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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SAO 76803
ReplyDeleteCalculated temperature: 4379K
There should be only 1 sig. fig. So it should be rounded to 4000K.
HD 31084
Calculated temperature: 6422K.
After rounding: 6000K
HD 33278
Calculated temperature: 6422K
After rounding: 6000K
HD 124320
Calculated temperature: 7139K
After rounding: 7000K
FEIGE 40
Calculated temperature: 7410K
After rounding: 7000K
HD 242908
Calculated temperature: 8028K
After rounding: 8000K
Relative luminosities according to SAO 76803:
Temperature of SAO 76803, the coolest one, is 4000K and its luminosity is assumed as 1.
Temperatures of HD 31084 and HD 33278 are 6000K and its relative luminosity will be around 5.(A little bit higher)
Temperatures of FEIGE 40 and HD 124320 are 7000K and their relative luminosity will be around 9.( A little bit higher)
Temperature of HD 242908 is 8000K and its relative luminosity will be 16.
Relative luminosities of HD 31084 and HD 33278 is
ReplyDelete5,0625.
Relative luminosities of FEIGE 40 and HD 124320 is 9,379.
Thank you Chris. Instead of a comma, there should be a decimal point.( 5.0625 and 9.379)
ReplyDeleteI used Wien’s law to compute for the estimate temperatures of the 6 stars. I reported my temperature to 2 SF
ReplyDeleteSAO76803 – 4400 K (coolest)
HD 33278 – 5900 K
HD 31084 – 6400 K
HD124320 – 7200 K
FEIGE 40 – 7400 K
HD242908 – 8300 K
I assumed that SAO76803 has a relative luminosity of 1 and the rest will be compared to this star.
I took the T^4 for all the stars and compared it to the T^4 of the coolest star
SAO76803 – 1.0
HD 33278 – 3.2
HD 31084 – 4.5
HD 124320 – 7.2
FEIGE 40 – 8.0
HD 242908 – 13
From Ria
ReplyDeleteThe peak wavelength of the spectrum of each star:
SAO76803 6600 angstrom
HD33278 4920 angstrom
HD31084 4450 angstrom
HD124320 4040 angstrom
FEIGE40 3880 angstrom
HD242908 3540 angstrom
Each of you will probably get different values for the max wavelength of the star. Post these wavelengths, calculate the temperature, explain how many sig figures are appropriate and then calculate the relative luminosities. Everybodies results will be a little different.
ReplyDeleteHD33278
ReplyDeleteTemperature 5800K
Relative Luminosity 3
HD124320
Temperature 7300K
Relative Luminosity 8
HD31084
Temperature 6400K
Relative Luminosity 5
HD242908
Temperature 9100K
Relative Luminosity 20
SOA76803
Temperature 4500K
Relative Luminosity 1
FEIGE40
Temperature 7300K
Relative Luminosity 8
correction to HD242908 My luminosity should have been 18 not 20
ReplyDeleteThe significant digits for the wavelength if based on the scale markings would be only 1 significant digit and the second digit would be an approximation. Is this a problem of not label the axis with more ticks? First, I found the wavelength with maximum and used Wein's law to calculate the temperature. I used Wein's temperature to rank the stars by luminousity. The used the temperature to determine the star class based on the table in stellar_spectroscopy.pdf. Second, I studied the traces and used the information on absorption in the same document to classify the star. Thus, I used two methods to determine the star class. Results are tabulated below:
ReplyDeleteCommas were used to separate the data because tabs see to be converted to spaces in the blog.
Name, Wavelength, Wein's Temp., Class vis Temp., Class Via Trace, Luminousity
SA0 76803, 6100A, 4749K, K, K, 1.0
HD-33278, 4700A, 6164K, A, F, 1.3
HD 31084, 4500A, 6438K, F, F, 1.4
HD 124320, 4100A, 7066K, F, A, 1.5
FEIGE 40, 3800A 7624K, B, A, 1.6
HD 242908, 3500A, 8277K, O, A, 1.7
Two out of the six agreed between the two methods. The other three (not including HD 242908) were very close agreement. I am not sure what is meant by smooth. HD 242908 did not have a peak so I used the highest possible value to determine the wavelength.
