Editing SSC/Structure/UniformDensity (section)

==Solution Technique 1==

Adopting [[SSCpt2/SolutionStrategies#Technique_1|solution technique #1]], we need to solve the integro-differential equation,

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{{Math/EQ_SShydrostaticBalance01}} 
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appreciating that,
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<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .
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For a uniform-density configuration, {{Math/VAR_Density01}} = <math>~\rho_c</math> = constant, so the density can be pulled outside the mass integral and the integral can be completed immediately to give,
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<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .
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Hence, the differential equation describing hydrostatic balance becomes,
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<math> \frac{dP}{dr} = - \frac{4\pi G}{3} \rho_c^2 r </math> .
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Integrating this from the center of the configuration &#8212; where <math>~r=0</math> and <math>~P = P_c</math> &#8212; out to an arbitrary radius <math>~r</math> that is still inside the configuration, we obtain,
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<math> \int_{P_c}^P dP = - \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math><br /> 
<math>\Rightarrow ~~~~ P  = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 </math>
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-->
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<math>~ \int_{P_c}^P dP </math>
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<math>~=</math>
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<math>~- \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math>
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<math>~ \Rightarrow ~~~ P </math>
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<math>~=</math>
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<math>~P_c - \frac{2\pi G}{3} \rho_c^2 r^2 \, .</math>
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We expect the pressure to drop to zero at the surface of our spherical configuration &#8212; that is, at <math>~r=R</math> &#8212; so the central pressure must be,

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

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