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Gravitational field in a hollow sphere

  1. Apr 19, 2004 #1
    Can someone please give me a qualitative justification for the gravitational field inside a uniform hollow sphere being zero? I'm having alot of trouble understanding this. Prof. said (in class) not to worry about the higher order polynomials involved, just be concerned with it "qualitatively" at this point. I'm lost.
    Thanks,
    m_d
     
  2. jcsd
  3. Apr 19, 2004 #2

    ZapperZ

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    You did not indicate what level of physics you already have. So, I have no idea if what I will say here makes any sense to you.

    There is a Gauss's law equivalent for gravitational field. So if you have seen gauss's law applied to electrostatic, you should understand the identical principle applied to gravitational field. So apply that.

    http://scienceworld.wolfram.com/physics/GausssLaw.html

    Zz.
     
  4. Apr 19, 2004 #3

    turin

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    Qualitatively, think of it like this: At any given point inside the sphere, there is x amount of mass to the right in the shape of a bowl, and 1-x amount of mass to the left in the shape of a shperical shell missing a bowl shaped cap. These two complementary portions have centers of mass at lets say rx and r1-x. Then, qualitatively, you can imagine that rx < r1-x by just such an amount that:

    x/rx2 = (1-x)/r1-x2.

    It would probably help to draw a picture.

    Qualitatively, you're supposed to realize that Fgrav is larger for larger amounts of mass and smaller at larger distances away from the mass. So, qualitatively, you can imagine that the effect of more mass (1-x) is canceled by the effect of further away (r1-x).
     

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    Last edited: Apr 19, 2004
  5. Apr 19, 2004 #4
    turin,
    Thanks, I think I can understand that. However, when I click on the link to your .gif, I get a "you need to login to view this" message--well, I am! Oh, well, I'll take that up with the admins.
    ZapperZ, FWIW, I'm in first year college physics for engineers (PHYS141 @ UoArizona). Next semester is my Electricity and Magnetism. So I have no idea what Gauss' law is, but I guess I'll find out.
    Thanks for the help!
     
  6. Apr 20, 2004 #5

    HallsofIvy

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    From any point inside the sphere, imagine a cone extending to a portion of the surface. Now extend that cone back to the opposite side of the sphere. If your point is not exactly at the center, your two cones will not intersect equal areas (and, so, not equal masses) because area is proportional to the square of the distance. HOWEVER, since gravitational force is inversely proportional to the square of the distance, the gravitational force from each of those portions of the surface will be the same. Since they are in opposite directions, the two equal but opposite forces cancel. The total gravitational force is 0.

    (The same thing is true of magnetic force as well.)
     
  7. Apr 20, 2004 #6

    turin

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    I like HallsofIvy's explanation better than mine. It more clearly incorporates the inverse square dependence.
     
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