Revisiting the Definition of Free Fall: Acceleration or Standing Still?

[marlon]

The thing that I don't understand is: where does the energy come from that can move many thousand of tons objects in an acceleration of 9.8m\s^2.

Good question. The answer : we do not know. The Newton formalism as well as general relativity describe the behaviour of gravity and allow us to make correct predictions when a certain set of initial conditions has been given. Theory and experiment correspond to each other. In Newtonian physics, one can somehow answer your question using the total energy conservation law : if an object falls down, the kinetc energy rises, if an object goes up, the kinetic energy lowers. The sum of the two is ALWAYS constant. That is the way you need to look at it. Ofcourse, you can ask the question : why is there conservation of total energy. Well, physics does not explains this because this conservation is a property (or axioma if you want) of nature.

Do you mean that mass in concentrations "creates" MORE gravity when they are together, than "piece by piece"? I don't mean the gravitation "created" by each piece, but the total "amount" of it. Like if 2 stars far away from each other, each resulting in a powerful gravity field. If you crushed one star into the other, would this result in a bigger gravitational field than the two stars did put together? (1+1=3?)

A pre-stellar gas cloud has a large amount of gravitational potential energy that causes the formation of stars only when there is a local fluctuation in the mass distribution. i mean, when, in the cloud, there is one region where the mass density is bigger than another region. If the mass would be uniformly distributed throughout the cloud, no gravitational contraction would occur.

Concerning your "(1+1=3)"-question : the potential energy U associated with gravity is U =- \frac{GmM}{r}
G : gravitational constant
m and M : two interacting masses
r : the distance between m and M


As you can see, this potential is negative which means that we are dealing with a "bound state". Indeed, once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape.

When U = 0, the two bodies are infinitely far away from each other and no gravitational interaction is occurring. Now, your 1+1=3 thing is CORRECT because the potential energy does not vary linearly (ie 1+1=2) when the two bodies approach each other. The U varies as \frac{1}{r} !

greets
marlon

edit : "the amount of gravity", as you write, is actually the potential energy U which is NOT equal to the gravitational force F. Intuitively, U expresses the amount of energy available for gravitational interaction. So, on earth, an object which is higher than another object will have a bigger U-value. Finally, the relation between F and U is : \vec {F} =- \vec {\nabla} \cdot U or in words : F equals the way U varies with respect to the distance r !

 

[disregardthat]

Ah, I understand now!

Except a few thing: why is the U negative in potential gravitational energy?
Just because objects need to bring this number over zero to escape from the objects gravitational field? Well, I need to know how to compare the gravitational potential energy from inertial acceleration energy.
If you don't understand I might have explained my self wrong.

If G has the units of Nm^2/kg^2 and is multiplied by 2 masses (kg*kg) divided by distance meter. Is the U unit gravitational potential energy in the units of: Nm/kg^4 ?

Would then: If two objects had the mass of 1 000 kg 1 000 meters away from each other then have the potential energy like this:?

U=- (G*m*M)\r
U=- ((6.6743*10^-11)100 000 000kg*100 000 000kg)/1 000m
U= -667.43Nm/kg^4

And it would be -6674.3Nm/kg^4 If the objects were ten times closer, right?

How would you describe that? -667.43 potential energy...

Edit: nevermind what i put in the quote! After rethinking I know that gravitational potential energy just is Nm. And Nm equals F right?

So U is kind of negative force? I get this thing now!

 

[marlon]

Except a few thing: why is the U negative in potential gravitational energy?

Like i said : The gravitational potential U is negative, which means that we are dealing with a "bound state". Indeed, once a mass is near a large body, it is trapped until something can provide enough energy to allow it to escape. In other words, the minus means that the two objects will "automatically" move closer to each other because of gravitation and it will COST energy if you want to move them away from each other. When r gets bigger (ie distance between m and M), the U value evolves toward 0. When r is infinite, U is 0 but U cannot exceed 0 !

Just because objects need to bring this number over zero to escape from the objects gravitational field?

