<p>Here are the results of some arbitrary cutoffs for the "core" based on $2010$ solar models calculated by <a href="http://www.ap.smu.ca/~guenther/evolution/solar_modeling.html" rel="noreferrer">Guenther et al</a>:</p>

<ul>
<li>Contributing $99\%$ of the total luminosity:<br>$R = 25.5\%$, $M = 9.88\times10^{29}\,\mathrm{kg}$ ($49.7\%$).</li>
<li>Nuclear reaction rate falls below $1\%$ of central rate:<br>$R= 27.2\%$, $M = 1.07\times10^{30}\,\mathrm{kg}$ ($54.0\%$).</li>
</ul>

<p>Overall, $\sim 1/4$ the radius ($\sim 1/64$ the volume) and $\sim 1/2$ the mass decently with some other sources. Unfortunately, there does not seem to be a clear dividing line demarcating the "core" like there is between the radiative zone and the convection zone.</p>

<p>Perhaps an alternative condition would be a local maximum of $\mathrm{d}(\ln P)/\mathrm{d}(\ln\rho)$, which occurs under $20\%$ of solar radius and is <em>roughly</em> constant until the convection zone.</p>

CC BY-SA 3.0
2014-10-24T01:40:20.260Z
Change in `what` over the change in density?
LDC3
CC BY-SA 3.0
2014-10-24T01:41:58.257Z
@LCD3 pressure.
Stan Liou
CC BY-SA 3.0
2014-10-23T23:46:37.793Z
2014-10-23T23:46:37.793Z
Stan Liou
<p>I'm getting a slightly different figure from <a href="http://www.space.com/58-the-sun-formation-facts-and-characteristics.html/" rel="nofollow">space.com</a> (not my favorite source, but a source nonetheless), which says</p>

<blockquote>
  <p>The core extends from the sun's center to about a quarter of the way to its surface. Although it only makes up roughly 2 percent of the sun's volume, it is almost 15 times the density of lead and holds nearly half of the sun's mass. </p>
</blockquote>

<p>I would expect it to be fairly dense because of the extreme conditions at the center, primarily the force of gravity pushing in on it. The "15 times the density of lead" is not constant throughout the Sun, and only applies in the <em>very center</em> of the core.</p>

CC BY-SA 3.0
2014-10-23T00:26:26.857Z
With 0.25 radius, the volume is 0.0156 the total volume, which is 1.56%. If 0.8% is 34% of the mass, than 1.56% should be 50% of the mass. Now all that is needed is a probe to determine if it is 20% or 25% of the radius.
LDC3
CC BY-SA 3.0
2014-10-23T00:29:47.050Z
I'm not sure I follow some of the numbers.
HDE 226868
CC BY-SA 3.0
2014-10-23T00:33:03.343Z
Which numbers are you confused about?
LDC3
CC BY-SA 3.0
2014-10-23T00:34:39.223Z
Are you justifying the data you found, or mine? If 0.8% is 34% of the mass, then (assuming constant density, which might not be true), 1.56% should be roughly 2/3 of the mass, not 50%.
HDE 226868
CC BY-SA 3.0
2014-10-23T00:36:32.313Z
Actually, I was justifying yours. But remember, the density between 20% and 25% is a lot less than that below 20% so the mass wouldn't double.
LDC3
CC BY-SA 3.0
2014-10-23T00:37:44.363Z
Right, which would be the fallacy in my hypothetical constant density argument. So I guess you're right with your rationale for 50%.
HDE 226868
CC BY-SA 3.0
2014-10-23T16:09:57.187Z
Why are you skeptical of the "roughly 2 percent"? That's consistent with the 1/4 radius figure because 1/4 cubed is 1.5625%, or "roughly" 2%. What you should be skeptical of is the "almost 15 times the density of lead" figure because $\frac 4 3 \pi (0.25 \, R_\odot)^3\,15\,\rho_{\text{lead}}$ is 1.9 solar masses. That "15 times the density of lead" is the density at the very center rather than the average density of the core (which is what the space.com article implies).
David Hammen
CC BY-SA 3.0
2014-10-23T17:28:43.707Z
@DavidHammen I hadn't yet done the calculations when I wrote that; I'll remove it.
HDE 226868
CC BY-SA 3.0
2014-10-22T22:48:45.023Z
2014-10-23T17:29:51.457Z
2014-10-23T17:29:51.457Z
HDE 226868
<p>I found out it is 34% of the Sun's mass on <a href="http://en.wikipedia.org/wiki/Solar_core" rel="nofollow">Wikipedia</a>.</p>

CC BY-SA 3.0
2014-10-22T06:08:16.770Z
2014-10-22T06:08:16.770Z
LDC3
<p>It is estimated that the sun's core is about 1/5 of the radius of the sun (from <a href="http://en.wikipedia.org/wiki/Solar_core" rel="nofollow">Wikipedia</a>). I know that the density of the plasma increases substantially the near the center, and that the volume of the core is 0.008 times the total volume. </p>

