Heat capacities are normally positive, but there is an important class of exceptions: systems of particles held together by gravity, such as stars and star clusters.
$\left(a\right)$Consider a system of just two particles, with identical masses, orbiting in circles about their center of mass. Show that the gravitational potential energy of this system is-$2$times the total kinetic energy.
$\left(b\right)$The conclusion of part $\left(a\right)$turns out to be true, at least on average, for any system of particles held together by mutual gravitational attraction:

${\overline{U}}_{\text{potential}}=-2{\overline{U}}_{\text{kinetic}}$

Here each $U$refers to the total energy (of that type) for the entire system, averaged over some sufficiently long time period. This result is known as the virial theorem. (For a proof, see Carroll and Ostlie ($1996$), Section $2.4$.) Suppose, then, that you add some energy to such a system and then wait for the system to equilibrate. Does the average total kinetic energy increase or decrease? Explain.

$\left(c\right)$A star can be modeled as a gas of particles that interact with each other only gravitationally. According to the equipartition theorem, the average kinetic energy of the particles in such a star should be $\frac{3}{2}KT$, where$T$ is the average temperature. Express the total energy of a star in terms of its average temperature, and calculate the heat capacity. Note the sign.
$\left(d\right)$Use dimensional analysis to argue that a star of mass $M$and radius $R$should have a total potential energy of $-G{M}^2/R$, times some constant of order $1.$
$\left(e\right)$Estimate the average temperature of the sun, whose mass is $2\times {10}^{30}kg$and whose radius is $7\times {10}^{8}m$. Assume, for simplicity, that the sun is made entirely of protons and electrons.

Step by step solution

01

Step: 1 Equating total kinetic energy: (part a)

Having two identical mass $m$in radius $r$circular orbit at centre mass.

Having that

$F_{\text{central}}=\frac{m{v}^2}{r};F_{grav}=\frac{G{m}^2}{4r^2}$

Equating centripetal and gravitational force

$\frac{m{v}^2}{r}=\frac{G{m}^2}{4r^2}$

The total kinetic energy is

$\begin{array}{l}K=K_1+K_2\\ K=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2\\ K=m{v}^2.\end{array}$

02

Step: 2 FInding potential energy: (part a)

The gravitational potential energy of system is

$V=-\frac{G{m}^2}{2r}$

The negative sign indicates the potential energy attractive.so,

$\begin{array}{l}m{v}^2=\frac{1}{2}\frac{G{m}^2}{r}\\ K=-\frac{V}{2}\\ V=-2K\end{array}$

03

Step: 3 Total energy: (part b)

The vital theorem for gravitational orbits,the energies as

$⟨V⟩=-2⟨K⟩$

The total gravitational energy system as

$\begin{array}{l}U=⟨K⟩+⟨V⟩\\ U=⟨K⟩-2⟨K⟩=-⟨K⟩\end{array}$

The average total kinetic energy is the total energy is negative, it is gravitationally bound and will not disintegrate over time. However, if we raise the system's energy by a sufficient amount to keep $U$negative, the kinetic energy must actually decrease.

04

Step: 4 Heat capacity: (part c)

The average kinetic energy of particle is

$U=-K=-\frac{3}{2}NkT$

The heat energy is negative.

$C=\frac{\mathrm{\partial }U}{\mathrm{\partial }T}=-\frac{3}{2}Nk$

The pressure is zero when the star expands as vacuum.

05

Step: 5 Potential energy: (part d)

The center of star as

$dV=-\frac{G{M}_{r}dm}{r}$

The mass and volume of sphere as

$$\begin{gathered} M_r=\rho \times V \\ dm=\rho \times dV \\ {M}_r=\frac{4}{3}\rho \pi r^3 \\ dm=4\rho \pi r^{2}dr \\ \begin{array}{l}dV=-\frac{G\left(\frac{4}{3}\rho \pi r^3\right)\left(4\rho \pi r^{2}dr\right)}{r}\\ \to dV=-\frac{16}{3}G{\rho }^2{\pi }^2{r}^{4}dr\end{array} \end{gathered}$$

06

Step: 6 Finding the value: (part d)

The potential energy as

$\begin{array}{l}V=\int dV\\ V=-\frac{16}{3}G{\rho }^2{\pi }^2{\int }_0^R r^{4}dr\\ V=-\frac{16}{3}G{\rho }^2{\pi }^2{\left[\frac{r^5}{5}\right]}_0^R\\ V=-\frac{16}{15}G{\rho }^2{\pi }^2{R}^5\\ \begin{array}{l}V=-\frac{16}{15}G{\rho }^2{\pi }^2\frac{R^6}{R}\\ V=-\frac{3G}{5R}{\left[\frac{4}{3}\rho \pi R^3\right]}^2\\ V=-\frac{3G}{5R}(M{)}^2\end{array}\end{array}$

07

Step: 7 Finding temperature of sun: (part e)

The average kinetic energy as

$\begin{array}{l}⟨K⟩=-\frac{1}{2}⟨V⟩\\ ⟨K⟩=-\frac{1}{2}\left[\frac{3G}{5R}(M{)}^2\right]\\ ⟨K⟩=\left[\frac{3G}{10R}(M{)}^2\right]=\frac{3}{2}NkT\\ T=\frac{G{M}^2}{5RNk}\end{array}$

The mass of electron compared to proton as

$\begin{array}{l}{N}_{\text{protons}}=\frac{\text{Mass of the sun}}{\text{Mass of proton}}\\ N_{\text{protons}}=\frac{2\times {10}^{30}}{1.67\times {10}^{-27}}\\ N_{\text{protons}}=1.197\times {10}^{57}\\ N=2\times N_{\text{protons}}\\ N=2\times 1.197\times {10}^{57}\\ N=2.394\times {10}^{57}.\end{array}$

From the above, the half beyond theSun's radius of core.

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