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Gravitation of non-uniform density

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A gaseous planet, with radius R, has a radially dependent density function
    [tex]\rho (r') = \rho_0 [\frac{r'}{R}]^2[/tex]
    where r' is the distance from the center planet. Find the magnitude of force for a mass m inside and outside of the planet.


    2. Relevant equations
    [tex] F = -\frac{GmM}{r^2} [/tex]
    [tex] F = -G m \int_V \frac{\rho(r') \hat{e_r}}{r^2}dv'[/tex]


    3. The attempt at a solution
    I'm pretty sure I did it right, but would like some confirmation. The mass inside the planet, taking a sample shell, would be

    [tex] M = \int_0^r 4\pi r'^2 dr' \rho(r')[/tex]

    which would give

    [tex] M = \int_0^r 4\pi r'^2 dr' \rho_0 (\frac{r'}{R})^2[/tex]

    and evaluates to

    [tex] M = \frac{4}{5} \pi r^5 \frac{\rho_0}{R^2}[/tex]

    such that the force inside will be

    [tex] F = \frac{-Gm r^3}{r^2}\rho_0 \hat{e_r}[/tex]

    Outside

    the same idea applies

    [tex] M = \int_0^R 4 \pi r'^2 dr' \rho(r')[/tex]

    and gives

    [tex] M = \frac{4}{5} \pi R^5 \frac{\rho_0}{R^2}[/tex]

    so that the force is

    [tex] F = \frac{-GmR^2}{r^2}\rho_0 \hat{e_r}[/tex]

    Both answers, more or less, make sense. Inside, as you move further away that the force will increase. Outside, the force will decrease as you move further away.
     
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2
    It's the same problem as a Gauss' law problem for a ball of non-uniform charge, and that's how the solution goes, so that all looks right.
     
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