All my talk about significant figures and I forgot to round correctly. Please excuse my mistake.
ReplyDeleteOOPS. I also used the blackbody simulator to determine the temperature. Most of the values for temperature were close except for HD242908. The Wein value was 8277K and the blackbody simulator was 8330K. Both were within the significant digits for temperature.
ReplyDeleteSince we are just the approximating the peak wavelength of each spectrum I believe that the temperature should be reported up to 2 significant figures:
ReplyDeleteSAO 76803 – 660nm - 4400 K
HD 33278 – 490nm- 5900 K
HD 31084 – 470 nm - 6200 K
HD 124320 – 400nm- 7200 K
FEIGE 40 – 380nm- 7700 K
HD 242908 – 330nm- 8900 K
The luminosity is defined mathematically by the equation:
L = 4 pi sigma (6.9E8)^2 (T^4).
I took the ratio of the T^4 of a given star relative to the T^4 of the coolest star (SAO76803)
SAO 76803 – 1
HD 33278 – 3.2
HD 31084 – 3.9
HD 124320 – 7.2
FEIGE 40 – 9.4
HD 242908 – 16.7
On the Intensity vs. Wavelength graphs, the horizontal axis is marked to the nearest thousand angstroms, so I can estimate to the nearest hundred angstroms. This will limit the estimation of temperature to two significant figures.
ReplyDeleteThese are the peak wavelengths I estimated for each star
HD33278-4900 A
HD124320-4100 A
SAO76803-6600 A
FEIGE40- 3900 A
HD31084-4600 A
HD242908-3600 A
Wein’s Law can be used to find the temperature.
HD33278
Sample: T = (3 x 10^7)/(max WL)
(3 x 10^7)/(4900 A) = 6100 K
Here are the estimated temperatures for each star listed from coolest to hottest
SAO76803-4500 K
HD33278-6100 K
HD31084-6500 K
HD124320-7300 K
FEIGE40- 7700 K
HD242908-8300 K
Using the Steffan-Boltzman Law, E is proportional to T^4. If we define the luminosity of SAO76803 to be 1.0, then we can use proportional reasoning to estimate the relative luminosity of the other stars. I took the ratio of the temperatures first, then raised it to the fourth power. The estimates are again limited to two significant figures, since this calculation is based on the peak wavelength estimate made earlier. Below are my relative luminosities for each star.
SAO76803-1.0
HD33278-3.4
HD31084-4.4
HD124320-6.9
FEIGE40- 8.6
HD242908-12
I used a ruler to measure the estimated peak wavelength:
ReplyDeleteSAO76803-6625 angstrom
HD33278-4896 angstrom
HD31084-4500 angstrom
HD124320-4042 angstrom
FEIGE40- 3833 angstrom
HD242908-3650 angstrom
Using Wien's law I calculated the temperature of each star. I used 2 significant figure for the Temperature since I just estimated the peak wavelength of each star
SAO76803-4100 K
HD33278- 5600 K
HD31084- 6200 K
HD124320-6900 K
FEIGE40- 7300 K
HD242908-7700 K
The Luminosity is directly proportional to the temperature raised to 4. The coolest star has a luminosity of 1. I compared the T^4 of each star to the T^4 of the coolest star
SAO76803-1.0
HD33278-3.5
HD31084-5.2
HD124320-8.0
FEIGE40- 10.
HD242908-12.
I used two significant figure for the Temperature since I just estimated the peak wavelength of each spectrum. The luminosity is directly poroportional to the temperature raised to 4th power.
ReplyDeleteSAO 76803
Peak Wavelength:660 nm
Temperature: 4400K
Relative Luminosity: 1
HD 33278
Peak Wavelength:490 nm
Temperature: 5900K
Relative Luminosity: 3.2
HD 31084
Peak Wavelength:450 nm
Temperature: 6400K
Relative Luminosity: 4.5
HD 124320
Peak Wavelength:405 nm
Temperature: 7200K
Relative Luminosity: 7.2
FEIGE 40
Peak Wavelength:380 nm
Temperature: 7600K
Relative Luminosity: 8.9
HD 242908
Peak Wavelength:360 nm
Temperature: 8000K
Relative Luminosity: 11
I also used 2 SF since I only approximate the peak of the emission line.
