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?
I) Let us work in units where $G=1=\rho$.
In this answer, we will additionally assume that the massive object is star-shaped. We can then use spherical coordinates $(r,\theta,\varphi)$. The surface profile is then given as
$$ {\bf r}~=~r{\bf n}~=~f({\bf n}){\bf n}, \qquad r~=~f({\bf n})~=~\sum_{\ell m} c_{\ell m} Y_{\ell m}({\bf n})~\geq~0 ,$$ $$ c_{\ell m}~=~\int_{S^2}\!d^2n ~Y^{\ast}_{\ell m}({\bf n})f({\bf n}),\tag{1}$$
where ${\bf n}\in S^2$ is a unit vector. In eq. (1) we have used DumpsterDoofus' idea to expand in spherical harmonics. The volume
$$ V[f]~:=~ \int_V \! d^3r ~=~\frac{1}{3}\int_{S^2}\!d^2n ~ f({\bf n})^3 .\tag{2}$$
is a functional of the surface profile $f({\bf n})$.
$$ \delta V ~=~\int_{S^2}\!d^2n ~ f({\bf n})^2\delta f({\bf n}),\qquad \delta^2 V ~=~2\int_{S^2}\!d^2n ~ f({\bf n})\delta f({\bf n})^2.\tag{3}$$
The volume is kept fixed by a constraint
$$ V[f]~=~V_0.\tag{4}$$
Potential:
$$ -\Phi({\bf r})~:=~\int_{V} \! \frac{d^3r^{\prime}}{|{\bf r}-{\bf r}^{\prime}|} ~=~\int_{S^2}\!d^2n^{\prime}\int_0^{f({\bf n}^{\prime})}\! \frac{r^{\prime 2}dr^{\prime}}{|{\bf r}-r^{\prime}{\bf n}^{\prime}|} .\tag{5}$$
$$ -\delta \Phi({\bf r}) ~=~\int_{S^2}\!d^2n^{\prime} \frac{f({\bf n}^{\prime})^2\delta f({\bf n}^{\prime})}{|{\bf r}-f({\bf n}^{\prime}){\bf n}^{\prime}|}.\tag{6}$$
Field:
$$ -{\bf g}({\bf r})~:=~ \nabla \Phi({\bf r}) ~=~\int_{V} \!d^3r^{\prime} \frac{{\bf r}-{\bf r}^{\prime}}{|{\bf r}-{\bf r}^{\prime}|^3} .\tag{7}$$
Radial field:
$$ \begin{align}-g_r({\bf r})~=~&\frac{\partial\Phi({\bf r})}{\partial r} ~=~{\bf n}\cdot \nabla \Phi({\bf r}) ~=~\int_{V} \!d^3r^{\prime} \frac{r-{\bf n}\cdot{\bf r}^{\prime}}{|{\bf r}-{\bf r}^{\prime}|^3}\cr ~=~&\int_{S^2}\!d^2n^{\prime}\int_0^{f({\bf n}^{\prime})}\!r^{\prime 2}dr^{\prime} \frac{r-{\bf n}\cdot{\bf n}^{\prime}r^{\prime}}{|{\bf r}-r^{\prime}{\bf n}^{\prime}|^3}.\end{align}\tag{8}$$
II) Potential energy:
$$ \begin{align} U[f]~:=~& -\iint_{V\times V} \! \frac{d^3r~ d^3r^{\prime}}{2|{\bf r}-{\bf r}^{\prime}|} ~=~ \frac{1}{2}\int_V \! d^3r ~\Phi({\bf r}) \cr ~=~& -\int_{S^2}\!d^2n\int_{S^2}\!d^2n^{\prime} \int_0^{f({\bf n})}\int_0^{f({\bf n}^{\prime})} \frac{r^2dr~ r^{\prime 2}dr^{\prime}}{2|r{\bf n}-r^{\prime}{\bf n}^{\prime}|}.\end{align}\tag{9}$$
$$ \delta U ~=~\int_{S^2}\!d^2n ~ f({\bf n})^2\delta f({\bf n})~\Phi(f({\bf n}){\bf n}).\tag{10}$$
$$ \begin{align}\delta^2 U~=~&-\int_{S^2}\!d^2n\int_{S^2}\!d^2n^{\prime}~ \frac{f({\bf n})^2\delta f({\bf n})~ f({\bf n}^{\prime})^2\delta f({\bf n}^{\prime})}{|f({\bf n}){\bf n}-f({\bf n}^{\prime}){\bf n}^{\prime}|}\cr &-\int_{S^2}\!d^2n~ f({\bf n})^2 ~\delta f({\bf n})^2g_r(f({\bf n}){\bf n})\cr &+2\int_{S^2}\!d^2n~f({\bf n})\delta f({\bf n})^2 \Phi(f({\bf n}){\bf n}).\end{align}\tag{11}$$
III) To treat the constraint (4) we use Lagrange multiplier method. We should minimize the functional
$$ E[f]~:=~U[f] +\lambda (V_0-V[f]).\tag{12}$$
The first variation is $$ \delta E ~=~\int_{S^2}\!d^2n ~ f({\bf n})^2\delta f({\bf n}) \{\Phi(f({\bf n}){\bf n})-\lambda\}.\tag{13}$$
We conclude that
The surface potential $\Phi(f({\bf n}){\bf n})=\lambda<0$ of a stationary shape $f$ is a constant, i.e. independent of the unit vector ${\bf n}$.
