Comment on how the pencil appears to move with respect to a background object. Now, repeat this but with the pencil at a different distance from your face. How does the apparent shift of the pencil change?
Send a screen shot or copy of this plot to Chris. Comment on your plot. Do you see a relationship? Compare it with a published HR diagram: for example, see Wikipedia, Hertzsprung-Russell Diagram
Sort the data by B-V. (Remember, this is a measure of temperature.) How does B-V compare with the spectral types in the table? Make a table to express the relation. Make a third column for temperature. How does your table compare with the axes in the published HR diagram from Wikipedia?
Using the distance modulus equation, what distance do you derive?
1) There is a negative relationship between absolute magnitude and B-V color difference. As absolute magnitude decreases (an increase in the y axis), which means that it is brighter the color difference between B-V filters is getting smaller (a decrease in the x axis). In other words the fainter the star is the more difference on the color when B-V filters are employed.
ReplyDeleteWhen I compared my plot with the published Hertzsprung - Russell diagram it is an example of main sequence in which diagram goes from the upper-left (hot and bright) to the lower-right (cooler and less bright).
2) According to the table; as B-V index decreases, temperature increases.
When I compare axes, I conclude that B-V index between -0.3 to 0 corresponds to B star class, 0.01-0.3 corresponds to A star class, 0.3-0.6 corresponds to F star class, 0.6-1.0 corresponds to G star class and 1.01-1.8 corresponds to K star class. I have seen that the range for star classes` B-V index is increasing as the temperature decreases. The range for B, A and F stars are 0.3, the range for G star is 0.4 whereas the range for K star class is 0.8.
B-V Class Temperature
-0.23 Bstar 9500-30000
-0.23 Bstar 9500-30000
-0.19 Bstar 9500-30000
-0.18 Bstar 9500-30000
-0.18 Bstar 9500-30000
-0.18 Bstar 9500-30000
-0.15 Bstar 9500-30000
-0.15 Bstar 9500-30000
-0.13 Bstar 9500-30000
-0.13 Bstar 9500-30000
-0.11 Bstar 9500-30000
-0.11 Bstar 9500-30000
-0.09 Bstar 9500-30000
-0.09 Bstar 9500-30000
0 Bstar 9500-30000
0 Bstar 9500-30000
0.03 A star 7000-9500
0.03 A star 7000-9500
0.1 A star 7000-9501
0.13 A star 7000-9502
0.13 A star 7000-9503
0.13 A star 7000-9504
0.14 A star 7000-9505
0.18 A star 7000-9506
0.21 A star 7000-9507
0.28 A star 7000-9508
0.31 F star 6000-7000
0.34 F star 6000-7000
0.43 F star 6000-7000
0.45 F star 6000-7000
0.46 F star 6000-7000
0.49 F star 6000-7000
0.62 G star 5200-6000
0.71 G star 5200-6000
0.72 G star 5200-6000
0.72 G star 5200-6000
0.76 G star 5200-6000
0.79 G star 5200-6000
0.8 G star 5200-6000
0.88 G star 5200-6000
0.88 G star 5200-6000
0.92 G star 5200-6000
1 G star 5200-6000
1 G star 5200-6000
1.06 K star 3900-5200
1.09 K star 3900-5200
1.17 K star 3900-5200
1.36 K star 3900-5200
1.41 K star 3900-5200
1.41 K star 3900-5200
1.44 K star 3900-5200
1.5 K star 3900-5200
1.55 K star 3900-5200
1.6 K star 3900-5200
1.74 K star 3900-5200
3) For HD 331246, B value is 9.0 and V value is 9.0.
Known:
B-V index will be 0 which is class B star according to my Hertzsprung-Russell diagram.
Absolute magnitude (m) = V=9.0
B-V index=0
Mv=0.2 which is derived from the graph.
Formula: V − Mv = 5log10(d ) -5
9.0-0.2=5log10(d ) -5
8.8=5log10(d ) -5
3.8=5log10(d )
0.76= log10(d )
d= 10^0.76 parsec
d= 3.26x10^0.76 ly
Parallax Experiment results
ReplyDeleteFirst, my arm was extended, my right eye lined up with the pencil and light switch. Next, I closed by right eye and opened my left eye. The pencil moved to the right from my point of view. Then, I bent my arm, my right eye lined up with the pencil and light switch. Next, I closed my right eye and opened my left. The pencil moved to the right further than before. I repeat the experiment for a third time with the pencil closer to my face. The pencil move right even further.
Parallax Experiment: When my arm was fully extended there was a shift in the apparent position of the pen I was holding. When I moved the pen closer the pen seemed to shift a greater distance. As distance decreased the apparent shift increased. Or, as distance increased the apparent shift decreased. So, there is an inverse relationship between these variables.
ReplyDeleteWhen I held my index finger out as far as the length of my arm so that it covered the door jamb with both eyes open, then with the left eye closed the finger shifted only slightly to my right. When the right eye was closed, the finger shifted a greater distance to the right. When I bent my arm to approximately one half of its length, then the shift was greater (to my right).
