[VernonNemitz]

I mentioned the gravitational slingshot when we were talking about gravity and work, Vernon. See http://www.thenakedscientists.com/forum/index.php?topic=27444.msg291021#msg291021.

So, you are indulging in hypocrisy? to claim that energy can transfer during a gravitational slingshot, but not transfer for other events?  Tsk, tsk.  Such inconsistency makes your argument even more worthless.

A better example for "never actually touch" would have been a magnet.

Wrong again.  It is not a better example because the point of my last message was about how "touching" is actually only an illusion.  Thus, to insist that energy can only be transferred when objects are in contact is to be wrong, since most interacting and energy-transferring objects are not actually engaging in the kind of contact you are imagining (a prime exception would be chemical reactions, where two molecules combine, but not relevant here).

The reason a nail is attracted to a magnet is because its in a magnetic field, which is local to the nail. It's similar for a gravitional field. A field has energy, and is a non-uniform disposition of the space/energy caused by one object that has some effect upon another.

Agreed, yet when a proton and an electron start out separated, and approach and interact and merge to form a hydrogen atom (radiating some energy), not only do physicists recognize that the atom has less mass-energy than the separated particles, they also are confident that the two particles maintain their mass ratio throughout the event.  Ditto for initially-separated proton and neutron combining to form a deuteron; their mass ratio stays constant.  I apply this to gravitation, keeping things consistent; you don't, because of your unwillingness to accept that energy can be undetectably tranferred across distance in more cases than those involving magnets or gravitational slingshots.

All I can do is reiterate that when you fire a cannonball up into the sky at 11.2 km/s it escapes the Earth's gravitational field, which is then diminished.

All you can do is repeat the same nonsense as before (but, I see, with less detail than before).  Yes, the velocity and kinetic energy of the cannonball is reduced during an escape from a gravitational field, and potential energy increases somewhere in the system while that happens.  Your feeble attempt to confuse the description of that system is noted and rejected.  We most certainly do not combine the mass of the cannonball with the mass of the main gravitating body, when computing the strength of the gravitational field that you are intending the cannonball to escape.  The system consists of separate masses (else, by definition, the cannonball would not be escaping!).

You really should look at the example I gave which explains why the energy goes into the cannonball rather than the far more massive Earth which has no detectable recoil. 

If I hadn't seen that or an equivalent example months ago, I wouldn't have started this discussion with you, pointing out the flaws in that explanation.  For example, the "no detectable recoil" is a red herring, since it is irrelevant; we are talking about gravitational potential energy being converted into an immeasurably-small (with current instruments) amount of mass, after all!  Who cares if planetary or stellar recoil is also immeasurable with current instruments?!!  I don't, and neither should you.  What matters in a hypothesis is consistency, throughout its own details, and also with other known observations.  Which you haven't got, not at all, in your hypothesis.

[Geezer]

Can we consider the question in terms of the Earth and an object of significant mass, for example, 10% of the mass of the Earth?

Although the forces involved might be a little greater, I would not think the underlying physics would differ in the slightest. A model that only works for cannonballs of negligible mass (relative to the mass of the Earth) does not strike me as terribly viable.

[VernonNemitz]

Can we consider the question in terms of the Earth and an object of significant mass, for example, 10% of the mass of the Earth?  Although the forces involved might be a little greater, I would not think the underlying physics would differ in the slightest. A model that only works for cannonballs of negligible mass (relative to the mass of the Earth) does not strike me as terribly viable.

In general, though, it is "accepted" to initially construct models that are not complicated --complications should be added after one is convinced that a basic notion works.  In this case the basic notion is that gravitational potential energy can be described as part of the overall mass of a System (not just "mass-energy" but actual simple mass).  Given 400+ years of associating potential energy with height-in-a-gravity-gradient, you perhaps can see some wisdom in introduce Change in a small way rather than a big way.  So, the change I've been proposing is specifically designed to be compatible-with/similar-to descriptions of potential energy in terms of the other Forces known to Physics.  And yes, if a proton and neutron, having similar masses, can maintain their mass ratio from the moment of an initial Strong Force interaction when separate, to final merging to form a deuteron, and releasing a significant amount of "binding energy", then I have no qualms about talking similarly about planetary-magnitude masses interacting similarly (though merely gravitationally).  The main difference will be a smearing of said planetary bodies during the final moment of the merge, something the proton and neutron don't do!  Which is why such a version of the event might be called "hairy"  [O8)] and offer a rationale to stick with the simple cannonball or plate, for now.

