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Re: [free_energy] Re: E!=mgh for the Earth

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  • Phil Karn
    ... *Which* proof, for Chrissakes? I ve lost count! Here. I ll do it again just for the exercise. We ll take teeny baby steps. Pay close attention. We will
    Message 1 of 106 , Oct 1, 2004
      David Thomson wrote:
      > Hi Phil,
      >>Several people on this list have proved that E=.5mv^2 several different
      >>ways. You just refuse to accept that they're right and you're wrong.
      > Point to the proof, Phil.

      *Which* proof, for Chrissakes? I've lost count!

      Here. I'll do it again just for the exercise. We'll take teeny baby
      steps. Pay close attention.

      We will start with some hopefully noncontroversial equations and rules
      with which you already agree:

      1. F = m*a [force is mass times acceleration - Newton's second law of

      2. v = a*t [velocity is acceleration times time]

      3. d = v*t [distance traveled is velocity times time, for constant velocity]

      4. E = F*d [energy is force times distance]

      5. Energy is conserved [first law of thermodynamics]. So the kinetic
      energy added to an object by accelerating it with a force is equal to
      the energy expended on the object by the accelerating force.

      I will first derive a formula for the total distance d traveled by an
      object that starts at rest at time 0, and is then placed under constant
      acceleration a. We want to know the distance traveled at time t. Since
      the velocity continually increases, we can't just use equation 3 as is.
      We have to take equation 2, the velocity as a function of time, and
      integrate it over time.

      d = definite integral from 0 to t [v(T)] dT
      = definite integral from 0 to t [a * T] dT
      = a * definite integral from 0 to t [T] dT [ok because a is constant]
      = a * 1/2 * t^2
      = 1/2 * a * t^2

      [I know integral calculus is far beyond your limited mental abilities,
      Dave, but it *is* the correct way to solve this problem. Besides, this
      particular integral is listed in all the math tables, *and* this
      specific formula is listed in the physics books. If you don't know
      calculus, learn it. If you *still* don't believe this answer, repeat
      Galileo's experiment and time a bunch of object falling different
      distances to the ground.]

      Now let's go back to the specific problem. Starting with equation 4, the
      energy expended on accelerating the object, which becomes the object's
      kinetic energy because it starts with none:

      E = F*d

      substituting equation 1 for F:

      E = m*a*d

      substituting our derived equation for d, and rearranging:

      E = m * a * 1/2 * a * t^2
      = 1/2 * m * a^2 * t^2
      = 1/2 * m * (a*t)^2

      But according to equation 2, a*t is just the final velocity, v, because
      a is constant. Substituting v for at:

      = 1/2 * m * v^2


      I guess your problem is really with integral calculus, because that's
      where that pesky factor of 1/2 first appears. Tough. Learn calculus.
      Accept it. Ignorance of the law is no excuse.

    • Phil Karn
      ... Good point; the kinetic energy of an object *does* depend on the inertial reference frame you use to measure the object s velocity. That s true in both
      Message 106 of 106 , Oct 4, 2004
        mintowheel wrote:

        > Part of Thomson's confusion about kinetic energy perhaps relates to
        > the need to define a reference frame when assigning the specific
        > value of an object's kinetic energy. Of course we must use the same
        > reference frame for both the velocity and kinetic energy (an obvious
        > fact to most of us). When I tried to explain this to him early on,
        > he seemed unable to comprehend this, and immediately rejected my
        > comments.

        Good point; the kinetic energy of an object *does* depend on the
        inertial reference frame you use to measure the object's velocity.
        That's true in both Newtonian and Einsteinean relativity. For objects on
        or near the earth, the obvious choice is the earth-centered-earth-fixed
        reference frame. That's the one in which the earth is stationary, though
        strictly speaking it's not inertial because of the earth's rotation.

        Away from the earth, things can get confusing. So you just state that
        whenever you measure kinetic energy, you have to specify the reference
        frame you're using.

        Perhaps Dave, in his confusion, felt that kinetic energy should only be
        defined in the context of a collision because the energy released in
        such a collision would be invariant to an observer's reference frame.
        But that's giving him way too much credit.

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