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

CC BY-SA 4.0
2024-08-04T12:36:26.083Z
2024-08-04T12:36:26.083Z
ProfRob
<p>Consider a small column at a distance <span class="math-container">$r$</span> from the center of a star with a height of <span class="math-container">$\Delta r$</span> and a base area of <span class="math-container">$\Delta S$</span>, a density within the column approximated to be uniform for a value <span class="math-container">$\rho(r)$</span>, and a mass of <span class="math-container">$M(r)$</span> within a radius of <span class="math-container">$r$</span> of the star. The conditions for hydrostatic equilibrium are:
<span class="math-container">$$
\Delta p = p(r + \Delta r) - p(r)
$$</span>
<span class="math-container">$$
\Delta p \Delta S = -\frac{GM(r)}{r^2}\rho(r) \Delta r \Delta S
$$</span>
Get:
<span class="math-container">$$
\frac{\Delta p}{\Delta r} = -\frac{GM(r)\rho(r)}{r^2}
$$</span>
Considering the whole Sun as a shell layer, and setting the pressure at the center <span class="math-container">$r = 0$</span> to be <span class="math-container">$p_0$</span>, then <span class="math-container">$\Delta p = -p_0$</span> and <span class="math-container">$\Delta r = R_\odot$</span>, the density of the shell is the whole Sun's density, we have:
<span class="math-container">$$
\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{GM(r)}{r^2} \frac{3M_\odot}{4\pi R_\odot^3}
$$</span></p>
<hr />
<p>Here is the problem: some books like <em>Astronomy - A Physical Perspective (Marc Leslie Kutner, 2ed)</em> will subsitute <span class="math-container">$r=R_\odot$</span> instead of <span class="math-container">$r=0$</span> (center) like this:</p>
<p><a href="https://i.sstatic.net/fpZYr76t.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fpZYr76t.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$
\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{GM(R_\odot)}{R_\odot^2} \frac{3M_\odot}{4\pi R_\odot^3}
$$</span></p>
<p><span class="math-container">$$
p_0 \sim \frac{GM_\odot^2}{R_\odot^4}
$$</span></p>
<p>But if we use <span class="math-container">$r = 0$</span> and it will be:
<span class="math-container">$$
\frac{-p_0}{R_\odot} = -\frac{GM(r)\rho_\odot}{r^2} = -\frac{G \times 0}{0^2} \frac{3M_\odot}{4\pi R_\odot^3}
$$</span>
which is a <span class="math-container">$0 \div 0$</span> paradox.</p>

CC BY-SA 4.0
2024-08-04T09:02:14.003Z
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.
Polaris5744
CC BY-SA 4.0
2024-08-04T09:06:59.857Z
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 $\Delta r = 0$
Firestar-Reimu
CC BY-SA 4.0
2024-08-04T09:09:22.920Z
$dr \approx \Delta r = R_s - R_c$, where $R_c = 0$ (core situation), so the $dr = R_s$
Polaris5744
CC BY-SA 4.0
2024-08-04T09:18:39.600Z
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)
Firestar-Reimu
CC BY-SA 4.0
2024-08-04T09:22:13.113Z
No $M(r) = M(R_s)$ because we are considering the star as a shell and again, the radius of the shell is $R_s$
Polaris5744
CC BY-SA 4.0
2024-08-04T09:47:34.093Z
the mass of M(r) is **within** a radius of r of the star?
Firestar-Reimu
CC BY-SA 4.0
2024-08-04T11:41:25.907Z
Yes, $M(r)$ is the mass within radius $r$
Polaris5744
<star><stellar-astrophysics>
Estimate the core pressure of a star
Firestar-Reimu
# Estimate the core pressure of a star

(Asked by: username_0 on 2024-08-04)

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:

[![enter image description here](https://i.sstatic.net/fpZYr76t.png)](https://i.sstatic.net/fpZYr76t.png)

$$
\frac{-p\_0}{R\_\odot} = -\frac{GM(r)\rho\_\odot}{r^2} = -\frac{GM(R\_\odot)}{R\_\odot^2} \frac{3M\_\odot}{4\pi R\_\odot^3}
$$

$$
p\_0 \sim \frac{GM\_\odot^2}{R\_\odot^4}
$$

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.

**Tags:** star, stellar-astrophysics
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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 $\Delta r = 0$  
> **username_1**: $dr \approx \Delta r = R_s - R_c$, where $R_c = 0$ (core situation), so the $dr = R_s$  
> **username_0**: 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)  
> **username_1**: No $M(r) = M(R_s)$ because we are considering the star as a shell and again, the radius of the shell is $R_s$  
> **username_0**: the mass of M(r) is **within** a radius of r of the star?  
> **username_1**: Yes, $M(r)$ is the mass within radius $r$  

---

## 1 answers

### Answer by username_2 on 2024-08-04. Score: 5

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*.


---