[Geezer]

Thanks Vernon. But surely it's the same system whether it's a cannonball or a moon (and I think it was you that pointed out it is a system). If the system is somehow "different" because the masses involved happen to be different, I have a really hard time believing we understand any of this.

EDIT: Come to think of it, if we need a simplifying assumption, let's assume the object has the same mass as the Earth. The original question imposed no limitations on the relative masses.

[Farsight]

A 10% mass is better Geezer. Consider some impulse that separates a 100kg object m₁ and a 10kg object m₂. Conservation of momentum means m₁v₁ = m₂v₂ so we know that the velocity v₂ of m₂ from the point of origin is ten times v₁. Let's say they are 100 m/s and 10 m/s. We can then work out the kinetic energy of each object using KE = ˝mv²:

For m₁ we get: ˝ * 100 * 10  * 10  = 5000  Joules
For m₂ we get: ˝ * 10  * 100 * 100 = 50000 Joules

The energy is not evenly distributed. It's very simple stuff, and we do understand it. When you alter the mass ratios such that one is very much bigger than the other, very little of the kinetic energy goes into the larger mass. See http://www.physicsclassroom.com/Class/energy/U5L1b.cfm re gravitational potential energy - there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height. 

[VernonNemitz]

When you alter the mass ratios such that one is very much bigger than the other, very little of the kinetic energy goes into the larger mass. See http://www.physicsclassroom.com/Class/energy/U5L1b.cfm re gravitational potential energy - there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height. 

Tsk, tsk, the mechanism is called "gravity".  AND you are being deceptive; it is not kinetic energy that transfers so much as energy that transfers; please remember that we are talking about a conversion process, here!  Thus, kinetic energy of the smaller body gets converted to potential energy which appears as mass of the gravitating body.  If the two interacting bodies were more equal in mass, then the larger body would also have notice-able momentum and kinetic energy, and it would notice-ably slow, also, as the distance between the two bodies reaches a maximum.  Its kinetic energy becomes potential-energy-in-the-form-of-mass that is added to the mass of the smaller body (because, obviously, both bodies are pulling each other toward a final stop; it is only fair and balanced that if the large body sucks energy out of the smaller one, then the smaller one is doing the same to the larger one).  The net result is that the mass ratio of the two bodies remains constant throughout the event, just like other interacting masses do with respect to the other Forces in Physics.  So, Farsight, you are being silly to claim there is "no mechanism" for energy transfer, since there most certainly is for the Strong, Weak, and ElectroMagnetic forces.  Why shouldn't it work for Gravitation, also?  Or are you going to start claiming that the proton-to-neutron mass ratio is different inside a deuterium nucleus than when the two particles are widely separated?  Good luck convincing the Quantum ChromoDynamics particle physicists that have been measuring masses to high accuracy (and matching to theory) for decades!

Not to mention there is one other side-effect of claiming that the potential energy of an object is only part of its own mass.  This side-effect relates to neutron stars, and the accumulation of mass upon them.  It can be shown that in order for a falling body to reach 70+% of lightspeed, the terminal velocity of a typical neutron star, most of the original mass of the falling body will have to be converted into kinetic energy.  For a bigger neutron star, say 90%c terminal velocity, even more of the original mass of the falling body will have to be converted into kinetic energy.  After impact and energy-dissipation/escape, only a small amount of original mass actually gets added to the totality of the star.  Note that the bigger the star gets, the less mass gets added!  The thing called a "black hole" would be impossible to reach, an "asymptotic" endpoint of the curve of accumulation of ever-smaller arriving masses.  But black holes do exist.  Therefore you are wrong, Farsight, one more way (your hypothesis leads to a prediction that does not match observation).

Geezer, perhaps I wasn't clear enough in my previous message.  Two bodies with significant gravitation tend to merge to become a single body.  This makes it difficult to continue to think of them as being two bodies, when comparing before-and-after properties.  That's my only objection to using two large masses.

[Geezer]

there is no mechanism by which the kinetic energy of a vertically-fired cannonball is transferred to the earth or the earth's gravitational field as the cannonball slows down and reaches its final height. 

Farsight, I'm afraid that is incorrect. Just because the changes in KE and PE of the Earth are negligible when firing a cannonball, that does not mean they do not exist.

