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B What is the need for introducing the concept of G.P.E?

  1. Aug 24, 2016 #1
    What is the need of introducing the concept of gravitational potential energy in physics? Can't we simply say that when an object is lifted up it falls down because of the force of gravity acting on it instead of saying that it has G.P.E and so it is falls down?
  2. jcsd
  3. Aug 24, 2016 #2


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    It is a useful tool in a lot of situations. For example, if you roll a ball down a ramp (or drop it off a cliff), you can easily figure out the final speed with only the height of the ramp (ignoring rotational inertia, drag and other losses).
  4. Aug 24, 2016 #3
    Yes you are right, if a body is lifted up, it falls down due to gravitational pull.

    I think GPE is introduced to explain the work done in the process.
    A body cannot lift itself up on its own. You have to do work against the gravitational force to lift the body up. Now, whenever work is done, there has to be some kind of energy involved somewhere (workdone = change in energy).
    The work you did should be stored in the object which is the GPE.
  5. Aug 27, 2016 #4
    G.P.E. is so usefull because the mass of objects is so small compared to the mass of the Earth that all the energy in the gravitational interaction comes back to the object, so, when a particle returns to its original possition (let's said it the zero height), all the potential energy is converted into kinetic energy.

    This is not the case if we calculate the G.P.E. between two bodies with a mass of the same order of magnitude.
  6. Aug 28, 2016 #5


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    Let me give an example where GPE helps you where Force is very inconvenient. A trolley rolls down a frictionless ramp that has a wavy profile (like some children's slides). You want to know its speed at the bottom.
    To work it out using forces you need to calculate the component of gravity that's acting down the variable slope at every point and to integrate the acceleration over the whole trip.
    Using GPE considerations, all you need to do is to equate the KE at the bottom with the lost GPE.
    Half m vsquared = mgh
    Solve for v and you're done. No calculus. Just two lines of working.
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