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(D) The gravitational potential energy $U$ of a uniform spherical star of mass $M$ and radius $R$ is given by $U = -\frac{3GM^2}{5R}$.The volume $V$ o

Vedclass · Accepted Answer

$P \propto V^{-4/3}$

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(D) The gravitational potential energy $U$ of a uniform spherical star of mass $M$ and radius $R$ is given by $U = -\frac{3GM^2}{5R}$.The volume $V$ of the star is $V = \frac{4}{3}\pi R^3$,which implies $R = (\frac{3V}{4\pi})^{1/3}$.Substituting $R$ into the expression for $U$,we get $U = -\frac{3GM^2}{5} (\frac{4\pi}{3V})^{1/3} = -\frac{3GM^2}{5} (\frac{4\pi}{3})^{1/3} V^{-1/3}$.The gravitational pressure $P$ is related to the change in potential energy by the relation $dW = P dV = -dU$,which gives $P = -\frac{dU}{dV}$.Differentiating $U$ with respect to $V$: $P = -[ -\frac{3GM^2}{5} (\frac{4\pi}{3})^{1/3} \cdot (-\frac{1}{3}) V^{-4/3} ]$.Simplifying this,we find $P = -\frac{GM^2}{5} (\frac{4\pi}{3})^{1/3} V^{-4/3}$.Since $G, M,$ and the constants are fixed,$P \propto V^{-4/3}$.

$A$ star is modeled as a uniform spherical distribution of matter,keeping the mass of the star constant. How does the gravitational pressure on the surface depend on the volume of the star?

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