= Solution
Assume $K>0$ and a positive central <mass density>. For this <polytrope of index one>, the spherical <hydrostatic pressure support equation> becomes
$$
2K\rho\frac{d\rho}{dr}=-\frac{Gm\rho}{r^2},\qquad
r^2\frac{d\rho}{dr}=-\frac{Gm}{2K}.
$$
Differentiate the second equation and use mass conservation. With $k^2=2\pi G/K$, this gives
$$
\frac1{r^2}\frac{d}{dr}\left(r^2\frac{d\rho}{dr}\right)+k^2\rho=0.
$$
The regular central solution, with $\rho(0)=\rho_c$ and $\rho'(0)=0$, is
$$
\rho(r)=\rho_c\frac{\sin kr}{kr}.
$$
This is also the index-one solution of the <Lane-Emden equation>. The <mass density> remains positive up to its first zero, so the free surface is at $kR=\pi$, giving
$$
\boxed{R=\frac\pi k=\left(\frac{\pi K}{2G}\right)^{1/2}.}
$$
Taking a later zero would include a region of negative <mass density> and would not describe a physical star.
Integrating the mass gives
$$
M=\frac{4\pi\rho_c}{k^3}\int_0^\pi \xi\sin\xi\,d\xi
=\frac{4\pi^2\rho_c}{k^3}.
$$
Consequently
$$
\boxed{\frac{\overline\rho}{\rho_c}=\frac{3M}{4\pi R^3\rho_c}=\frac3{\pi^2}.}
$$
The radius is independent of the central <mass density>, whereas the mass is proportional to it; this is the special <polytropic mass-radius relation> at index one.
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