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ToSeek
2004-Apr-21, 02:05 PM
Someone on GLP is claiming that gravity violates the law of conservation of energy because it keeps doing work without running down. This is one of those things that's almost too simple to deal with. It seems obvious to me that his claim is wrong, but I can't put it into words. Would one of our local physics experts like to take a shot at it? Thanks.

Glom
2004-Apr-21, 03:01 PM
For starters, he doesn't even know which law he means. It wouldn't violate the first law, it would violate the second law of thermodynamics.

In order for gravity to be inefficient, there would need to be a means of gravitational potential energy to be dissipated without the object moving through the gravitational potential field. That doesn't happen. Inherent inefficiency is due to dynamic factors. Just sitting at a point in a gravitational potential field isn't dynamic.

ToSeek
2004-Apr-21, 03:12 PM
For starters, he doesn't even know which law he means. It wouldn't violate the first law, it would violate the second law of thermodynamics.


Not from where he's coming from, I don't think. My impression is that what he's trying to say is, to give an example, if you're a long distance away from the Earth with an infinite number of rocks, you can keep dropping those rocks forever, and the Earth's gravity doesn't get used up. It doesn't make sense to me that it would, but I really can't explain why.

swansont
2004-Apr-21, 03:14 PM
Someone on GLP is claiming that gravity violates the law of conservation of energy because it keeps doing work without running down. This is one of those things that's almost too simple to deal with. It seems obvious to me that his claim is wrong, but I can't put it into words. Would one of our local physics experts like to take a shot at it? Thanks.

"Just existing" isn't doing any work. You're right that it's a simple thing that's tough to deal with. I think it's a notion that "everything runs down because of friction" that people who aren't familiar with Newton's laws have, and don't really know how "work" is defined. An object held to the surface of the earth has no work done on it by gravity, since there is no displacement along the direction the force. No energy is dissipated. (planet rotation is perpendicular to the force, so no work there either)

The same notion comes into play with magnets - that there must be an energy source because it keeps sticking to the fridge - and leads some into trying to get perpetual motion. Magnets don't run down in that way because magnetic forces do no work.

Glom
2004-Apr-21, 03:18 PM
Not from where he's coming from, I don't think. My impression is that what he's trying to say is, to give an example, if you're a long distance away from the Earth with an infinite number of rocks, you can keep dropping those rocks forever, and the Earth's gravity doesn't get used up. It doesn't make sense to me that it would, but I really can't explain why.

Yes, but those rocks have to be lifted up through the G-field first.

Ut
2004-Apr-21, 05:28 PM
Not from where he's coming from, I don't think. My impression is that what he's trying to say is, to give an example, if you're a long distance away from the Earth with an infinite number of rocks, you can keep dropping those rocks forever, and the Earth's gravity doesn't get used up. It doesn't make sense to me that it would, but I really can't explain why.

Yes, but those rocks have to be lifted up through the G-field first.

Yeah. No matter how far away from the Earth you are with your rocks, you're still in the Earth's gravitational field, and you have gravitational potential energy due to your position in the field. The rocks have energy, too. Releasing the rocks simply causes their potential energy to be converted into kinetic energy. The rocks will not gain energy, simply change what kind they have.

There, I think I said that in enough different ways for a GLPer to at least understand what's being said.

Incidently, using an "infinite number of rocks = infinite energy" argument will probably spawn if your example is being used over there. If it does, I want to see the fallout.

Ricimer
2004-Apr-21, 06:01 PM
Yeah, its basically what glom said.

If the gentleman isn't going to violate conservation of energy in the first place (by magically "appearing" the rocks) then he has to drag those rocks from the earths surface, up to wherever he drops them.

The energy he expends getting the rocks there is retrieved when he drops the rock and it goes back.

If he just appears it there, that violates conservation, in and of itself. So you better end up with a net energy gain.

eburacum45
2004-Apr-21, 08:12 PM
I might be wrong here, but this is the way I see it;

The attractive force of gravity is a result of (and in opposition to) the original expansion of the universe, and if all the objects in the universe were to fall back together into a single point mass once again then gravity would have finally exhausted itself.

However this 'big crunch' scenario looks extremely unlikely, as the force of expansion in the universe is much greater than that of gravity/contraction for some reason, and apparently is increasing in strength.

