Timeline for What differentiates a "group of rogue stars" from a galaxy?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2017 at 17:09 | history | edited | Pablo | CC BY-SA 3.0 |
added 1 character in body
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| Jul 18, 2017 at 21:12 | answer | added | PERFESSER CREEK-WATER | timeline score: -4 | |
| Jul 15, 2017 at 2:14 | history | tweeted | twitter.com/StackAstronomy/status/886046215044857856 | ||
| Jul 14, 2017 at 14:32 | comment | added | zephyr | Not necessarily. If it's velocity wrt to the galaxy is greater than the escape velocity of that galaxy, then we say it is not gravitationally bound. For example, the escape velocity of the Milky Way is several hundred $km/s$ (how much it is exactly depends on a variety of factors). Basically, if the star's motion away from the galaxy can never be slowed to a stop by that galaxy's gravitational pull, we consider it gravitationally unbound. Certain hypervelocity stars are certainly traveling too fast to be bound and can become "rogue" once they leave their host galaxy. | |
| Jul 14, 2017 at 14:27 | comment | added | Pablo | No problem. Anyway, I think the main definition is missleading too. Any star is at least very weakly gravitationally bound to the closest galaxy, isnt it? or how does that work? | |
| Jul 14, 2017 at 14:22 | comment | added | zephyr | I tried to clean up your question to have better links and quotes. I believe I maintained the spirit and meaning of your question but if not, feel free to roll back. | |
| Jul 14, 2017 at 14:22 | history | edited | zephyr | CC BY-SA 3.0 |
Cleaned up formatting, grammer, and spelling mistakes.
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| Jul 14, 2017 at 10:27 | vote | accept | Pablo | ||
| Jul 14, 2017 at 6:48 | answer | added | ProfRob | timeline score: 6 | |
| Jul 13, 2017 at 22:39 | history | edited | Pablo | CC BY-SA 3.0 |
edited title
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| Jul 13, 2017 at 22:28 | history | asked | Pablo | CC BY-SA 3.0 |