ReplyDeleteIn this first plot of the nearby stars of B-V vs Mv there does not appear to be a relationship because the independent variable is not defined until the B-V numbers are sorted.
By sorting by the B-V color then we can see a relationship between color and absolute magnitude. The brighter the star, the hotter it is. When the scale of the Y-axis is increased, then this table compares to the diagonal line of the main sequence stars adequately in spite of many outliers.
I selected the star HD331241 to calculate the distance in parsecs using the distance modulus equation. It was a F5 D star so I estimated its absolute magnitude from the HR diagram on wikipedia to be 1. This choice seemed like it might have less error as in this range the plot was slightly horizontal around an absolute magnitude of 1.
Thus V= 10.73. 10.7-1.0 = 9.7
9.7 +5= 5log (d)
14.7/5 = log (d)
2.94 = log (d)
0.496 parsecs = distance
After, adventures in reading about luminosity and playing with the blackbody simulator, I arrived at a solution for the determination of the distance. The star used for the determination was HD331066. It has a b=3.93 and v=3.70 and the difference b-v is 0.23. I was also able to obtain the same value for b-v using a plot of temperature versus b-v. The b-v values were obtained using the blackbody simulator for b and v and then subtracting the values. The temperature of HD331066 was 7400K and this was used to find b-v from the graph. WOW! The value from the graph was the same for b-v as the database value. The b-v was used to located Mv on the HR Diagram. The value was between 1.5 and 2.0. Using the distance modulus formula the distance was between 21.9 pc and 27.5 pc. The values on my table were not as useful as the graph of temperature vs b-v.
ReplyDeleteHD 331085 T = 4800K, Class K (SIMBAD Class K B 10.47 V 9.12)
ReplyDeleteB-V: 10.47-9.12 = 1.35
9.12 – 1.3 = 5 log10(d) – 5
12.82/5 = log10 (d)
2.564 = log10 (d)
0.409 parsecs = d
Correction:
ReplyDeleteB-V: 10.47-9.12 = 1.35
9.12 – 8.2 = 5 log10(d) – 5
5.92/5 = log10 (d)
1.184 = log10 (d)
15.3 parsecs = d
When I closed my left eye and then opened the right eye, the shift of my index finger’s position was to the right. And then I tried the other eye and I observed that the shift of my finger’s position was to the left. Then I tried the same procedure with my finger closer to my eyes. I observed a greater shift of my index finger’s position. I can say that the relative shift is greater for a closer object relative to the observer than the farther object
ReplyDeleteI used the simulator to determine the temperature of each star. I tabulated all my data and obtained the following temperature ranges:
Class B stars: 11600K – 12500K
Class A stars: 7800K – 9400K
Class F stars: 5800K – 7000K
Class G stars: 4200K – 5300K
Class K stars: 3400K – 4300K
Class M stars: 3000K – 3100K
After correlating all the data, I notice that the more negative is the B-V the greater is the temperature of the star and the more positive is the B-V the cooler is the star.
After plotting B-V vs Mv, I obtained a graph similar to Hertzsprung–Russell diagram .
Using the modulus equation, the distance of the star HD331085 is 6.31 parsecs or 20.5809248 light year
Closing my left eye then the right eye will give me a shift of any obeject's position to the right and closing the right eye then the left will give a shift of position to the left. The closer the object to the obeserver the greater is the shift.
ReplyDeleteFrom the Excel Data and the simulator in module 2, I got the following temperature range
.
B 10,000 - 15000K
A 7,500 - 10,000 K
F 6,000 - 7,500 K
G 5,000 - 6,000 K
K 3,500 - 5,000K
M < 3,500 K
The distance of star HD331054 is 8.87156012037961 parsecs (28.9358022 lightyear)
My graph looks like the HR diagram.
The vertical axis represents the star’s luminosity or absolute magnitude.
The horizontal axis represents the star’s surface temperature (not the star’s core temperature. On the diagram, the higher (hotter) temperatures are on the left, and the lower (cooler) temperatures are on the right.
A star in the upper left corner of the diagram would be hot and bright. A star in the upper right corner of the diagram would be cool and bright. The Sun rests approximately in the middle of the diagram. A star in the lower left corner of the diagram would be hot and dim. A star in the lower right corner of the diagram would be cold and dim
The apparent image shifting in the direction of the closed to opened eye. When I closed my left eye then opened the right, the image shifted to the right. As I did the opposite procedure, the image shifted to the left.
ReplyDeleteThe relative amount of shift was greater for a close object than a far one.
After some data analysis, I got the following:
B 10,500 to 15,000
A 7,400 to 10,500
F 6,000 to 7,400
G 4,700 to 6,000
K 3,500 to 4,700
M <3,500
My graph resembles the shape of the HR diagram. The majority of stars fall along a curving diagonal line called the main sequence but there are other regions where many stars also fall.
For Star HD331059, its distance using modulus equation V-Mv=5log10(d)-5
ReplyDeleteis 3.63 parsecs or 11.8397396 lightyears
When the pencil is further away, it does not appear to shift as much with respect to the background when I open one eye and close another.