The second variation is
$$ \begin{align}\delta^2 E~=~&-\int_{S^2}\!d^2n\int_{S^2}\!d^2n^{\prime}~ \frac{f({\bf n})^2\delta f({\bf n})~ f({\bf n}^{\prime})^2\delta f({\bf n}^{\prime})}{|f({\bf n}){\bf n}-f({\bf n}^{\prime}){\bf n}^{\prime}|}\cr &-\int_{S^2}\!d^2n~ f({\bf n})^2 \delta f({\bf n})^2g_r(f({\bf n}){\bf n})\cr &+2\int_{S^2}\!d^2n~f({\bf n})\delta f({\bf n})^2 \underbrace{\{\Phi(f({\bf n}){\bf n})-\lambda\}}_{=0}.\end{align}\tag{14}$$
For systematic reasons, the second variations (14) should be consistent with the constraint (4) to first order.
IV) We finally consider a ball $f({\bf n})=R$. We then have
$$V~=~\frac{4\pi}{3}R^3, \qquad {\bf g}({\bf r})~=~-\frac{4\pi}{3}{\bf r}, \qquad \Phi({\bf r})~=~2\pi\left(\frac{r^2}{3}-R^2\right), $$ $$ \Phi(R)~=~-\frac{4\pi}{3}R^2, \qquad U~=~-\frac{16\pi^2}{15}R^5.\tag{15}$$
We want to show that the ball is a stable stationary shape. We have to show that the second variation (14) is semipositive definite. Eq. (3) simplifies to
$$ \frac{\delta V}{R^3} ~=~\int_{S^2}\!d^2n ~\delta f({\bf n}) ~=~\sqrt{4\pi}\delta c_{00},\tag{16}$$
$$\frac{\delta^2 V}{2R^3}~=~\int_{S^2}\!d^2n ~|\delta f({\bf n})|^2 ~=~\sum_{\ell m}|\delta c_{\ell m}|^2.\tag{17}$$
Since the constraint (4) should be maintained to first (but not necessarily second) order, we must demand that the zeromode vanishes
$$\delta c_{00}~=~0.\tag{18}$$
The second variation (14) simplifies to (see e.g. Ref. 1)
$$ \begin{align}\frac{\delta^2 E}{R^3} ~=~&-\int_{S^2}\!d^2n \int_{S^2}\!d^2n^{\prime}~\frac{\delta f({\bf n})~\delta f({\bf n}^{\prime})^{\ast}}{|{\bf n}-{\bf n}^{\prime}|} -\frac{g_r(R)}{R}\int_{S^2}\!d^2n~\delta f({\bf n})^2 \cr ~=~&-\int_{S^2}\!d^2n\int_{S^2}\!d^2n^{\prime}~\delta f({\bf n})~\delta f({\bf n}^{\prime})^{\ast}\sum_{\ell m} \frac{4\pi}{2\ell+1}Y^{\ast}_{\ell m}({\bf n})Y_{\ell m}({\bf n}^{\prime})\cr &+\frac{4\pi}{3}\int_{S^2}\!d^2n~|\delta f({\bf n})|^2 \cr ~=~&4\pi\sum_{\ell m}\left(\frac{1}{3}-\frac{1}{2\ell+1}\right)|\delta c_{\ell m}|^2 ~\geq~0,\end{align}\tag{19}$$
which is non-negative since the zeromode (18) is absent. So the ball is a stable stationary shape.
References:
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.