PhantomWolf
2004-Apr-22, 01:11 AM
Actually the closer you are to the center of gravity, the less energy you have, so the objects falling to get as close to that point as possible are losing energy and thus obeying the 2nd Law.

darkdev
2004-Apr-22, 05:34 AM
"Just existing" isn't doing any work. You're right that it's a simple thing that's tough to deal with.
I like this one. Picture gravity as a mountain. It does not work at all. If you happen to be at the top, you have the potential to fall down. If you are at the base, you have no potential to fall down, unless you are drunk...

And I know for a fact alcohol effects gravity and angular velocity ;)

Toutatis
2004-Apr-22, 11:16 AM
Kind friends,

I have noticed several threads Re: gravitation lately --- While most replies are ‘spot on’ I feel greater understanding might be lent via broader scope than that ‘invited’ by the OPs --- Please note: the following discussion is greatly simplified – but with an eye to perspicuity NOT condescension… :-)


Q) Why doesn't gravity run down?

A) Succinctly, gravity in not energy - it is, instead, a REACTION to chronological linear acceleration. Matter, owing to its 'deflection' of space-time, engenders chronological linear acceleration (note: linear acceleration, in the strictest sense, implies alteration of linear velocity. Linear velocity is a vector quantity (where Magnitude = rate of linear displacement while Direction relates to the line along which said displacement occurs...) Hence alteration in either or both of rate or direction constitutes acceleration...

As a workaday example consider the 'spatial' linear mass ‘forces’ experienced by an 'observer' while changing the rate and/or direction of his/her automobile --- the mass of the observer's body 'attempts' to proceed along the 'original' line at the 'original' rate however the motion of the Auto, having been ‘adjusted’ in regards to said parameters, 'interferes' creating the illusion of a force... Do you see? - Mass 'forces' aren't really forces at all - but rather reactions (That said: the conservation laws are not violated by *true* [static] forces [e.g. static magnetism, electrostatic attraction/repulsion, ponderomotive effects [where applicable], etc…) A static force cannot perform work (net) and, hence, is not representative of energy...

But to return to gravitation:

Comprehension of gravitation at the intuitive/qualitative level requires that one ‘get their mind round’ the fact that time is a dimension of space --- Inasmuch as mass alters chronological velocity -- (all other influences aside) distinct quantities of mass will tend toward convergence --- Please think really, really hard about this (bearing the more conventional mass forces in mind) then *WHAM* you'll comprehend gravitation at the ‘visceral’ level --- *I PROMISE* http://www.badastronomy.com/phpBB/images/smiles/icon_biggrin.gif

Of course such comprehension is not necessary so long as one has a quantitative grasp of the subject --- BUT IT’S LOTS MORE FUN!!! :-)

Very best regards
D. Sarandon

Eroica
2004-Apr-22, 11:43 AM
My impression is that what he's trying to say is, to give an example, if you're a long distance away from the Earth with an infinite number of rocks, you can keep dropping those rocks forever, and the Earth's gravity doesn't get used up.
When the rocks are falling, they do take energy from the Earth. But when they land and come to rest, they give it back.:-k

Glom
2004-Apr-22, 11:47 AM
As the rocks are lifted through the potential field, they gain potential energy. As they fall, this potential energy is transferred to kinetic energy. When it hits the surface, the kinetic energy is released as heat and sound and kinetic energy in the ejecta.

Eroica
2004-Apr-22, 11:52 AM
As the rocks are lifted through the potential field ...
What if the rocks come from Mars? :D

Toutatis
2004-Apr-22, 12:26 PM
Ok here's one for ya...

Suppose we lift a rock off and away from Earth (thus 'storing' [potential] energy)

Then convert all Earth's mass to energy --- What happens to the potential energy 'invested' when we extracted the rock???

Hint: If you're pondering "Could it be that conservation of energy is inapplicable to potential energy???" - Careful!!! (To wit: not quite!!!)

There is an answer to this -- however I think I'll let ya sweat awhile http://www.rainbowpirates.com/forum/images/smiles/icon_twisted.gif

I may make you feel but I can't make you think -- Anderson

TTFN
Sarandon

George
2004-Apr-22, 03:11 PM
As the rocks are lifted through the potential field, they gain potential energy. As they fall, this potential energy is transferred to kinetic energy. When it hits the surface, the kinetic energy is released as heat and sound and kinetic energy in the ejecta.

