Whats is volume of mass at the centre of the sun

Click For Summary
SUMMARY

The volume of mass at the center of the Sun is calculated to be approximately 3.805 × 1025 m3, corresponding to a radius of 0.3 R. Within this active fusion core, the proton–proton chain reaction occurs at a rate of 9.2 × 1037 fusions per second, resulting in a mass fusion rate of 6.155 × 1011 kg/s. This core represents one-third of the Sun's total volume, highlighting the significant density and activity within this region.

PREREQUISITES
  • Understanding of solar physics and stellar structure
  • Familiarity with the proton–proton chain reaction
  • Basic knowledge of volume and mass calculations
  • Awareness of solar radius measurements (R)
NEXT STEPS
  • Research the implications of solar core fusion rates on stellar evolution
  • Explore advanced calculations of stellar volumes using different radii
  • Study the effects of solar mass loss on the solar system
  • Investigate the role of hydrogen nuclei in stellar fusion processes
USEFUL FOR

Astronomers, astrophysicists, and students of solar physics will benefit from this discussion, particularly those interested in the dynamics of stellar cores and fusion processes.

karen_lorr
Messages
63
Reaction score
0
Thank you
 
Last edited:
Astronomy news on Phys.org

Whats is volume of mass at the center of the sun?

Total solar radius:
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.

Active solar core fusion radius:
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}
I am trying to get a visualization of dimension of a certain amount of mass within the core of the sun?
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}}}
But if you could "see" it, how much space would this lot take up.
One-third the present volume of the Sun.
[/Color]
Reference:
Solar - core - Wikipedia
 

Attachments

  • core01.jpg
    core01.jpg
    7.8 KB · Views: 509
Last edited:
Simple nitpick: one third radius = 1/27 volume, or .3 radius -> .027 volume.
Still pretty large. I would have guessed much smaller.

Jim Graber
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K