Free fall = standing still [Archive]

devino

Dec24-06, 05:07 PM

Do you mean our weight from the 'pull' of earth's gravity can be seen as a gravitational acceleration? Like a 1g acceleration from a rocket is equal and indistinguishable from a 1g 'pull' you feel on the earth's surface.

The acceleration from free fall is quite different from the force of gravity while standing on the surface of the earth. This has me woundering if they are two different forces. Gravity is inversly proportional to its distance squared. Logically I see that in free fall you are weightless. How can this be? Could you weigh an object in free fall to test Newton's Inverse proportionality theory? And I don't mean measure its inertia either.

If you are satisfied with the standard model of gravity and motion then thats great, A satisfaction I fail to enjoy. The more time I spend reading and learning about physics, motion and in particular gravity the more I find questions, contradictions and confusion. And somehow I enjoy it all the more.

disregardthat

Dec25-06, 07:33 AM

See someone who understands me :)

Anyway, doesn't this mean anything? Gravitational acceleration is not the same thing as inertial acceleration since you feel the G powers. So wouldn't it be most correct to compare 0 g in the universe with an acceleration of 9.8m\s^2 on earth? Since both is "free fall" and 0 g. Just that one of them is accelerating more than the other compared to eachother.

D H

Dec25-06, 12:23 PM

A body in free-fall is accelerating. It is not "standing still". Strictly speaking, a body is in free-fall when gravity is the only force acting on the body. Relaxing this strict definition a bit, one could say a body is in free-fall if the gravitational force dominates all other forces. A sky diver who jumps off a hovering helicopter reaches terminal velocity after falling a 1000 feet or so. The sky diver feels his full weight after reaching terminal velocity.

An spacecraft in low-Earth orbit is in near free-fall conditions. The spacecraft is not standing still; it is moving at more than 17,000 miles per hour. Because the exoatmosphere is so thin, atmospheric drag is several orders of magnitude small than the gravitational acceleration.

You do not feel the acceleration due to gravity. When you are standing on the surface of the Earth, the net force on your body is zero. So why do you feel anything? You don't feel gravity pulling every part of your body toward the center of the Earth. You do feel the upward mechanical force of the Earth on your feet that exactly the downward gravitational force. You don't feel gravity pulling your arms down. You feel the tensile forces on your bones, tendons, muscles, and skin that keep your arm from falling off of your body. As with the force on your feet, this tensile force is directed upward and exactly counterbalances the gravitational force.

ShawnD

Dec25-06, 12:42 PM

If you are in free fall you would be standing still, with no acceleration right?
So in a way you are decelerating(or accelerating in the opposite direction) in 9.8m\s^2, when you are standing on the earth

I think I know where the confusion is coming from. The number of 9.8m/s^2 is actually a derived value.
http://en.wikipedia.org/wiki/Law_of_universal_gravitation

That formula is actually meant to solve for the force of gravity between 2 masses, which would be you and earth. Solving for it, you get your force of gravity, F. Another law of physics is that force is related to mass and acceleration,
F = ma
If you move the terms around you get:
a = F/m
If you fill in your body weight (using a scale) for F, and fill in your body mass (using a balance) for m, you get a = 9.8m/s^2. This is how fast you will accelerate in free fall, relative to the earth.

Now for where the confusion comes in. You're saying something to the effect of "if I'm standing still, how am I accelerating? this is total BS" and you are correct. When you are standing still, you are not accelerating. That acceleration is a derived value under the condition that you are falling. When you are not falling, it does not apply, and you would use the univeral gravitation formula to find your weight when standing still.

disregardthat

Dec25-06, 02:19 PM

Well, you can't deny that gravitational acceleration and inertial acceleration is different? Inertial acceleration requires and external force pushing the object. And the accelerating object feels G power.

In gravitational acceleration an object does not feel G power. (i am not including air resistance here) it accelerates all right, but no force other that gravity "drags" the object.