No, but what you mean here is that it will cost energy to move the objects away from each other and escape their mutual gravitational field.

Well, I need to know how to compare the gravitational potential energy from inertial acceleration energy.
If you don't understand I might have explained my self wrong.

First of all, we don't use "inertial acceleration energy" but just potential energy associated with some force. If you have an equation for that force, the potential energy is calculated like THIS. You will also find another explanation of the "-" sign . Go check it out.

Edit: nevermind what i put in the quote! After rethinking I know that gravitational potential energy just is Nm. And Nm equals F right?

Nm are the units of U which is the POTENTIAL ENERGY, NOT THE FORCE. Again, the relation between the two is explained in the reference i gave you.

Potential energy is the energy of the system's configuration, positions , etc etc. We apply a force onto that system. You see the difference ?

So U is kind of negative force? I get this thing now!

Just to be clear, U is NOT a force but the gravitational potential energy !

regards
marlon

edit : and the mass in your example is 1000 kg NOT the 100 000 000kg you used !

 

[disregardthat]

Yeah, I get it now. Thanks I will check the link

Sorry, 100 000 000 was 1000*1000... I multiplied the answer with the answer...

But here it is not mentioned the radius of the objects that have mass! Isn't this a significant factor when the objects are closing in?

Like an object of 1000 kg in one km cm will act differently on an object that has 1000 kg over a cubed km! Doesn't it?

 

[marlon]

But here it is not mentioned the radius of the objects that have mass! Isn't this a significant factor when the objects are closing in?

Actually, NO. Newton has proven that we can treat the planets as point particles. Only when we use the law of gravitation on the earth, we need to look at the actual radius of our planet. But that is exactly how gravitation "becomes" gravity where g = \frac{GM_{earth}}{R_{earth}^2}

marlon

 

[disregardthat]

Well, what about black holes, they don't have more mass than the star had when it collapsed. Still, it's gravitational potential energy is extreme just before the event horizen (not to mention the even horizon, but that is something that doesn't fit in the theory of relativity, If I understood it right)

And I think because this works for black holes, it should work for other objects too. That radius does matter.

 

[marlon]

Well, what about black holes, they don't have more mass than the star had when it collapsed.

So ?

Besides, the behaviour of black holes is NOT desrcibed by classical physics. Remember that.

Still, it's gravitational potential energy is extreme just before the event horizen (not to mention the even horizon, but that is something that doesn't fit in the theory of relativity, If I understood it right)

Nope, it DOES fit general relativity. Ever heard of the Schwarzschild radius ?
Only the singularity does not fit general relativity.

And I think because this works for black holes, it should work for other objects too. That radius does matter.

I don't understand what you mean by this. Besides, what does this have to do with what we have been talking about ?

marlon

 

[disregardthat]

Well, what i meant with my question about the radius of the object, and why it didnt matter. I came up with the black hole. You say that it DOES fit in general relativity, (i will check Schwarzschild radius up) but why does an object reach singularity, even though there is not added any more mass to the object that once was a star. The gravitational force have obviously risen, and only because of the radius of the object...

So i guess that the effect also would occur with an object that had very low density over a huge volume, if it was compressed to a very tiny volume.

If the Earth's mass was compressed to a cubed meter(assuming that it isn't enough to create singularity, which i doubt), it would not create a stronger gravitational field?

 

[marlon]

Well, what i meant with my question about the radius of the object, and why it didnt matter. I came up with the black hole. You say that it DOES fit in general relativity, (i will check Schwarzschild radius up) but why does an object reach singularity,

"An object does not reach singularity", the singularity is a property of the black hole where the laws of physics no longer apply.

So i guess that the effect also would occur with an object that had very low density over a huge volume, if it was compressed to a very tiny volume.

yes but in that case you have a mass distribution that is varyin over a certain distance. The earth, and planets in general do not have this property in classical physics. Treating them as point particle, as proven by Newton (go check this out !) gives an accurate description of what's going on.

