User:Tohline/SSC/UniformDensity

Uniform-Density Sphere (structure)

Solution Technique #1

Adopting solution technique #1, we need to solve the integro-differential equation,

<math>~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}</math>

appreciating that,

<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .

For a uniform-density configuration, <math>~\rho</math> = <math>\rho_c</math> = constant, so the density can be pulled outside the mass integral and the integral can be completed immediately to give,

<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .

Hence, the differential equation describing hydrostatic balance becomes,

<math> \frac{dP}{dr} = - \frac{4\pi G}{3} \rho_c^2 r </math> .

Integrating this from the center of the configuration — where <math>r=0</math> and <math>P = P_c</math> — out to an arbitrary radius <math>r</math> that is still inside the configuration, we obtain,

<math> \int_{P_c}^P dP = - \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math>
<math>\Rightarrow ~~~~ P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 </math>

We expect the pressure to drop to zero at the surface of our spherical configuration — that is, at <math>r=R</math> — so the central pressure must be,

<math>P_c = \frac{2\pi G}{3} \rho_c^2 R^2 = \frac{3G}{8\pi}\biggl( \frac{M^2}{R^4} \biggr)</math> ,

where <math>M</math> is the total mass of the configuration. Finally, then, we have,

<math>P(r) = P_c\biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] </math> .

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation