# Explain how scaling of the inverse square law breaks down at a stars surface

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?

• A star doesn't have a solid surface. See en.wikipedia.org/wiki/Photosphere – PM 2Ring Mar 30 '20 at 9:02
• What does this correspond to in the question? @PM2Ring – User1997 Mar 30 '20 at 9:03
• 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. – PM 2Ring Mar 30 '20 at 9:33

For the answer to the question a), just use a Taylor expansion in the parameter $$x= R/d \ll 1$$, so that $$$$(1-x^2)^{3/2} \simeq 1- \frac{3}{2} x^2$$$$ and then you obtain a inverse square law in $$d$$ $$$$P_{rad}= \frac{2\sigma T^4}{c} \frac{R^2}{d^2}$$$$
$$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}}$$