If the Earth's mass was compressed to a cubed meter(assuming that it isn't enough to create singularity, which i doubt), it would not create a stronger gravitational field?

In classical physics ? No. In general relativity, there would indeed be a stronger wrap in the space time continuum localized at one "point".

greets
marlon

 

[disregardthat]

But I thought Newton didn't insert relative factors in his equations, making them incorrect in enormous scales. So if we inserted a relative factor here would the radius be a significant factor?

In general relativity, there would indeed be a stronger wrap in the space time continuum localized at one "point".

what i got from this was an 'yes' to that question.

And I have read about the U=-GMm/r in 'hyperphysics', I understand that, at least reasonably. I guess and assume it is correct, and I am not saying I am correct, I am just questioning in because I thought radius would matter in this.

yes but in that case you have a mass distribution that is varyin over a certain distance. The earth, and planets in general do not have this property in classical physics.

What 'propety' don't they have, I didn't excactly get that.

 

[D H]

yes but in that case you have a mass distribution that is varyin over a certain distance. The earth, and planets in general do not have this property in classical physics. Treating them as point particle, as proven by Newton (go check this out !) gives an accurate description of what's going on.

Off-topic,

Marlon, you are now speaking outside your area of expertise. Your statement would be correct if the mass distributions of the planets was spherical. However, the Earth, the Moon, and Mars cannot be treated as point masses because their mass distribution is not spherical. One needs to account for the deviations from spherical mass distribution to attain "accurate description of what's going on". We use spherical harmonics to model the static gravitational field of a planet and a rather ad-hoc thingy called the "tidal Love number" to model how the plasticity of a planet effects the gravitational field.

The http://www.csr.utexas.edu/grace/" is an ongoing experiment to develop an accurate model of the Earth's gravitational field.


Back on topic,

And I have read about the U=-GMm/r in 'hyperphysics', I understand that, at least reasonably. I guess and assume it is correct, and I am not saying I am correct, I am just questioning in because I thought radius would matter in this.

I am assuming you meant the planet's radius when you said "radius". That only comes into play if the mass distribution is not spherical. In Newtonian physics, an object with a spherical mass distribution exerts the exact same gravitational field as does an equally massive point mass.

The Earth does not have a spherical mass distribution. The point mass approximation (for which U=-GMm/r) is correct to "zeroth order".

 

[marlon]

What 'propety' don't they have, I didn't excactly get that.

Well, that mass varies locally throught a planet's volume. Planets are considered to be point particles in classical physics and that works just fine.

marlon

 

[marlon]

Off-topic,

Marlon, you are now speaking outside your area of expertise. Your statement would be correct if the mass distributions of the planets was spherical. However, the Earth, the Moon, and Mars cannot be treated as point masses because their mass distribution is not spherical. One needs to account for the deviations from spherical mass distribution to attain "accurate description of what's going on". We use spherical harmonics to model the static gravitational field of a planet and a rather ad-hoc thingy called the "tidal Love number" to model how the plasticity of a planet effects the gravitational field.

True, but that does NOT change the fact that planets (like the moon and earth) are considered to be point particles in classical physics. I know this is an approximation but it is an approximation that gives correct results, so no problem. Even F=mg assumes that the Earth is a perfect sphere and all results with this law are correct. So, within our "earthly" observations and their required accuracy this approximation works just fine. That is all i am saying.

marlon

 

[Doc Al]

True, but that does NOT change the fact that planets (like the moon and earth) are considered to be point particles in classical physics.

For many purposes it's OK to treat planets as point particles, but for other purposes it would be ludicrously inaccurate. (Try explaining the tides using a point particle model of Earth and moon.) I doubt you mean it this way, but someone can take your words as implying that treating planets as point particles is some fundamental assumption of classical physics, which of course it is not. What you are pointing out is that many of the standard "results" used in elementary physics are based on this practical approximation--which is certainly true.

 

[disregardthat]

Well, so basically you mean that a sphere with a radius of x and a mass of y, has the same gravitational effect on objects (treating everythign as point particles) as a sphere of x\1000 with a mass of y? Unless the mass is concentrated enough to create an even horizon.

