A spherically symmetric star of total mass $M_{s}$ has pressure $P(r)$ and mass density $\rho(r)$, where $r$ is the radial distance from its centre. These quantities are related by the equations of hydrostatic equilibrium and mass conservation:

$$\begin{aligned} \frac{d P}{d r} &=-\frac{G M(r) \rho}{r^{2}} \\ \frac{d M}{d r} &=4 \pi \rho r^{2} \end{aligned}$$

where $M(r)$ is the mass inside radius $r$.

By integrating from the centre of the star at $r=0$, where $P=P_{c}$, to the surface of the star at $r=R_{s}$, where $P=P_{s}$, show that

$$4 \pi R_{s}^{3} P_{s}=\Omega+3 \int_{0}^{M_{s}} \frac{P}{\rho} d M,$$

where $\Omega$ is the total gravitational potential energy. Show that

$$-\Omega>\frac{G M_{s}^{2}}{2 R_{s}}$$

If the surface pressure is negligible and the star is a perfect gas of particles of mass $m$ with number density $n$ and $P=n k_{B} T$ at temperature $T$, and radiation pressure can be ignored, then show that

$$3 \int_{0}^{M_{s}} \frac{P}{\rho} d M=\frac{3 k_{B}}{m} \bar{T},$$

where $\bar{T}$ is the mean temperature of the star, which you should define.

Hence, show that the mean temperature of the star satisfies the inequality

$$\bar{T}>\frac{G M_{s} m}{6 k_{B} R_{s}}$$