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Set $A={\Phi }_0{R}_0^{\beta }<0$, $q={\lambda }^2$ and $s=R^2+q{z}^2$. The flattened power-law gravitational potential is $\Phi =A{s}^{-\beta /2}$. In axisymmetric cylindrical coordinates, the Poisson equation for Newtonian gravity givesThe first derivatives are ${\partial }_R\Phi =-\beta AR{s}^{-(\beta +2)/2}$ and ${\partial }_z\Phi =-\beta Aqz{s}^{-(\beta +2)/2}$. Differentiating again and collecting powers givesTherefore the required mass density isThis is a formal density for arbitrary $\lambda$, but a physical dark matter distribution must be nonnegative. The density positivity for a flattened power-law potential condition isNecessity follows by evaluating on the midplane and symmetry axis; sufficiency follows because both numerator coefficients are then nonnegative. In this range the origin is a locally integrable density cusp: the mass enclosed near radius $r$ scales as $r^{1-\beta }$ and tends to zero, so no point mass needs adding. The scale-free distribution has infinite total mass at large radius and represents an idealized background, not a finite isolated halo.

For the cold, non-self-gravitating disk, radial balance of a circular orbit is $R{\Omega }^2={\partial }_R\Phi (R,0)$. ThusThe displayed positive root chooses the rotation orientation; the opposite orientation has the negative of this angular frequency.

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

Write $q=|r-r^{\prime }|$. Away from the source, the radial Laplacian gives ${\nabla }^2(-1/(4\pi q))=q^{-2}{\partial }_q[q^2/(4\pi q^2)]=0$. Through a small source-centred sphere, the outward flux of its gradient isThis identifies a unit Dirac delta at the source. More explicitly, for a smooth compactly supported test function $\psi$, apply Green's second identity outside the radius-$\epsilon$ ball. The inner outward normal of that punctured domain is $-e_q$. The term $H{\partial }_n\psi$ has integral $O(\epsilon )$, while $-\psi {\partial }_{n}H$ tends to $\psi (r^{\prime })$. Thus $\int H{\nabla }^2\psi dV=\psi (r^{\prime })$, proving the distributional identityWhenever the integral converges, convolution with this Green function of the Laplacian solves the Poisson equation for Newtonian gravity:The harmonic term represents the homogeneous freedom and is fixed by boundary conditions. For an isolated finite mass distribution, regularity away from the prescribed sources and the normalization $\Phi \to 0$ at infinity select $h=0$.

For a point mass at the origin, the delta integral gives the Newtonian potential of a point mass,

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.

For a self-consistent Gaussian angular-momentum suppression in a stellar distribution, put $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $s=\log (r/r_a)$. At large radius the Poisson equation for Newtonian gravity reduces to $y_{ss}+y_s\sim -\lambda e^y$, where $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. The asymptotic balance $y_s\sim -1/s$ gives $e^y\sim 1/(\lambda s)$, the displayed density, and circular speed $v_c^2\sim {\sigma }^2/\log (r/r_{\ast })$. A sufficiently small regular core can have an intermediate isotropic singular isothermal sphere envelope with $\rho \simeq {\sigma }^2/(2\pi G{r}^2)$ and $v_c^2\simeq 2{\sigma }^2$. At the actual regular centre, however, $\rho \to {\rho }_c$ and $v_c\propto r$; the intermediate power law must not be mistaken for a central limit.

A plane-symmetric, self-gravitating stellar population with constant vertical velocity dispersion satisfies the vertical Jeans equation and the Poisson equation for Newtonian gravity. With ${\Phi }^{\prime }(0)=0$, their solution isIts surface density is $\Sigma =2{\rho }_0{z}_0$, so $z_0={\sigma }_z^2/(\pi G\Sigma )$. Its outer density tail is exponential with scale height $z_0/2$; its central profile is not exponential.

Self-gravity is the gravitational acceleration produced by a system's own mass density. It is obtained by solving the Poisson equation for Newtonian gravity for that density, rather than treating the Newtonian gravitational potential as an externally fixed field. It competes with pressure and rotational support in gravitational instability of an astrophysical disk.

For a mean-field gas of parallel infinite rods with total mass per unit length $M=Nm$ confined to radius $R_e$, put $q=GM\beta /2$, $\beta =m/(kT)$ and $F=1-q+q{R}^2/R_e^2$. The isothermal solution is $\psi =\psi (R_e)-2\log F/\beta$ with mass density $\rho =M(1-q)/(\pi R_e^2{F}^2)$ and enclosed line mass $M(<R)=M(R^2/R_e^2)/F$. Direct differentiation verifies the cylindrical Poisson equation for Newtonian gravity. Regularity requires $q<1$, or $T>T_{\min }$; equality is a singular core-concentration limit, not a regular equilibrium.