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I'm working on a problem in Thorne & Blandford's Modern Classical Physics regarding the shape of a constant density, spinning planet (Exercise 13.5). It asks you to argue that the gravitational potential inside a slowly spinning, uniform density planet can be written in the form $$ \Phi = \frac{2\pi G \rho r^2}{3}+Ar^2P_2(\mu)$$ where A is a constant and $P_2$ is the Legendre polynomial with argument $\mu = \sin(latitude)$. The first part of the question asked you to show that the spatially variable part of the gravitational potential for a non-rotating planet of uniform density was equal to the first term in the above formula.

The best answer I can come up with is that a slowly rotating planet will have an equatorial bulge and therefore the potential inside the planet will no longer be that of a uniform sphere, so the second term arises from the planet's ellipticity. However, the solution associates this second term to the general solution of Laplace's equation. What is the relationship between a planet's ellipticity and the solutions to Laplace's equation? Why is Laplace's equation relevant in this context given that inside the planet, the gravitational potential should obey Poisson's equation, not Laplace's?

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I suspect what they're doing here is expanding the potential $\Phi$ in some kind of formal power series: $$ \Phi = \Phi^{(0)}+ \Phi^{(1)} + \cdots $$ where $\Phi^{(0)}$ is the potential of a non-rotating planet and $\Phi^{(1)}$ is a "small" correction to account for its rotation. This whole power series should satisfy $\nabla^2 \Phi = 4 \pi G \rho$.

But the first term already obeys Poisson's equation with density $\rho$: $\nabla^2 \Phi^{(0)} = 4 \pi G \rho$. This means that the second term must satisfy $\nabla^2 \Phi^{(1)} = 0$, i.e., Laplace's equation.

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  • $\begingroup$ I think this makes sense. Would that mean that the angular velocity is hidden somewhere in A, such that if we were to take that velocity to zero then the $\Phi^{(1)}$ term would vanish? $\endgroup$ Jul 26, 2022 at 2:30
  • $\begingroup$ @GianfrancoGrillo: That would make sense to me, but I'll admit that I don't have the textbook or the problems statement in front of me so I may be missing some context. $\endgroup$ Jul 26, 2022 at 11:31

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