity.

 

The main question asked if objects fell at the same speed when outside a vacuum. [g] is the gravitational acceleration that an object experiences. Speed is related to velocity, although [g] still acts the same on both objects, the objects speeds will differ. If this were not true then somebody jumping out of a plane with a parachute would meet a terrible fate. A lot of research has gone into the study of drag which does rely on [g].

 

I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics.

 

A vacuum scenario only illustrates a part of physics which proves that objects experience the same gravitational acceleration. It has been further developed to explain why we can jump out of a plane with a parachute and reach the ground with a constant terminal velocity that we can survive. If you disagree with this then take a peice of cardboard and paper with the same dimentions and drop them. I did say that SA of the objects is a main component when determining drag. When determining how much force needs to be applied by air resistance to let the objects reach terminal velocity is dependent on its mass.

 

We deal with more than the force of gravity on Earth. There are many forces that we interact with and we need to take this into consideration. If we look at a specific case of falling in a vacuum it is impractical. If you can prove to me that mass does not play a role in determining the speed of an object that is dropped outside of a vacuum I will be impressed.

 

Please note that I did not deny [g] but am mearly stating that Fg=mg is the amount of force needed to act against the falling object upwards to cause it to reach terminal velocity. I stand by my comment that heavier objects will fall faster than lighter objects when comparing their terminal velocities. If you compare the amount of force needed to make a heavier object reach terminal velocity to that of a lighter object there will be a big difference. The force, SA, air density and drag coefficients will determine how long it would take an object to reach its terminal velocity.

 

I did specify that in a vacuum they will have the same acceleration and velocity. I did not write all of this for a vacuum, I include air resistance. The study of aerodynamics would be meaninless if we only considered objects falling in a vacuum.

 

The question itself requires more definition. Of course air resistance affects the rate of fall. Terminal velocity is a function of the atmosphere we fall through and the surface area of the object it self. This puts an entire different question up. It would be beneficial to determine why objects fall at all. What attracts objects? You seem to believe that "gravity" attracts objects and "gravity" is the controller of all objects with mass. Like mass generates a force all its own. Admittedly there is a force that attracts objects. But mass does not generate that force. We acquire mass initially from the definition [mg] and [ma]. Where mass is conserved. Both [mg] and [ma] are subjective measurements of force as we have defined it. [F=mg] and [F=ma}. Where [mg] and [ma] can be objectively measured, calling them force is universalizing what we feel and because what we feel can be quantified as [mg] and [ma] does not justify a force beyond that which we feel and have set equivalent to objective measurements of "resistance".

 

Axioms wrote> "I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics."

 

This assumes that the attraction is caused by mass. So the further mass is from another mass the weaker the attraction. But you are treating mass as equivalent to matter. Where mass is a measure of resistance and while we feel an equal and opposite resistance to the force we apply, this does not make the resistance a function of what we feel, namely force. And because we can calculate the objective quantities we feel [mg] and [ma] and call them force does not make the universe a function of the force we feel. Both [mg] and [ma] function within the parameters of least action motion. The mathematics functions best with least action motion. So as long as we have quantities like mass [m] and acceleration [a] and [g] and velocity as [v], all of which function within least action motion quantitatively still goes awry because we treat these quantities as though they represent a force beyond the force we apply. And we use mass as a proportional amount of matter beyond its origin which is in classical planetary surface object action.

 

So when we lift an object we say that the planet applies an equal and opposite force that we feel. Whereas the object we lift has a resistance that we can quantify in units of [mg] anywhere we apply the force we feel. The planet exacts a uniform attraction on non-uniform atoms. We define this action as gravitational force where Einstein dealt with it as a uniform gravitational field, still relying on the force we apply to the resistance we work against, to define that resistance we feel as gravity.

 

[F=mg] can also be written as [F=Nnmgx] where [N]=Avogadros Number and [n] = the number of moles and [mg] = the relative weight of a single atom [x]. This applies to the measure of a single isomer of an element but the principle can be applied to all elements. The action we call force then becomes a function of the resistance of a number of atoms where the attraction is uniform on non-uniform atoms. And what we feel is precisely defined.

 

This is much more general of an answer to the original question but it helps to precisely define the terms we use. Have a good time.

johnreed

[DrRocket]

Saying all 3 will fall at the same rate in air is stupid because empirically we can prove this is not true. Saying air resistance as a function of surface area is the only opposing force to gravity is wrong because all 3 have the same surface area.

