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If the radiation pressure at distance $d>R$ from the center of an isotropic black body star is found to be $$P_{rad}=\large{\frac{4\sigma T^4}{3c}}\left[1-\left(1-\frac{R^2}{d^2}\right)^{\frac{3}{2}}\right],$$

a) How do I show that $P_{rad}$ obeys an inverse square law for $d \gg R$?

b) Why does the inverse square law scaling break down close to the stars surface?

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  • $\begingroup$ A star doesn't have a solid surface. See en.wikipedia.org/wiki/Photosphere $\endgroup$ – PM 2Ring Mar 30 '20 at 9:02
  • $\begingroup$ What does this correspond to in the question? @PM2Ring $\endgroup$ – User1997 Mar 30 '20 at 9:03
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    $\begingroup$ As that article says, the glowing surface of a star isn't actually a surface, it's like a translucent glowing fog. The Sun's photosphere is around 100 kilometers thick. $\endgroup$ – PM 2Ring Mar 30 '20 at 9:33
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For the answer to the question a), just use a Taylor expansion in the parameter $x= R/d \ll 1$, so that \begin{equation} (1-x^2)^{3/2} \simeq 1- \frac{3}{2} x^2 \end{equation} and then you obtain a inverse square law in $d$ \begin{equation} P_{rad}= \frac{2\sigma T^4}{c} \frac{R^2}{d^2} \end{equation}

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As about (b):

Why does the inverse square law scaling break down close to the stars surface?

$$ P_{\,d\approx R} = \lim_{d \to R} {\frac{4\sigma T^4}{3c}}\left[1-\left(1-\frac{R^2}{d^2}\right)^{\frac{3}{2}}\right] = {\frac{4\sigma T^4}{3c}} $$

In other words, when you are at the star surface - you get as much radiation flux as possible, thus radiation density depends just on star surface temperature.

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  • $\begingroup$ Thank you @AgniusVasiliauskas $\endgroup$ – User1997 Mar 30 '20 at 11:21

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