My point actually was: It is more correct to say that an object falling due to gravity and accelerating is actually standing still, since it is only trying to find it's correct place in space.

So if an object experience 0 G, it won't move. If a star comes around and drags it towards it, the object still experience 0 G's even though it's moving. So why don't you just agree with me that gravity cannot be seen upon as a force like an object pushing another, attracting magnets and charges connecting. :)

disregardthat

Dec25-06, 02:28 PM

ah,

"So what people standing on the surface of the Earth perceive as the 'force of gravity' is a result of their undergoing a continuous physical acceleration caused by the mechanical resistance of the surface on which they are standing."

- Wikipedia, General Relativity

Another statement that proves my point that we are decelerating (accelerating in the opposite direction of the earth):

"Similarly, two balls initially at rest with respect to and above the surface of the Earth (which are parallel paths by virtue of being at rest with respect to each other) come to have a converging component of relative velocity as both accelerate towards the center of the Earth due to their subsequent free-fall".

- Wikipedia, General Relativity

devino

Dec25-06, 04:08 PM

Inertial Mass is the resistence to change of motion, I see it as a time delay. 2 masses one is half the size of the other and with the same force applied it would take the larger mass twice as long to react, accelerate and travel the same discance as the smaller one. This does not hold true for free fall and Jarle brings up a good point that free falling bodies don't feel a G force with gravitational acceleration. I had not thought of that.
When you are standing on the surface of the Earth, the net force on your body is zero. I have a problem with this statment and its not a problem with the posters reply but that this is what is taught in physics today. The net force on your body is not zero. I don't mean to argue but 'come on' there is work being done even if the work equation is showing zero work due to zero distance. This equation was desgined to measure the amount of work being done, and this is where it fails. I believe the reason why we are taught that 'net force is zero while standing on the earth' is that it might make gravity seem like perpetual energy. So dismissing it is way easier then opening up that can of worms.

You do not feel the acceleration due to gravity. D H, think of this, how do you know when your leaning to far to the left and starting to fall over while your standing up? It is your ability to detect gravity, or more precisely detect accelertation, in you inner ear. A sensory organ that can detect acceleration (I think of it as our sixth sence).

I also have trouble with imagining the upward force that holds objects from falling through the earth. Again general physics here, a stack of books sitting on a table are pushing down with 'x' force and the table is pushing up with equal opposite 'x' force. so If we can magically take the books away fast enough does the table fly up due to this upward force? I see it as weight displacement, no opposing force here unless you see displacement as equal and opposite resistance.

Jarle, I look at it as all objects are in motion and the term 'at rest' is realative but not absolute. I remember reading a quote from einstein and it caused me to think of gravity as an acceleration....perhaps an acceleration into time. I am really stretching this though here but it helps me understand time dilation due to mass.

D H

Dec25-06, 05:14 PM

You do not feel the acceleration due to gravity.
D H, think of this, how do you know when your leaning to far to the left and starting to fall over while your standing up? It is your ability to detect gravity, or more precisely detect accelertation, in you inner ear. A sensory organ that can detect acceleration (I think of it as our sixth sence).

The otoliths in our ears and the accelerometers in a spacecraft or fighter jets sense linear acceleration due to all forces but gravitation. No device that directly senses the acceleration due to gravity can exist. True free-fall is indistinguishable from inertial motion per the equivalence principle. General relativity went one step further: Per general relativity, free-fall is inertial motion.

Think of how this relates to what you said at the start of your post:
Jarle brings up a good point that free falling bodies don't feel a G force with gravitational acceleration.

------

I also have trouble with imagining the upward force that holds objects from falling through the earth. Again general physics here, a stack of books sitting on a table are pushing down with 'x' force and the table is pushing up with equal opposite 'x' force. so If we can magically take the books away fast enough does the table fly up due to this upward force? I see it as weight displacement, no opposing force here unless you see displacement as equal and opposite resistance.