Rerun your test with two equal masses and you will discover that they both behave identically. Only the magnitudes of the effects vary with mass. The physics is the same, regardless of the relative masses.

Furthermore, enough with pejorative comments like "it's very simple stuff".

[Farsight]

Vernon, the kinetic energy becomes potential energy, but it's cannonball potential energy. It isn't transferred out of the cannonball. I've never claimed that black holes do not exist, just that they're better described by the Weinberg Field Interpretation rather than the Misner/Thorne/Wheeler Geometric Interpretation. This doesn't invalidate what I've been saying, not at all. Note that what I'm saying is no hypothesis, it's in line with educational establishments such as http://www.physicsclassroom.com/Class/energy/U5L1b.cfm.   

Geezer, the mass of the earth is 5.9742 × 10²⁴ kilograms. Run the numbers and you'll see that the cannonball gets the lion's share of the kinetic energy. There really is no mechanism by which this is  transferred to the earth or the earth's gravitational field. If you beg to differ, please explain it or provide some educational reference that does. The "simple stuff" comment wasn't perjorative, the arithmentic of momentum and kinetic energy really is straightforward.

[VernonNemitz]

Vernon, the kinetic energy becomes potential energy, but it's cannonball potential energy.

Nope; the cannonball is part of a system, and so is the potential energy part of that system.  Geezer has one important thing right about requesting us to talk about a system in which the masses are more nearly equal to each other; you will be unable to assign the potential energy to just one part of it.

It isn't transferred out of the cannonball.

Just because you say so, that does not mean you are right.  Neither of us is that good.  But at least my position is consistent with respect to the other three Forces in Physics, and all the experimental observations about how they act.  (The advantage is that the other forces are much stronger than gravitation; therefore much more potential energy and potential-energy-as-mass can be involved in their interactions --enough to be measured, such as the proton-neutron mass ratio I've previously mentioned.)  You have yet to explain why picking an inconsistent position makes sense.  The only thing you appear to have claimed is, basically, "I must be right because I don't understand the mechanism that makes any alternative explanation possible."  Which is a worthless argument.

I've never claimed that black holes do not exist, ...

Nor did I claim that you made any such claim.  What I did claim (in different words) was that your hypothesis has logical consequences that do not match reality, and therefore your hypothesis must be erroneous.

Note that what I'm saying is no hypothesis, it's in line with educational establishments such as http://www.physicsclassroom.com/Class/energy/U5L1b.cfm.  

Hardly!  I took a look at that page, and it clearly presents the conventional view that gravitational potential energy is defined in terms of a two-body system, not in terms of just one of those bodies.  Equally clearly, it is a hypothesis on both our parts to take the view that gravitational potential energy actually takes the form of mass, and to take the view that the field-gradient-of-one-body-and-height-of-second-body description is a workable and convenient bookkeeping trick, but not actually true (kind of like Feynmann's proposal that antiparticles move backward in time is a workable and convenient bookkeeping trick, but nobody who uses it seriously thinks it is actually true).  We even need that conventional explanation to compute how much potential-energy-as-mass we are talking about!  We only disagree on how that mass gets distributed in the system....

[Geezer]

Geezer has one important thing right about requesting us to talk about a system in which the masses are more nearly equal to each other; you will be unable to assign the potential energy to just one part of it.

Thanks Vernon. Yes, that's precisely my point.

Interactions between massive bodies make it very apparent that they are part of a common gravitational system. When the the ratio of the masses becomes very great we can disregard the KE and PE effects on the greater mass, which is the way we teach terrestrial mechanics/dynamics (as in the link that Farsight provided).

The measureable effects on the larger mass may asymptotically approach zero, but that does not mean they do not exist, it just means they are difficult to measure.