This is how I see it, too. The energy released leaves the system and the net gravitational field will not be the result of the sum of the orginal two masses since energy (which counts as mass) has left the system. The final two masses + energy release should equal 0. Entropy gets involved, too, no doubt, as the turn of events has gone "down hill".

rsa
2004-Apr-22, 04:10 PM
Ok here's one for ya...

Suppose we lift a rock off and away from Earth (thus 'storing' [potential] energy)

Then convert all Earth's mass to energy --- What happens to the potential energy 'invested' when we extracted the rock???

My guess... the sudden change in the gravitational field would cause the potential energy to be paid back via the release of gravity waves.

???

Glom
2004-Apr-22, 04:40 PM
Ok here's one for ya...

Suppose we lift a rock off and away from Earth (thus 'storing' [potential] energy)

Then convert all Earth's mass to energy --- What happens to the potential energy 'invested' when we extracted the rock???

According to relativity, energy also warps spacetime and so in effect casts a gravitational field. Energy and mass are equivalent. As long as the energy is still where it is, there'll be no change.

Wiley
2004-Apr-22, 09:36 PM
Ok here's one for ya...

Suppose we lift a rock off and away from Earth (thus 'storing' [potential] energy)

Then convert all Earth's mass to energy --- What happens to the potential energy 'invested' when we extracted the rock???

According to relativity, energy also warps spacetime and so in effect casts a gravitational field. Energy and mass are equivalent. As long as the energy is still where it is, there'll be no change.

This reminds me of the problem, "Which weighs more a compressed spring or an uncompressed spring?" The answer: a compressed spring.

Dropping a rock is similar, but a closed system. The total mass + energy of the Earth-rock system is constant. As the rock gains kinetic energy, a little bit of mass is lost.

Glom
2004-Apr-22, 10:13 PM
Dropping a rock is similar, but a closed system. The total mass + energy of the Earth-rock system is constant. As the rock gains kinetic energy, a little bit of mass is lost.

No, the mass remains constant because in a conservative system, the energy remains constant. The energy is changed between kinetic and gravitational potential, but it's still there and hence the mass equivalence is still there.

George
2004-Apr-22, 10:29 PM
Dropping a rock is similar, but a closed system. The total mass + energy of the Earth-rock system is constant. As the rock gains kinetic energy, a little bit of mass is lost.

No, the mass remains constant because in a conservative system, the energy remains constant. The energy is changed between kinetic and gravitational potential, but it's still there and hence the mass equivalence is still there.

I agree with Glom (once again). But I am curious if the falling rock becomes hotter due solely to the falling process (excluding other reasons such sunlight and friction from gas particles,etc). For some reason, I want to say entropy increases so the rock will get hotter as it falls but will it? #-o

constible
2004-Apr-23, 01:11 PM
When one drops a rock from a mountain, the earth moves towards the rock and the rock moves towards the earth. It's just that the earth is much more massive so there's no noticeable effect on it.

The energy used by the rock is the same as the energy used by the earth, just in opposite directions, cancelling each other out.


That is a simple way of looking at it.

swansont
2004-Apr-23, 01:45 PM
When one drops a rock from a mountain, the earth moves towards the rock and the rock moves towards the earth. It's just that the earth is much more massive so there's no noticeable effect on it.

The energy used by the rock is the same as the energy used by the earth, just in opposite directions, cancelling each other out.


Note, however, that energy is not a vector.

Eroica
2004-Apr-23, 03:57 PM
An apple falls from a tree. Its rest mass when it's lying on the ground is less than its original rest mass when it was hanging from the tree. Right? :-k

milli360
2004-Apr-23, 04:00 PM
An apple falls from a tree. Its rest mass when it's lying on the ground is less than its original rest mass when it was hanging from the tree. Right?
Why?

Eroica
2004-Apr-23, 04:19 PM
Well, I thought that objects fall because they're trying to attain a lower energy state. Electrons occupy the lowest energy level available, and if there's anything to the Unified Field Theory, something similar should apply to gravity.

When the apple is hanging from the tree all its energy is rest-mass, therefore its gravitational potential energy is part of its rest mass. In lifting it up to the tree its rest-mass would have increased by the amount of potential energy it gains. When it falls that extra energy is converted into kinetic energy. When it hits the ground that kinetic energy is released as heat or some form of mechanical energy. So its rest-mass is now less than it was when it was on the tree. :-k

milli360
2004-Apr-23, 04:31 PM
When it hits the ground that kinetic energy is released as heat or some form of mechanical energy.
When I took Physics 101, our labs included an experiment where we would take lead shot and load it into a two meter tube, and seal it at both ends. There was a hole for a thermometer, to measure the heat of the lead shot. Then, we would turn the tube over and over, essentially allowing the shot to fall hundreds of meters. Then we measured the temperature again--it corresponded pretty well to the conversion.