The mechanical resistance is a true force. What would happen if you replaced the table with springs? The springs will fly up you take the books away fast enough. The table is nothing more than an extremely stiff spring.

disregardthat

Dec25-06, 05:17 PM

I just read the article about general relativity, or at least half of it on wikipedia. It got me thinking. If we are accelerating at the speed of 9.8m\s^2 as this quote means "people standing on the surface of the Earth perceive as the 'force of gravity' is a result of their undergoing a continuous physical acceleration caused by the mechanical resistance of the surface on which they are standing.", don't we get a higher velocity? let's say we accelerate at 9.8m\s for a long time, let's say thirty years, as most people live at this earth, wouldn't we get a velocity higher than speed of light? I know we wouldn't but we would be very near the limit since further adding of N force would decrease amount of acceleration. But that would mean that we had a very very high weight, since an object moving very close to speed of light, experience a extreme weight.

I know that gives little sense, but i didnt think logically, I just thought the effect of an acceleration of 9.8m\s^2 over thirty years, and that would give that effect. and the GR points out that we ARE accelerating 9.8m\s^2 AWAY from the earth at all times when we are standing on it. I know everything is relative. but wouldn't a perosn feel that he is moving at 9.9999999% at the speed of light, in the way of measuring the force needed to be added to reach a higher acceleration? (it's humonguos!)

And the table are not pushing the books upwards. the atoms are only sharing the weight of the books, and the earth's atoms is sharing the weight of the table. When the books are removed, the atoms have freedom to organaize as they were when the books werent there, this may cause a slight "jump", but of course a string's atosm would organize in a way more dramatically way than the table as D H says.

D H

Dec25-06, 10:35 PM

The table most certainly is pushing the books upward. There is no "sharing of atoms". At the interface between the book and table, the electrons orbiting the nuclei at the face of the book and the electrons orbiting the nuclei at the top of the table repel each other. The upward force is electromagnetic.

Your interpretation of "So what people standing on the surface of the Earth perceive as the 'force of gravity' is a result of their undergoing a continuous physical acceleration caused by the mechanical resistance of the surface on which they are standing (http://en.wikipedia.org/wiki/General_relativity#Treatment_of_gravitation)" is too three-dimensional. It is spacetime, not just space, that gravity curves. The mechanical resistance of the Earth against the people standing on it, and of table against the books piled on it, are real forces that counteracts the spacetime curvature. Without those mechanical forces, the books would follow a geodesic in spacetime. In other words, they would fall through the table.

disregardthat

Dec26-06, 09:31 AM

Well, if you look at it that way, then yes. The table are "pushing" the books at the same force as the books are "pushing" the table. But the electrons are only resisting the force of the "pushing" books, isnt that right?

Well, the quotes are taken directly from wikipedia about the general relativity which indeed are working with space time, a four dimensional space... There is no inertial force that are pushing the books. The gravity are letting the books "fall" into the earth. Or at least that is what it want. The table is resisting this acceleration. And the mechanical forces is excactly what I am saying that is decelerating for example the books from falling through the table. I didnt deny the mechanical forces, they was what i was thinking of....

Anyway, the main point was that an object in inertial movement is moving along the coordinates, the geometrical 2 dimensial area we can look at spacetime as. In free fall, the object is standing still in this geometrical area, it is not moving along any coordinates. At least that was what I ended up with when i was finished reading that article.

Ok, i don't know if this is right, it just seems logical at the moment: Look at it like this: If we only hypothetical lived in a 2 dimensial world, moving upwards wont give any meaning. Moving upwards in a three dimensional world would give meaning There was no force that moved a hypothetical object upwards, but if you wanted to move sideways, it needs force. Falling due to gravity would not give any meaning in a three dimensional world, but we are not living in a three dimensional world, we are living in Spacetime a four dimensional world, where free fall give a meaning. we are changing our position in this 4d world and see this as accelerating towards a point. If we only lived in a three dimensional world we would be standing still (with no external force added)

D H

Dec26-06, 11:13 AM

Jarle, you appear to have some misconceptions. What do you mean by "inertial force", "inertial acceleration", "inertial movement", and "standing still"?