[Farsight]

Working through the numbers with equal masses does not assist with understanding here. Let's work the numbers with a 1kg cannonball fired upwards at 11,200 m/s. We’re ignoring air resistance, so this is escape velocity. The mass of the earth is 5.9742 × 10²⁴kg, so conservation of momentum means:

m₁v₁                   = m₂v₂
m₁v₁                   = 1 * 11,200
5.9742 × 10²⁴ x v₂= 11,200
v₂                       = 11,200 / 5.9742 * 10²⁴
v₂                       = 11.2 / 5.9742 * 10²¹
v₂                       = 1.8747 * 10^-21
 
Working out the kinetic energy via KE = ˝mv²:

For Earth we get: ˝ * 5.9742 × 10²⁴ * 1.8742 * 10^-20 * 1.8742 * 10^-20 =  1.0492 * 10^-17  Joules
For the cannonball we get: ˝ * 11,200 * 11,200 = 6.72 x 10⁷ Joules

Please check my arithmetic, but this says that the Earth acquires roughly one 10^-25th of the energy acquired by the cannonball. Since the cannonball has escape velocity it leaves the earth’s gravitational field and escapes the system. Conservation of energy tells us the cannonball takes the energy away with it, which can also deduce by imagining a new cannonball poised ready to fall to earth. Alternatively we can envisage the earth as a large collection of cannonballs being fired upwards in succession. If these fired cannonballs did not remove their kinetic energy, and if instead this kinetic energy was transferred to the residual cannonballs, the collective gravitational field of the latter would have to rise in proportion to their mass. This is because energy causes gravity, and mass only causes gravity because of the energy content - a rapidly spinning mass would exert more gravity than one that was motionless. Ergo the kinetic energy of the fired cannonball is converted into cannonball potential energy, and is then removed into the vastness of space.   

[VernonNemitz]

Since the cannonball has escape velocity it leaves the earth’s gravitational field and escapes the system. Conservation of energy tells us the cannonball takes the energy away with it ...

Your error is obvious. The kinetic energy of the cannonball approaches zero, due to gravity-caused slowing of the cannonball's velocity as its height increases, before the cannonball escapes the system.  Conservation of energy therefore involves the system, and not the cannonball only.  Thus, as kinetic energy converts to potential energy, the system acquires that potential energy, not the cannonball only.  If we choose to think that that potential energy appears as mass, then we are now free to divide up that mass between both bodies of the interacting system, in the manner that makes the most sense, in terms of other and similar observations --such as a kinetically-energized electron climbing out of the electrostatic field of a proton, and escaping.  Quantum ElectroDynamics, the most accurately-measured/supported theory in Physics, indicates that the overall mass ratio of the two bodies does not change throughout the event, so we have no reason to say things must work differently with respect to gravitation.

[Geezer]

Farsight, I think I see the source of the problem.

You said:

Since the cannonball has escape velocity it leaves the earth's gravitational field and escapes the system.

The problem with that statement is that, even though the cannonball has escape velocity, it does not mean it has left the Earth's gravitational field.

Gravitational fields don't stop at some arbitrary boundary. They just get weaker.

At escape velocity the cannonball and the Earth are still mutually attracting each other, but the KE of the cannonball is great enough that the Earth is not able to "pull it back".

Consequently, it's not valid to say that the cannonball has escaped the system. In fact, I don't think it's possible to define any point at which the cannonball has left the system. No matter how hard we try, there is always going to be some interaction. It may be an incredibly weak interaction of course, but it will not go to zero.

You also said, assertively, "Working through the numbers with equal masses does not assist with understanding here."

Why not? Because you said so? If you are going to dismiss another posters suggestion, unless you subscribe to "proof by loud assertion", it is customary to provide some reasoning.

[VernonNemitz]

Gravitational fields don't stop at some arbitrary boundary. They just get weaker.  At escape velocity the cannonball and the Earth are still mutually attracting each other, but the KE of the cannonball is great enough that the Earth is not able to "pull it back".  Consequently, it's not valid to say that the cannonball has escaped the system. In fact, I don't think it's possible to define any point at which the cannonball has left the system. No matter how hard we try, there is always going to be some interaction. It may be an incredibly weak interaction of course, but it will not go to zero. 

Your theory is exactly correct, but what matters in the human realm is the practicality of things.  That means we indeed can specify a place where for all practical purposes we can say the Earth is not significantly affecting an object that has left its immediate vicinity.  I'm going to arbitrarily pick 2 Astronomical Units for this distance (the Asteroid Belt is about 1.8 AU from Earth).  If we want to talk about an object that has "escaped", then the velocity of this object can be computed to be (initial velocity) minus (escape velocity), pretty simple, don't you think?  Certainly an escaped object will always have at least that velocity...and at a distance from Earth such as 2 AU, the actual velocity will be close enough to that simply-computed velocity so as not to make a practical difference, over the short term at the very least, and likely also over the middle-term.