(This is excactly what i mean, It is the concentration of mass in a little volume that gives the black hole the extreme gravity (at least at small distances from it) so I just can't see why spheres with a lower readius but same amount of mass would not act in the same way)

 

[marlon]

For many purposes it's OK to treat planets as point particles, but for other purposes it would be ludicrously inaccurate. (Try explaining the tides using a point particle model of Earth and moon.)

Agreed

I doubt you mean it this way, but someone can take your words as implying that treating planets as point particles is some fundamental assumption of classical physics, which of course it is not.

Indeed, but that's why i wrote "approximation" each time i mentioned this.

What you are pointing out is that many of the standard "results" used in elementary physics are based on this practical approximation--which is certainly true.

Exactly.

marlon

 

[Hootenanny]

Well, so basically you mean that a sphere with a radius of x and a mass of y, has the same gravitational effect on objects (treating everythign as point particles) as a sphere of x\1000 with a mass of y?

In classical mechanics yes, in relativity no.

(This is excactly what i mean, It is the concentration of mass in a little volume that gives the black hole the extreme gravity (at least at small distances from it) so I just can't see why spheres with a lower readius but same amount of mass would not act in the same way)

I will say this again (as other have said). In classical physics, any sphere can be treated as a point mass provided you are outside the sphere, this turns out to be a good approximation for most uses.

Apologies for butting in Marlon et al.

 

[marlon]

(This is excactly what i mean, It is the concentration of mass in a little volume that gives the black hole the extreme gravity (at least at small distances from it) so I just can't see why spheres with a lower readius but same amount of mass would not act in the same way)

Ok, got it. Essentially you are right : you would evolve to a "black hole type situation" if an object has a lot of mass contained in a smaller volume or if you would take a certain amount of mass but lower the volume in which it is contained. But again, my point is that black holes require a tremendous amount of gravity, and thus a very big collapse of mass. This kind of behaviour is not described by classical physics (ie the formula's we have been discussing). I wanted to point out that there is a distinction to be made here : classical physics versus general relativity. One cannot just take a general relativity phenomenon and expect it to be properly described in terms of classical physics. That's all.

marlon

 

[marlon]

In classical mechanics yes, in relativity no.

Indeed, that's what i have been trying to say as well.

I will say this again (as other have said). In classical physics, any sphere can be treated as a point mass provided you are outside the sphere, this turns out to be a good approximation for most uses.

Actually, i don't understand why "people" make such a fuss about this. This "system" or approximation makes life a lot easier and gives correct results, so what's the problem ? If it ain't broken, don't fix it

Apologies for butting in Marlon et al.

No problem.

marlon

 

[disregardthat]

I see, I see.

As I have understood: We can use the same equation for the force of the Earth's gravitational field, if it is a perfect sphere and the same density all over, and will get the same correct answer undependent on the radius of the object.

 

[D H]

In classical physics, any sphere can be treated as a point mass provided you are outside the sphere, this turns out to be a good approximation for most uses.

If it ain't broken, don't fix it

It depends on what you mean by "most uses". Orbital mechanics is used primarily to track and predict the location of the objects mankind has placed in orbit around the Earth. Secular effects arise from the Earth's non-spherical mass distribution; in other words, the point mass model is broken. We have worried about these effects since the beginning of the space age.

 

[disregardthat]

D H, the point mass is not broken, it just cannot be used on the Earth since it is not a perfect sphere. At least that's what Marlon's saying.

 

[Hootenanny]

D H, the point mass is not broken, it just cannot be used on the Earth since it is not a perfect sphere. At least that's what Marlon's saying.

It can be used on earth, depending on the degree of accuracy required. For example, if you wish to calculate the trajectory of a bullet fired from a gun do you think that because we treat the Earth as a point mass and ignore that fact that it is non a uniform sphere will make a significant difference? How about calculating the terminal velocity of a falling object on earth?

