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Calculating the Gravitational Potential of a Spehrical Shell

  1. Jul 24, 2007 #1
    I'm going through The Feynman Lectures On Physics. I've taken a few classical physics classes before, however Feynman's detailed explanations are much better than anything I've gotten from my teachers. However, I'm having a little difficulty following the mathematics when he tries to show that a hollow spherical shell acts as though it has all its mass located at the center.

    I will read the paragraph previous to the math I am having difficulty with, and then read off the math.

    "We now demonstrate the correctness of this miracle. In order to do so, however, we shall consider a thin uniform shell instead of the whole earth. Let the total mass of the shell be m, and let us calculate the potential energy of a particle of mass m' a distance R away from the sphere and show that the potential energy is the same as it would be if the mass m were at the center. If we call x the distance of a certain plane section from the center, then all the mass that is in a slice dx is at the same idstance r from P, and the potential due to the ring is -Gm'dm/r. How much mass is in the small slice dx? An amount


    Here, a is the radius I believe. I don't know where he pulled y from. Mu is the density of a piece of the spherical shell.

    What I can't understand is how he is doing the math for all this. Where does the sin theta come from? I'm slightly lost here.

    This is in the Feynman Lectures, by the way.
    Last edited: Jul 24, 2007
  2. jcsd
  3. Jul 24, 2007 #2
    Can you tell me the chapter it's in, I have to see it in the book. I'll try to see if I can figure it out.
  4. Jul 24, 2007 #3
    It's in chapter 13-4. Thanks.
  5. Jul 24, 2007 #4

    Doc Al

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    Staff: Mentor

    To understand what he's doing, refer to the diagram next to the equations. The diagram shows the meaning of x and y.

    dm refers to the mass of the ring-shaped slice of the sphere. Yes, a is the radius of the sphere. y is the perpendicular (vertical) distance of the ring from the line O-P. Since ds is the length of the side of the ring, the thickness of the ring along the x-axis (which is the line O-P) is [itex]dx = ds \sin\theta[/itex].
    Book I, chapter/page 13-9.
  6. Jul 24, 2007 #5
    Ahh. That makes perfect sense now. Thanks a lot.
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