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For a fixed polytropic index $n>0$ and fixed equation-of-state constant $K$, let ${\xi }_1$ be the first zero of the regular Lane-Emden equation solution. A finite-radius model requires such a zero; for the usual nonnegative indices this holds for $n<5$. The surface radius and mass follow from $r=\alpha \xi$ and $\rho ={\rho }_c{\theta }^n$:The last equality integrates the Lane-Emden equation; define the positive Lane-Emden surface mass constant ${\omega }_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. Substituting $\alpha =C_1{\rho }_c^{(1-n)/(2n)}$ givesThe central-density exponents cancel when the required powers are taken. Thus the polytropic mass-radius relation isIt is independent of ${\rho }_c$, with $n,K$ and composition held fixed. For $n=1$ the radius is fixed; for $n=3$ the mass is fixed. These limiting powers need no division by a vanishing exponent.

For $n=0$, the literal pressure-density power and the supplied expression for $\alpha$ are singular. Interpret this case separately as an incompressible planetary interior with constant density. Then $M=4\pi \rho R^3/3$, or $R\propto M^{1/3}$ at fixed density. A weakly compressed rocky body, or a rough uniform-density approximation to Earth, is an example; realistic terrestrial planets are stratified and compressible.

For $n=3/2$, $P\propto {\rho }^{5/3}$. This describes a cold nonrelativistic degenerate electron gas of fixed composition, or a fully convective monatomic ideal gas at fixed entropy. Examples are a nonrelativistic white dwarf, a sufficiently cooled partly degenerate brown dwarf as an approximation, and an approximately fully convective low-mass star for the ideal-gas version. The fixed-$K$ scaling is $R\propto M^{-1/3}$. It must not be applied to an entire main-sequence stellar sequence with different entropies; the entropy dependence of a polytropic mass-radius relation explains the distinction.

For an incompressible planetary interior of density $\rho$, the enclosed mass and local gravity are $M(r)=4\pi \rho r^3/3$ and $g(r)=4\pi G\rho r/3$. Hydrostatic equilibrium therefore givesIntegrating inward from negligible surface pressure at $r=R_p$ yields the uniform-density planetary pressure profile:The surface gravity is $g_s=GM/R_p^2=4\pi G\rho R_p/3$. Eliminate $\rho R_p$ to obtainThe units of $g_s^2/G$ are pressure. For a nonzero imposed surface pressure, add $P_s$ throughout and interpret the boxed central value as $P_c-P_s$. The constant-density approximation is crucial; real centrally concentrated rocky planets need a different profile and generally a larger central pressure at the same mass and radius.

For a spherical incompressible planetary interior, $M(r)=4\pi \rho r^3/3$ gives $dP/dr=-4\pi G{\rho }^{2}r/3$. With $P(R)=0$, $P(r)=2\pi G{\rho }^2(R^2-r^2)/3$. The surface gravity $g_s=4\pi G\rho R/3$ then yields $P_c=3g_s^2/(8\pi G)$. Add an imposed surface pressure to the entire profile if needed.