An object in "free fall" is moving along a geodesic in curved spacetime. It is not "standing still".

disregardthat

Dec26-06, 11:36 AM

:\ I am not so into the definitions of the terms...

I think I meant inertial acceleration.

An object in "free fall" is moving along a geodesic in curved spacetime. It is not "standing still".

Yes, but isn't that the system used to coordinate the four dimensional spacetime we live in? If we strictly think if a coordinatal field in a three dimensional world we would be standing still? I am pretty sure i read about it in the article. I think what I meant was the three dimensional Euclidean space, that we are standing still in this three dimensional system. Not in Spacetime, the four dimensional geodesic space. Am I right, or is this way out into the wilderness?

devino

Dec27-06, 01:55 AM

Whats important for me is to try and understand the thoughts here. Because I said I had a problem with the upward force explination did not mean I thought it was incorrect, I just visualize it in a different way. I have been going over the replys here and I still need help with D H's explination of that force, upward force seems unrealistic. Maybe I am looking at it in the wrong manner.

The definition of terms used is important to help get our thoughts across as I am interested in understanding your ideas here and I hope you feel the same.

MeJennifer

Dec27-06, 02:17 AM

The table are "pushing" the books at the same force as the books are "pushing" the table. But the electrons are only resisting the force of the "pushing" books, isnt that right?

No, I don't think that is correct.

The books are not pushing the table and neither are the electrons resisting anything. It is the opposite the electrons are pusing the books up!

But of course this view is general relativity not Newtonian.

The key in understanding this leads to the core of general relativity, namely the equivalence principle.

MeJennifer

Dec27-06, 02:26 AM

If we strictly think if a coordinatal field in a three dimensional world we would be standing still?
One of the conclusions we can draw from relativity is that it is useless to say if something is standing still or is in constant motion. It is the same, the laws of physics are invariant to inertial motion.
We are only moving relative to something else, and that something else must have mass. But whether we accelerate or not is absolute and something we can measure.

A comet approaching the earth is in free fall (if we exclude atmospheric resistance for a moment). Now is it standing still or moving? It does not matter, if we say it is moving we always have to provide a refererence as to what it is moving in relation to. Is it accelerating? Well in a general relativity sense no, the movement is inertial. While movement is a relative concept in relativity acceleration is not. So when something stops the comet or slows it down, for instance the earth's atmosphere, then we can say that the electro-magnetic forces in the air accelerate this comet, accelerate away from the direction of the center of the earth.

disregardthat

Dec27-06, 10:01 AM

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.

An object could be in parallell acceleration in outspace with an object on earth, with the object in space moving by inertial force, and the object on earth moving because of the gravity. The difference is that the object on earth DOES NOT feel G powers, it is like it is just standing still(standing still=constant velocity, I may have explained myself a bit wrong here), even though it is accelerating.

If find that pretty impossible, that's why i need to (in my head) differ the two kind of movements. One is inertial motion\acceleration, and one is gravitational motion\acceleration.

I thought since the object on earth accelerationg is not feeling any G's they can't be "moving" in a 3 dimensional space, but in a 4 dimensional space, yes. If a object moved upwards in a 2 dimensional space it would be standing still in the observers eyes. If it moved upwards in a three dimensional spave it WOULD move in the observers eyes too.

That's why I mean that there can be a coherence between this when we add another dimension, the fourth. That the object in free fall are "standing still" in the three dimensional space, but not in the fourth.

.......or what?

marlon

Dec27-06, 10:28 AM

You are saying that acceleration and a constant velocity is not the same,

Ofcourse it's NOT the same.

and I assume it is because in acceleration you experience G force.

