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Gravitational Potential Energy and force

  1. Jun 15, 2005 #1
    The gravitational potential energy ([tex]U[/tex]) of an object, from some reference point - such as the surface of the earth, is defined as the amount of work required to move the object from the reference point to its position.

    The force of gravity is given by:

    [tex]
    F = -\frac{GMm}{r^2}
    [/tex]

    where [tex]r[/tex] is the distance of the object from the reference point. Example: the distance of a satellite from the earth's surface.

    If you make a graph of force versus distance [tex]r[/tex] you would see it begin at [tex]-\infty[/tex] and decrease exponentially towards zero as [tex]r[/tex] increases.

    If you marked the horizontal axis for the reference point and the distance of the object, the gravitational potential energy would be represented by the area bounded by the graph and the horizontal axis, between the two marked points. You can calculate this with the integral:

    [tex]
    U = -\int_a^b \frac{GMm}{r^2} dr
    [/tex]

    where [tex]a[/tex] is the reference point and [tex]b[/tex] is the position of the object.

    If you solve this you end up with:

    [tex]
    U = -GMm(\frac{1}{a} - \frac{1}{b})
    [/tex]

    However, the forumula you find in physics books is this:

    [tex]
    U = -\frac{GMm}{r}
    [/tex]

    where [tex]r = b - a[/tex]

    I remember my physics teacher saying that the latter formula is used to make things simpler - when [tex]r=\infty, U=0[/tex].

    I can't figure out how they managed to get from the first formula to the second one, to me they don't seem to be interchangeable... :grumpy: ...???
     
    Last edited: Jun 15, 2005
  2. jcsd
  3. Jun 15, 2005 #2
    The last formula you cite is the gravitational potential energy with respect to a specific point, and the first gives you the GPE difference between two points. For

    [tex]U = -\frac{GMm}{r}[/tex] with r = b - a, taking a = 0 gives you the GPE of an object in M's gravitational field where the distance between them is b.

    For the other case, if you take

    [tex]U = -GMm\left(\frac{1}{a} - \frac{1}{b}\right)[/tex]

    and multiply out you get:

    [tex]U = \frac{-GMm}{a} - \frac{-GMm}{b}[/tex]

    which is just the difference in GPE for an object m in M's gravitational field in two different positions. It is the initial potential at radius a, minus the potential at radius b. The negatives end up doing a little happy dance and cancelling.
     
  4. Jun 15, 2005 #3
    Potential energies are usually (and this holds for GPE) classically defined as the energy to take a particle from infinity to a certain point. So, following this, [itex]a=\infty[/itex] and b=r. It all drops out from that.
     
  5. Jun 16, 2005 #4
    whozum, you can't set [tex]a = 0[/tex] because the this gives you an improper integral. This also implies that the two objects, if allowed to "fall" toward each other, could become infinitely close, which is impossible. To find the GPE of the object at some distance, you have have some reference point greater than zero, which is present in all real life applications. For example, if an object is some distance from a planet, the planet surface is the reference. The integration limits would be from the planets radius to the distance of the object from the center of the planet.

    James, that doesn't seem logical to me. If you move an object from the surface of the earth to some arbitrary height and then release it, it will fall and convert its potential energy to kinetic energy. The instant before it hits the ground, all the potential energy will have been transformed to kinetic. How much kinetic energy? The same amount of energy that was required to move it to that height. This is equal to the amount of work done on the object from it's starting point - the earth's surface, to whatever height. How could the force of gravity at a one million mile radius have anything to do with an object that is only at a 1 mile radius?
     
    Last edited: Jun 16, 2005
  6. Jun 16, 2005 #5

    pervect

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    The formula you find in physics books is not r=b-a, it's r=a.

    The gravitational potential energy is then just -GmM/r + constant.

    The value of constant is arbitrary, and usually set to zero. The reason the constant is arbitrary is that when you calculate the potential difference between two locations, you subtract the potential at location b from location a. The constant then doesn't matter.
     
  7. Jun 16, 2005 #6

    dextercioby

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    Cosider a system of coordinates in which the source of the gravitational field is located at the point of position vector [itex] \vec{a} [/itex].

    In this system of coordinates,the gravitational potential energy of interaction between a body of mass "m" situated at the point of position vector [itex]\vec{r} [/itex] and the source of the field of mass "M" is

    [tex] U=-\frac{GmM}{\left|\vec{r}-\vec{a}\right|} [/tex]

    Daniel.
     
  8. Jun 16, 2005 #7
    Felix - it doesn't matter where you define the potential from, as it's potential differences that matter. It is a standard definition to take GPE (and electrical potential, etc etc) as the energy required to move the object from infinity to the point.

    When you talk of raising an object above the earth, the potential at each point is defined by this, subtract the two, and you have the potential difference (close to the earth, flat field, this will equal mgh).
     
  9. Jun 16, 2005 #8
    You misunderstood me. The model that I set up explained the gravitational PE for a point mass' field. The latter model, the difference in potential between two points with respect to a field would give you the potential between, say, 100m above the surface and the surface.
     
  10. Jun 16, 2005 #9
    whozum, I see what you're saying now.

    James, I understand now. My physics teacher didn't do a great job of clarifying. Also, I misread some of the class notes about it. The first formula I gave is the correct formula for difference in GPE, [tex]\Delta U[/tex]. The second one is for the specific value of GPE at that point, using [tex]\infty[/tex] as a reference. It all makes sense now, but my teacher never really pointed out that [tex]\infty[/tex] was the reference.

    Thanks for the responses.
     
    Last edited: Jun 16, 2005
  11. Jun 16, 2005 #10
    Ah! That would explain it then... Glad you've got your head around it.
     
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