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

[disregardthat]

"Ofcourse it's NOT the same."
That's why I said it...


"Huh ? So, you conclude from this that acceleration and constant velocity are the same ? Explain, please."
Uhhh, no I don't, I didn't conclude that acceleration and constant velocity is the same thing. If you read what i wrote, you would understand that I just said that under acceleration you experience G force, at least when something is pushing on the object.


"What do you mean by that ?"
That 2 objects have the same acceleration in the same direction with same (rising) velocity.


"1) What's parallel acceleration"
read above
"2) What's "outspace""
Outspace = outer space


"With respect to what frame of reference ? Without defining that, you cannot be speaking about "inertial forces"."
By inertial forces I mean forces that are result of mass' effect on each other.

"Now THAT is an inertial frame."
...


"You mean g force, right ? The ratio of acceleration to gravitational acceleration. Or the force to weight ratio. 6 G's means that the force on an object is 6 times it's weight."
G force, G power whatever, you get too hugned up in details that people could think themselves too. See? you understood that i mean g force.



"1) standing still is NOT the same as having constant velocity."
Since everything is relative, I mean standing still = no G force. Acceleration = G force (just not gravitational acceleration, and that's basically what i don't understand)
"2) standing still eventhough it is accelerating ? What the ... ? With respect to what reference frames are you talking here ?"
... Standing still = not feeling G force(like constant velocity in outer space), gravitational acceleration = not feeling any G force.(free fall)
And by G force i mean in the accelerating\moving object's frame.
If something is pushed on Earth it will feel the G force. If something is standing still in the Earth's frame it will feel G force. But if it falls, no G force.



"Nonsense, check the definition of G-force again."
I will

"Jarle, this is just rubbish. Pardon the French but do YOU even know what you are saying here ?"
Both me and you are talking english.

"Look, if you want to have a constructive debate here you need to change your attitude and the way you present your questions. Try to answer my questions as a start. They will make this topic a whole lot clearer. If you talk about inertial forces you ALWAYS need to specify the frame of reference. Be more accurate with that. Refrain from using ill concieved concepts like "outspace", "parallel acceleration" and constant velocity with acceleration. THINK ABOUT WHAT YOU ARE WRITING DOWN."
Explain what is wrong instead of just pointing out how bad I have understood this... And I am not that good in the necessarily needed definitions of everything I say.

"Also Jarle, i am sure that you have read the PF Guidelines. They clearly state how you should present your questions on this forum. The way you are doing it right now is a big NONO. Please, put in some extra effort, we will be patient enough with you but we need to see progress on that. If not, actions WILL be taken. So, take this as constructive criticism and learn from it."
... I am only saying what I believe, and hope that someone would correct me in a friendly way...

 

[marlon]

Therefore, the gravitational field we feel at the surface of the Earth is really a fictitious force like those of other non-inertial frames of reference.​

This is EXACTLY what GR deals with !

Don't worry, these are difficult concepts that require a thorough study. One cannot expect to learn this from Wikipedia. Actually, this is the very reason why i jumped into this thread because many people are just using hollow words but do not know what they are talking about. Just to be clear, i read your posts and aside my two objections they are completely correct.

regards
marlon

 

[marlon]

"Ofcourse it's NOT the same."
That's why I said it...

Ok, so they are NOT the same. That's clear now.

In post 25 you wrote "You are saying that acceleration and a constant velocity is not the same, and I assume it is because in acceleration you experience G force. The thing that doesn't say click in my head then, is gravity."

What i don't get is "in acceleration you experience G force". How does this prove that acceleration and constant velocity are NOT the same ? What i ask is this : why do you need G force to prove this ?

I just said that under acceleration you experience G force, at least when something is pushing on the object.

Correct.

"What do you mean by that ?"
That 2 objects have the same acceleration in the same direction with same (rising) velocity.

Ok then, that's clear.

"2) What's "outspace""
Outspace = outer space

Ok, got it.

