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Clarification on determining the Potential of a body

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the gravitational potential both inside and outside a spherical shell of inner radius b and outer radius a?


    2. Relevant equations

    [tex]\Phi = -2\pi*G*\rho*\int (r'*r') \int (sin(\theta)/r) d\theta[/tex]

    Integration range for the first integral is from a to b, while the integration range for the 2nd integral is (R+r' to R-r') for [tex]\Phi[/tex] outside of the shell and from (R-r' to R+r') inside the shell

    3. The attempt at a solution

    This is actually an example in the book, and I am able to work out the example found in the book, but only through taking the second integral in the above equation for granted. I do not quite understand why

    A) we need a double integral,
    B) how we identify it as "[tex](sin(\theta)/r) d\theta[/tex], and
    C) how the bounds are determined to be 0 to [tex]\pi[/tex].

    I find the first integral very easy to understand, since we are looking for the 'whole' area of the shell, coupled with the fact that if we multiplied the 2pi factor in, it would look like the area of a circle (Or in this case, of the shell)

    Any clarification would be greatly appreciated!
     
  2. jcsd
  3. Apr 8, 2010 #2

    vela

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    I'm not sure what you're saying about the limits.

    In spherical coordinates, the volume element is [itex]dv=r^2 \sin\theta d\theta d\phi dr[/itex], so the potential is given by

    [tex]\Phi(\textbf{r}) = -\int \frac{G\rho dv'}{|\textbf{r}-\textbf{r}'|}=-\int_a^b \int_0^{2\pi} \int_0^\pi \frac{G\rho r'^2 \sin\theta d\theta d\phi dr'}{|\textbf{r}-\textbf{r}'|}[/tex]

    If you pull the gravitational constant and density out front and integrate over phi to get the factor of 2 pi, you get the result you have (if I'm interpreting what the variables in your expression stand for correctly).
     
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