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Conservation of ME and stable orbit

  1. Mar 1, 2010 #1


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    Imagine a planet in elliptical orbit around a star, without any perturbation from third bodies and moving in perfect vacuum. The only force acting in the system is gravity.

    As the planet gets closer to the star, gravity acts in the direction of its motion, so it gains speed and KE but loses PE. As it gets away from the star, gravity slows it down and it loses KE but gains PE. Hence mechanical energy is conserved.

    But if we now consider that the planet is a macroscopic object with a significant volume and a certain composition, doesn’t this have an impact on the analysis? The side closer to the star will be subject to a stronger force than the opposite side. It seems to me that the difference gets wider as the planet gets farther from the star, so the planet is stretched. And such difference gets narrower as the planet gets closer to the star, so in this leg of the trip it compresses. And if the composition of the planet does not enable it to be perfectly elastic and recover its original size, can’t there dampening arise, i.e. a supplementary loss of KE that may make it finally collapse against the star?

    Is it so? And if so, it is because the principle of conservation of mechanical energy only applies to ideal point masses? Or, rather, am I misunderstanding the principle and mixing things that have nothing to do with each other?
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  3. Mar 1, 2010 #2
    Ignoring the many strange ideas you have about gravity, if you want an example of gravitational influence on a planet, consider the tides. Does the temporary displacement of water destroy the ocean? Of course, the earth being so much more massive (and so close) to our moon means that we can cause moon-quakes...

    Your idea that gravity becomes "broad" is not correct; it becomes weak. It becomes weak very very quickly with distance, so the level of deformation of a planet that you describe doesn't occur in a planet with a stable orbit.

    Your notions of KE and ME seem... random. As for any idea that gravity "stretches" a planet... no, that's not how it works.
  4. Mar 1, 2010 #3
    Yes, that is how it works. The earth's tides are the result of the stretching effect (tidal forces) caused by the moons gravity. http://en.wikipedia.org/wiki/Tidal_force And these forces could destroy a planet or moon - if it came too close to a massive object. I'm not sure, but I think Saturn's rings were formed that way.
    Last edited: Mar 1, 2010
  5. Mar 2, 2010 #4


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    Hmm... Did I say anything strange or wrong in the introduction? If you point at any wrong conception and explain why it is wrong, I will be willing to accept it, but I do not see the reason for those general statements.

    Well, that is precisely what the question is about, obviously, as Turtle Mister rightly pointed. Can tidal effects in a macroscopic body be responsible for loss of KE?

    By mentioning and describing this effect, consequently, you seem to be accepting the basic idea proposed in the OP. What are ocean tides but a bulging of the planet, that is to say, a streching?

    It seems, then, that you are only objecting the magnitude of the effect under particular circumstances. That is not the question. Even assuming that a stable orbit is eternal (perpetual motion?), the question would still be valid in other circumstances. Consider objects in rectilinear free-fall, with large size, from a far distance. When they collide, the collision may not be fully elastic and entail loss of KE. But the purely gravitational side of the interaction, is it perfectly elastic? Why?

    Even in the case of an elliptical orbit (of which the free-fall case is just a subset, a degenerate ellipsis), one tends to think that perfect stability is just an ideal that may not hold eternally true, due to, particularly what I am mentioning. Nothing is perfectly stable, unfortunately... or fortunately...
  6. Mar 2, 2010 #5


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    Thanks, Turtle. Yes, tidal effects (call that "difference of strength of the gravitational field across distance", if you judge from an inertial frame or a "fictitious force", if you judge from the concerned accelerated frame) mean stretching/compression and hence deformation. What is deformed may elastically return to its original size or not; in the latter case, it may deform permanently or even, as you point out, beyond elasticity point, break into pieces.

    In other words, a system, being subject only to internal forces, cannot "self-accelerate" (the velocity of its center of mass remains the same) but it can "self-deform". In fact, it seems that is what happens with every system in real life. And a system where gravity is the sole force in action cannot be an exception.

    The question is then: when you study that "a system that is only subject to gravity conserves mechanical energy", how does this principle link with the above realization? Two possibilities:

    (a) The principle of conservation of ME still absolutely holds. I am missing something that preserves it, in all cases, under all circumstances.

    (b) The principle only strictly applies to ideal point masses. For real macroscopic bodies, in normal circumstances, it still applies as a most accurate approximation but it fails under extreme circumstances.

    What is the answer?
  7. Mar 2, 2010 #6
    Ok, I get your question. The answer is that energy isn't lost from the system, and you're asking a question which has been asked an answered here before, "Where is the energy stored in a gravitational field?". Not the best worded question, but the discussion was decent.

    As an idealized problem, why not do a bit of research on the 2 body problem? It's more than you're asking, but I think more in the spirit of what you're asking.

    Oh, and you can call tidal forces or deformation "stretching", but this often makes people imagine stretching towards the source of gravity, which confuses them in relation to Earth's tides... etc...

