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Two related questions (visual magnitude + Uranus)
Hi group!
As you can imagine, a side-effect of the thread I initiated last week on "elusive Uranus", I got now extra-eager and impatient trying to see Uranus. So, I learned that under reasonably dark sky, it is a naked eye object, and very obvious with binoculars (should be the case with mine, a 10x50 pair). Ok, but what about under full Moon? (well, or near-full, like in the past couple of days?) I normally would think not, but I ask because (and this leads me to my other "related" question) on my front yard, in a heavily light- polluted area --- right on the border between white (big-city level light- pollution) and red, as per light-pollution maps http://www.jshine.net/astronomy/dark_sky/ I went out just to try and see the moon through the binoculars to see if it was worth the effort to take the telescope to shoot a couple of pictures, and *surprise* !!! I could see Andromeda (the galaxy!) *naked eye* (I had to see through the binoculars to double check that it wasn't my imagination). Even *yesterday*, I could see it (even though yesterday I would say it was *truly barely* visible naked eye --- but through the binoculars it was like incredibly obvious. (I'm sure this is not at all news for most experienced astro-enthusiasts out there --- but it did take me by surprise!!). I tried locating Uranus through binoculars, but without success --- nothing seemed like a reasonable suspect. So, I see that M31's visual magnitude is 3.4 (according to Wikipedia). That's a lot, and actually I would expect a star of 3.4 to be visible in downtown NYC or some *heavily light-polluted* area (am I over- estimating??). So, this suggests that the 3.4 is sort of the integral over the area, to obtain the "total amount of light" that the area covered by Andromeda emits (well, that we receive, I should say). The funtion being integrated is somehow the "density" or the "true" value of the luminous intensity, as in amount of light per unit of surface; with stars, the thing works more or less like the function being a Dirac's delta, where its weight is precisely the visual magnitude of the given star. So, *finally* the question: is there a measurement of the brightness of a celestial object that could be more directly related to the level of light pollution at which the object becomes visible? Intuitively, it seems like this would be closely related (if not being the exact same measure) to the "density" that I was describing above --- more specifically, the maximum value of this density of the given object (for objects that occupy an area, like M31) IOW, I would like to be able to determine whether I *should* be able to see some target (in this case Uranus, of course) simply by the fact that some other target is visible --- I mean, it seems to me (pure intuition and rough estimate) that if Andromeda is visible under some particular light-pollution conditions, then a 5.6 object *should* be visible as well (possibly even obvious). But then, I don't think I can conclude that from the number 3.4 that I see for Andromeda; I do not think the 3.4 reflects how faint Andromeda really is. Hope I'm making some sense and that you will be able to understand what I'm asking, and shed some light (I know, I know, this must be the second lamest pun, after the one we all know for Uranus :-\ :-) ) Thanks, Carlos -- |
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Two related questions (visual magnitude + Uranus)
On Sep 10, 11:44 am, wrote:
Hi group! As you can imagine, a side-effect of the thread I initiated last week on "elusive Uranus", I got now extra-eager and impatient trying to see Uranus. So, I learned that under reasonably dark sky, it is a naked eye object, and very obvious with binoculars (should be the case with mine, a 10x50 pair). Ok, but what about under full Moon? (well, or near-full, like in the past couple of days?) I normally would think not, but I ask because (and this leads me to my other "related" question) on my front yard, in a heavily light- polluted area --- right on the border between white (big-city level light- pollution) and red, as per light-pollution mapshttp://www.jshine.net/astronomy/dark_sky/ I went out just to try and see the moon through the binoculars to see if it was worth the effort to take the telescope to shoot a couple of pictures, and *surprise* !!! I could see Andromeda (the galaxy!) *naked eye* (I had to see through the binoculars to double check that it wasn't my imagination). Even *yesterday*, I could see it (even though yesterday I would say it was *truly barely* visible naked eye --- but through the binoculars it was like incredibly obvious. (I'm sure this is not at all news for most experienced astro-enthusiasts out there --- but it did take me by surprise!!). I tried locating Uranus through binoculars, but without success --- nothing seemed like a reasonable suspect. So, I see that M31's visual magnitude is 3.4 (according to Wikipedia). That's a lot, and actually I would expect a star of 3.4 to be visible in downtown NYC or some *heavily light-polluted* area (am I over- estimating??). So, this suggests that the 3.4 is sort of the integral over the area, to obtain the "total amount of light" that the area covered by Andromeda emits (well, that we receive, I should say). The funtion being integrated is somehow the "density" or the "true" value of the luminous intensity, as in amount of light per unit of surface; with stars, the thing works more or less like the function being a Dirac's delta, where its weight is precisely the visual magnitude of the given star. So, *finally* the question: is there a measurement of the brightness of a celestial object that could be more directly related to the level of light pollution at which the object becomes visible? Intuitively, it seems like this would be closely related (if not being the exact same measure) to the "density" that I was describing above --- more specifically, the maximum value of this density of the given object (for objects that occupy an area, like M31) IOW, I would