Sunday, January 30, 2011
Module 3, Assignment 5
Post a list of the stars (identifying them by their HD, SAO or Feige numbers) your estimates of temperature (in degrees K), and the number of significant figures you believe are appropriate. Calculate the relative luminosity, compared to the coolest star in your sample, for each of the other 5 stars. To do this, select the coolest star in your sample and define its luminosity as 1, and scale the remaining stars to this one.
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Assignment 5
ReplyDeleteOther stars
Star Peak(A) Temp(2 sig figs) relative temp Relative luminosity
HD33279 4800 6000 1.354 3.37
HD124320 4000 7290 1.645 7.33
HD31084 4500 6460 1.458 4.52
HD 242908 3600 8090 1.826 11.12
SAO76803 6500 4430 1 1
FEIGE 40 3900 7390 1.668 7.66
Sorted by temperature
Star Peak(A) Temp(2 sig figs) relative temp Relative luminosity
SAO76803 6500 4430 1 1
HD33279 4800 6000 1.354 3.37
HD31084 4500 6460 1.458 4.52
HD124320 4000 7290 1.645 7.33
FEIGE 40 3900 7390 1.668 7.66
HD 242908 3600 8090 1.826 11.12
At first I had forgotten that 1nm(10-9 m) = 10 Angstroms(10-10 m)
There seems to be plenty of error in these numbers.
Estimating the peak is not easy and gets more difficult as you go from hot stars(+/- 50A) to cold stars(+/- 500A). Also is seems that a change in 1A in wavelength is equivalent to a change of 1K in star temperature. For this reason temperatures should be quoted to 2 significant figures.
As we used the coolest star as our reference point I am sure that the relative luminosities have large errors.
One problem that Chris and I grapple with is the linear nature of this blog: in other words, how can I make sure everyone sees a given post, given that people work at different speeds? So I'm posting this ahead of most of your posts, but hope everyone reads it.
ReplyDeleteI just want to remind you that one carrot we hold out in this class is a visit to Kitt Peak to observe with me at a research telescope. The opportunities are primarily in the spring and fall: summers are monsoon season, and nobody wants to observe clouds! There will be an opportunity in April, if you have completed the majority of the modules: Saturday night, April 23 (I will point out that this is the day before Easter) If you think this would work for your schedule, please let me know: I need to arrange for a room in the dorm at Kitt Peak if you are coming. I will, of course, send lots more details about this if you are coming, but basically you can expect to arrive (your own transportation) about noon or a little later on Saturday, observe until VERY LATE (astronomers observe until dawn, but I understand you may not make this) then sleep late in the morning, and leave when you wish.
I have corrected my data so that is shows the temperature to 2 significant figures
ReplyDeleteOther stars
Star Peak(A) Temp(2 sig figs) relative temp Relative luminosity
HD33279 4800 6000 1.354 3.37
HD124320 4000 7300 1.645 7.33
HD31084 4500 6500 1.458 4.52
HD 242908 3600 8100 1.826 11.12
SAO76803 6500 4400 1 1
FEIGE 40 3900 7400 1.668 7.66
Sorted by temperature
Star Peak(A) Temp(2 sig figs) relative temp Relative luminosity
SAO76803 6500 4400 1 1
HD33279 4800 6000 1.354 3.37
HD31084 4500 6500 1.458 4.52
HD124320 4000 7300 1.645 7.33
FEIGE 40 3900 7400 1.668 7.66
HD 242908 3600 8100 1.826 11.12
At first I had forgotten that 1nm(10-9 m) = 10 Angstroms(10-10 m)
There seems to be plenty of error in these numbers.
Estimating the peak is not easy and gets more difficult as you go from hot stars(+/- 50A) to cold stars(+/- 500A). Also is seems that a change in 1A in wavelength is equivalent to a change of 1K in star temperature. For this reason temperatures should be quoted to 2 significant figures.
As we used the coolest star as our reference point I am sure that the relative luminosities have large errors.
I believe I can measure the peak wavelength to within 50 angstroms, which translate to 2 significant digits in the temp. The hottest star is about 10 times hotter than the coolest star.
ReplyDeleteStar Temp Relative Luminosity
SAO 76803 4800 1
HD 33278 6300 3.0
HD 31084 6500 3.4
HD 124320 7400 5.6
FEIGE 40 7800 7.0
HD 242908 8600 10.3
I listed my results as: star, temperature, luminosity.