<p>$$\frac {V_{core}}{V_{sun}}=\frac {4/3\pi (0.2)^3}{4/3\pi (1)^3}$$
$$\frac {V_{core}}{V_{sun}}=\frac {(0.008)}{(1)}$$</p>

<p>What is the mass of the core compared to the total mass?</p>

CC BY-SA 3.0
2014-10-22T05:54:15.737Z
Can you give references for these numbers?
Py-ser
<the-sun><mass><core>
Mass of sun's core
LDC3
# Mass of sun's core

(Asked by: username_0 on 2014-10-22)

It is estimated that the sun's core is about 1/5 of the radius of the sun (from [Wikipedia](http://en.wikipedia.org/wiki/Solar_core)). I know that the density of the plasma increases substantially the near the center, and that the volume of the core is 0.008 times the total volume.

$$\frac {V\_{core}}{V\_{sun}}=\frac {4/3\pi (0.2)^3}{4/3\pi (1)^3}$$
$$\frac {V\_{core}}{V\_{sun}}=\frac {(0.008)}{(1)}$$

What is the mass of the core compared to the total mass?

**Tags:** the-sun, mass, core
> **Comments:**
> **username_1**: Can you give references for these numbers?  

---

## 3 answers

### Answer by username_2 on 2014-10-23. Score: 5

Here are the results of some arbitrary cutoffs for the "core" based on $2010$ solar models calculated by [Guenther et al](http://www.ap.smu.ca/~guenther/evolution/solar_modeling.html):

* Contributing $99\%$ of the total luminosity:  
  $R = 25.5\%$, $M = 9.88\times10^{29}\,\mathrm{kg}$ ($49.7\%$).
* Nuclear reaction rate falls below $1\%$ of central rate:  
  $R= 27.2\%$, $M = 1.07\times10^{30}\,\mathrm{kg}$ ($54.0\%$).

Overall, $\sim 1/4$ the radius ($\sim 1/64$ the volume) and $\sim 1/2$ the mass decently with some other sources. Unfortunately, there does not seem to be a clear dividing line demarcating the "core" like there is between the radiative zone and the convection zone.

Perhaps an alternative condition would be a local maximum of $\mathrm{d}(\ln P)/\mathrm{d}(\ln\rho)$, which occurs under $20\%$ of solar radius and is *roughly* constant until the convection zone.

> **Comments:**
> **username_0**: Change in `what` over the change in density?  
> **username_2**: @LCD3 pressure.  

---

### Answer by username_3 on 2014-10-22. Score: 3

I'm getting a slightly different figure from [space.com](http://www.space.com/58-the-sun-formation-facts-and-characteristics.html/) (not my favorite source, but a source nonetheless), which says

> The core extends from the sun's center to about a quarter of the way to its surface. Although it only makes up roughly 2 percent of the sun's volume, it is almost 15 times the density of lead and holds nearly half of the sun's mass.

I would expect it to be fairly dense because of the extreme conditions at the center, primarily the force of gravity pushing in on it. The "15 times the density of lead" is not constant throughout the Sun, and only applies in the *very center* of the core.

> **Comments:**
> **username_0**: With 0.25 radius, the volume is 0.0156 the total volume, which is 1.56%. If 0.8% is 34% of the mass, than 1.56% should be 50% of the mass. Now all that is needed is a probe to determine if it is 20% or 25% of the radius.  
> **username_3**: I'm not sure I follow some of the numbers.  
> **username_0**: Which numbers are you confused about?  
> **username_3**: Are you justifying the data you found, or mine? If 0.8% is 34% of the mass, then (assuming constant density, which might not be true), 1.56% should be roughly 2/3 of the mass, not 50%.  
> **username_0**: Actually, I was justifying yours. But remember, the density between 20% and 25% is a lot less than that below 20% so the mass wouldn't double.  
> **username_3**: Right, which would be the fallacy in my hypothetical constant density argument. So I guess you're right with your rationale for 50%.  
> **username_4**: Why are you skeptical of the "roughly 2 percent"? That's consistent with the 1/4 radius figure because 1/4 cubed is 1.5625%, or "roughly" 2%. What you should be skeptical of is the "almost 15 times the density of lead" figure because $\frac 4 3 \pi (0.25 \, R_\odot)^3\,15\,\rho_{\text{lead}}$ is 1.9 solar masses. That "15 times the density of lead" is the density at the very center rather than the average density of the core (which is what the space.com article implies).  
> **username_3**: @DavidHammen I hadn't yet done the calculations when I wrote that; I'll remove it.  

---

### Answer by username_0 on 2014-10-22. Score: 0

I found out it is 34% of the Sun's mass on [Wikipedia](http://en.wikipedia.org/wiki/Solar_core).


---