like to be able to determine whether I *should* be able to see some target (in this case Uranus, of course) simply by the fact that some other target is visible --- I mean, it seems to me (pure intuition and rough estimate) that if Andromeda is visible under some particular light-pollution conditions, then a 5.6 object *should* be visible as well (possibly even obvious). But then, I don't think I can conclude that from the number 3.4 that I see for Andromeda; I do not think the 3.4 reflects how faint Andromeda really is. Hope I'm making some sense and that you will be able to understand what I'm asking, and shed some light (I know, I know, this must be the second lamest pun, after the one we all know for Uranus :-\ :-) ) Thanks, Carlos Your question(s) are good ones. Have a read of this paragraph, note visual magnitudes and how amateurs can use 'the little dipper' as a seeing reference, http://en.wikipedia.org/wiki/Ursa_Minor#Stars The four stars constituting the "bowl" of the little dipper are unusual in that they are of second, third, fourth and, fifth magnitudes. Hence, they provide an easy guide to determining what magnitude stars are visible, useful for city dwellers or testing one's eyesight. It's very convenient as Polaris is easy to find. If you keep notes, log the dimmest star you can see, inverted vision is cheating ;-). Cheers Ken |
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Two related questions (visual magnitude + Uranus)
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Two related questions (visual magnitude + Uranus)
The four stars constituting the "bowl" of the little dipper are unusual in that they are of second, third, fourth and, fifth magnitudes. Hence, they provide an easy guide to determining what magnitude stars are visible Huh ... Interesting. However, the caveat (working against me, as per Murphy's Law :-) ) ... I can see only two of them with zero effort, the third one *barely*, and not at all the fourth one (the one with magnitude 5, that is). With the binoculars, however, all of them are extremely obvious. But that's looking to the north, which is opposite to where the Moon is, and also, more or less opposite to where the light pollution is coming (as I'm to the north-west of the city, north/north-west is the darkest side of the sky from where I live --- the Moon is now being added to the light pollution, and Pisces and Uranus are precisely around there ...) *sigh* ... anyways, next weekend!!! However, the Moon shots, prime focus with a 1000mm focal distance f/5 Newtonian, are not half bad --- I'm pretty sure they're my personal best; maybe some of the past eclipse pictures might have come out a bit more impressive, not being so excessively bright. If you're curious, two of them are he http://www.mochima.com/personal/DSC_5578.jpg http://www.mochima.com/personal/DSC_5580.jpg Thanks, and thanks Chris for your message! Carlos -- |
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Two related questions (visual magnitude + Uranus)
On Sat, 10 Sep 2011 15:01:46 -0600, Chris L Peterson
wrote: Correct. The magnitude of extended objects is determined by the total light emitted, so large objects often have low magnitudes, when in fact, they are very hard to see. Correct. Two examples: the integrated magnitude of the Andromeda Galaxy (M31) is around +3 and the magnitude of the Milky Way is around -7 which is brighter than Venus. A stellar object of mag -7 would be easily visible even under the most light polluted skies and even in daytime, while the Milky Way, which extends across the sky, can quickly be washed out even by moderate light pollution. |
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Two related questions (visual magnitude + Uranus)
Carlos wrote:
So, I learned that under reasonably dark sky, it is a naked eye object, and very obvious with binoculars *(should be the case with mine, a 10x50 pair). Ok, but what about under full Moon? *(well, or near-full, like in the past couple of days?) It should be quite possible to *see* Uranus with a full Moon (unless the Moon is right next to Uranus--dang, the 13-year-old in me is giggling like a madman), but it's non-trivial to identify which dot you're looking at is Uranus. Uranus's apparent size is 3-4 arcseconds, and when magnified 10x is indistinguishable from a star. (At least, I certainly can't tell the difference.) One really wants a chart with Uranus's position plotted day by day, or the computer equivalent thereof. Something that goes down to seventh magnitude or so is nice, as those stars are generally easily visible through binoculars. Tilt the map so that it aligns with what you see in the sky; it may help to draw out circles on the map that match the binoculars' field of view (usually about 5 or 6 degrees at that magnification), to give you an idea of what is likely to appear in the binoculars along with Uranus. So, *finally* the question: *is there a measurement of the brightness of a celestial object that could be more directly related to the level of light pollution at which the object becomes visible? As others have indicated, surface brightness, measured in magnitudes per solid angle (square arcsecond or square arcminute), is the beast you want. It should be pointed out that it's not exactly what you're describing, because the visibility of an object depends not only on the light pollution, but also (among other things) on the instrument you use to observe it--both aperture and magnification. But for general purposes, surface brightness is useful. You may find the discussion at this page http://mysite.verizon.net/vze55p46/id18.html somewhat useful. It also discusses the notion of peak brightness, which as its name implies tells you how bright the brightest part of the object is. -- Brian Tung (posting from Google Groups) The Astronomy Corner at http://www.astronomycorner.net/ Unofficial C5+ Page at http://www.astronomycorner.net/c5plus/ My PleiadAtlas Page at http://www.astronomycorner.net/pleiadatlas/ My Own Personal FAQ at http://www.astronomycorner.net/reference/faq.html |
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