 

You have missed the point, and the effect of bending in bchanging the drag coefficient is both secondary and unimportant. The bending of a sheet of paper will reduce the drag coefficient, but the steel plate will still fall faster in air.

 

The net downward force on the falling square will be, to a standard ballistic approximation

 

[math] f= mg - C \rho v^2A[/math]

 

where [math]A[/math] is the surface area presented [math]\rho [/math] is the air density and [math] C[/math] is a shape dependent drdag coefficient which can be taken as constant over the small range of velocities that will be encountered in this situation (all speeds are well below mach 1 for instance). Neglecting any bending the constant [math]C[/math] will be the same for paper, cardboard or steel.

 

So the acceleration will then be

 

[math] a = \frac {f}{m} = g - \frac {\rho v^2 A}{m}[/math]

 

which increases with increasing [math]m[/math] showing clearly why steel falls faster than paper or cardboard in air -- assuming that you don't make the steel so thin that its mass is not appreciably greater than that of the cardboard or paper sheets.

Edited October 31, 2011 by DrRocket

[johnreed]

Sorry,what do you mean? g is acceleration, and in this context of newtonian mechanics it's constant acceleration regardless of location... Also, are you suggesting the atoms are falling = the air is falling?

 

I'm not quite clear as to what you're saying.... can you explain?

 

Hello: I mean that barring theoretical locations based on our notion of gravitational force like area in around and near theoretical gravity based black holes, wherever we place a balance scale the factor [g] in the product [mg] will be identical on both balance pans. [g] itself can vary on the moon and other planets and it can vary on planet Earth depending on location. However in any case [g] will be constant at any given location that we can balance objects. However [g] acts on the balance scale as well and both pans. So it is uniformly acting on all objects we measure on a balance pan. [g] is therefore a consequence of location and varies according to location but yes it is constant at any given location. However it is different in magnitude on the moon for example where mass remains the same magnitude anywhere you can use the balance scale. So the magnitude of [g] is a consequence of location. Therefore all objects must fall at [g]. So the rate of fall depends on [g] and not on mass. Yes heavier objects will fall faster than feathers any place an atmosphere can influence the rate of fall. As far as atoms falling goes all atoms fall at the same rate but the atmosphere can remain in layers according to relative mass and density like a cork will float on the water where a rock will sink. All atoms fall but some atoms fall farther than others. The air has already ascended to its proper level like a cork released at the bottom of the sea will rise to the top. Weight is the human measure they acquire from the balance scale but the balance scale only compares mass as a function since [g] is a consequence of where the measure of mass is taken. I am trying not to wave my hands here. Does this suffice to explain it. Thank you for the question.

[DrRocket]

I am trying not to wave my hands here.

 

Better try harder.

[Axioms]

The question itself requires more definition. Of course air resistance affects the rate of fall. Terminal velocity is a function of the atmosphere we fall through and the surface area of the object it self. This puts an entire different question up. It would be beneficial to determine why objects fall at all. What attracts objects? You seem to believe that "gravity" attracts objects and "gravity" is the controller of all objects with mass. Like mass generates a force all its own. Admittedly there is a force that attracts objects. But mass does not generate that force. We acquire mass initially from the definition [mg] and [ma]. Where mass is conserved. Both [mg] and [ma] are subjective measurements of force as we have defined it. [F=mg] and [F=ma}. Where [mg] and [ma] can be objectively measured, calling them force is universalizing what we feel and because what we feel can be quantified as [mg] and [ma] does not justify a force beyond that which we feel and have set equivalent to objective measurements of "resistance".

 

Axioms wrote> "I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics."

 

This assumes that the attraction is caused by mass. So the further mass is from another mass the weaker the attraction. But you are treating mass as equivalent to matter. Where mass is a measure of resistance and while we feel an equal and opposite resistance to the force we apply, this does not make the resistance a function of what we feel, namely force. And because we can calculate the objective quantities we feel [mg] and [ma] and call them force does not make the universe a function of the force we feel. Both [mg] and [ma] function within the parameters of least action motion. The mathematics functions best with least action motion. So as long as we have quantities like mass [m] and acceleration [a] and [g] and velocity as [v], all of which function within least action motion quantitatively still goes awry because we treat these quantities as though they represent a force beyond the force we apply. And we use mass as a proportional amount of matter beyond its origin which is in classical planetary surface object action.

