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STF 2351 distance?

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#1 Astrodj

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Posted 26 October 2012 - 01:34 PM

On the evening of the 24th I observed STF 2351, an equal brightness double in Lyra. Besides being a lovely white pair of 7.7 magnitude stars separated by 5.1", I was wondering two things.

1. Being so close in our sky to the "Double Double", does this maybe get neglected by double star folks more than other DEB's?

2. Can anyone tell me how far away this system is? I would like to be able to mentally place it where it actually is in the galaxy.

Thanks,
DJ

#2 fred1871

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Posted 26 October 2012 - 07:45 PM

I don't know how 'neglected' STF 2351 is, but it's common enough that if you're in the neighbourhood of a famous showpiece, you might get overlooked. :shocked:

Distance of the pair? Hipparcos has a parallax, but it has a large error bar. 4.77mas +/-1.72. In other words, the distance measured is approximate, unsurprising given the small parallax. 4.77mas translates to ~210 parsecs, which is ~680 light years. Taking the error bar literally, it tells us the pair is somewhere in the 500-1,000 ly range, with the 680 figure being likely closer to correct than the extremes.

Most doubles we see - except for a few super-bright in absolute magnitude that are also very wide apart in real separation - are in our local region of the galaxy. Paul Couteau, in his book Observing Visual Double Stars, says that "the country of double stars with computable orbits has its frontier about 1,200 ly away, with our current means of observation" - that's using refractors up to 1-metre (40-inches) aperture.

Of course we can see other pairs further off, especially among the dimmer pairs in terms of apparent magnitude, with stars that are type B in particular, to some extent type A, in spectral classification, because these are intrinsically brighter stars. But "within a few thousand light years" covers most visual doubles. Even at 5,000 ly we're still limited to a modest section in width of our Milky Way galaxy, though we've more than covered the thickness.

We see STF 2351 separated about 5" in our line of sight, despite its distance. If it were 10 times further away (~6800 ly), it would appear ten times closer (0.5"), and 100 times dimmer (inverse-square law on brightness). That's 5 magnitudes fainter, so you'd have a pair of stars around mag 12.5 at 0.5" separation. That's getting to the limits of what's visually detectable with any telescope, as the super-tight pairs fainter than 10th magnitude become extremely difficult even in large telescopes. Tycho found some pairs nearly this difficult,though about a magnitude brighter - such as TDT 724, mags 11.01 and 11.16 at 0.5" - which had been missed by the surveys using large refractors.

I did a quick scan of some dim super-close pairs in part of the WDS, and most examples like this one are Tycho (or Hipparcos) doubles. A few were picked up by Aitken, with the Lick 36-inch refractor, by Couteau with the 30-inch refractor at Nice, and especially by Rossiter with the 27-inch Lamont-Hussey refractor in South Africa, but Tycho and Hipparcos found more, no doubt helped by perfect seeing in space. :cool:

An example of one of Rossiter's toughest finds is RST 968 - stars of mag 11.13 and 11.50 at about 0.4" when discovered in 1930. He found a number of 11th mag pairs below 0.5", but that's about the limit in terms of faintness.

Leaving aside the much-too-hard-for-ordinary-telescopes category, we can enjoy the relatively nearby pairs such as STF 2351, a very nice double I agree, and easily seen even with small telescopes.

#3 drollere

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Posted 26 October 2012 - 10:11 PM

the alternative approach is to use spectroscopic parallax, which i'll demonstrate.

WDS gives visual magnitudes 7.6 and 7.64, and spectral types A1V and A0V. those have morgan-keenan absolute magnitudes (from the table HERE) of 1.6 and 1.4.

then the distance modulus or "delta" (visual mag. - absolute mag.) is 7.6-1.6=6.0 and 7.64-1.4=6.24.

and the distance (in parsecs) is found as = 10^((delta/5)+1) or for our two stars, 10^((6.0/5)+1) and 10^((6.24/5)+1). 10^2.2 = 158 and 10^2.25 = 177, so we get a distance of 158 or 177 parsecs.

obviously the stars are closer to each other than 60 light years, but the estimates bracket a likely distance, and suggest the amount of error the estimate contains.

fred is right to use the astrometric parallax if it is available and seems trustworthy. note that the "large" confidence bound means the parallax could be as large as 4.77+1.77 mas = 0.00654 arcseconds, which gives a minimum distance of 1/0.00654" = 153 parsecs. so the two estimates overlap, and suggest the astrometric estimate errs (if at all) on the long side. 180 parsecs seems like a reasonable best guess to me.

that would make the minimum separation between the two stars rather large -- more than 1200 AUs -- which would be plausible because A type stars are rather massive. it also makes it highly likely that one or both stars has a much closer or spectroscopic companion.

i'm not here to undercut fred's excellent post, but to point out that there is more than one way to get to a distance estimate, especially for stars farther than a few hundred parsecs. also, there is a lot of uncertainty in any estimation method, so you need to take estimates of distance, mass, brightness, etc., etc. with a grain of salt. salt is not usually approved in a diet, but in astronomical data you can lay it on pretty thick and still eat healthy.

