# Estimate the core pressure of a star

Consider a small column at a distance $$r$$ from the center of a star with a height of $$\Delta r$$ and a base area of $$\Delta S$$, a density within the column approximated to be uniform for a value $$\rho(r)$$, and a mass of $$M(r)$$ within a radius of $$r$$ of the star. The conditions for hydrostatic equilibrium are: $$\Delta p = p(r + \Delta r) - p(r)$$ $$\Delta p \Delta S = -\frac{GM(r)}{r^2}\rho(r) \Delta r \Delta S$$ Get: $$\frac{\Delta p}{\Delta r} = -\frac{GM(r)\rho(r)}{r^2}$$ Considering the whole Sun as a shell layer, and setting the pressure at the center $$r = 0$$ to be $$p_0$$, then $$\Delta p = -p_0$$ and $$\Delta r = R_\odot$$, the density of the shell is the whole Sun's density, we have: $$\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{GM(r)}{r^2} \frac{3M_\odot}{4\pi R_\odot^3}$$

Here is the problem: some books like Astronomy - A Physical Perspective (Marc Leslie Kutner, 2ed) will subsitute $$r=R_\odot$$ instead of $$r=0$$ (center) like this:

$$\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{GM(R_\odot)}{R_\odot^2} \frac{3M_\odot}{4\pi R_\odot^3}$$

$$p_0 \sim \frac{GM_\odot^2}{R_\odot^4}$$

But if we use $$r = 0$$ and it will be: $$\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{G \times 0}{0^2} \frac{3M_\odot}{4\pi R_\odot^3}$$ which is a $$0 \div 0$$ paradox.

• The question stated that we consider the star as one shell, this shell thus has a radius of $R_s$, taking $r=0$ makes no sense here. Commented Aug 4 at 9:02
• No I think that the shell is $r = r \sim r+\Delta r$ and $r = 0$, $\Delta r = R_s$. If you take $r=R_s$ then $\Delta r = 0$ Commented Aug 4 at 9:06
• $dr \approx \Delta r = R_s - R_c$, where $R_c = 0$ (core situation), so the $dr = R_s$ Commented Aug 4 at 9:09
• But you should admit that $M(r)=M(R_c)$, as the gravity comes from the inner core (the outer shell provides 0 gravity to the inside due to the Shell theorem) Commented Aug 4 at 9:18
• No $M(r) = M(R_s)$ because we are considering the star as a shell and again, the radius of the shell is $R_s$ Commented Aug 4 at 9:22

Your formula for $$\Delta p/\Delta r$$ is an approximation. It only becomes exact as $$\Delta r \to 0$$.
Clearly, to use the approximation you have to make an arbitrary decision about what values of $$M(r)$$, $$\rho$$ and $$r$$ you are going to use. Using $$M_\odot$$, $$\bar{\rho}$$ and $$R_\odot$$ is one choice. It leads to an incorrect answer because the density in the Sun isn't uniform.
Choosing $$r=0$$ makes no sense at all because it gives an indeterminate result. Choosing some other value of $$r$$ would require us to know what the mass is within that radius and the average density within that radius.