 

So when we lift an object we say that the planet applies an equal and opposite force that we feel. Whereas the object we lift has a resistance that we can quantify in units of [mg] anywhere we apply the force we feel. The planet exacts a uniform attraction on non-uniform atoms. We define this action as gravitational force where Einstein dealt with it as a uniform gravitational field, still relying on the force we apply to the resistance we work against, to define that resistance we feel as gravity.

 

[F=mg] can also be written as [F=Nnmgx] where [N]=Avogadros Number and [n] = the number of moles and [mg] = the relative weight of a single atom [x]. This applies to the measure of a single isomer of an element but the principle can be applied to all elements. The action we call force then becomes a function of the resistance of a number of atoms where the attraction is uniform on non-uniform atoms. And what we feel is precisely defined.

 

This is much more general of an answer to the original question but it helps to precisely define the terms we use. Have a good time.

johnreed

 

 

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. If we say that Fg=mg what exactly is generating the force of gravity? There must be a gravitational field but what generates this? We know from Newtons laws that Fg= G*m1m1/d^2. The propotional constant G= 6.678*10-11 is known as the universal gravitational constant. This force tends to pull objects towards each other. 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. 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.

 

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

 

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.

 

This is so far off the topic where heavier objects will fall faster than lighter objects on Earth outside a vacuum. The question was very specific and it was defined properly. I don't understand why you have ventured off in this direction. I dont know what you are trying to argue here but picture this situation:

 

You are sitting in a test. You are given the mass of the paper sheet, cardboard sheet and steel sheet with their dimensions being equal. They state it is not in a vacuum. They tell you it is dropped from 50m above the ground. Which will hit the ground first? Explain why this is the case?

 

Who has the right answer here? I can prove why the steel sheet hits the ground first with mathematics (I have above in my previous post). Your initial statement suggested that they hit the ground at the same time. This is based on the fact that the gravitational field has equal effects on the objects so it must be true that their speeds are equal? Wrong. You need to consider what the question implys and answer accordingly. If you venture off into trying to define what makes it fall to the ground you have missed the point of the question. It is a practical question not theoretical one.

[johnreed]

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. If we say that Fg=mg what exactly is generating the force of gravity? There must be a gravitational field but what generates this? We know from Newtons laws that Fg= G*m1m1/d^2. The propotional constant G= 6.678*10-11 is known as the universal gravitational constant. This force tends to pull objects towards each other. 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. 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.

 

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

 

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.

 

This is so far off the topic where heavier objects will fall faster than lighter objects on Earth outside a vacuum. The question was very specific and it was defined properly. I don't understand why you have ventured off in this direction. I dont know what you are trying to argue here but picture this situation:

 

You are sitting in a test. You are given the mass of the paper sheet, cardboard sheet and steel sheet with their dimensions being equal. They state it is not in a vacuum. They tell you it is dropped from 50m above the ground. Which will hit the ground first? Explain why this is the case?

 

Who has the right answer here? I can prove why the steel sheet hits the ground first with mathematics (I have above in my previous post). Your initial statement suggested that they hit the ground at the same time. This is based on the fact that the gravitational field has equal effects on the objects so it must be true that their speeds are equal? Wrong. You need to consider what the question implys and answer accordingly. If you venture off into trying to define what makes it fall to the ground you have missed the point of the question. It is a practical question not theoretical one.

 

 

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

 

jr writes> Gravity [mg] is a force that we feel. Mass [m] is a conserved comparative resistance of "objects" or "bodies" or "blobs" we can quantify. If our definition of mass derives from the force that we feel [mg] then objects, bodies and blobs are all equally specific.

 

 

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

 

jr writes> If you are saying that [Fg=mg] you are saying that [mg^2=mg]. How do you define [g] as unity? On what basis?

 

There must be a gravitational field but what generates this?

 

Jr writes> Your logic for there "must be" a field based on what we feel, is over my head. Since all "blobs" fall at the same rate then atoms also fall at the same rate. It is difficult to weigh one atom on the balance scale at a time so we use Avogadro's number and moles to do the job. So atoms are legitimate substitutes for bodies, objects and blobs. If the attraction is on atoms then our gravitational field becomes a construct we use to navigate through a type of electromagnetic field that we don't feel in the way we feel the electromagnetism we are familiar with which we directly feel when we are being electrocuted. Such a field can derive from a form of electromagnetism that acts on all blobs rather than only objects with internally and externally alligned atoms. It depends on the internal dynamics of stars and some planets and maybe some objects that qualify as moons. Which dynamics as we have projected them are based on the force we feel and call gravity.

