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Can we extract energy from a gravitational field ?

  1. Apr 2, 2013 #1
    hey there

    MTW mention that the mass energy of the Earth moon system is less than the mass energy that the system would have if the 2 objects were at infinite separation.
    MTW say that it is due to gravitational forces.
    Please correct me if I'm wrong : The fact that the moon follows a geodesic of the Earth's gravitational field, means that the Earth is exerting a force on the moon which means Earth is losing energy due to that. The gyroscope precession also is due to the rotation of the Earth and the dragging of the spacetime => means the gravitational force is responsible for that.
    As long as Earth has mass, it bends spacetime => it will always have a gravitational field force .
    Can we extract a 'free' energy from that ?

    Thanks,

    A lone navigator in the MTW jungle.
     
  2. jcsd
  3. Apr 2, 2013 #2

    mfb

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    No. Why should it (if we neglect gravitational waves)?
    No. We can extract energy from gravity if we lower stuff from space to earth, but then we have it on earth and not in space any more. The energy is not free.

    What is MTW, by the way?
     
  4. Apr 2, 2013 #3

    WannabeNewton

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    Misner Thorne and Wheeler - Gravitation. It is a GR bible of sorts. What section is this in the book anyways zn5252? I'm interested in reading it. I'm curious to see how you concluded that an object following a geodesic would be subject to forces when by definition the motion of such a particle satisfies ##u^{a}\nabla _{a}u^{b} = 0##.
     
  5. Apr 2, 2013 #4
    Thanks for your reply. This is section 20.4, I mean chapter 20, section 4
     
  6. Apr 2, 2013 #5
    I meant that it is subject only to gravitational forces...Imagine we tie a rope to an asteroid which is coming from infinity towards the sun. Gently, due to the curvature of the sun, the stiffness of the rope will change due to the Sun's gravitational forces...
    Now if we attach a spring to the asteroid with the other end at infinity. If the asteroid gets closer to the sun, the spring stiffness will change, and then the astroid will be pulled back and so forth... is this reasonable ?
     
  7. Apr 2, 2013 #6

    mfb

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    Changing the stiffness of the spring at different lengths will require (or give) energy, depending on the position and the stiffness change. You cannot get free energy. The setup does not matter, it is possible to prove this mathematically with the current ("known") laws of physics. You would have to observe some violation of those laws.
     
  8. Apr 2, 2013 #7

    Because Earth is creating a spacetime curvature, which means there is energy in that location. Is this energy infinite ? does it get lost ? wasted ? somehow ?
    If is exerting a gravitational force, then that means, there is energy being 'consumed' ? otherwise how would it maintain this force ?
    Sorry but I'm confused !
     
  9. Apr 2, 2013 #8

    mfb

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    No
    No
    No
    "Maintaining" a force does not consume energy. Moving something against a force requires energy.

    I edited my post while you were posting, the addition could be interesting.
     
  10. Apr 2, 2013 #9
    Are you talking about stiffness of the spring, as in Energy = kx^2/2, if k = F/x ? Or were you talking about the deflection of the spring associated with increasing force?
     
  11. Apr 2, 2013 #10

    Jonathan Scott

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    For practical purposes, gravitational energy is conserved locally, like all other forms of energy. It works like the Newtonian approximation, in that a particular configuration of masses has a certain amount of potential energy, and if the configuration changes that energy can be turned into other forms, such as kinetic energy, or some stored form (for example mechanical, electrical or chemical, all of which are effectively special cases of electromagnetic field energy).

    A simple case is when a test mass is falling freely in a gravitational field. In that case, the rest mass of the object can be considered to be effectively multiplied by the local gravitational time dilation factor, which decreases it by the relevant potential energy, but at the same time the kinetic energy of the object increases by the same amount, so that the total energy is constant in free fall.

    However, this semi-Newtonian model is very limited, and does not work when two or more source masses are considered, as the total time dilation effect the masses would have on one another's energy is actually twice the potential energy of the system. This suggests that there needs to be the same amount of gravitational energy stored somewhere "in the field", to bring the overall energy back to what it should be, and it is possible to use an analogy to the Coulomb energy of the electric field to model this situation.

    In the GR way of looking at things, the actual energy density in a vacuum is zero, and GR makes it clear that in the general case, gravitational energy cannot be assigned a unique location, as even its existence is dependent on the observer's coordinate system. However, although there is no global conservation of energy in GR, it is possible to describe energy flow using particular "pseudotensors" which work from specific points of view.

    In very extreme gravitational situations, where fields are changing rapidly, such as in very close binary star systems, it is possible for some of the gravitational energy to be radiated away as gravitational waves in space, and in this case the system gradually loses overall energy. The well-known case of the Hulse-Taylor binary pulsar is an example where the orbital energy of the system is gradually being lost at a rate which is consistent with the expected rate of gravitational radiation according to GR.
     
  12. Apr 2, 2013 #11
    indeed, f=-kx
     
  13. Apr 2, 2013 #12
    Yes then if these forces do not consume energy, then why the mass energy of the Earth moon system is less than the mass energy that the system would have if the 2 objects were at infinite separation (MTW Chapter 20 section 4)? this is non intuitive at all for me at least up to my very limited current knowledge of GR...
    Is this related to the potential energy curve ?
     
  14. Apr 2, 2013 #13

    Bill_K

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    Sounds like you need to brush up a bit on Newtonian physics before tackling GR.
     
  15. Apr 2, 2013 #14
    indeed Bill, it has been a while...
     
  16. Apr 2, 2013 #15

    Nugatory

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    Well, there are tidal electrical generators that work by putting a turbine in the tidal flow. Of course the ultimate source of the energy is the kinetic energy of the earth-moon system, but it's transferred by work done on the water by gravitational forces.
     
  17. Apr 2, 2013 #16

    mfb

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    I think you are mixing the concepts of energy and power here.
     
  18. Apr 2, 2013 #17

    Dale

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    Consider the earth and moon at rest wrt each other and at infinite separation. Now, consider the moon to be falling towards the earth from its initial position at infinity until it gets to its actual current orbital distance from earth.

    What is its speed when it gets to the final position? What is its final KE? How do those values differ from the moon's actual speed and KE? What can you infer about the energy at infinite separation compared to the actual current energy?
     
  19. Apr 2, 2013 #18
    Are they simple to explain and understand?

    I don't know the concepts of energy and power, but would like to.
     
  20. Apr 2, 2013 #19

    Mentz114

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  21. Apr 3, 2013 #20
    Indeed. It takes work to separate them : this solves the problem. They are gravitationally bound together. If this was not so they would have continued their paths unhindered...

    I just wanted an explanation in terms of the spacetime curvature....not Newtonian physics.

    this reminds me of to the potential energy binding for chemical bonds.
     
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