[Geezer]

Vernon, of course, for practical purposes what you say is quite true. But for something to "leave a system" we need to be able to define a boundary where the influences exerted by the system no longer exist. I don't believe that's possible in this case.

Just because the cannonball gets farther from Earth, that does not mean its potential energy with respect to Earth is eliminated.

If we temporarily relocate Earth to a region of space where all other gravitational influences are negligible and transport the cannonball to any distance from Earth, then "park" it in space, I maintain it will return to Earth. It might take a long time to get there, but if there are no other gravitational influences to redirect it, I'm pretty sure that's where it's going to go.

Other than the minor inconvenience associated with temporarily relocating Earth, this seems like a perfectly practical demonstration of the effect.

[VernonNemitz]

Geezer, what I was trying to say was that if we can specify a place where for practical purposes a cannonball or other object is not being significantly gravitationally affected by the Earth, then we can also say that the object, in that place, has for practical purposes escaped the system that had previously included both the Earth and the cannonball or other object.  I agree that technically there is never any escape outside of infinite distance, but since that is not achievable, a practical distance will have to do in its place --and can do in its place, over the short-term or middle-term.

[Geezer]

Vernon, I think that depends on what we are trying to do.

If we only want to calculate practical effects, then I agree with you. However, if we are trying to describe a general case where we are allowed to use terms like "leaves the system", we have to be able to unambiguously define when something crosses some boundary and actually "leaves the system". If we cannot create an unambiguous definition, the term has no meaning.

As we know of no limit to the extent of gravitational fields, I do not understand how it we can possibly ever create such a definition.

To say that something has effectively left the system, or has virtually left the system, reminds me of the girl who said to her father,

"But Daddy, I'm only a wee bit pregnant."

[Farsight]

Vernon, there is no mechanism by which the cannonball kinetic/potential energy leaves the cannonball. You might claim it does, but you have no mathematics and no logic to support that hypothesis. Quantum ElectroDynamics and "the mass ratio of the two bodies" is not appropriate when the cannonball gets 10²⁵ times the energy that the earth gets. 

Geezer: when an object has escape velocity, as it gets further away its velocity reduces because the Earth's gravity pulls it back. But the velocity reduces less than the gravity reduces. So it always gets further away. It never comes back.

Now come on guys, the kinetic energy of the cannonball is vastly greater than the kinetic energy of the earth. This is no proof by assertion, it's a mathematical proof. And there is absolutely no logic and no evidence to support the assertion that the energy you gave to the cannonball is somehow transferred to the earth or the earth's gravitational field. The cannonball gets away, and whilst that kinetic energy is converted into potential energy, it gets away too. It's obvious that it does, because a billion years later the cannonball finds a new earth, and falls down at 11.2km/s. It gives this new earth all the energy lost to the old one.

I don't know what morte I can say to explain this. I've given you the logic, the references, the arithmetic. There's nothing else I can give.

[yor_on]

I wonder if I should open a new thread about a plane and a treadmill here?
Or maybe this question is already answered

'Potential energy' will differ on the choice of system, or frames you compare against. There is always frames of rest relative 'potential energy' and there will always be a 'potential energy' to objects at rest with each other when arbitrarily compared to something outside that frame.

That means that every object you can think of, at all times, will have a variety of 'potential energy' simultaneously depending on your choice of 'system', don't it?

Now if we changed that to 'gravitational impact' instead

[Geezer]

Geezer: when an object has escape velocity, as it gets further away its velocity reduces because the Earth's gravity pulls it back. But the velocity reduces less than the gravity reduces. So it always gets further away. It never comes back.

Farsight, please go back and read what I said.

In the case where the object achieves escape velocity, I agree with you, it never comes back, and I did not say otherwise. But that does not mean it gave up all of its potential energy. That's the critical distinction.

To prove that it has not given up its potential energy, I proposed my slightly impractical experiment. In that experiment, we eliminated the kinetic energy of the object and put it in some stationary position (with respect to the Earth). We then wait for a long time and observe what happens. I believe it will accelerate towards Earth. Why would it not?

If it does, it still has potential energy with respect to the Earth. If it does not, it has no potential energy with respect to the Earth.

Anyway, back to my point. When an object achieves escape velocity, does that mean, or does that not mean, that it has given up its potential energy? I believe that is where you are confused. Let's try to answer one question at a time.