 

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

 

jr writes> Please explain the expression [mg^2].

Newton is my main idol and Feynman is number 2. So it is with great defernce and respect that I point out the following. Newton's first law gave us linear and single object spin angular momentum. It did not give us two body orbit angular momentum. Newton derived two body orbital angular momentum by applying perfectly circular spin angular momentum to Kepler's law of areas. They are both least action consistent and rely on our measure of comparative mass which with respect to Newton conjectured "If it is true here it is true everywhere". This was in the time of Dante' and Newton was a believer.

 

Newton's third law gave us the equal and opposite idea for force. Since what we lift can be quantified as resistance [mg] and is equal to a force we feel [F] by definition [F=mg]. We can call the force we feel [mg] a force that is generated by the planet. However we are alive and animate and have the propensity to "feel" through our tactile sense. The planet feels nothing I assume. So the planet can be acting uniformly on non-uniform atoms which we exert an effort to lift.

 

The propotional constant G= 6.678*10-11 is known as the universal gravitational constant.

 

This constant of proportionality is taken from the comparative measure of the behavior of planet surface object mass. Initially using a device that measured the torque between two planet surface objects. All natural motion is least action consistent. Mass operates independent of celestial object motion so it can be applied to that motion with impunity based on our conjecture that if it is true here it is true there. I give you that. In fact I already gave you that.

 

This force tends to pull objects towards each other. 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.

 

jr writes> If I accept the notion that if it's true here it's true there I would have to agree altho' I still have difficulty with [mg^2/m = [g]. Again please explain.

 

This illustates that the force of gravity is dependent on the mass of the object.

 

jr writes> I have shown on the balance scale that [m] and [g] are distincly separate where [g] controls and [m] obeys without effect until collision which is where we come into it.

 

It is easy to think of examples. When we landed on the moon, which has less mass than the Earth, its gravitational fie

ld 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 did not aid in determining the strength of the force that we feel. I did say that gravity is a force that we feel and that mass is the comparative measure of pans full of non-uniform atoms. So that the attraction is a uniform attraction on non-uniform atoms which is a more massive form of electromagnetism that acts comparatively weakly on ALL atoms. Einstein worked out a notion for a uniform gravitational field based on our perception of visual events.

 

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

 

jr writes> He predicted that light bends near large objects and the Englishman (his name is momentarily lost to me) eagerly confirmed the displacement. Our line of vision is easily displaced by many things. Rising heat from the desert displaces our horizontal vision but has no effect on our vision from space. On the other hand strong electromagnetic fields can displace our line of vision in a quantifiable manner. Attributing strong electromagnetic fields to the force we feel is just another form of defining the universe after our own image.

 

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.

 

jr writes> I won't argue on paper. I will explain on paper.

 

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 said that mass is quantifiable and derived from the objective quantifiable measure of [mg]. This was long before atoms were an accepted item which were not accepted until Einstein's paper on Brownian motion coupled with his 4 other paper in 1905. Mass was appropriated to and not originated from atomi science.

 

This is so far off the topic where heavier objects will fall faster than lighter objects on Earth outside a vacuum. The question was very specific and it was defined properly. I don't understand why you have ventured off in this direction. I dont know what you are trying to argue here but picture this situation:

 

You are sitting in a test. You are given the mass of the paper sheet, cardboard sheet and steel sheet with their dimensions being equal. They state it is not in a vacuum. They tell you it is dropped from 50m above the ground. Which will hit the ground first? Explain why this is the case?

 

jr writes>

The steel will of course hit the ground first. It is not subject to wind drift like the paper and cardboard Nor will it veer very much from its straight path and it might even hit the ground at the same time as a steel sheet thrown like a frisbee at 50m. I use the balance scale to demonstrate my point. It operates the same in or out of a vacuum. I suspect that it might also have a limited function at the bottom of the sea.

 

Who has the right answer here? I can prove why the steel sheet hits the ground first with mathematics (I have above in my previous post). Your initial statement suggested that they hit the ground at the same time.

 

jr writes>

I was not refering to such a modified specific set of circumstances. "Do all objects fall at the same rate? No." was the question and answer I attempted to address.

