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Gravitational potential energy

  1. Sep 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Why is the gravitational potential energy of a ball a distance r from the center of the Earth negative?
    2. Relevant equations
    U_\text{grav}(r) = - GMm/r

    (To me, this makes sense because gravity is an attractive force and bodies will want to minimize the distance between them if only gravity is acting.)

    3. The attempt at a solution

    The force of gravity is given by Newton's universal law, so I'm thinking the potential energy due to this force is the negative of the work done on the ball by gravity over a distance, or
    \Delta U = -W = - \int_{r_1}^{r_2} \; \mathbf{F} \cdot \mathbf{dr}

    since the force and the displacement are in the same direction,

    \Delta U = -W = + \frac{GMm}{r}\vert^{r_2}_{r_1}

    if r1 is at infinity and r2 is equal to r,
    \Delta U = +\frac{GMm}{r}

    What did I do wrong?

    Last edited: Sep 1, 2015
  2. jcsd
  3. Sep 1, 2015 #2


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    First decide whether the force vector should be written as ##\mathbf{F} = + \frac{GMm}{r^2} \hat{\mathbf{r}}## or as ##\mathbf{F} = - \frac{GMm}{r^2} \hat{\mathbf{r}}##.

    Then think about whether ##\hat{\mathbf{r}} \cdot \mathbf{dr} = dr## or ##\hat{\mathbf{r}} \cdot \mathbf{dr} =-dr##
  4. Sep 1, 2015 #3
    Oh...minus sign! That makes since because the force of gravity is directed to the center of the Earth, in the -r^ direction. Thanks!
  5. Sep 2, 2015 #4


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    I would have answered the stated question in a completely different way.
    The zero level in a potential field is, in general, arbitrary. For practical purposes, all that matters is the potential differences between points. The choice of where to set the zero is either up to the individual or a matter of convention.
    For gravitational potential, the convention is that it is zero at infinity. I leave you to complete the explanation from there.
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