Question

Ques. Derive the expression for the mean temperature in a star:
< T > M^2/3 < P > ^1/3

Answer 1

Answer on Question #69702, Physics / Astronomy | Astrophysics

Ques. Derive the expression for the mean temperature in a star: $< T > M^{\wedge}2 / 3 < P >^{\wedge}1 / 3$

Answer:

Internal energy $E_{\mathrm{i}} = -E_{\mathrm{g}} / 2$

C $E_{\mathrm{g}}$ is Gravitational Energy.

$$\mathrm{d}P / \mathrm{d}m = -G m / (4\pi r^4)$$

Where $P$ is the pressure, $m$ is the mass enclosed in the spherical surface of radius $r$.

Gravitational Energy of the star

$$E_{\mathrm{g}} = -G M^2 / R$$

Where, $M =$ mass of the star and $R =$ radius of the star

Let,

$$\rho = M / [4\pi R^3 / 3]$$ $$R = [3 M / (4\pi \rho)]^{\wedge} 1 / 3$$

Internal Energy

$$E_{\mathrm{i}} = 1 / 2 * G M^2 * [3 M / (4\pi \rho)]^{\wedge} - 1 / 3 = 1 / 2 * G M^{\wedge} (5 / 3) \rho^{\wedge} (1 / 3) * (3 / 4\pi)^{\wedge} 1 / 3$$

Internal Energy (of a mono-atomic ideal gas or gas in the form of ions)

$$E_{\mathrm{i}} = 3 / 2 k T N = 3 / 2 k T [M / \mu m_{\mathrm{H}}]$$

Where, $k =$ Boltzmann's constant, $T =$ average temperature of the star, $N =$ number of molecules/particles of gas, $M =$ mass of the gas, $m_{\mathrm{H}} =$ mass of the particle of gas basically N is proportional to the mass M of star

We get:

$$E_{\mathrm{i}} = M^{\wedge} (5 / 3) * \rho^{\wedge} (1 / 3) = T * M$$ $$T = M^{\wedge} (2 / 3) * \rho^{\wedge} (1 / 3)$$

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