 

This is based on the fact that the gravitational field has equal effects on the objects so it must be true that their speeds are equal? Wrong.

 

jr writes> I was not considering a "gravitational field". I do not recognize the reality of such a field. It is functional for us and enables us to navigate the universe as we understand it but it is subject to a force we "feel" and one we have assigned to inanimate matter by defining the objective quantifiable resistance we act on [mg] as a generated force rather than a "resistance" that we apply a generated force to. We feel the force we generate. Inanimate objects feel nothing. A moving resistance has momentum. A moving resistance has force. These quantites function mathematically as [mv] and [ma or mg]. We live in a least action consistent universe where comparative resistance [mass] is independently conserved in the classical planet surface frame. So we think that it is proportional in the celestial frame to the magnitudes we measure locally (refering to areas on celestial surfaces and the space between all the celestial objects).

 

I chose a balance scale to show that [g] is a consequence of location where mass is a measure of comparative resistance and not a generator of force. The balance scale does not require a vacuum. It requires any location within which it can function. We use a balance scale to compare weight [mg]. Weight is no more objective than the word "heavy" or the word "effort". However the mathematical formulation for weight is quantitatively effective because we work against resistance which varies according to mass and its rate of travel be it [g], [a] or [v] or just straight speed [s/t]. The balance scale compares the resistance of objects at any location it can operate. We call this mass but we think it generates and derives from force or weight which is what we feel.

[johnreed]

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.

[matty]

The presence or absence of a vacuum does not affect the gravitational acceleration. Two objects in a vacuum will fall with the same acceleration. It's been tested.

 

Here's a crude experiment

http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html

 

The list of tests of the weak equivalence principle is fairly long

http://en.wikipedia.org/wiki/Equivalence_principle#Tests_of_the_weak_equivalence_principle

 

Apparently, given all of the forces acting on these experiments, i.e., leakage and a host of other issues, they're all pretty crude.

 

I know you guys have a habit of indulging in the supportive nature of fondling generalities but I'm a lover of specifics. If you break away from the comfort zone in generalities, this becomes far more spurious a debate, and, yes, this to date was brought in as an issue for and carried out as a debate and not a 'discussion,' however lofty that sounds.

 

 

That's not the vacuum we're talking about, matty.

 

http://www.oregon.gov/ODA/MSD/mass.shtml

 

I found it curious a device had been conjured which renders all objects in any space the same, so I did want to know about it if there was one, it's very interesting, indeed, in that case. But an impartial reader following through these passages in the link above certainly gets another, much less cut and dried impression, all the way to the paragraph, "...Thus, the lifting force on the object and that on the standard are not the same and the object will no longer appear to have the same mass as the standard. The mass that the object now appears to have is the 'conventional mass'..."

 

And, anyway, if we're measuring "true mass", just how null and void can it all have been rendered to begin with??

 

Worse yet, when I dug into that link, no offense intended at all, it isn't your own, you were just trying to catch me up, I getit--but I got a good giggle when in the link I read that generally liquids can't be pulled. Heh, Guess they weren't aware of modernday hydraulics at the time--or your basic medical syringe--or even a plunger??, now that one's just a shame, I swore we might be able to expect to find any one of these applications in some context in science venues abound...

 

 

 

 

 

!

Moderator Note

Don't take the discussion in that direction. Asking questions and learning is what this forum is for.

 

I didn't even read dude's post there, but I thought Mods weren't to be too eager to take action within the context of threads they're participating in??

 

!

Moderator Note

The first thing to get the hang of is the rules.

 

Lol, too funny, maybe you oughta take a gander back at them yourself, it seems Ophiolite has a more respectable handling on it all than a couple of you do, eh. Nice, that's some posturing under the haughty guise of 'scientific' pursuit... Hmf~

Edited November 8, 2011 by matty

[swansont]

I didn't even read dude's post there, but I thought Mods weren't to be too eager to take action within the context of threads they're participating in??

If you had, then you might have realized that the posts were outside of the context of the thread. I had not interacted with that poster in this thread, so there should be no appearance of any conflict of interest.

 

In any event, I am never eager to have to intervene, because that means a rule has been broken. Once it's required, though, I will do so.

[Klaynos]

!

Moderator Note

I would strongly recommend remaining on topic and reviewing the forum rules.

Comments and replies to this modnote or previous ones will be removed.

