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How is - (gravitational) potential energy and free fall compatible?

  1. Nov 23, 2013 #1
    I was having a few problems with Negative gravitational potential energy, I wasn't able to put my finger on why because it worked so well with everything else I had learned.

    I tried to say it made sense because in space you don't free fall, but that's not actually true, though the gravitational energy is smaller you still accelerate (just very slightly) towards a planetary body.

    If Ugrav is your store of potential KE, how can it be negative even in cases when you are accelerating towards a planetary body? This question has realy slowed my studying down, help would be greatly appreciated.
  2. jcsd
  3. Nov 23, 2013 #2

    Doc Al

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    Staff: Mentor

    All that matters is the change in gravitational PE. When you fall towards a planet,
    you lose PE and gain KE. (The actual value of PE at any point is arbitrary and depends on what you chose as your reference point.)
  4. Nov 23, 2013 #3
    I thought about this after I asked the question but I suspected that I had made another mistake

    Do you mind confirming something for me.

    The equation is derived (the way I learned it) by comparing the Ugrav of two positions and then making the second position (the final one) infinitely far from the center of the planetary object, does this mean that U is only 0 when r is infinity because there is no longer a difference between the Ugrav of the initial position and the final one?

    And is U (as in Ui) always negative because the final position is made the reference position even though it's in the opposite direction of the centre if the planetary object.

    It's cool that this equation has its reference point choosen as it forms, though this confused me as I thought the reference point was simply the centre of the Earth given the fact that the "work done" half of the equation was formulated using the centre as the ref.

    Seems like there are two reference points one where U = minus infinity (r = 0) and U = 0 (r = infinity) however Ui being equal to minus infinity only means that it is inivity less than Uf right?

    So for future reference the Ugrav of two or more objects are measured on a scale of 0 to infinity i.e relative to the position of the FINAL position from the deriving equation.

    God help me.

    Also KE, when converted from U will move the object toward the Earth in the opposite direction of the reference point, thus it has potential to do negative work relative to the reference point.
    Last edited: Nov 23, 2013
  5. Nov 23, 2013 #4


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    You can choose any reference point and any finite value of U at that point. Choosing U = 0 when r = infinity is convenient for orbital mechanics, finding escape velocity, etc.

    A reference point if "U = infinity when r = 0" doesn't really mean anything, because you can't do any sensible arithmetic with "infinity" (well, not without using some more devious math than you probably know about yet).

    On the other hand, if you are dealing with motion close to the surface of the earth and assuming gravitational acceleration is independent of altitude, it's simpler to take U = 0 at the surface of the earth (or any other convenient altitude). That is where the formula "mgh" comes from, of course.
  6. Nov 24, 2013 #5
    Thanks for all your help.
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