"With respect to what frame of reference ? Without defining that, you cannot be speaking about "inertial forces"."
By inertial forces I mean forces that are result of mass' effect on each other.

That's not the definition of an inertial force.

G force, G power whatever, you get too hugned up in details that people could think themselves too. See? you understood that i mean g force.

No, YOU need to be more correct in what you say.

"1) standing still is NOT the same as having constant velocity."
Since everything is relative,

I mean standing still = no G force.

Well that is NOT the same as standing still in an inertial frame or with respect to any other frame.

Acceleration = G force (just not gravitational acceleration, and that's basically what i don't understand)

That is NOT the definition of g force. Didn't i already tell you this ?

"2) standing still eventhough it is accelerating ? What the ... ? With respect to what reference frames are you talking here ?"
... Standing still = not feeling G force(like constant velocity in outer space), gravitational acceleration = not feeling any G force.(free fall)

Forget the g force and think in terms of inertial mass and gravitational mass. THAT IS THE MAIN TOPIC HERE.

And by G force i mean in the accelerating\moving object's frame.

But what if that frame is not an inertial frame ? I urge you to study the content of the equivalence principle !

If something is pushed on Earth it will feel the G force. If something is standing still in the Earth's frame it will feel G force. But if it falls, no G force.

I agree but clearly you see that the above contradictis with "your" definition of standing still ("Standing still = not feeling G force"). This is exactly what i wanted to point out. You need to be talking about reference frames as well and DO IT CORRECTLY.

"Nonsense, check the definition of G-force again."
I will

I really hope so.

"Jarle, this is just rubbish. Pardon the French but do YOU even know what you are saying here ?"
Both me and you are talking english.

Yes, but you use the wrong definitions and terminology so things get, err, confusing to say the least

Explain what is wrong instead of just pointing out how bad I have understood this... And I am not that good in the necessarily needed definitions of everything I say.

I try to help you in finding the right way out. First i just wanted to show you what you were doing wrong. Now, we can talk about solutions. But, "solutions" also requires that YOU study and go look for the correct definitions. On this forum, we do not just spoonfeed you because experience has taught us that it serves no purpose. Besides, this is clearly written in the PF Guidelines.

... I am only saying what I believe, and hope that someone would correct me in a friendly way...

Well, that is NOT the way science works (nor does this forum). You need to come up with facts and correct definitions. How can we help someone who does not speak the correct language ?

So, i suggest you rewrite your original question BUT keep in mind what we have been talking about concerning : g force, inertial frames, inertial mass and gravitational mass.

regards
marlon

 

[disregardthat]

All right, I understand...

But the very reason I am posting at this forum is excactly because I don't understand. My main source is Wikipedia, and everything there is not that understandable, especially since english is my main language.

I see now that I have misunderstood the concept of inertial movement. But the definitions given on wikipedia I don't understand. So couldn't you insted of criticize me, give the correct definition?

All right with this said, I want to know what really is different from gravitational acceleration from acceleration due to "pushing or pulling"(which I understood as inertial acceleration... You pointed that it is not correct, but what is correct then?)

Both of the "forces" result in movement in our world. But acceleration due to pushing and pulling ALWAYS(as I have understood) result in a higher G force effect on an object. So that's one difference.
In my mind I thought that maybe gravitational acceleration, or "force" is not movement in the same way as 'movement due to pushing and pulling'(what is the word for that if it is not inertia? (don't hit me)) I began to think of the coordinate system I read about in the general relativity article. Movement or acceleration due to pushing and pulling, and movement due to gravity makes you move in this coordinate system(which I believe is fourdimensional, is that correct?)

Then it popped to me, that maybe movement due to pushing and pulling makes you move in the three dimensional system, and gravitational movement didn't. I thought that gravitational movement was only movement in the four dimensional system. So, I am asking you, is this correct? Or what is the correct way of seeing it? What I meant as "standing still", wasn't in the relative way of thinking, not standing still in any other's frame. I meant that it was standing still in a different coordinate system than is present in our weird world. (3 dimensional system, no curving) If this doesn't make sense then explain why...

