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Perpetuum Mobile and Gravitation

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  1. Sep 17, 2014 #1
    I have fundamental question about what is called the “law of conservation of energy”.
    We all hear about the tidal power stations which using the tidal power. The source of the tidal power came from the changes in the gravity field between the moon and the earth. Allegedly, because of the law of conservation of energy this influence must cause to some energy lose in the moon or the earth. And indeed we know that the moon orbit get longer and slower over time. My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal?
    According to the general relativity theory the gravitation is the time space curve effect of a big object. This curve is not “energy consuming”, which means basically - two objects can spin around each other in space forever even that such a spin is a change in momentum that should consume energy according to the classic theory. The question is did the tidal effects caused by that “miracle” eternity momentum changes are indeed “energy consuming”?
    Lest imagine that instead of the moon there is black hole and the earth is spinning around it. This can cause to tidal power effects exactly like happen by the moon. This energy coming from the black hole which means that the black hole mass must be reduce according to the equation of e=mc^2. This is against what we know about black holes which are never losing any mass.
    But if the answer is “no” that mean in other words that the tidal power stations are kind of “Perpetuum Mobile” - creating energy from nothing. This is of course a weirder conclusion.
     
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  3. Sep 17, 2014 #2

    Nugatory

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    We do if we want to understand what's going on here - it is a necessary part of the physical explanation. You also may have a basic misunderstanding about how to apply conservation of energy here:
    Classical theory does not say that; the classical theory says that in the absence of friction the circular orbit will last forever. The direction of the moon's momentum changes, but its kinetic energy (which is a function of the magnitude of the momentum, not its direction) does not.

    You are also misunderstanding black holes. Whether an object is a black hole or not, the gravitational effects are the same everywhere above its surface (physical surface of a planet, event horizon of a black hole). So the physics of the rotating earth orbiting a rotating black hole are the same as the physics of the earth orbiting an ordinary star of the same mass. We can extract energy from the system using tidal power stations, and the power comes not from inside the black hole but from the kinetic energy of the rotating moving objects.

    By the way... questions of this sort sometimes degenerate into argments about whether teh current physical understanding of these systems is correct, If that's where this thread is going, expect it to be locked pretty quickly - we're here to explain what the science says and how it works, not to argue its correctness with people who are unwilling to understand either.
     
  4. Sep 17, 2014 #3

    jbriggs444

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    Can you firm that statement up?

    The tides cause the Earth to slow its rotation and cause the moon to increase its distance from the Earth. The change increases the gravitational potential energy of the moon with respect to the Earth, making it less and less negative [assuming a zero point at infinity].

    If "changes in the gravity field" is the same thing as "change in gravitational potential energy", the above statement gets the sign wrong. The change in gravitational potential energy is a drain on net tidal power, not a contribution.
     
  5. Sep 17, 2014 #4

    CWatters

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    We don't have any reason to believe conservation of energy wouldn't apply.

    Proving it by measurement would be hard. Pretty sure there are other things that effect the moons orbit.
     
  6. Sep 17, 2014 #5
    Well, you probably must invest some energy in order to change the direction of the moon's momentum . Where do you think this energy is coming from?
     
  7. Sep 17, 2014 #6

    A.T.

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    Changing the direction doesn't require energy.
     
  8. Sep 17, 2014 #7
    Let's say I want to change the direction of some moving object, how can I do it?
     
  9. Sep 17, 2014 #8

    A.T.

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    By applying a force perpendicular to the object's velocity.
     
  10. Sep 17, 2014 #9

    Drakkith

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    If energy is expended, then that energy must go somewhere. In the case of a perfectly circular orbit, the velocity, and thus the kinetic energy, is the same at all points in the orbit. So where did the energy go? The answer is nowhere. No energy was expended.

    Consider that energy is the potential to do work, and that work is: w=fd, where f is force and d is displacement. This only works if the force is applied along the same axis that the displacement takes place in. In the case of a circular orbit, the force is ALWAYS perpendicular to the displacement, so no work is performed on the orbiting object and no energy is expended.
     
  11. Sep 17, 2014 #10

    russ_watters

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    I'll try:

    Power generators are often named after the carrier of the energy, not the original source of the energy. Wind power and hydroelectric power originate from the sun, for example (also: "steam engine" doesn't tell you what is powering it). Similarly, "tidal power" is only carried by the movement of the tides: the energy that is being turned into electricity is the rotational kinetic energy of the earth. Tidal friction and tidal power plants slow the rotation of the earth.

    The lunar recession is an additional consequence of that, but my understanding (less certain...) is that energy is conserved in that interaction: the moon is gaining momentum in its orbit, but the total energy of the orbit (kinetic and potential) stays constant.

    http://www.talkorigins.org/faqs/moonrec.html
     
  12. Sep 17, 2014 #11

    jbriggs444

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    That's clearly incorrect. The kinetic energy of an object in [circular] orbit is half of its escape energy. Worded slightly differently, it is half of its potential energy deficit. The higher the orbit, the lower the potential energy deficit and the higher the total energy.

    The moon is losing momentum in its orbit, going slower and slower but orbitting higher and higher. The net impact is that its angular momentum is increasing over time.

    Edit: clarified that the above applies for a set of different circular orbits. Russ may be recalling that for a single elliptical orbit the sum of potential and kinetic energy is a constant.
     
    Last edited: Sep 17, 2014
  13. Sep 17, 2014 #12
    The force of the tidal bulges on the moon is in the direction of the velocity of the moon, so the total energy of the moon must increase.

    Combining mv^2 = GMm/r^2 with E = (1/2)mv^2 - GmM/r will get you:

    E = - (1/2)GmM/r for the total energy. (with 0 at infinity)
     
  14. Sep 17, 2014 #13
    Gravitational waves do carry energy. We've never measured gravitational waves directly, but there's strong indirect evidence from study of binary stars. The orbit gradually decays to a smaller, faster orbit. http://www.space.com/17346-gravity-waves-binary-stars-speed.html

    AFAIK, all orbits decay this way, but the decay is very, very slow for the Earth/moon, that other effects are dominant.
     
  15. Sep 17, 2014 #14

    davenn

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    interesting article, but it doesn't account for the earth-moon distance increasing, rather than decaying

    from wiki ... Due to tidal acceleration, the orbit of the Moon around Earth becomes approximately 2.2 cm more distant each year.

    Dave
     
    Last edited: Sep 17, 2014
  16. Sep 18, 2014 #15
    So you are basically saying I can open any screw without spending any energy? this is good news.
     
  17. Sep 18, 2014 #16
    If there's no friction, this is indeed possible.
     
  18. Sep 18, 2014 #17
    Why the friction changs the fact that the force "is ALWAYS perpendicular to the displacement"?
     
  19. Sep 18, 2014 #18

    Drakkith

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    No, you have a force in the direction of displacement (up). When you turn the screw, the grooves exert a force on the material the screw is in that forces the screw up and out of the material. Friction drastically increases the amount of energy required to get the screw out.
     
  20. Sep 18, 2014 #19

    A.T.

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    Because friction is parallel to displacement.
     
  21. Sep 18, 2014 #20
    If we ignore the "up" direction (let's say our screw does not have grooves or we just try to rotate a wheel or change the course of some asteroid) what you saying is that we not have to invest energy to do so?
    Anyway the moon’s orbit is not a perfect round.
     
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