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Homework Help: Force due to gravity inside planet

  1. Nov 27, 2012 #1
    Here is a simple problem in classical gravitation.

    Consider a spherical planet of radius R, and let the radial coordinate r originate from the plant's center. If the density of the planet is ρ from 0 ≤ r < R/2 and ρ/3 from R/2 < r < R, then my work tells me that the maximum force due to gravity inside the planet is at r = R/2, not at r = R as one might expect.

    [tex]0\leq r\leq R/2,\qquad F_G = \frac{Gm\left(\rho \frac{4}{3}\pi r^3\right)}{r^2} = \frac{4\pi}{3}Gm\rho r \\
    R/2 \leq r \leq R,\qquad F_G = \frac{Gm\left[ \rho\frac{4\pi}{3}\left(\frac{R}{2} \right)^3 + \frac{\rho}{3}\frac{4\pi}{3} \left( r^3 - \left(\frac{R}{2} \right)^3 \right) \right]}{r^2} = \frac{4\pi}{3}\frac{Gm\rho}{r^2}\left[ \left(\frac{R}{2} \right)^3 + \frac{1}{3}r^3 - \frac{1}{3}\left(\frac{R}{2} \right)^3 \right][/tex]

    My work is above. Is this be correct, that the maximum force due to gravity would be at r = R/2? Thanks for your time!
     
  2. jcsd
  3. Nov 27, 2012 #2

    mfb

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    Neglecting 4pi/3 Gmρ as prefactor:

    In the inner part: ##F \propto r##
    In the outer part: ##F \propto (\frac{R}{2})^3/r^2 + 1/3 (r-\frac{R}{2}) = \frac{1}{r^2} ((\frac{R}{2})^3+\frac{1}{3}r^3-\frac{1}{3}(\frac{R}{2})r^2)##

    I get a different third term for the force.

    ##\frac{dF}{dr}=-\frac{R^3}{4r^3} + \frac{1}{3}## which is 0 at ##4r^3=3R^3##, it has a minimum in the less dense region.

    F(R/2) > F(R), so I can confirm your result.
     
  4. Nov 27, 2012 #3
    Thanks mfb. Regarding our difference, I think your third term might be mistaken. If we hollow out the planet from 0 ≤ r < R/2, then the force due to gravity in the domain R/2 ≤ r ≤ R is

    [tex] F_G = \frac{Gm\frac{\rho}{3}}{r^2}\left[ \frac{4\pi}{3}r^3 - \frac{4\pi}{3}\left(\frac{R}{2}\right)^3\right] [/tex]

    By the shell theorem, it as if all the volume enclosed is contained in a point at the center. Do you agree?
     
  5. Nov 27, 2012 #4

    mfb

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    Oh, you are right.
     
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