I didn't mean to express my thoughts as statements, but merely as thinking I have done. I didn't mean to say that it is like that, but that I think it might have been something in that direction.

---------------------------------------------------------------------

I included the G force because I thought it was a direct effect of acceleration. I thought that since we doesn't feel G force when we were accelerating because of gravity, we couldn't be moving the same "way" as movement due to pushing and pulling, which I have explained some paragraphs above.

I read about G force in Wikipedia, and I understand that we only feel the G force when we are standing still in the center of the Earth's frame.(approximately 9.8m\s^2) But most of the explanations I don't understand fully. I also understand that G force is the unit of acceleration. Which brings me back to parts of my very question, why don't we feel G force when we accelerate towards the earth?

 

[Mindscrape]

G-force is just a useless scaling factor for force, and has absolutely nothing to do with relativity.

Are you asking why we don't feel a force in free fall (other than drag force)? The answer is inertia.

 

[D H]

No instrument, including our inner ears, can feel the "G force" in free drift. This is the very heart of the equivalence principle. If you are in an enclosed box in space, there is no experiment you could conduct in that box to determine whether you accelerating solely due to gravitation or simply drifting in flat space.

This is very important for spacecraft equipped with inertial navigation systems. The accelerometers can feel all forces exerted on the vehicle except for gravity. The gravitational force dominates all other forces on the spacecraft , and yet it is the one force that cannot be measured directly. To accurately propagate the vehicle's state based on those measured forces, the spacecraft navigation system must compute the gravitational force.

 

[disregardthat]

Well, how would you explain and define inertia?

 

[marlon]

All right, I understand...

But the very reason I am posting at this forum is excactly because I don't understand. My main source is Wikipedia, and everything there is not that understandable, especially since english is my main language.

First, don't use a public website to learn physics. Use proper course material. If you want i can recommend you some great books.

I see now that I have misunderstood the concept of inertial movement. But the definitions given on wikipedia I don't understand. So couldn't you insted of criticize me, give the correct definition?

I am not criticising you, i am trying to convince you that you understand the mistakes you made. It is not this forum's task to just give you some definitions. Again, science does NOT work like that.

All right with this said, I want to know what really is different from gravitational acceleration from acceleration due to "pushing or pulling"(which I understood as inertial acceleration...

Ahh, so what is the difference between inertial acceleration and gravitational acceleration. This is equivalent to asking the difference between inertial and gravitational mass.

The inertial mass is defined as the m in Newton's second law: m.a = F, or, in other words, the m in the formula for momentum: p = m.v. This mass is the constant of proportionality between the applied force (we are not running in circles here...) and the acceleration of the object.

To study the dynamics of any interaction in classical physics, we will need this m for Newton's second law.


Now, Newton was the first guy to discover a very special interaction: gravitation. The force of gravitation is written as F = -G.M1.M2/r^2

Here, G is Newton's gravitational constant and M1 and M2 are DEFINED as "gravitational mass". The reason why Newton made a distinction between the M's and the inertial mass m is very obvious. We need to apply a force of m to make it "move" while gravity is out of our hands. THAT IS ALL.

Suppose that m is the mass of a car and M1 is the gravitational mass of that SAME car, M2 is the gravitational mass of another object like the earth.

Now, experimentally, we found out that M1=m. That is the equivalence principle ! Also, if you would apply Newton's second law on m you get:

ma = -G.M1.M2/r^2 and m=M1

a = -G.M2/r^2 and the car stands on the Earth's surface so, r= the Earth's radius. If you would fill in the Earth's mass and the erath's radius in that equation, you get a = 9.81 m/s² ! THIS IS A CORRECT VERSION OF REALITY and thus a good theoretical model.