Huh ??? So, you conclude from this that acceleration and constant velocity are the same ? Explain, please.

The thing that doesn't say click in my head then, is gravity.

What do you mean by that ?

An object could be in parallell acceleration in outspace with an object on earth,

1) What's parallel acceleration
2) What's "outspace"

with the object in space moving by inertial force,

With respect to what frame of reference ? Without defining that, you cannot be speaking about "inertial forces".

and the object on earth moving because of the gravity.

Now THAT is an inertial frame.

The difference is that the object on earth DOES NOT feel G powers,

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.

it is like it is just standing still(standing still=constant velocity, I may have explained myself a bit wrong here), even though it is accelerating.

1) standing still is NOT the same as having constant velocity.
2) standing still eventhough it is accelerating ? What the ... ? With respect to what reference frames are you talking here ?

I thought since the object on earth accelerationg is not feeling any G's

Nonsense, check the definition of G-force again.

they can't be "moving" in a 3 dimensional space, but in a 4 dimensional space, yes. If a object moved upwards in a 2 dimensional space it would be standing still in the observers eyes. If it moved upwards in a three dimensional spave it WOULD move in the observers eyes too.

Jarle, this is just rubbish. Pardon the French but do YOU even know what you are saying here ?

Look, if you wanna 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.

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.

regards
marlon

marlon

Dec27-06, 10:41 AM

The books are not pushing the table and neither are the electrons resisting anything. It is the opposite the electrons are pusing the books up!

But of course this view is general relativity not Newtonian.

So you say that this electron behaviour is described by general relativity ? I would very much disagree with that because last time i checked, electrons are not really heavy mass objects :wink:

Also, Jarle is using electrons to explain gravity. We should be explaining him the concept of physical regimes. To explain the book-stuff, WE DON'T NEED TO BRING IN ELECTRONS. Electrons are atomic scaled entities and thus described by QM ! I know what he is trying to say (he is actually referring to the exclusion principle) but we first need to set some things straight here. Starting with : using the correct language that corresponds to mainstream physics. If we don't do that, this thread will degenerate (which is already going on) into a bunch of useless posts full of misconceptions and mistakes.

Let's get back on the right track, people, or this thread will be closed.

marlon

MeJennifer

Dec27-06, 10:45 AM

So you say that this electron behaviour is described by general relativity ? I would very much disagree with that because last time i checked, electrons are not really heavy mass objects :wink:

No I am not, and I strongly suspect that you very wel know I do not say that at all.
So please be so kind and stop twisting what I am writing here.

marlon

Dec27-06, 10:48 AM

No I am not, and I strongly suspect that you very wel know I do not say that at all.
So please be so kind and stop twisting what I am writing here.

Twisting ? I am just asking. So, what do you mean by the last sentence in :

"The books are not pushing the table and neither are the electrons resisting anything. It is the opposite the electrons are pusing the books up!

But of course this view is general relativity not Newtonian."

Even a book and a table are not heavy mass objects :wink:

marlon

D H

Dec27-06, 11:01 AM

An important thing to remember is that the correspondence principle applies to general relativity and to quantum mechanics. That is, those theories must yield results commensurate with that which we see/measure in our ordinary world.

For example, we see an object that is subject only to gravitational forces (i.e., in free-fall) accelerating in a Newtonian inertial frame. General relativity must yield an commensurate answer. It does just that. Gravitation is not a force in general relativity. The massive object geometrically distorts space and time. An object follows a straight line in this distorted space-time. In our distorted view, we see the object as accelerating.

We see a book standing still on a table. In Newtonian mechanics, the net force on the book must be (and is) zero. The downward gravitational force is counterbalanced with an upward mechanical force. Quantum mechanics explains the upward force quite nicely. It is a real force. We know how the force arises and what carries the force.

