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Grav field fun!

  1. Apr 13, 2005 #1
    [tex] \vec{g}(x,y,z) = -kG((x^3 y^2 z^2)\hat{e_x} + (x^2 y^3 z^2)\hat{e_y} + (x^2 y^2 z^3) \hat{e_z})[/tex] given this grav field (k is constant)

    find the mass density of the source of this field, and what is the total mass in a cube of side 2a centered about the origin?

    hmmm well we all know...[tex]\int \int \vec{g} \bullet d\vec{a} = 4 \pi G m_{enclosed}[/tex]

    and density [tex] p = \frac{m}{V} [/tex] at least the overall density of it is (non-differential)

    sooooo...[tex]\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}[/tex]

    now the limits i made a cube of side 2a, because the flux through a box is easier when g is given in cart coords....any way i get...

    [tex]m_{source} = \frac{-2}{3 \pi}G k a^9[/tex] how do I get a neg mass (unless this is dark matter which it very well could be) and I'm thinking i missed something about density cause why would it ask that first and then the mass enclosed second?...a lil help?
     
    Last edited: Apr 13, 2005
  2. jcsd
  3. Apr 13, 2005 #2

    dextercioby

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    Are u sure u set the integratiing limits from -a to +a for all three integrals?

    Daniel.
     
  4. Apr 13, 2005 #3
    yes...i am sure
     
  5. Apr 13, 2005 #4

    dextercioby

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    You forgot about the minus in the RHS of Gauss's theorem for the gravitostatic field...

    [tex] \oint\oint_{\Sigma} \vec{g}\cdot d\vec{S}=\mbox{-}4\pi G m_{\mbox{enclosed by}\ \Sigma} [/tex]

    Daniel.
     
    Last edited: Apr 13, 2005
  6. Apr 13, 2005 #5
    i see so that just answers the negative mass problem, now how is it that this grav field increases as you move farther away from the "source" this tells me that the "mass" enclosed is ever increasing i suppose so i guess having the mass enclosed as a function of the flux box side is alright what do you think?
     
  7. Apr 14, 2005 #6
    anyone? anyone? buler buler?
     
  8. Apr 14, 2005 #7

    dextercioby

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    Why did u conclude it increases?

    Daniel.
     
  9. Apr 14, 2005 #8
    ok if [tex]\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}[/tex] then...ill just do it and show you

    [tex]\nabla \bullet \vec{g} = -9kGx^2y^2z^2[/tex]
    [tex]\frac{9k}{4 \pi }\int^a_{-a}\int^a_{-a}\int^a_{-a} (x^2y^2z^2) dxdydz = \frac{2}{3}k \pi a^9 = m_{source/enclosed}[/tex] so as my "flux cube" getts larger obviously it isn't limiting at some value as the function is ever increasing so, the mass should always be increasing yes?
     
  10. Apr 14, 2005 #9

    dextercioby

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    Of course the mass increases,but you said about the field.It's not the same thing...:wink:


    Daniel.
     
  11. Apr 14, 2005 #10
    alright cool thanks man all that for yes u did it right except a negative sign lol
     
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