ReplyDeleteHD 33278 6100K, 3.3 W/m2 (I think these are the correct units)
HD 124320 7200K, 6.9 W/m2
HD31084, 6400K, 4.4 W/m2
HD242908 9100K, 17 W/m2
SAO 76603, 4500K, 1 W/m2
Feige40, 7400K, 7.7 W/m2
The error in these measurements are +/- 100A. The scale on the graph is by 1000. You can estimate 1 place past what you can measure for certain. I used Wien's Law to convert 100A to temperature, to determine the error in K, but I'm not sure this is the right way, since you get +/- 2.897 X 105
I used Wien’s law to compute for the estimate temperatures of the 6 stars. I used the simulator to confirm my answers:
ReplyDeleteSAO76803 – 4400 K (coolest)
HD 33278 – 5900 K
HD 31084 – 6400 K
HD124320 – 7200 K
FEIGE 40 – 7400 K
HD242908 – 8300 K
For the relative luminosity, I took the 4th power of each temperature. Compare the other star’s T^4 to the SAO76803’s T^4
SAO76803 – 1.0
HD 33278 – 3.2
HD 31084 – 4.5
HD 124320 – 7.2
FEIGE 40 – 8.0
HD 242908 – 13
I computed the peak wavelength for each star using ratio and proportion. I measured the division for each interval and computed for the wavelength of the peak. Using the simulator, I determine the temperature of the stars. I used 2 significant figures for my values:
ReplyDeleteSAO76803 –(6604 angstrom)- 4390K -- 4400K
HD33278 - (4917 angstrom)- 5910K -- 5900K
HD31084 – (4450 angstrom)- 6440K -- 6400K
HD124320 – (4042 angstrom)- 7240K –- 7200K
FEIGE40 – (3875 angstrom)- 7430K –- 7400K
HD242908 – (3540angstrom) - 8280K -- 8300
Relative luminosity is proportional to T^4. The coolest star has a relative luminosity equal to 1. Scaling other temperature relative to the coolest star. I got the following values (using 2 SF):
SAO76803 – 1
HD33278 – 3.23 round off to 3.2
HD31084 – 4.48 round off to 4.5
HD124320 – 7.17 round off to 7.2
FEIGE40 – 8.0
HD242908 – 12.66 round off to 13
WOW, I did all the work and then checked my numbers against the blog, I found that my relative luminosities were exactly the same as Reny and Beatriz.
ReplyDeleteI estimated where the peak was in angstroms (Hard to get the estimate without digital data to analyze, wish I could use Image J to find the peak). I then used the applet to convert wavelength (in nm) to temperature and rounded to 2 sig figs. Raised the temperature to the ^4 and compared it to the coolest star. These are my results.
Peak Wavelength
HD242908 3500
FEIGE40 3900
HD124320 4000
HD31084 4500
HD33278 4900
SAO76803 6600
Temperature
HD242908 8300 K
FEIGE40 7400 K
HD124320 7200 K
HD31084 6400 K
HD33278 5900 K
SAO76803 4400 K
Luminosity and Relative Luminosity
HD242908 4745832100000000 12.6
FEIGE40 2998657600000000 8.0
HD124320 2687385600000000 7.2
HD31084 1677721600000000 4.5
HD33278 1211736100000000 3.2
SAO76803 374809600000000 1.0
These were all based on the estimation of the peak wavelength.
I used the simulator to determine the temperatures of the stars at a specific peak wavelength. I used 2 significant figures to estimate the temperatures of the star since the graphs are not the spectra of a perfect blackbodies
ReplyDeleteHottest to coolest:
Star Temperature
HD 242908 - 8300 K
FEIGE 40 - 7600 K
HD 124320 - 7100 K
HD 31084 - 6400 K
HD 33278 - 5900 K (sun-like Star)
SAO 76803 - 4400 K (coolest among the 5 stars)
Star Relative Luminosity:
Luminosity is directly proportional to the temperature raised to 4th power. I determine the relative luminosity of the other stars by setting the luminosity of the coolest star by 1.
SAO 76803 ( 3.75E14) - 1
HD33278 (1.21E15) - 3.2
HD 31084 (1.68E15) - 4.5
HD124320 (2.54E15) - 6.8
FEIGE40 (3.34E15) - 8.9
HD242908 (4.75E15) - 13
I used the blackbody simulator to determine the temperatures of the stars at a specific peak wavelength. I used 2 significant figures to estimate the temperatures of the stars. The luminosity of the stars is computed to T raise to 4. I am presenting the results in column (stars (coolest to hottest), temp. using the simulator, 2 sig figs, and luminosity)
ReplyDeleteSTARS TEMPERATURE 2 SIG FIGS LUMINOSITY
(blackbody
simulator)
SAO76803 4380K 4400K 1
HD 33278 5900K 5900K 3.2
HD 31084 6440K 6400K 4.5
HD 124320 7170K 7200K 7.2
FEIGE 40 7490K 7500K 8.4
HD 242908 8190K 8200K 12
Based on the above data the hottest star is almost 2x hotter as the coolest star. The higher the temperature, the greatest the luminosity.