[swansont]

http://www.oregon.gov/ODA/MSD/mass.shtml

 

I found it curious a device had been conjured which renders all objects in any space the same, so I did want to know about it if there was one, it's very interesting, indeed, in that case. But an impartial reader following through these passages in the link above certainly gets another, much less cut and dried impression, all the way to the paragraph, "...Thus, the lifting force on the object and that on the standard are not the same and the object will no longer appear to have the same mass as the standard. The mass that the object now appears to have is the 'conventional mass'..."

 

And, anyway, if we're measuring "true mass", just how null and void can it all have been rendered to begin with??

 

That's a metrology link and they are discussing the small but very real effect of buoyancy of the air which is important if you want to do a precision measurement. I don't see the connection to the topic at hand.

 

Worse yet, when I dug into that link, no offense intended at all, it isn't your own, you were just trying to catch me up, I getit--but I got a good giggle when in the link I read that generally liquids can't be pulled. Heh, Guess they weren't aware of modernday hydraulics at the time--or your basic medical syringe--or even a plunger??, now that one's just a shame, I swore we might be able to expect to find any one of these applications in some context in science venues abound...

 

In those systems/examples you are pushing the liquid.

[matty]

I don't know jack about metrology and the difference and so then I guess this would be moving away from context so, now that I think about it, the latter is semantics, since pumps are what we're referring to then and they're set at lowest points for most pushing force achievable.--But I never thought about it like that, I suppose. I tend to think of their draw but, yeah, there is a lot more going on than that when I refresh myself.

 

Not so complicated in out of context syringe, I don't think, you draw back and by suction, pull in liquid.--Anyway, that just sounded ridiculous to me but not if you, in fact, break hydraulics down.

 

-You win, So what.~

Edited November 9, 2011 by matty

[HiThere]

heres my view. first post n all..

your question contradicts itself slightly. In the first sentence youve stated "In a vacuum" but further down youve wrote " ...have more force to "push" air molecules out of the way". So to answer both questions (in a vacuum and in atmosphere) in a vacuum, the only particles there, are the ones you put there. so because there is no mass to create any opposing force then they would fall at a uniform rate (unless the objects were ridiculously huge enough to cause its own gravity but thats going too far lol) so because theres no atmosphere, aerodynamics pale into insignificance because there is no 'aero' for the masses to be 'dynamic' through. they would simply fall at a rate set by whatever mass is causing the gravity, namely a planet

In somewhere with an atmosphere, then mass counts for a great amount, BUT! that doesnt necessarily mean a heavier thing falls faster than a lighter thing. eg a tank vs a tank with a parachute. you are ignoring a lot of forces that must be considered. is it windy? do the objects have the same aerodynamic properties? are they of similar shape? My overall guess would be that it is entirely dependant on what/who you throw... and what you throw it off

[Joshmckenna]

Okay,

 

Lets say two stones hit the ground at the same time. However, this is not necessarily true. The reason is rooted in the concept of air resistance. In the case of extreme air resistance, where a falling object experiences a drag force equal in magnitude to the gravitation force, the object reaches a terminal velocity. Terminal velocity is the result of air resistance slowing the stone's acceleration to zero.

 

Vt = Root(2mg/CpA)

 

Even if the objects do not reach their terminal velocities before they hit the ground, the denser stone still should hit first, because air resistance is unable to hold it back as much as air resistance can hold back the less dense stone. Air resistance depends on the contact area of the object with the medium. When the object has a high mass and low contact area, then the weight is more significant than air resistance. Conversely, when the object has a low mass and high contact area, then the weight is comparable to air resistance. This means that the stone that hits the ground first will be theone that has the highest mass-to-contact area ratio.

Hopefully this clears up any questions if this was occuring outside a vacuum.

Edited May 24, 2013 by Joshmckenna

[xyzt]

The "time to collision" in VOID is given by:

 

[math]t_i=\frac{D^{3/2}}{\sqrt{2G(M+m_i)}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math]

 

where [math]R[/math] is the radius of the massive body and [math]r_i[/math] is the radius of the test probe.

If the two test probes have the same radius [math]r_1=r_2[/math], then the "time to collision" depends only on the mass of the respective test probes, the larger the mass, the shorter the time. The reason is that the object attracts the Earth, just as the Earth attracts the object. More massive bodies make the Earth close the distance faster.

Edited June 14, 2014 by xyzt