So if m=M1 --> a=g and also if a=g --> m=M1 : HERE YOU SEE THE ACTUAL EQUIVALENCE !

Finally, some words on general relativity, although this has NOTHING to do with your original question.

The principle of equivalence is the corner stone of Einstein's general relativity, because Einstein realized that, if inertial and gravitational mass are the same, that the "force of gravity" can be an entirely geometrical effect (and not a genuine force), because no property of the object proper is needed to determine the trajectory (which can hence be a "bending of spacetime" itself, and not simply a trajectory of a specific object, related to its properties).

I thought that gravitational movement was only movement in the four dimensional system. So, I am asking you, is this correct?

No, it is not correct. Gravitation can be 1,2 or 3 dimensional but never 4 dimensional. At least not in classical physics.

regards
marlon

 

[marlon]

Well, how would you explain and define inertia?

Check out Newton's first Law :http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html

marlon

 

[disregardthat]

I think I understood that, but why don't gravitational movement need any energy to make an object move? When you say gravitational and inertial mass is the same, i guess you meant that inertial and gravitational movement is the same also, if I didn't misunderstand.

I trust you on this equilance principal thing, although I don't udnerstand it all fully.

About the Newton law: thanks.

 

[marlon]

I think I understood that, but why don't gravitational movement need any energy to make an object move?

Well, indeed gravity just "seems to be happening" and that is exactly why gravity and it's associated mass and acceleration seem to be different than inertiam mass and acceleration. But again, this is just the way it is. Don't worry if you don't understand where gravity comes from because Newton also did not get that and WE STILL DO NOT KNOW !

That being said, the formalism behind gravity allows us to accurately describe this interaction and make correct predictions with it. That is all there is to it and i hope you understand the value of this formalism. Not knowing where gravity comes from does not equal the fact that we cannot describe it.

When you say gravitational and inertial mass is the same, i guess you meant that inertial and gravitational movement is the same also, if I didn't misunderstand.

If you mean by movement "acceleration" than YES !

I trust you on this equilance principal thing, although I don't udnerstand it all fully.

What is it you do not get ?

marlon

 

[D H]

Gravitation is not so special that it is immune from concepts of energy and work. A pre-stellar gas cloud has a large amount of gravitational potential energy because the gas that comprises the protostar is so diffuse. The potential energy drops drastically as the gas cloud collapses gravitationally. The temperature (kinetic energy) increases with the decrease change in potential energy. The star eventually ignites.

That inertial and gravitational mass are the same does not mean that inertial movement and gravitational movement are the same. Inertial and gravitational movement are not the same in Newtonian mechanics. Inertial movement in Newtonian mechanics is movement with no net force.

The equivalence principle is axiomatic in Newtonian mechanics. In Newtonian mechanics, this principle means the $m$ in $F=ma$ is the same as the $m$ in $F=-\frac{GMm}{r^2}$.

Marlon is one of the PF experts on relativity. I will let him speak on how GR treats inertial frames and the equivalence principle. I will keep my foot firmly planted on the Earth (and out of my mouth).

I asked you some time ago to clarify what you meant by "inertial acceleration" and "inertial movement". This, I think, is the heart of your problem.

 

[disregardthat]

:) wow, an expert on relativity?

If you mean by movement "acceleration" than YES !

yeah, I did.

But when you say that the gravitational and inertial mass and acceleration is the same, what do you have to say to that gravitational acceleration accelerates an object at the same speed, undependent on it's mass(weight), but 'push and pull'(what is the scientific word for that?) acceleration depends on the mass(weight)

I understand it's like this:
Gravitational acceleration: (x)N\kg = (x)m\s
Push and pull acceleration: xN = (y)m\s

Or at least that gravitational acceleration does not depend on the mass of object, but 'push and pull' acceleration does. How can one say that it's the same thing?

