# Whats is volume of mass at the centre of the sun

1. May 19, 2012

### karen_lorr

Thank you

Last edited: May 19, 2012
2. May 19, 2012

### Orion1

$$R_{\odot} = 6.955 \cdot 10^8 \; \text{m}$$
Inside 24% of the Sun's radius, 99% of the power has been generated, and by 30% of the radius, fusion has stopped nearly entirely.

$$r_f = 0.3 R_{\odot}$$
Average active fusion core volume:
$$V_f = \frac{4 \pi r_f^3}{3} = 3.805 \cdot 10^{25} \; \text{m}^3$$
$$\boxed{V_f = 3.805 \cdot 10^{25} \; \text{m}^3}$$
The proton–proton chain occurs around 9.2×10^37 times each second in the core of the Sun. This reaction uses four free protons (hydrogen nuclei).

Solar core fusion reaction rate:
$$\frac{dn}{dt} = 9.2 \cdot 10^{37} \; \frac{\text{fusions}}{\text{s}}$$
Solar mass fusion reaction rate:
$$\frac{dm}{dt} = 4 m_p \frac{dn}{dt} = 6.155 \cdot 10^{11} \; \frac{\text{kg}}{\text{s}}$$
$$m_p - \; \text{proton mass}$$
$$\boxed{\frac{dm}{dt} = 6.155 \cdot 10^{11} \; \frac{\text{kg}}{\text{s}}}$$
One-third the present volume of the Sun.

Reference:
Solar - core - Wikipedia

#### Attached Files:

• ###### core01.jpg
File size:
7.8 KB
Views:
77
Last edited: May 19, 2012
3. May 26, 2012

### jimgraber

Simple nitpick: one third radius = 1/27 volume, or .3 radius -> .027 volume.
Still pretty large. I would have guessed much smaller.
Best,
Jim Graber