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Potential Energy and gravity

  1. Jun 21, 2009 #1


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    I am now reading Rosenblum's and Kuttner's book "Quantum Enigma". In one of the first chapters they talk about the different forms of energy. Regarding Potential Energy, they give the example of a rock held in the air having potential energy because of the force of gravity. They go on to state that when rock hits the ground it transfers ALL of its potential energy to the ground, and at that point is has no remaining potential energy.

    I am confused with the difference with being held in the air, and resting on the ground. In reality, for the rock on the ground, if an earthquake were to occur and the ground were to separate under the rock, it could actually fall further.

    So, is it really correct to state a rock lying on flat ground really has zero potential energy?

  2. jcsd
  3. Jun 21, 2009 #2
    You can set the zero for your potential energy anywhere that is convenient. Usually that is the lowest available position, but it doesn't have to be. Potential energy can be positive or negative, and only differences matter. For a rock in a constant gravitational field, the potential energy is [tex]mg(h-h_0)[/tex] where [tex]h_0[/tex] is your arbitrarily chosen zero point for potential energy. (Or, equivalently, one can choose to place the origin of their coordinate system anywhere that is convenient.) Only differences will matter, and when you take the difference in potential energy between two positions, then the h_0's will cancel. They have chosen a particular origin for the convenience of their illustration where the PE of a rock on the ground is zero. If the earthquake were to occur and the rock fell into the earth, then its PE would become negative.
  4. Jun 21, 2009 #3
    In reality , the only way an object on the earth can have zero potential energy is when it is at the centre of the earth , for then it has no tendency to fall further. However , the concept of zero potential energy on the earth's surface is a misleading one. Potential energy is a relative term that physicist use for convenience in describing physical systems , i.e , when newton's laws prove to be just too complicated. Potential energy is relative . In the given situation , the ground has been assumed to be the reference point , i.e , it has arbitrarily been assigned zero potential energy , and with respect to that , the potential energy of different heights has been calculated .
  5. Jun 21, 2009 #4


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    Agreed. Actually can't one say no object in the universe has ABSOLUTELY zero potential energy? That rock at the center of the earth is still being gravitationally pulled by every other object in the universe, although in a VERY small way.

    Now that makes sense.

  6. Jun 21, 2009 #5
    1) Get a rock; heavier is better.
    2) Stick your foot out forward; right foot if your are right footed; otherwise left.
    3) Rest the rock on your toe.
    4) If it does not hurt, then lift the rock above your head.
    5) Drop the rock on your toe.
    6) Calculate potential energy mgh for the rock.
  7. Jun 21, 2009 #6


    Staff: Mentor

    How do you determine if you are right- or left-footed?
  8. Jun 21, 2009 #7
    If you write with your left foot you're left-footed, and vice-versa.
  9. Jun 21, 2009 #8
    This statement is not quite right. You can define any object in the universe to have ABSOLUTELY zero potential energy. The only thing you need to do is define that object as the zero potential.

    On earth, the zero potential is usually taken at ground level, just because it's convenient. You could take the zero potential on the moon, and every calculation would yield the same answer. The only problem is that you would then need to know the exact distance between your object and the moon, and that distance varies.

    Here is an example. Suppose you hold a rock at a height h above the ground. The potential energy of this rock, relative to the ground (which means, the ground is the zero potential), is just:
    [tex]P_1 = mgh[/tex]

    The potential energy of the same rock, relative to the center of the earth (which means, the ground potential is the center of the earth), is:
    [tex]P_2 = mg(r+h)[/tex]
    where r is the radius of the earth (or the distance between the center of the earth and the ground under your feet).

    Yes, those two potential energies are very different, and they do describe the same rock.

    For example, let's calculate how much kinetic energy the rock gains when you drop it from your hand onto the ground. This energy is the difference between the potential energy where you release the rock and the potential energy where it lands.

    In the first case, we have:
    [tex]E_1 = P_1(y = h) - P_1(y = 0) = mgh - 0 = mgh[/tex]

    In the second case, we have:
    [tex]E_2 = P_2(y=h) - P_1(y = 0) = mg(r+h) - mg(r+0) = mgr + mgh - mgr = mgh = E_1[/tex]

    The answers are the same!!

    This example shows that it does not matter where you take the zero potential, as it will always cancel if you look at any potential difference. And you always look at potential differences in physics. A single value of the potential energy doesn't say much, and is indeed arbitrary if you don't define the zero potential.
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