I tried the Wien's law and the simulator to estimate the temperatures of the stars. I have the same results for both methods. My precision of my temperature is up to 2 SF since I only estimate the peak wavelength in a 1000 angstrom division. I will try to analyze later the absortion lines to get the accurate temperature of the 6 stars. For the mean time, these are my results:
ReplyDeleteSAO76803 – 4400 K
HD33278 – 5900 K
HD31084 – 6300 K
HD124320 – 7200 K
FEIGE40 – 7600 K
HD242908 – 8300 K
Luminosity = (surface area of star) × flux of energy through its surface. It is also equal to [4pi×(star's radius)^2] × [sigma(star's surface temperature)^4].
I assumed that the the diameters of the stars
are all the same and I can use the temperature to determine the relative luminosity of each star. The star SAO 76803 is the coolest star among the 5 stars so its relative luminosity equals 1:
SAO76803 – 4400 K - 1
HD33278 – 5900 K - 3.2
HD31084 – 6300 K - 4.2
HD124320 – 7200 K - 7.2
FEIGE40 – 7600 K - 8.9
HD242908 – 8300 K - 13
I just tried to use the SIMBAD and determine B-V (measure of temperature) then I used the simulator to check if I have the same temperature. These are my data:
ReplyDeleteSAO76803 – 3190K
HD33278- 3980K
HD31084 – 5790K
HD124320 – 8160K
FEIGE40 – 9760K
HD242908 – 7550K
I got different answers. Any comments???
From John Webster
ReplyDeleteIn doing this Module, the entering data comes from estimating the wavelength from the graphs. We can only estimate the λ to 100Å. This means we are only doing 2 sig figs.
After estimating the wavelength I calculated the following temperatures using Wien’s Law and relative luminosity using Stefan-Boltzmann’s Law :
SAO 76803 4457° K 1
HD 33278 5912° K 3
HD 31084 6438° K 4.5
HD 124320 7242° K 7
FEIGE40 7428° K 8
HD 242908 8277° K 12.6
I used my wavelength estimates and entered them into the Blackbody simulators to check my calculations to make sure they were in the ballpark. Using the intensity data from the simulator, my results were just slightly different:
SAO 76803 1
HD 33278 3
HD 31084 4.3
HD 124320 7
FEIGE40 7.7
HD 242908 11.8
I amcurrently operating on not nearly enough sleep (took 140 kids to Magic Mountain on Friday - a whopping 2 hours of sleep Friday noght before the dogs went crazy at first light - sheesh!). But I think I have this done sort of correctly:
ReplyDeleteCoolest: SAO 76803 peak 650 nm; 4400 K assigned luminosity 1
HD33278 peak 480 nm; 60000 K; luminosity 3.3
HD 31084 peak 450 nm; 6500 K; luminosity 4.5
HD 124320 Peak 410nm; 7100 K; luminosity 6.5
FEIGE 40 Peak 390 nm; 7500 K; luminosity 8.3
Hottest: HD 242908 Peak 350 nm; 8200 K; luminosity 12
The estimates of the peak give at best 2 SF, but I used all the digits indicated by the simulator to complete the calculations, rounding only at the end.
Below I listed the star, peak wavelength, temperature (calculated using Wien’s equation), and relative luminosity. Since the wavelength was given by thousands, we can estimate to one more place. Since wavelength has two sig figs, each subsequent calculation can only have two sig figs.
ReplyDeleteHD 33278, 4500 A, 6400 K, 4.5
HD 124320 4100 A, 7100 K, 6.8
HD 31084 4500 A, 6400 K, 4.5
HD 242908 3200 A, 9100 K, 18.3
SAO 76803 6600 A, 4400 K, 1
FEIGE 40 3700 A, 7800 K, 9.9
On each of the graphs, the scale is in 1000 Angstroms so I think that the accuracy of estimating the peak wavelength is 10% of that or 100 Angstroms. Therefore our calculations should be rounded to two significant figures.
ReplyDeleteThe following are the stars listed from coolest to hottest and includes the wavelength in Angstroms, calculated temperature in Kelvin and relative luminosity.
SAO76803 6500A 4500K 1
HD 33278 4800A 6000K 3.2
HD 31084 4600A 6300K 3.8
HD124320 4100A 7100K 6.2
FEIGE 40 3800A 7600K 8.1
HD242908 3600A 8000K 10
Great work, everyone!I'm trying to send individual comments if I see particular confusion, but overall we are very pleased. You should have gotten an individual email from me asking for you SSN so I can have a check cut for you, for the first 3 modules: still need to hear form a few of you...
ReplyDeleteWell I estimated the temperatures of the stars as the following:
ReplyDeleteSAO76803 ~ 4800 K
HD33278 ~ 6400 K
HD31084 ~ 6700 K
HD124320 ~ 7300 K
FEIGE40 ~ 7500 K
H242908 ~ 9400 K
This corresponds to an adjusted luminosity of
SAO76803 ~ 1
HD33278 ~ 3.2
HD31084 ~ 3.8
HD124320 ~ 5.4
FEIGE40 ~ 6.0
H242908 ~ 15