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Gravitational potential of a point within a hollow sphere

  1. Jun 7, 2006 #1

    eep

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    Hi,
    I know that if you have a hollow sphere of some finite mass, when you place an object inside it, it experiences no gravitation force. Does this mean that the point also has no gravitational potential energy?

    thread title should be "Gravitational" not "Gravitional"
     
  2. jcsd
  3. Jun 7, 2006 #2

    rcgldr

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    A "point" is an abstract concept, it's infinitely small and has no mass. A real object would have mass and create a gravitational field on it's own, regardless of whether or not it was inside a hollow sphere.
     
  4. Jun 7, 2006 #3

    eep

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    Well say you've got some tiny object inside of a hollow sphere. What's the gravitation potential energy of the object due to the surrounding mass of the hollow sphere?
     
  5. Jun 7, 2006 #4

    rcgldr

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    Edit: correcting my posts in this thread.

    The potential energy within the hollow sphere is constant, regardless of the objects position within the sphere.

    It's the net gravitation forces on the objects that sum up to zero, regardless of the objects position within the shpere.
     
    Last edited: Jun 7, 2006
  6. Jun 7, 2006 #5

    eep

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    But how can it's gravitation potential energy due to the spherical shell be 0 if the potential is defined as

    [tex]
    \frac{-GMm}{|r|}
    [/tex]

    How do you add a bunch of negative numbers together and get 0?
     
  7. Jun 7, 2006 #6

    rcgldr

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    Edit: update to reply:

    If you add up all the forces on the object, the forces sum up to be zero at any location within the sphere. If there is no net force on the object, then the potential energy should be zero, or at least a constant independent of position.

    I don't understand how potential energy from multiple sources could be summed without taking direction (location of those sources) into account. It seems to me that you'd need to sum up vectors, not scalars.
     
    Last edited: Jun 7, 2006
  8. Jun 7, 2006 #7
    The potential has to be a constant (since [itex] -\nabla V = \vec{F} = 0[/itex])but, need not be zero. You have to calculate the potential as two parts...from the reference point (which, in this case, is usually taken at infinity) to the surface of the sphere, and the other from the surface to the point under consideration. In each part (an integral) you have to use the form of the field (which is essentially the force) which exists in that particular region.The second integral will vanish since the field is zero inside, and the first part will come out in such a way that it is a constant.
     
  9. Jun 7, 2006 #8

    vanesch

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    As the previous poster said, you do not have to get 0. You only have to get a constant, independent of position.
     
  10. Jun 7, 2006 #9
    This assumes one is outside the sphere, not inside.
     
  11. Jun 7, 2006 #10

    eep

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    Potential energy is a scalar, not a vector.
     
  12. Jun 7, 2006 #11

    pervect

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    As others have mentioned, the condition for zero force is only that the gravitational potential be constant. It does not matter to the force what the magnitude of the potential is, the force is proportional to the rate of change of potential (dV/dx).

    In classical mechanics, the "zero point" of potential is arbitrary in any event - one can add or subtract any constant number from the potential function without changing the physics.

    A good way of looking at this: the escape velocity inside a hollow sphere will be the same as the escape velocity on the surface of a hollow sphere. An object will be at escape velocity when it has just enough kinetic energy to reach infinity.
     
  13. Jun 7, 2006 #12

    rcgldr

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    If you add up all the forces on the object, the forces sum up to be zero at any location within the sphere. If there is no net force on the object, then the potential energy should be zero, or at least a constant independent of position (within the sphere).

    I don't understand how potential energy from multiple sources (in this case, the integral sum of all the points of the hollow shpere) could be summed without taking direction (location of those sources) into account. It seems to me that you'd need to sum up vectors, not scalars.

    Edit: The object experiences no gravitation force. Original and incorrect reply: The object inside a hollow shpere will experience a pull in all directions, expanding the object slightly (assuming that the object isn't decompressable).

    What I don't know for sure is if the object inside a hollow shpere isn't spherical, if the potential energy is still independent of where in the sphere that the object is located. If you divide the object into smaller and smaller components, then as the size of the components approach being points, then each point's potential energy is independent of it's location, so it would seem that the integral sum of all these points would be independent as well.
     
    Last edited: Jun 7, 2006
  14. Jun 7, 2006 #13

    Doc Al

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    Since the gravitation field within the hollow sphere is zero, the object will experience no gravitational force at all. It will not be pulled in all directions; it will experience no stress or strain, and will not be stretched or expanded.
     
  15. Jun 7, 2006 #14

    rcgldr

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    Agreed, I just realized that when I considered the integral sum of all the points that make up the object inside the sphere. Each point experiences zero force, so the entire object experiences zero force. I need to take more time in composing replies instead of editting them afterwards, in the words of Ronald Reagon, I'll try to avoid "shooting from the hip".
     
    Last edited: Jun 7, 2006
  16. Jun 7, 2006 #15

    rcgldr

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    This was brought up before but never clarified. What gravitational effect would a third object, outside the sphere, have on an object inside the sphere? My intution is that it would be the same as if the hollow sphere wasn't there at all.
     
  17. Jun 7, 2006 #16

    Hootenanny

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    Indeed, this would the case, as long as the object inside the hollow sphere did not collide with the hollow sphere, note that the gravitational field of the third object will also infuence the hollow sphere.
     
  18. Jun 8, 2006 #17

    Doc Al

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    If you are adding up the gravitational field contributed by each mass element, then of course you'd need to add them as vectors. But if, instead, you added up the potential contributed by each mass element, only distance, not direction, would matter since the potential is a scalar. (That's one of the advantages of using potential--it's a scalar.) And if you set up the calculation and do the integral you'll find that the potential at any point is independent of position within the hollow sphere.

    Since field and potential are intimately related (given the potential function, one can derive the field), either method must lead to the same conclusion.

    Exactly. The field contributions from all masses just add up. Since the sphere contributes nothing, all you have is the field from that third object.
     
  19. Jun 8, 2006 #18

    reilly

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    A pedantic point or two: first, you must assume that the mass distribution of the sphere is spherically uniform, it can be radially nonuniform. Second, you must assume that the inside mass has no influence on the mass distribution of the sphere -- unlike, say, the same problem for a charged spherical shell, isolated or grounded. And yes, the charge outside will give a -GmM/s portential for the inside charge, where s is the distance between the two masses. But, is this the correct total potential?

    If the shell has mass M', what is the potential for the outside charge?

    Regards,
    Reilly Atkinson
     
  20. Jun 9, 2006 #19

    rcgldr

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    My point about potential energy from multiple sources was meant to be more general than the hollow sphere.

    Take an object that is between the Earth and the Moon right at the point where it does not fall towards either. In this case, the potential energy is probably indeterminate. If the object moves and falls towards the Moon it's potential energy is far less than if the object falls and moves towards the Earth. How is this situation handled with scalar math?
     
  21. Jun 9, 2006 #20

    Doc Al

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    The gravitational field at that point is zero; the potential has a local maximum; an object placed there will be in unstable equilibrium.
    Not at all. See below.
    This is true.
    Very simply. The gravitational potential at a point outside a spherical object (assume the earth and moon are spherically symmetric--close enough) is given by:
    [tex]-G\frac{M}{R}[/tex]
    where, as is typical, we take the potential as zero at infinity, and R is the distance to the center of the gravitating object.

    So the potential at some point due to the gravity of Earth and Moon is just the sum of each contribution:
    [tex]-G (\frac{M_{earth}}{R_{earth}} + \frac{M_{moon}}{R_{moon}}) [/tex]
     
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