The inertial mass is defined as the m in Newton's second law: m.a = F, or, in other words, the m in the formula for momentum: p = m.v. This mass is the constant of proportionality between the applied force (we are not running in circles here...) and the acceleration of the object.

I know that m=mass a=acceleration and f= force. But what does p stand for. And i assume v=velocity?
And I have read that this formula (m*a=f) is not correct in relativity, (with high amounts of mass, acceleration and force)

Gravitation is not so special that it is immune from concepts of energy and work. A pre-stellar gas cloud has a large amount of gravitational potential energy because the gas that comprises the protostar is so diffuse. The potential energy drops drastically as the gas cloud collapses gravitationally. The temperature (kinetic energy) increases with the decrease change in potential energy. The star eventually ignites.

So you are saying:
If you have 100 atoms in different place, and calculate their total amount of gravity, that this result would be LESS than if you put the 100 atoms together and calculated the amount of gravity they "created" together?? :O

 

[marlon]

:) wow, an expert on relativity?


yeah, I did.

But when you say that the gravitational and inertial mass and acceleration is the same, what do you have to say to that gravitational acceleration accelerates an object at the same speed, undependent on it's mass(weight), but 'push and pull'(what is the scientific word for that?) acceleration depends on the mass(weight)

Well, let's review the equivalence principle again. Basically it says that if you stand in an elevator on a scale that measures your weight, you cannot distinguish whether your weight comes from the elevator standing on the Earth because of gravity or whether it comes from the elevator moving up with an acceleration a = -g. In other words, your weight in both cases is the same so : mg=m'a (in magnitude). Now the equivalence says that if m=m' then a=g OR if a=g then m=m'.

So, this principle only says that if a=-g, your measured mass will be the same. This means that movement under gravity is equivalent to an accelerated movement upwards with a=-g.

By the way, the "push and pull" acceleration is the inertial acceleration

I understand it's like this:
Gravitational acceleration: (x)N\kg = (x)m\s
Push and pull acceleration: xN = (y)m\s

Or at least that gravitational acceleration does not depend on the mass of object, but 'push and pull' acceleration does.

I am not sure i get this. Both the inertial acceleration a = F/m and the gravitational acceleration g = F/m' So what is the problem ? Both of them depend of mass m and m'. Ofcourse, the value of g is independent of the gravitational mass m' but that has nothing to do with the equivalence principle as i outlined in the first paragraf. The clue is that the product of ma nd m'g are equal ! So ma=m'g, if a=g then m=m', that's all.

Here is another example : suppose your mass is 80kg, now if you jump out of a window you still have that mass but yet you feel like you have no mass/weight ? This is exactly the same as what i wrote above with the elevator example: if a=g then m=m' !

I know that m=mass a=acceleration and f= force. But what does p stand for. And i assume v=velocity?

p is linear momentum : $$\vec{p} = m \vec{v}$$

And I have read that this formula (m*a=f) is not correct in relativity, (with high amounts of mass, acceleration and force)

Indeed it is not.

So you are saying:
If you have 100 atoms in different place, and calculate their total amount of gravity, that this result would be LESS than if you put the 100 atoms together and calculated the amount of gravity they "created" together?? :O

I don't get what you mean by "their total amount of gravity" .

marlon

 

[disregardthat]

Ok thanks, I understand the the gravitational mass and inertial mass is the same. 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.

A big tanker falls just as fast as a little stone on earth. Where does this energy come from? The amount of force needed to accelerate a tanker in space at 9.8m\s^2, is enormous, especially in camparison of the little rock.

About the atoms:

Gravitation is not so special that it is immune from concepts of energy and work. A pre-stellar gas cloud has a large amount of gravitational potential energy because the gas that comprises the protostar is so diffuse. The potential energy drops drastically as the gas cloud collapses gravitationally. The temperature (kinetic energy) increases with the decrease change in potential energy. The star eventually ignites.

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?)

Amount of gravity=gravitational force

 

[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.