There is very little the this universe which is perfectly spherical (or perfect in any geometric sense), all that is important is that the error due to any assumptions is very small in comparison to the required accuracy.

As an aside and further to DH's comment; we don't used classical physics to track orbiting satellites, we must use relativistic physics so we obviously don't treat the Earth as a point mass.

 

[disregardthat]

Ah, finally. So what's the realitivistic equation for the force of gravity? It must be something else than F=GMm/r^2

 

[Hootenanny]

Ah, finally. So what's the realitivistic equation for the force of gravity? It must be something else than F=GMm/r^2

There is not one equation, but ten known as the Einstein field equations. Basically they describe the curvature of space time which results in the observed gravitational force. They can be written down as a single tensor equation. There are a few tutorials available on the Internet, you should search for the "Einstein Field Equations".

 

[D H]

As an aside and further to DH's comment; we don't used classical physics to track orbiting satellites, we must use relativistic physics so we obviously don't treat the Earth as a point mass.

We do not use relativistic physics to track orbiting satellites. I ought to know. I have been working with satellites (guidance, navigation, and control) for 26 years. The errors induced by using the Newtonian approximation are tiny, much smaller (for example) than the errors induced by imperfectly modeling exoatmospheric density.

We do need to account for relativity for one thing. The clocks on any satellite run at a slightly faster rate than those on the Earth (38 microseconds per day or about 1 millisecond per month). That effect is negligible on most satellites; a calibrated quartz clock is accurate to a few seconds per month. The GPS satellites, however, need very accurate timing and thus carry atomic clocks. The relativistic effect on those clocks must be taken into account. The state (position and velocity) of a GPS satellite is still propagated using Newtonian mechanics.

 

[Hootenanny]

We do not use relativistic physics to track orbiting satellites. I ought to know. I have been working with satellites (guidance, navigation, and control) for 26 years.

My apologies, I did not intend my comment as a personal attack in any way.

We do need to account for relativity for one thing. The clocks on any satellite run at a slightly faster rate than those on the Earth (38 microseconds per day or about 1 millisecond per month). That effect is negligible on most satellites; a calibrated quarz clock is accurate to a few seconds per month. The GPS satellites, however, need very accurate timing and thus carry atomic clocks. The relativistic effect on those clocks must be taken into account. The state (position and velocity) of a GPS satellite is still propagated using Newtonian mechanics.

This is what I meant by my comment, I should have been more specific (i.e. specified that relativity is used to correct the clock times); not to model the position of an orbiting satellite.

 

[marlon]

, it just cannot be used on the Earth since it is not a perfect sphere. At least that's what Marlon's saying.

No, that is NOT what i have been saying. What i said is that the earth, although not being a perfect sphere, is considered to be a point particle (just like any other planet) in classical physics. C'mon, i have been saying this over and over again.

marlon

 

[marlon]

It depends on what you mean by "most uses". Orbital mechanics is used primarily to track and predict the location of the objects mankind has placed in orbit around the Earth. Secular effects arise from the Earth's non-spherical mass distribution; in other words, the point mass model is broken.

This is very true yet it does not change the fact that spherical objects (like planets, or at least like the idialization of them) are considered to be point particles in classical physics. Hootenanny already gave some examples, which i am sure, you will not deny. In light of Jarle's original comments on gravity, i wanted to make sure he understood this. That is all.

marlon

 

[D H]

This is very true yet it does not change the fact that spherical objects (like planets, or at least like the idialization of them) are considered to be point particles in classical physics.

I agree. There is no difference between the gravitational potential for a point mass and a spherical mass in classical physics for all points outside of the spherical object.

 

[disregardthat]

Well, I agree, the Earth could be looked upon as a point particle. BUT, what I meant that that does not makes the equation 100% correct due to it's non-sphere like form... But it is probably insignificant...