ReplyDeleteStar Peak Wavelength (angstrom)
SAO 76803- 6600
HD 33278 - 4900
HD 31084 - 4500
HD124320 - 4100
Feige 40 - 3800
HD 242908- 3600
I used Wiens law to compute for the temperature of the stars.
Star Temp (in 2 SF)
SAO 76803 4391K - 4400K
HD 33278 5914K - 5900K
HD 31084 6439K - 6400K
HD124320 7068K - 7100K
Feige 40 7626K - 7600K
HD 242908 8049K - 8000K
Using Stefan-Boltzmann law, I computed for the ratio of T^4 of each star relative to the T^4 of the coolest star
Relative Luminosity in 2 SF (Relative to SAO76803)
SAO 76803 - 1.0 - 1.0
HD 33278 - 3.23 - 3.2
HD 31084 - 4.48 - 4.5
HD124320 - 6.78 - 6.8
Feige 40 - 8.90 - 8.9
HD 242908 - 10.93- 11
My values for luminosity were not calculated correctly. Below are my revised values based on luminosity proportional to T**4.
ReplyDeleteSAO 76803 1
HD33278 3
HD31084 3
HD 124320 5
FEIGE 7
HD 242908 9
From the spectrum, I measured the peak wavelength and used the equation below to get the temperature of the stars:
ReplyDeleteT = 2.9 x 10^6 / wavelength peak in nanometer.
Star Temperature in 2 SF
HD 242908 - (365nm)- 8000 K
FEIGE 40 - (383nm)- 7600 K
HD 124320 - (404nm)- 7200 K
HD 31084 - (450nm)- 6400 K
HD 33278 - (490nm)- 5900 K (sun-like Star)
SAO 76803 - (662nm)- 4400 K
Each star's temperature is raised to the 4th power and then divided each of these Steffan-Boltzmann Law values by that of the coolest star (SAO76803).
Star Relative Luminosity
SAO 76803 - 1
HD 33278 - 3.23
HD 31084 - 4.48
HD 124320 - 7.17
FEIGE 40 - 8.90
HD 242908 - 10.93
I used the blackbody applet to estimate the temperature of each of the stars based upon an estimate of the peak wavelength given in the graphs of luminosity to wavelength.
ReplyDeleteHD 31084,6500 K @ 4400 A; luminosity ratio 5.6
HD 242908, 8200 K ,3600 A ; had the highest luminosity ratio of 16.
Coolest star SAO 76803, 4200 K @6900 A, ratio 1.
FEIGE 40, 7200 K @ 4000 A, ratio 8.5
HD 33278, 5700 K @ 5100 A, ratio 3.3
HD 124320, 7200 K @ 4000 A , ratio 8.5
HD 31084, 6500 K @4400 A, ratio 5.6
I calculated the relative luminosity of each of the other stars to SAO 76803 using the Stefan-Boltzmann Law (luminosity is proportional to the 4th power of the temperature).
I agree with 2 significant figures due to estimation of the wavelength in the applet.
Using Wien's law : T = 2.9E6/wavelength peak.
ReplyDeleteI got the following temperature using 2 SF
Star Temperature
SAO 76803 at 660nm peak – 4400 K
HD 33278 at 490 nm peak – 5900 K
HD 31084 at 450 nm peak – 6400 K
HD 124320 at 400nm peak – 7200 K
FEIGE 40 at 380 nm peak – 7700 K
HD 242908 at 330nm peak – 8900 K
Assigning SAO 76803 to relative luminosity of 1, then I raised each star's temperature to the 4th power and then divided each of these Steffan-Boltzmann Law values by that of the coolest star (SAO76803).
SAO 76803 – 1.0
HD 33278 – 3.2
HD 31084 – 4.5
HD 124320 – 7.2
FEIGE 40 – 9.4
HD 242908 – 17