So, maybe the apple is just hotter? :)

JohnOwens
2004-Apr-23, 07:09 PM
When one drops a rock from a mountain, the earth moves towards the rock and the rock moves towards the earth. It's just that the earth is much more massive so there's no noticeable effect on it.

The energy used by the rock is the same as the energy used by the earth, just in opposite directions, cancelling each other out.

Note, however, that energy is not a vector.
Worse than that, it's actually the momenta (which are vectors) that exactly cancel each other. And since p = m*v, and E = m*v^2/2, E = p*v/2. Since the v's are very different for Earth and a rock but the p's must be the same, the E's will be very different, i.e. tiny for Earth, big for the rock.

Added:
Well, I thought that objects fall because they're trying to attain a lower energy state. Electrons occupy the lowest energy level available, and if there's anything to the Unified Field Theory, something similar should apply to gravity.

When the apple is hanging from the tree all its energy is rest-mass, therefore its gravitational potential energy is part of its rest mass. In lifting it up to the tree its rest-mass would have increased by the amount of potential energy it gains. When it falls that extra energy is converted into kinetic energy. When it hits the ground that kinetic energy is released as heat or some form of mechanical energy. So its rest-mass is now less than it was when it was on the tree. :-k
Or look at it with GR. The apple closer to Earth will act much like an apple under constant acceleration, i.e. time passes slower for it from the higher apple's point of view. Would this mean it also seems to have a greater mass, if measured by the higher apple? These two lines of reasoning seem to give opposite results. Perhaps they just cancel out, energy from the one way of thinking goes to the other, and the rest mass remains the same? :-k

Toutatis
2004-Apr-25, 12:11 PM
Just a quick note to advise all interested parties that I haven't neglected my (attempted) ‘brain teaser’ (back yonder) --- :-) --- I'll be on it like a duck on a June bug just as soon as time permits the thorough ‘treatment'! --- Please note that the question presumes conservation of *potential* energy - sans insistence such presumption is valid...

Some of your responses imply that (Re: gravitation) the seat of 'potential energy' lies within the masses involved - An assertion which would seem to require increment in mass paralleling (but not necessarily proportional to) increased 'potential energy' http://www.badastronomy.com/phpBB/images/smiles/icon_eek.gif - (No comment... yet ;) )

Hint #2 Where *TRUE* force fields are involved (e.g. magnetism) the 'potential energy' is 'stored' in the field (think ideal elasticity) -- HOWEVER, as discussed in another post, gravitation is merely reaction not *true* force... (What does this suggest? ;) )

BTW nothing here should be construed as suggestive that any response received thus far is correct OR incorrect http://www.badastronomy.com/phpBB/images/smiles/icon_biggrin.gif

Very best regards
Sarandon

poorleno
2004-Apr-25, 07:05 PM
I'm sorry; but i didn't read the entire thread. The first 15 posts or so mention nothing about the duality of this "magic energy", however. Just thought i'd mention that even though it seems that the apple is falling toward the ground, the ground is also falling toward the apple with respect to its relative mass.
The way I think about it, is that the end result is the same as the start : Sure the apple moves more, but the focus point of the sum of these two masses stays at the same place, does it not?

PhantomWolf
2004-Apr-25, 11:52 PM
It seems that people are forgetting their Phyiscs 101.

While lifting a rock will increase its potential energy, but it isn't gained from the Earth, it is gained from the WORK that the lifter is doing to raise position and therefore its potential. Because you are doing work on the rock, it allows of an increase of energy, quite within the 2nd Law. Dropping the rock will cause the Gravitational potenital energy to attempt to dimish itself by forcing the rock downwards and thus transfer the potential energy into kenetic energy resulting in a total energy loss. This means when dropped that no Work has been done on the rock, it's just obeying the 2nd Law and attempting to get to a lower energy state.

Glom
2004-Apr-26, 01:50 AM
This means when dropped that no Work has been done on the rock, it's just obeying the 2nd Law and attempting to get to a lower energy state.

Ooh. Wicked in its omniscience.

