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Gravitational Force on a Particle in Jupiter's Core

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Jupiter has a core of liquid metallic hydrogen, with uniform density $\rho_c$, with radius $R_c$. This is surrounded by a gaseous cloud $R_g$, where $R_g>R_c$. Assume the cloud is of uniform density $\rho_g$.

    The problem also specifies that we are to assume both regions of Jupiter are spherical (so it's two spheres, one inside the other)

    What is the gravitational force on an object of mass $m$ located a distance $r$ from Jupiter's center? Consider the following cases:

    A) $r<R_c$ (i.e., within the core)
    B) $R_c<r<R_g$
    C) $r>R_g$

    2. Relevant equations

    $$F=\frac{Gm dM}{|d-d'|^2}\hat{d-d'}$$

    (With the hat denoting unit vector)

    3. The attempt at a solution

    I think what we're supposed to do here, since the density is not given as a continuous function, is to calculate the gravitational force due to the core, and the gravitational force due to the surrounding gas, and then superpose them. Since the density is uniform, we know that in a small chunk of mass dM (I think, this is one of the main thing's I'm not sure about):

    $$dM=\rho P^2 \sin\theta d\theta d\phi dP$$

    In spherical coordinates, $(P, \phi, \theta)$, that is. The $\rho$ is either $\rho_c$ or $\rho_g$, depending on which integral we're trying to do. The distance term will be given by:

    $$\vec{d-d'}=(P \cos \phi \sin \theta - r_1) \hat{i} + (P \sin \theta \sin \phi-r_2) \hat{j} + (P \cos \theta - r_3)\hat{j}$$

    Where $(r_1,r_2,r_3)$ are the coordinates of the mass located a distance $r$ from the center. This gives:

    $$|d-d'|=\sqrt{P^2 - 2r_1P\cos(\phi)\sin(\theta) - 2r_2 P \sin \theta \sin \phi - 2r_3 P cos\theta + r^2}$$

    (Note that $r_1^2 + r_2^2 + r_3^2 = r^2$ from the way we defined them)

    I'm not sure this is right, especially because doing this produces the obscenely complicated integral for the force from the core region:

    $$F=\int_0^{2\pi}\int_0^\pi \int_0^{R_c} \frac{mG\rho_c P^2 sin \theta}{P^2 - 2r_1P\cos(\phi)\sin(\theta) - 2r_2 P \sin \theta \sin \phi - 2r_3 P cos\theta + r^2} dP d\theta d\phi$$

    Which looks too complicated to be right, especially since it contains three constants that weren't given. Could somebody either confirm I'm on the right track, or give me some guidance as to where I went wrong/how to get back on the right track? Thanks.
  2. jcsd
  3. Apr 6, 2015 #2

    Doc Al

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    Staff: Mentor

    Way too complicated! Start by answering this problem: A mass m sits on the surface of a spherically symmetric planet with mass density ##\rho## and radius R. What gravitational force does the planet exert on the mass?

    No calculus required! Just Newton's law of gravity.
  4. Apr 6, 2015 #3

    Doc Al

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    Staff: Mentor

    Hint: Take advantage of spherical symmetry.
  5. Apr 7, 2015 #4
    Thanks for the reply. I think I see what you're saying. Since we have uniform density, we just have:

    $$M=4\pi R^2 \rho$$

    And since it's on the surface, the distance is just R

    So $$F=4mg\rho \pi$$

    (Note I'm not sure that's exactly right but I woke up a few minutes ago so I can't quite get my brain to work well enough to check)

    How do I deal with the outer shell, though? That seems more complicated.
  6. Apr 7, 2015 #5

    Doc Al

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    Staff: Mentor

    Careful. You want the volume of the sphere, not the surface area.

    Understood. Give it another try. :wink:

    Actually, it's just as easy---even easier. First take a guess as to what the field inside a uniform spherical shell would be. Then look up Newton's Shell Theorems.
  7. Apr 7, 2015 #6
    Aha! I suspected t was something like that (I was thinking Gauss' Law), but I couldn't quite make the connection. I will attempt that solution when I get home, thank you!

    And yeah I realize that I should've plugged (4/3)pi R^3 in there, like I said, I had just woken up and was not in a state of full coherence, haha.
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