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(A) For a spherical shell of radius $r$ and thickness $dr$,the weight of the layer is balanced by the pressure difference across the layer.$	herefore

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$1 / R^{4}$

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(A) For a spherical shell of radius $r$ and thickness $dr$,the weight of the layer is balanced by the pressure difference across the layer.$	herefore$ Weight $= \{p - (p + dp)\} 4 \pi r^{2} = -dp \cdot 4 \pi r^{2}$Also,weight $= (dm)g = (ho \cdot 4 \pi r^{2} dr) \cdot \frac{GM(r)}{r^{2}}$,where $M(r) = ho \cdot \frac{4}{3} \pi r^{3}$.So,$-dp \cdot 4 \pi r^{2} = ho \cdot 4 \pi r^{2} dr \cdot \frac{G ho \cdot \frac{4}{3} \pi r^{3}}{r^{2}}$$-dp = \frac{4}{3} \pi G ho^{2} r dr$Integrating from $r=0$ to $R$,where $p(R)=0$ and $p(0)=P_{center}$:$P_{center} = \int_{0}^{R} \frac{4}{3} \pi G ho^{2} r dr = \frac{2}{3} \pi G ho^{2} R^{2}$Since $ho = \frac{M}{\frac{4}{3} \pi R^{3}}$,we have $ho^{2} = \frac{M^{2}}{\frac{16}{9} \pi^{2} R^{6}}$.Substituting $ho^{2}$ in the pressure expression:$P \propto \frac{M^{2}}{R^{6}} \cdot R^{2} = \frac{M^{2}}{R^{4}}$Thus,the average gravitational pressure $P_{av} \propto \frac{1}{R^{4}}$.

$A$ star of mass $M$ and radius $R$ is made up of gases. The average gravitational pressure compressing the star due to the gravitational pull of the gases making up the star depends on $R$ as

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