User:Tohline/SSC/UniformDensity
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Uniform-Density Sphere (structure)
Solution Technique #1
Adopting solution technique #1, we need to solve the integro-differential equation,
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<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 </math>.
Finally, then, we have,
<math>P(r) = \frac{2\pi G}{3} \rho_c^2 R^2 \biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] = \frac{GM \rho_c}{2R} \biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr]</math> ,
where <math>M</math> is the total mass of the configuration.
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