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Confusion about Gravitational Potential

  1. Jun 11, 2004 #1

    dcl

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    Heya's
    Im a bit confused reguarding gravitational potential energy.
    I've seen 2 different calculations

    U(h) = mgh
    where h is the height above the ground.

    and
    U = -GMm/r (off the top of my head, havnt got my notes here with me right now)

    Could anyone help clear this up for me? Why does one expression have a negative potential?
     
  2. jcsd
  3. Jun 11, 2004 #2
    The former, U(h), applies to a uniform gravitational field where the zero of potential is at h = 0. The later, U = -GMm/r, applies to the gravitational field of a spherical body where the zero potential is at r = infinity. Plot each on a graph and you'll notice that each is a decreasing function of r, only the shape of the curve is different.

    Keep in mind that only changes in potential are physically meaningful. You can always add a constant without changing the physics. The constant is usually chosen to make the math as simple as possible.

    Pete
     
  4. Jun 11, 2004 #3

    dcl

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    I see, makes sense. Thanks for clearing that up.
     
  5. Jun 11, 2004 #4
    Just to add to what pmb_phy said, we use mgh much of the time in calculations because it is a good approximation to the change in the gravitational potential (given by U = -GMm/r) for small distances. i.e. when h << r.

    Matt
     
  6. Jun 11, 2004 #5

    Gokul43201

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    [tex]r=R+h [/tex]

    So,

    [tex]U =\frac {-GMm} {r} = \frac {-GMm} {R+h} = \frac {-GMm} {R} (1 + h/R)^{-1}[/tex]

    For h<<R, you can expand the last term binomially, and neglect terms of second order and up. So,

    [tex]U = \frac {-GMm} {R} (1 - h/R) =\frac {-GMm} {R} + \frac {GMmh} {R^2}[/tex]
    [tex] = Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh [/tex]

    Since we are interested only in changes in potential, we can throw away the constant term.
     
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