#4 Astrodj

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Posted 27 October 2012 - 05:52 PM

Fred,

Thank you for your thorough explanation concerning distances of visual double stars. The 1200ly frontier is interesting. Most of the naked eye stars lie within the same local bubble, or there abouts. After reading your explanation, it makes sense that for most binaries to be telescopically visible they would need to be relatively nearby, galactically speaking.

Even though I have been somewhat aware of the fact (for many years) that most of what we can visually see of our own galaxy with modest equipment is very much "local", I find I am still getting used it. :o

#5 Astrodj

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Posted 27 October 2012 - 06:05 PM

Bruce,

An excellent explanation of an alternative method. Thank you. I was aware of the use of spectroscopy for distance determination but must admit I had a pretty hazy understanding of it.

Your explanation of the method, and how it is used in combination with astrometric paralax where applicable is very understandable.

It is very interesting to understand more fully how distance estimates are derived.

#6 GlennLeDrew

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Posted 28 October 2012 - 11:00 PM

Fred and Bruce,
Excellent discourses on two perspectives!

Bruce,
For the benefit of other readers, could you outline the method for finding the minimum separation?

Why do you suppose it likely that one of the stars is a very close binary? I don't see any difficulty in imagining two single, early A stars of this separation being reasonably securely bound. Or if their separation along the line of sight is considerable, they could be unbound, now sharing common space motion.

#7 Ed Wiley

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Posted 31 October 2012 - 06:06 PM

VI/135/table15 All-sky spectrally matched Tycho2 stars (Pickles+, 2010)

200 and 205 parsecs, but no errors listed and the distances do not appear in the details page.

Ed

#8 drollere

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Posted 01 November 2012 - 03:53 PM

Bruce, For the benefit of other readers, could you outline the method for finding the minimum separation?


certainly.

1. distance modulus is delta = (m - M), where M is the estimated absolute magnitude and m is the visual magnitude.

2. then distance D in parsecs = 10^((delta/5)+1).

3. then minimum separation S in astronomical units = D*10^(log(rho)+0.13), where rho is the visible angular separation in arcseconds, and 0.13 is a correction devised by paul couteau to compensate for the fact that most binary orbits are tilted to our direction of view, so the apparent separations are foreshortened.

in the example, if D = 180 pc and rho = 5", then

S = 180 * 10^(log(5)+0.13) = 180 * 10^0.83 = 1216.95 AU

Why do you suppose it likely that one of the stars is a very close binary? I don't see any difficulty in imagining two single, early A stars of this separation being reasonably securely bound. Or if their separation along the line of sight is considerable, they could be unbound, now sharing common space motion.


simple probabilities. the probability that a star is in a binary system increases with mass: for A type stars (at 2-3 solar masses) the estimated multiplicity fraction is over 80%. doesn't the AB pair count? yes, but the fact that two A stars have joined implies they were formed in a mass dense environment, which makes close companions more likely. note, for example, that the B stars in the orion trapezium are gravitationally bound, and all except D have close visual companions, and all also have very close spectroscopic companions. "likely" is the keyword.

#9 drollere

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Posted 01 November 2012 - 05:21 PM

Even though I have been somewhat aware of the fact (for many years) that most of what we can visually see of our own galaxy with modest equipment is very much "local", I find I am still getting used it.

a concrete way to evaluate the problem is to pick any arbitrary visual magnitude as "the faintest binary i am able to resolve" with your telescope, then use absolute magnitude to derive the distance limit of your view.

if your resolution limit is magnitude 10, which includes about 95% of all primary stars listed in 19th century double star catalogs, then the maximum distance of a star of absolute magnitude M is (distance modulus again):

D = 10^((10-M/5)+1)

the Sun has an absolute magnitude of 4.83, so the farthest distance at which you would resolve a solar type binary system is:

D = 10^((10-4.83/5)+1) = 10^2.03 = 107 parsecs = 349 light years.

couteau's observational limit of 1200 light years (~370 parsecs) is based on computable orbits (couteau, "observing visual double stars" p.182) -- not only on intrinsic brightness but also observable separation.

his criterion for computable orbit is a visual magnitude of at least 10 and a period of at most 200 years. using kepler's third law, binary period implies separation, if we make assumptions about the combined mass of the two stars. couteau's limit case is an A5 binary. A5 stars are roughly 2 solar masses, so the combined mass of an A5 binary system is 4 solar masses. with a period of 200 years, the separation R is the cube root of (P^2)*M, or (200^2)*4 = 160,000^0.33 or 54 AU.

at 370 parsecs 1" will span 370 AU, so 54 AU will equal a visual angle of 54/370 = 0.15". and finally, 0.15" is roughly couteau's limit number because it is approximately the resolution limit of the largest refracting telescopes (couteau's standard for double star observations). this limit says nothing about intrinsically brighter type O or B binaries, or more widely separated binaries with periods greater than 200 years. these can be seen at much greater distances -- they just won't yield "computable orbits".

#10 GlennLeDrew

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Posted 01 November 2012 - 07:54 PM

Thanks for all the details, Bruce. Kind of makes a mini toolkit of double star astrophysics.






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