George
2004-Apr-26, 03:24 AM
This means when dropped that no Work has been done on the rock, it's just obeying the 2nd Law and attempting to get to a lower energy state.

But the rock is driven by the force acting on it as it is in a gravitational field. Therefore, the gravitational field is doing work on it as it travels down. It is the same amount of work it will take to move it back (excluding irreversible considerations such as air friction).

PhantomWolf
2004-Apr-26, 03:49 AM
This means when dropped that no Work has been done on the rock, it's just obeying the 2nd Law and attempting to get to a lower energy state.

But the rock is driven by the force acting on it as it is in a gravitational field. Therefore, the gravitational field is doing work on it as it travels down. It is the same amount of work it will take to move it back (excluding irreversible considerations such as air friction).

I must admit my Physics is rusty, but the way I understand it, you could argue for a -negative- amount of work, but gravity isn't doing any work on it at all. It is more like electrons attempting to lower their energy state, in the same way the rock lowers its energy state by travelling towards position of lowest possible energy, until something stops it. Because it is losing energy it is not having work done on it, you only can have work when something is gaining energy. At least that's the way I was taught it, so if my Physics professeor got it wrong....

Eroica
2004-Apr-26, 11:35 AM
While lifting a rock will increase its potential energy, but it isn't gained from the Earth, it is gained from the WORK that the lifter is doing to raise position and therefore its potential...
I would regard the lifter as part of the Earth. After all, the lifter's mass contributes to the gravitational field that "pulls" the falling apple back down to the ground.

George
2004-Apr-26, 04:31 PM
Whew, you would think ToSeek's question wouldn't be this hard. #-o

It may be a matter of convention as to how we describe the behavior of the rock since we don't know exactly what gravity is. However, we can nicely describe it's behavior in normal circumstances. Any force acting on a body (rock) that causes it to move a given distance can equate to a value we call "work". It seems fair to say that gravity is not the actual force but it may be fair to say that it is the gravitational field that is responsible for the force we can measure acting upon it.

Some will say it is the rock we see "falling" that is really at rest, and it is Earth that is rising up to it due to the equivalence principle which says gravity and acceleration are the same. Thereby, it was Newton who was accelerating up to the apple and not the apple falling down on him (did that actually happen?). In engineering school, we only had one class that dealt with deeper physics which we called "science fiction 101". It seemed superfluous as the Schroedinger and Einstein equations were impractical in a our normal macro and relatively slow (pun intended) work environment.

I really am not good at this stuff, but I would guess the work calculations, based on the force (gravitational) on the rock and the distance traveled, will accurately represent the rock's real world situation. The conversion of it's potential energy to kinetic energy, as well as momentum, are also straight forward and easy (excluding friction, etc.).

Maybe once we learn the real "guts" in gravity we can see it more clearly as we do our understanding of electromagnetism.

I would still like to know if, and/or how, entropy would apply for a falling rock (no friction issues). #-o

PhantomWolf
2004-Apr-26, 10:42 PM
While lifting a rock will increase its potential energy, but it isn't gained from the Earth, it is gained from the WORK that the lifter is doing to raise position and therefore its potential...
I would regard the lifter as part of the Earth. After all, the lifter's mass contributes to the gravitational field that "pulls" the falling apple back down to the ground.

I guess this is were we differ. ;)

As far as I can tell any lifter, be it a tree taking nutrients, a person carrying it, or a crane, has to do the lifting with their own energy, not the Earth's. If you pick up a rock, then it is your Muscles doing the work, and they do that by breaking down sugars which store your energy, so you transfer some of that energy into the rock, energy from you, not from the Earth itself.

I'd also argue that gravity doesn't "pull" things but that objects move towards gravity by the field converting their gravitational Potential Energy into Kenetic Energy in as a way of lowering their overall energy state (something that all things are wanting to do under the 2nd Law.) This does allow for an interesting prediction. If the Rock is the object moving towards the gravity source, rather than the gravity applying force and therefore Work, not only do we get the observed energy decrease (if gravity was doing work on the rock it should increase its energy but we see an overall decreace, not increase) then we could claim that gravity is doing -Negative- work on the rock. This would increase the Gravity Source's energy and -INCREASE- the Gravity of the source, rather then decrease. Because the rock is now part of the gravity source then this is is true. It would be interesting to see if the gravity source actually increase in mass by greater the the rock's mass too, but I some how doubt that we can measure it if it does. :)