Timeline for What's the safe distance from a hypernova?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2021 at 12:13 | comment | added | Anonymous | @ProfRob, I didnt realize that but thanks. Anyways i made another mistake of 3 magnitude so they partly cancel out, but I'll edit that soon. I searched on wikipedia and got that 10^-2 stars are at 10 times the suns mass and 10^-3 at 20, and the distribution gives 10^-3.5 for 30 times the mass, so Im 4 magnitude off. | |
| Oct 23, 2021 at 17:16 | comment | added | user177107 | not this again... | |
| S Oct 23, 2021 at 9:10 | history | suggested | Bad Chad | CC BY-SA 4.0 |
Improved punctuation, improved formatting, improved lazy usage of symbol substituents like "pi", fixed grammar ("Sun" is uppercase), added missing closing parenthesis to the stream-of-consciousness digress.
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| Oct 22, 2021 at 21:38 | review | Suggested edits | |||
| S Oct 23, 2021 at 9:10 | |||||
| Oct 22, 2021 at 12:51 | comment | added | ProfRob | Your numbers are confused/confusing. The stellar mass function goes as a power law something like $N(M) \propto M^{-2.3}$, it is not a "normal" distribution. The integral above a particular mass $M_0$ therefore goes as $M_0^{-1.3}$. i..e. the fraction of stars more massive than $8M_\odot$ that are also $>32 M_{\odot}$ is $4^{-1.3} = 0.16$, not $10^{-6}$ as you claim. | |
| Oct 22, 2021 at 9:05 | review | Late answers | |||
| Oct 22, 2021 at 13:34 | |||||
| Oct 22, 2021 at 8:52 | history | edited | Anonymous | CC BY-SA 4.0 |
added 120 characters in body
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| S Oct 22, 2021 at 8:46 | review | First answers | |||
| Oct 23, 2021 at 21:35 | |||||
| S Oct 22, 2021 at 8:46 | history | answered | Anonymous | CC BY-SA 4.0 |