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  #1291 (permalink)  
Old 17-December-2005, 04:57 AM
evanoconnor evanoconnor is offline
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A Metal Nose

Commonly mistaken for a metal tongue which was actually Jebediah Springfield's. Just Joking.
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  #1292 (permalink)  
Old 17-December-2005, 06:33 AM
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More precisely a silver nose.
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  #1293 (permalink)  
Old 17-December-2005, 06:56 AM
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According to Carl Sagans "Cosmos" page 59, it was a gold nose.

I guess since Henrick was closest, you go next.
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  #1294 (permalink)  
Old 17-December-2005, 07:26 AM
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I knew the answer before posting. I've read Sagan but also many Astronomy books say the same.
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  #1295 (permalink)  
Old 17-December-2005, 08:01 AM
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To stay with Tycho, what's this instrument for? Link to picture
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  #1296 (permalink)  
Old 17-December-2005, 01:10 PM
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Measuring the angular separation of two stars?
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  #1297 (permalink)  
Old 18-December-2005, 06:37 PM
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Yes
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  #1298 (permalink)  
Old 18-December-2005, 11:57 PM
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Quote:
Originally Posted by crosscountry
We had a similar "big array" here at UF which operated for many years ... there was a similar one in the Owens Valley in California. Ours had "only" 640 poles and it operated at 26.3 MHz. For some time at the UF Astronomy Department PhD was said to stand for Post hole Digger.
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  #1299 (permalink)  
Old 19-December-2005, 03:12 AM
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Quote:
Originally Posted by IMO
For some time at the UF Astronomy Department PhD was said to stand for Post hole Digger.
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Old 19-December-2005, 01:15 PM
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Given a close double star whose two components are of visual magnitude 3.085 and 5.088, what is the integrated magnitude (to three decimal places) of the double star?
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  #1301 (permalink)  
Old 19-December-2005, 02:06 PM
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Quote:
Originally Posted by Eroica
Given a close double star whose two components are of visual magnitude 3.085 and 5.088, what is the integrated magnitude (to three decimal places) of the double star?
2.926
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  #1302 (permalink)  
Old 19-December-2005, 02:24 PM
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I'm not doubting IMO's answer here, but wouldn't that be dependent on the position of the two stars? I mean, if one eclipses the other (say the brighter eclipses the duller one), wouldn't the visual magnitude then be 3.085?
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  #1303 (permalink)  
Old 19-December-2005, 02:31 PM
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Quote:
Originally Posted by Fram
I'm not doubting IMO's answer here, but wouldn't that be dependent on the position of the two stars? I mean, if one eclipses the other (say the brighter eclipses the duller one), wouldn't the visual magnitude then be 3.085?
But at that point, the other one would not have a visual magnitude of 5.088, right?

PS: I got 2.924 for the answer, slightly different from IMO's
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  #1304 (permalink)  
Old 19-December-2005, 02:44 PM
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Quote:
Originally Posted by Fram
I'm not doubting IMO's answer here, but wouldn't that be dependent on the position of the two stars? I mean, if one eclipses the other (say the brighter eclipses the duller one), wouldn't the visual magnitude then be 3.085?
a) Double star may not mean mutually orbiting stars ... such a system would usually be called a close binary star,
b) most binary systems, even close binaries, do not eclipse,
c) the magnitude during the eclipses depends on the surface temperatures and radii of the stars,
d) when asked for integrated magnitude one virtually always means "what will be the magnitude if the two stars are unresolved".
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  #1305 (permalink)  
Old 19-December-2005, 03:16 PM
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I just noticed that if you use 2.5 for 10^(.2), you get IMO's answer, but if you use 2.512 (or even 2.51), you get my answer.
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  #1306 (permalink)  
Old 19-December-2005, 04:10 PM
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Quote:
Originally Posted by hhEb09'1
I just noticed that if you use 2.5 for 10^(.2), you get IMO's answer, but if you use 2.512 (or even 2.51), you get my answer.
IIRC the formulae are
I/Io=10^(-0.4*dM) and dM=-2.5*Log10(I/Io)
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  #1307 (permalink)  
Old 19-December-2005, 04:15 PM
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Quote:
Originally Posted by IMO
IIRC the formulae are
I/Io=10^(-0.4*dM) and dM=-2.5*Log10(I/Io)
Yes, but that is only two digits precision, as (10^(2/5)) = 2.5118864315095801110850320677993..., so to get the precision asked for, you'd have to use 2.512

PS: I just noticed I typed 10^(.2) in my previous post when I meant 100^(.2)
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  #1308 (permalink)  
Old 19-December-2005, 04:22 PM
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Quote:
Originally Posted by hhEb09'1
Yes, but that is only two digits precision, as (10^(2/5)) = 2.5118864315095801110850320677993, so to get the precision asked for, you'd have to use 2.512
??? ... 5.0000000 magnitudes is a factor of 100 in I
(or 2.500000 magnitudes is a factor of 10)

calculate 2.5*Log10(100) ... = 5.0000000
or the other way, 10^(-.4*5)= 10^2 .... = 100.0000

It IS true that 1 magnitude is an intensity ratio of 2.512 but it is unclear to me what that has to do with this problem.
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  #1309 (permalink)  
Old 19-December-2005, 04:43 PM
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Quote:
Originally Posted by IMO
2.926
Correct.

I used my desktop calculator (1000.2 = 2.5118864315095801110850320677993) and a piece of paper to work it out, rounding off at the end. I got the same answer as IMO.
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  #1310 (permalink)  
Old 19-December-2005, 04:49 PM
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I'm with IMO, 2.926
[Edit: Sorry I didn't reload to see Eroica last post, just ignore me ]
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  #1311 (permalink)  
Old 19-December-2005, 04:49 PM
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