Is it possible to use the Lagrangian formalism involving physical terms to answer the question of why all planets are approximately spherical?
Let's assume that a planet is 'born' when lots of particles of uniform density are piled together, and constrained only by the Newtonian gravity of the $N$ particles with no external gravity.
My question is: Is it possible to use the variational calculus starting from this situation to prove that planets should be spherical?
Answer
I interpreted the question as asking whether it's possible to show that a sphere is the minimum energy shape of an object being acted on by its own gravity. I attempted to do this using spherical harmonics, but got stuck part of the way there; I'm posting this anyways just in case someone can figure out how to complete the last bit.
The gravitational self-energy of a matter distribution $\rho(\mathbf{r})$ which gives rise to a gravitational potential $V(\mathbf{r})$ is given by $$E=\frac{1}{2}\left\langle V,\rho\right\rangle=\frac{1}{2}\left\langle \nabla^{-2}\rho,\rho\right\rangle$$ where $\langle,\rangle$ represents the inner product and $\nabla^{-2}$ is the inverse Laplacian. If there is a change in the matter distribution $\rho=\rho_0+\delta\rho$, then the self-energy changes to $$E=\frac{1}{2}\left\langle \nabla^{-2}(\rho_0+\delta\rho),\rho_0+\delta\rho\right\rangle=\frac{1}{2}\left\langle \nabla^{-2}\rho_0,\rho_0\right\rangle+\left\langle \nabla^{-2}\rho_0,\delta\rho\right\rangle+\frac{1}{2}\left\langle \nabla^{-2}\delta\rho,\delta\rho\right\rangle \\ =E_0+\delta E+O(\delta^2)$$ where the self-adjointness of the Laplacian was used in the second to last step.
Thus it suffices to determine the sign of $\delta E=\left\langle V_0,\delta\rho\right\rangle$. Since $V_0$ is just the gravitational potential of a uniform-density sphere (which has simple closed form) this becomes tractable.
Let the matter density be $\rho_d$ and assume the planet is incompressible. The radius of the planet $R(\Omega)$ can be expanded in the real harmonics as $$R(\Omega)=R_0+\sum_{L=0}^\infty\sum_{m=-L}^L\delta C_{Lm}Y_{Lm}(\Omega)=R_0+\delta R(\Omega).$$ To conserve volume, note that $$V=\frac{4}{3}\pi R_0^3=\frac{1}{3}\int R(\Omega)^3\,d\Omega=\frac{1}{3}\int \left(R_0+\sum_{L=0}^\infty\sum_{m=-L}^L\delta C_{Lm}Y_{Lm}(\Omega)\right)^3\,d\Omega \\ =\frac{4}{3}\pi R_0^3+2\sqrt{\pi}R_0^2\delta C_{00}+R_0\sum_{L=0}^\infty\sum_{m=-L}^L\delta C_{Lm}^2+O(\delta^3) \\ \Rightarrow \delta C_{00}=-\frac{1}{2\sqrt{\pi}R_0}\sum_{L=1}^\infty\sum_{m=-L}^L\delta C_{Lm}^2+O(\delta^3). $$ Since the volume conservation term $\delta C_{00}$ scales as $\delta^2$ let's rewrite it as $\delta^2C_{00}$, giving $$R(\Omega)=R_0+\sum_{L=1}^\infty\sum_{m=-L}^L\delta C_{Lm}Y_{Lm}(\Omega)+Y_{00}\delta^2C_{00}=R_0+\delta R_1+\delta^2 R_2.$$ With a bit of visualization one can see that $$\left\langle V_0,\delta\rho\right\rangle=\int d\Omega R(\Omega)^2\int_{R_0}^{R(\Omega)}V_0(r)\rho_d\,dr \\ =\int d\Omega R(\Omega)^2\left[-\frac{G M R_1 \rho_d}{R_0}\delta+\frac{GM(R_1^2-2R_0R_2)\rho_d}{2R_0^2}\delta^2+\:...\right] \\ =\int d\Omega\left[-GMR_0\delta R_1\rho_d-\frac{1}{2}GM\rho_d\left(3(\delta R_1)^2+2R_0\delta^2R_2\right)\:...\right]$$ where in the last step terms in the integral have been arranged according to their order in $\delta$.
The first term vanishes due to symmetry of the real harmonics, and the second term simplifies due to orthonormality to become $$=-\frac{1}{2}GM\rho_d\left[3\sum_{L=1}^\infty\sum_{m=-L}^L\delta C_{Lm}^2-2\sum_{L=1}^\infty\sum_{m=-L}^L\delta C_{Lm}^2\right] \\ =-\frac{1}{2}GM\rho_d\sum_{L=1}^\infty\sum_{m=-L}^L\delta C_{Lm}^2<0.$$ Unfortunately, this predicts that the energy change is always favorable to deformation, rather than always unfavorable! So I must have an arithmetic error somewhere, but I am too tired to find out where it is (the units at least work out correctly to Joules). However, this gives a general idea of how one can proceed down this line of proof.
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