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On the website https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Physics_(Boundless)/5%3A_Uniform_Circular_Motion_and_Gravitation/5.5%3A_Newtons_Law_of_Universal_Gravitation#:~:text=Isaac%20Newton%20proved%20the%20Shell,inside%20of%20it%20is%20zero.

Isaac Newton proved the Shell Theorem, which states that: A spherically symmetric object affects other objects gravitationally as if all of its mass were concentrated at its center.

From this statement, it seems as though the shell theorem only requires the mass to be a perfect symmetrical sphere in order to apply. However, if a solid sphere has non-uniform density, can we still assume that all its mass is concentrated at its center?

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if a solid sphere has non-uniform density, can we still assume that all its mass is concentrated at its center?

Yes. The shell theorem depends on the spherical symmetry, not uniformity. Meaning that the density can vary as a function of $r$, as long as it is constant with respect to $\theta$ and $\phi$

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  • $\begingroup$ What do you mean by constant with respect to θ and ϕ? $\endgroup$
    – john
    Commented Mar 4, 2022 at 14:58
  • $\begingroup$ Also is there a reason why shell theorem still applies to non-uniform density? I thought when proving shell theorem, we assume uniform mass distribution? $\endgroup$
    – john
    Commented Mar 4, 2022 at 15:00
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    $\begingroup$ @john - each shell layer can have its own density, but has to be constant all the way around (so for all latitude/longitude angles). If all the mass is at the North Pole clearly the shell theorem does not apply. $\endgroup$
    – Jon Custer
    Commented Mar 4, 2022 at 16:29
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    $\begingroup$ @john What Dale means in this answer is that the shell theorem still applies even if density varies with radial distance from the center so long as density has no dependence on direction ($\theta$ and $\phi$). $\endgroup$
    – David Hammen
    Commented Mar 4, 2022 at 19:44

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