Gravitational potential energy of orbiting object vs. non-orbiting object

growingpains

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Dec 22, 2011
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    It seems as though gravitational potential energy is proportional to the distance from the massive object (e.g. Earth) if the suspended object is non-orbiting (e.g. a box being held up in the sky). PE = mgh

    However, if the object is orbiting, the gravitational potential energy is inversely proportional to the radius of the orbit. U = (-G x M x m) / r

    Could someone explain this difference?
     

    chiddler

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    Apr 6, 2010
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      I was reading about this yesterday. Wikipedia, for once, has a clear explanation:

      " If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount.

      Consider a book placed on top of a table. As the book is raised from the floor, to the table, some external force works against the gravitational force. If the book falls back to the floor, the "falling" energy the book receives is provided by the gravitational force. Thus, if the book falls off the table, this potential energy goes to accelerate the mass of the book and is converted into kinetic energy."
       

      growingpains

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      Dec 22, 2011
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        Thanks, that was helpful.

        Is there some property of orbiting objects that makes them exhibit an inverse relationship in regards to distance and gravitational potential energy? I just find it peculiar how the potential energy to distance/radius relationship happens to "flip" as the object changes from non-orbiting to orbiting.
         
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        MrNeuro

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        Mar 17, 2010
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          Well, like magnetic force, magnetic potential is R instead of R^2.
          which then relates back to atoms bonding energies and deriving distance (angstroms) from that....lowest potential energy is where the compound will be happy and if E=kqq/r (g=GMm/r) you can find out r knowing E and the charges on the atoms

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