In general relativity, the table frame is not an inertial frame. The gravitational "force" that appears to pull on the book is a fictitious force in general relativity, akin to the Coriolis force in a rotating reference frame in classical physics.

-----------

:\ I am not so into the definitions of the terms...
I think I meant inertial acceleration.

This is a contradictory term. Inertial motion is motion without acceleration: Newton's First Law of Motion.

I have been going over the replys here and I still need help with D H's explination of that force, upward force seems unrealistic. Maybe I am looking at it in the wrong manner.

Try turning things sideways if the upward force seems unrealistic. Instead of gravity pulling the book downward, imagine you are pushing the book horizontally. The book pushes back on you, does it not? Newton's Third Law: for every action, there is an equal and opposite reaction.

marlon

Dec27-06, 11:09 AM

We see a book standing still on a table. In Newtonian mechanics, the net force on the book must be (and is) zero. The downward gravitational force is counterbalanced with an upward mechanical force.

Agreed, but what exactly do you mean by "mechanical" force ?

Quantum mechanics explains the upward force quite nicely. It is a real force.

How ? The upward force is there because gravity avts on the book. That is Newton's third law. No problem. But if gravity is not acting on the book, or the book is not laying on a table, the upward force is NOT there. You say that QM explains this upward force. I politely ask you TWO THINGS :

1) how does QM explain the upward force exactly ?
2) what QM property is responsible for this force and in the case there is no upward force, is that QM entity no longer there because the force is not there ?

In general relativity, the table frame is not an inertial frame.

So ? Isn't general relativity all about "we no longer need inertial frames" ?

The gravitational "force" that appears to pull on the book is a fictitious force in general relativity,

WHAT ? A fictitious force in general relativity ? Why do you think GR exists in the first place ? What problem of special relativity is solved in GR ?

What you are saying here is not correct, sorry.

akin to the Coriolis force in a rotating reference frame in classical physics.

Now, THAT is correct.

marlon

marlon

Dec27-06, 11:19 AM

The table most certainly is pushing the books upward. There is no "sharing of atoms".

CORRECT.

At the interface between the book and table, the electrons orbiting the nuclei at the face of the book and the electrons orbiting the nuclei at the top of the table repel each other. The upward force is electromagnetic.

First of all, in QM, electrons do NOT orbit nuclei because that violates the HUP. Keep in mind that the Bohr model is not complete ! Check out our FAQ thread for that matter.

Secondly, you say that the upward force is electromagnetic because of repulsion and opposing gravity. I agree with the latter but concerning the former, please look at this :

Suppose we rotate the book + table 90° so that book and table are vertical now. We just put them together like this || (first | = book, second | is the table) and they are touching sides so there is " an interface". Now, the repulsion must still be there because the interface nuclei are still the same. Actually, the magnitude of this force is STILL equal to the opposing gravity force. Yet NOTHING happens : the book will NOT move away from the table !

You see ? Clearly, the opposing force is NOT electromagnetic. It has NOTHING to do with the interface nuclei what so ever because if that was the case, the opposing force must also depend on the material of the table.

I hope you get my point. Indeed, the opposing force can be explained using QM (and yes, there is some interfacial charge repulse as a REACTION to gravity, yet that is NOT the classical EM force but more a Van der Waals like quadrupole (local repulsion of electronic charge densities) interaction) but :

1) this is NOT necessary
2) has nothing to do with the actual nuclei or atoms at the interface !

My point : let's not use QM in classical physics. That just does NOT make any sense and THAT's why we have this little thingy called "physical regimes".

marlon

D H

Dec27-06, 11:50 AM

I agree that QM is not needed here. Then again, neither is GR. The Newtonian approximation for human-sized, low-speed, Earth-bound objects is far more accurate than the accuracy of all but the finest of instruments.

I brought in QM because some of the posters didn't see an upward force on the book on the table. The upward force exists because (a) gravity is pulling the book down, and (b) the book is laying horizontally on the top of the table. If there is a better name for this upward force than "mechanical", tell me. (No sarcasm here. Proper terminology is important.)