    Obviously I oversimplified.
  8. Mar 2, 2010 #7


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    What about the effect that tidal forces have on angular momentum, which affects the rate of rotation and orbital path in a two body system?
  9. Mar 2, 2010 #8
    The 2 body problem is unsolved... I know... a pal works on it for a living. That said, the net effect between 2 bodies is that orbit evolves, but energy isn't lost from the system. Lets take the tides again as an example; we know that SOME energy (however little) is going to end up as thermal energy as a result of friction of the water against the seafloor and waves breaking, right?

    Over time the rotation of a body under other transfers of energy and momentum is going to see orbital and rotational decay. It's not that energy is LOST from the system, but it's no longer doing the work that it used to. As the bodies near each other, remember that that GPE is the rest /2... so what happens? If you consider the closed system: 2 bodies and their related fields, etc... ignoring ejecta, conservation holds where it needs to.

    EDIT: good link: http://hyperphysics.phy-astr.gsu.edu/HBASE/gpot.html
  10. Mar 2, 2010 #9


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    Frame Dragger, thanks for taking the time to answer. But I didn’t find the thread you mentioned and am still confused.

    It's clear to me that the energy of the system, the same way that it cannot be created out of the blue, cannot be lost. No question about that. If *kinetic* energy is lost due to deformation, it must be because such energy is transformed into another form, namely internal energy, in whatever modality. The question is only whether *mechanical* energy is always conserved, under the exclusive action of gravity or in other words, whether gravity is fully conservative (of ME), in all circumstances, even when there are significant tidal effects involved.

    Let me explain my doubt with another example. A ball is shot from the ground of a planet, in vacuum. For this purpose, energy is supplied to it, say by a cannon, in quantity = 1. But during the interaction there will be (i) deformation and, since the ball is not perfectly elastic and it’ll not fully recover its natural size, (ii) some conversion of the supplied energy into internal energy. We somehow immediately measure the resulting KE and it is = 0.9. On the other hand, suppose that the ball’s volume is so large that there is a significant gravitational tidal effect: as the ball ascends, it noticeably stretches. Again, since the ball’s material is not perfectly elastic, this would entail a new loss of KE, which would have converted into internal energy. When it finally stops, we measure the height h and conclude that the ball has acquired GPE wrt the ground = - GMm/h = … what? - 0.8, for example? Or - 0.899 or whatever but less than 0.9? And then the ball descends and as it falls, it compresses. With a subsequent additional loss of KE? Would that mean that it reaches the ground with KE = 0.7 or 0.898 or whatever but less than the GPE acquired at the top? Or is the principle of conservation of *mechanical* energy (under the sole influence of gravity) somehow preserved so that the ball reaches the highest point with GPE = - 0.9 and returns to the ground precisely with KE = 0.9?
  11. Mar 2, 2010 #10
    Oh. You know, the issues I was talking about tend to think of idealized masses without those effects being present.

    Lets say that deformation is radiated as heat... that makes it even more confusing because it would seem that energy could not possibly be recovered as GPE or KE. You know what, I don't know the answer to your question, and now I have the same question. I've always thought of this in terms of a weight striking a spring such as this http://cnx.org/content/m15102/latest/ but although the math is there I'm still not seeing that it answers your question.

    I THINK the answer is that the case you're thinking of probably doesn't require that energy be conserved in that fashion, because I cannot imagine how your point wouldn't be valid otherwise. I'm definitely open to a physics advisor stepping in at this point, my pride is nothing. :rofl:
  12. Mar 3, 2010 #11


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    I read the link and it's very good. I had read and liked other articles from the same author, but missed this one, which is very appropriate.

    Somehow it gives the answer: conservation of mechanical energy is an "ideal process" that requires two conditions, namely (i) "the system is isolated" and (ii) "no change in thermal energy within the system", which can be summarised in one: no dissipation of mechanical energy into non-mechanical forms, either due to external or internal forces = no friction, either external or internal. Under these conditions, obviously, the principle must hold, because in the end what we are saying here is "mechanical energy will be conserved to the extent it has to..."

    What becomes more confusing and this text is not an exception is when one starts using the concept of "conservative force", as if there were forces that deserve that adjective and other that do not.

    If we talk at the level of fundamental forces, leaving aside strong and weak nuclear force, we are left with gravitational and electromagnetic force.

    If we then consider the "microscopic" level (loose language, yes, I don't know where to put the limit), it seems clear that they are both conservative: there is no friction in the interaction between an electron and a proton... or between two atoms interacting gravitationally.

    If we go up to macroscopic level, considering extended bodies, some degree of friction, even a tiny one, seems unavoidable in any interaction. But if we admit so and then say that contact force is non-conservative but gravity is, we are mixing apples with pears, non-fundamental with fundamental forces. I think we should say that both macroscopic contact and gravity force are non-conservative at macroscopic level, while their fundamental counterparts, electromagnetic and gravitational force, are, at microscopic level. But maybe I am just re-inventing the wheel. Is that maybe the difference between "gravity" and "gravitation"?
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