To prepare for the IOAA and to deepen my understanding of astrophysics, I have been systematically studying astrophysics. Today, I reached the chapter on “The Sun” and performed a simple estimation of the Sun’s central pressure and temperature.
Astrophysics provides several equations regarding stellar structure. Assuming a star is an isotropic sphere, we have:
\frac{dm(r)}{dr}=4\pi r^2 \rho(r) — Mass Equation
where m(r) is the total mass within a spherical region at a distance r from the center of the star, and \rho(r) is the density of the matter at distance r. We can also write this in integral form: m(r)=\int_0^R 4\pi r^2 \rho(r)dr where R is the stellar radius. This equation essentially represents the summation of the mass of each shell, so I will not derive it in detail.
Additionally, for the pressure of a star, the following equation is given:
\frac{dP(r)}{dr}=-\frac{Gm(r)}{r^2}\rho(r) — Hydrostatic Equilibrium Equation
Since \frac{Gm(r)}{r^2}=g(r), where g(r) is the gravitational acceleration at distance r from the center, the equation can also be written as: \frac{dP(r)}{dr}=-g(r)\rho(r)
The derivation of this equation is not difficult. A star can be approximated as a fluid. Within an infinitesimal change dr, g(r)=\frac{Gm(r)}{r^2} can be considered constant. For a shell of thickness dr, the mass is approximately 4\pi r^2 \rho(r)dr, and the gravitational force is 4\pi r^2 \rho(r)g(r)dr. The surface area is 4\pi r^2, so the pressure change is \Delta P = -g(r)\rho(r)dr. Note that dP is the difference between the pressure at (r+dr) and r, so dP should be negative. Therefore: dP=-g(r)\rho(r)dr
Now, we can attempt to estimate the Sun’s central pressure and temperature. We need to make two assumptions: 1. The Sun is uniform, meaning the density is constant at \rho = 1411 \text{ kg/m}^3; 2. The surface pressure of the Sun is 0. Both of these assumptions are incorrect, but they are sufficient for a simple estimation of the Sun’s parameters.
Assuming the Sun’s density is constant, we can integrate the mass equation to get m(r)=\frac{4}{3}\pi r^3\rho. Substituting this into the pressure equation, we have: dP=-\frac{4}{3}G\pi r\rho^2 dr
Integrating this yields: P=-\frac{2}{3}G\pi r^2 \rho^2+C. Substituting the boundary conditions (r=R, P=0), we find C=\frac{2}{3}G\pi R^2 \rho^2, which gives P=\frac{2}{3}G\pi\rho^2(R^2-r^2). Setting r=0 gives the pressure at the center: P_c=\frac{2}{3}G\pi R^2 \rho^2\approx 1.347 \cdot 10^{14} \text{ Pa}
How can we estimate the central temperature? For pressure, we have P=nkT, where n is the number of molecules per unit volume, k=1.38066 \times 10^{-23} \text{ J/K}, and T is the temperature. For an ideal gas, n=\frac{\rho}{m}, where m is the average mass of a single molecule. Since 94% of the Sun is hydrogen, m \approx 1.66 \times 10^{-27} \text{ kg}. From this, we can calculate: T_c \approx 11,480,000 \text{ K}
Error Analysis:
More precise theoretical calculations yield a central pressure of 3.4 \times 10^{16} \text{ Pa} and a central temperature of 15 million K. The pressure is nearly 200 times our estimate, while the temperature is approximately 1.3 times our estimate. It is evident that both results are underestimates. Beyond providing a basic understanding of solar research, these calculations help us realize that the Sun’s density is non-uniform (which is obvious...) and that the Sun is not a perfectly ideal fluid, among other things.
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