In the Newtonian model, the upward force must exist to make the net acceleration of the book zero. In the GR model, something impedes the book from following a geodesic through the table. In light of a better name, that something is a force.

marlon

Dec27-06, 11:55 AM

D H

Dec27-06, 12:10 PM

I understand the point you make and i agree with it.

But i was primarily talking about the content of post 31 :

1) you talking about fictitious forces in GR
2) the distinction you make, between fictitious and real forces in GR

Again, i just wanted to create some clarity here. That's all.

marlon

Oh crap.

A little knowledge is dangerous, and I was being very dangerous using wikipedia as a fount of knowledge. I was relying on this little tidbit from http://en.wikipedia.org/wiki/Introduction_to_general_relativity#The_equivalence _principle
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.

and I didn't bother to look at the discussion page for that article to see whether that statement is disputed.

disregardthat

Dec27-06, 12:13 PM

:uhh:

"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""
:uhh: 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 eachother.

"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 wanna 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

Dec27-06, 12:14 PM

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

Dec27-06, 12:37 PM

:uhh:

"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""
:uhh: 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 eachother.

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 :rolleyes:

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

Dec27-06, 01:06 PM

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 criticise 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 wierd 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

Dec27-06, 01:16 PM

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

Dec27-06, 01:21 PM

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

Dec27-06, 02:00 PM

Well, how would you explain and define inertia?

marlon

Dec27-06, 02:15 PM

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 criticise 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 :wink: 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

Dec27-06, 02:19 PM

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

Dec27-06, 02:29 PM

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

Dec27-06, 02:38 PM

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

Dec27-06, 02:59 PM

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

Dec27-06, 05:10 PM

:) 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

Dec28-06, 06:54 AM

:) 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 :wink:

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

Dec28-06, 11:30 AM

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 eachother, 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

Dec28-06, 02:27 PM

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 eachother, 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 occuring. 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

Dec28-06, 05:35 PM

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 eachother 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...

[B]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

Dec28-06, 06:45 PM

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 wanna 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 (http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pe). You will also find another explanation of the "-" sign :wink:. Go check it out.

[B]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

Dec28-06, 07:17 PM

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

Dec29-06, 06:37 AM

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

Dec29-06, 11:51 AM

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

Dec29-06, 12:26 PM

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

Dec29-06, 01:19 PM

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

Dec29-06, 02:05 PM

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

Dec29-06, 03:14 PM

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

Dec29-06, 06:31 PM

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 Gravity Recovery And Climate Experiment (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

Dec30-06, 06:44 AM

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

Dec30-06, 06:47 AM

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

Dec30-06, 07:13 AM

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

Dec30-06, 08:17 AM

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

Dec30-06, 08:29 AM

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

Dec30-06, 08:30 AM

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

Dec30-06, 08:34 AM

(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

Dec30-06, 08:37 AM

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 :wink:

Apologies for butting in Marlon et al.

No problem.

marlon

disregardthat

Dec30-06, 09:38 AM

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

Dec30-06, 10:36 AM

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 :wink:

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

Dec30-06, 03:48 PM

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

Dec31-06, 05:00 AM

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

Dec31-06, 05:02 AM

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

Hootenanny

Dec31-06, 05:17 AM

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

Dec31-06, 07:40 AM

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

Dec31-06, 07:45 AM

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

Jan3-07, 07:57 PM

, 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

Jan3-07, 08:00 PM

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

Jan3-07, 08:13 PM

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

Jan4-07, 10:17 AM

Hootenanny

Jan4-07, 10:48 AM

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, '' matters in [I]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

Jan4-07, 10:58 AM

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

Jan4-07, 03:10 PM

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

Jan6-07, 11:10 AM

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

Jan6-07, 11:15 AM

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...