Anyway, I believe that there must be another factor involved here, since a black hole is evidence that the smaller the volume the mass is spreaded on, is making a difference in the gravitational effect. Since the blakc holes density made the gravitational force exceed the speed of light (you know what I mean...) means that the density DOES matter, or what? The factor might be pretty insignificant when comparing planets to stars and such, but in the end, it would really matter. :\

I am not saying it works like this, I am saying that I believe it should work like this, even though it is not taken in consideration in classical physics.

 

[Hootenanny]

Well, I agree, the Earth could be looked upon as a point particle. BUT, what I meant that that does not makes the equation 100% correct due to it's non-sphere like form... But it is probably insignificant...

Anyway, I believe that there must be another factor involved here, since a black hole is evidence that the smaller the volume the mass is spreaded on, is making a difference in the gravitational effect. Since the blakc holes density made the gravitational force exceed the speed of light (you know what I mean...) means that the density DOES matter, or what? The factor might be pretty insignificant when comparing planets to stars and such, but in the end, it would really matter:\

I don't quite understand what your point is here. In the fields/applications where the effects of GR / non-uniformity of the Earth is not significant, then why should we over complicate calculations, when the result from a much simpler calculation would be sufficiently accurate? In the fields where such effects are significant, then they will already have been taken into account (Check DH's orbital mechanics). What your saying is in essence correct, '[insert assumption here]' matters in some applications. However, for many 'everyday' applications classical physics and the assumption that [insert assumption here] is more than sufficiently accurate (as is treating the Earth as a uniform sphere and point mass and ignoring GR). The reason we make assumptions is to simplify calculations, without losing any significant degree of accuracy.

 

[disregardthat]

Well, that was kind of what i said... That the calculations would indeed be accurate enough, even if we treat the planet as a point particle.

Anyway, it didn't have much to do with my other question.

I meant that if you put 1000 kg of material in 1m^3. This would create LESS gravitational energy for a point outside the sphere than if the volume was 1cm^3. I know it does not work that way in classical physics, but I thought that the black hole would be a good example of gravity working this way. but please, TRY to understand my point instead of just saying it doesn't mean anything if I haven't defined it totally correct.

 

[Hootenanny]

I meant that if you put 1000 kg of material in 1m^3. This would create LESS gravitational energy for a point outside the sphere than if the volume was 1cm^3. I know it does not work that way in classical physics, but I thought that the black hole would be a good example of gravity working this way. but please, TRY to understand my point instead of just saying it doesn't mean anything if I haven't defined it totally correct.

Yes, what you are saying is in essence correct. I would point out that Gravitational Energy in GR is a bit of a fuzzy subject (as far as I know Marlon feel free to chip in), so it is better to talk in terms of curvature of space-time.

 

[disregardthat]

Ah, the acknowledge I was waiting for! puh...

So, you say that this is correct, why did Marlon (as I presume knows a good deal about this) deny it?

 

[marlon]

So, you say that this is correct, why did Marlon (as I presume knows a good deal about this) deny it?

Clearly, you are not a good reader. It is not the first time i have been thinking this about you. In post 68 i wrote this as a response to your above question :

"Ok, got it. Essentially you are right : you would evolve to a "black hole type situation" if an object has a lot of mass contained in a smaller volume or if you would take a certain amount of mass but lower the volume in which it is contained. But again, my point is that black holes require a tremendous amount of gravity, and thus a very big collapse of mass. This kind of behaviour is not described by classical physics (ie the formula's we have been discussing). I wanted to point out that there is a distinction to be made here : classical physics versus general relativity. One cannot just take a general relativity phenomenon and expect it to be properly described in terms of classical physics. That's all."


marlon, sighs bigtime...well...whatever...

 

[disregardthat]

All right, sorry, I understood now fairly enough after intensive reading. Well, I guess my point is proven then...

 

[marlon]

Well, I guess my point is proven then...

And your misconceptions have been replaced by proper physical knowledge

marlon

 

[disregardthat]

Misconceptions and misconseptions...

I was asking why it wasn't like so and so, not saying that it WAS like that. You see the difference?

 

[Hootenanny]

I can see this thread takes circular path... I don't think we need to be so pedantic about the language, do we chaps?