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Confused about gravitational potential energy?

  1. Feb 4, 2014 #1

    MrBillyShears

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    There is several things I am confused about with gravitational potential energy. So, first of all, shouldn't U=mgh always be written ΔU=mgΔh, because isn't that equation only dealing with differences of potential energies when close to the surface of earth?
    Second, with the equation U=-GMm/r, there is a +K for the constant of integration at the end. What role does that play? Does that have something to do with reference frame, or is that just some random constant. I know that with differences of potential energies (ΔU=U(f)-U(i)) it will subtract itself and become irrelevant, but when you are just dealing with U(r) what significance does it have? Do you always just assume it's 0? Otherwise, wouldn't it make the single quantity U(r) useless if it could be anything?
    Mind you it is rather late and I wrote this in pure confusion so some of the things I wrote may not make sense.
     
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  3. Feb 4, 2014 #2

    olivermsun

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    Differences require a baseline. ΔU = U - U0 for some U0 which you define. Similarly, Δh = h- h0. You combine the definitions by assuming U = 0 when h = 0. It doesn't really matter because you only ever care about ΔU, e.g., an object fell this height so it gained this much energy...
     
  4. Feb 4, 2014 #3

    rcgldr

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    U = mgh assumes U = 0 at h = 0. It's an approximation for objects close to the surface of the earth. U = -GMm/r assumes U = 0 at r = ∞.
     
  5. Feb 5, 2014 #4
    At some level, what matters is what we can measure. In this case it is the force (basically energy difference), [itex]\mathbf{F}=-\nabla U[/itex]. On integration, we are free to add any integration constant to U. So what matters is the functional form of U(r), not its value. If you study more classical mechanics (i.e. some Lagrangian and Hamiltonian mechanics), you will see that kind of thing over and over again.
     
    Last edited: Feb 5, 2014
  6. Feb 5, 2014 #5

    jtbell

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    It determines the location where U = 0.

    If we let K = 0, then U = -GMm/r, and U = 0 at r = ∞.

    If we let K = GMm/R (where R is the radius of the earth), then U = GMm(1/R - 1/r), and U = 0 at the earth's surface.

    Either way, we get the same ΔU between two different r's, so they give the same physical results. K = 0 gives us a simpler formula, so we usually make that choice instead of the second one.

    However, there is an interesting application for the second choice. Let r = R + h, where h is the distance above the earth's surface. Then
    $$U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)$$
    I leave it as an exercise to show that for h << R (i.e. close to the earth's surface), U ≈ mgh. Hint: using the gravitational force law you can show that g = GM/R2.
     
  7. Feb 5, 2014 #6

    MrBillyShears

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    Thanks guys, I get it now.
     
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