ReplyDelete1) The plot of Mv vs B-V shows that as the absolute magnitude value increases (or the dimmer the star), the B-V value increases. The Hertzsprung-Russell diagram shows the same pattern, it just has the axes flipped from what I plotted in Excel
2) Negative B-V values correlate with class B stars, which are relatively hot. The lower B-V values correlate with higher temperature star types, such as Class B and Class A. Higher values correlate with cooler star types, such as Class K and Class M. This matches the Hertsprung-Russell diagram very nicely.
3) Star HD 331078. B value is 10.39 and V value is 10.05. This star has a B-V value of 0.34, This should correspond with an absolute magnitude of approximately 2.5 on the Hertzsprung-Russell diagram. Using the distance modulus equation,
10.05-2.5=5log10(d) -5
2.51 = log10(d)
324 parsecs = d
Correction:
ReplyDeleteFrom Simbad
HD331085 is a K0 star
Its B value is 10.47 and its V value is 9.12
Its B-V value is 1.35 and absolute magnitude is approximately equal to 8.2
Using modulus equation:
9.12 - 8.2 = 5log10(d)-5
d = 15.275660582380723 parsecs or 49.8236485 lightyears
Parallax is an apparent displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines. In the activity described in module 6, the left and right eye correspond to the two different line of sights. When I close my left eye and open the right eye, the shift of the pencil's position is to the right. And if I close my right eye and open the left eye, the shift of the pencil's position is to the left.
ReplyDeleteThe closer the pencil to my face the greater is apparent displacement and the farther the object to the observer the lesser is the apparent displacement
The plot I had is similar to the HR diagram. From the B-V, I used the simulator to obtain the temperature of each star. These stars are classified from B-A-F-G-K-M (hottest to coolest).From the graph, I noticed that the more negative is the B-V the greater is the temperature.
For the star HD331241:
B= 11.30 and V= 10.73
B-V=0.77
Apparent magnitude = 10.73
Absolute magnitude= 4.6
Using the distance modulus equation, the distance of the star HD331241 is 168 parsec or 549 lightyear
Correction on my distance calculation for HD 331078: With a B-V value of 0.34, I am estimating an absolute magnitude of 3.0. Using the distance modulus equation,
ReplyDelete10.05-3.0=5log10(d) -5
2.51 = log10(d)
260 parsecs = d
The easiest way to see parallax is to do this activity. I closed my right eye and noted where on the background object the image of my finger appeared. Then I opened my right eye and closed the left one without moving my finger. The image of my finger appeared to have jumped to a different spot on the background wall.
ReplyDeleteIt is possible to tell how far away your finger is from your eyes just by measuring the distance between your eyes and the distance that the image appeared to move in degrees of arc.
The closer is the object from the observer the greater is the apparent displacement.
For HD 331055 (class F8 star)
B=10.97 V=10.39
B-V = 0.58
Absolute magnitude = 2.6
Apparent Magnitude = 10.39
using distance modulus equation, HD331055 has a distance of 361.41 parsec or 1 178.79 lightyear
When I shut my left eye then open the right, the pencil moves to the right. If I shut my right eye then the left, the pencil moves to the left. The closer the pencil to the eye the greater is the shift in position
ReplyDeleteFrom B-V values, I used the simulator and set the value of B-V for each star then I tabulated my temperature in another column. These are my temperatures:
Temp Spectral Class
9750 A1
9640 A2
8250 A3
8260 A4
8160 A5
7550 A6
7800 A7
12500 B0
14700 B1
13200 B2
14700 B2
13500 B2.5
13200 B3
12500 B3
13200 B3
12000 B5
12000 B7
11200 B8
11600 B9p
6830 F0
6640 F2
6130 F2mF5
6030 F6
5840 F6
4670 G1
4960 G5
4920 G8
4790 G8
4140 G9
3930 K0
3770 K0
4440 K2
3290 K3
4000 K5
3210 K5
3420 K7
3140 M0
3340 M1
3070 M3
3000 M4
The spectral classification is in this order B-A-F-G-K-M (increasing temperature)
My scattered graph is similar to HR diagram.
From Simbad:
HD331085 is a K0 star. Its B=10.47 and V=9.12. The absolute magnitude is 8.2 and apparent magnitude is 9.12 Using the formula for modulus equation, the distance of this star is equal to 15.28 parsec(49.8 ly)
The pencil has a greater shift when I move it closer to my eyes. The shift of the pencil is to the right when I close my left eye and open the right eye. The shift will be left when I close my eye right eye and open the left eye.
ReplyDeleteB-V Temperature Range
-0.23 to -0.09 10,500 to 15,000 (B star)
0 to 0.21 7,400 to 10,500 (A star)
0.28 to 0.49 6,000 to 7,400 (F star)
0.62 to 0.8 4,700 to 6,000 (G star)
0.79 to 1.36 3,500 to 4,700 (K star)
1.41 to 1.74 <3,500 (M star)
The more negative is the B-V the hotter is the star. My plot is similar to HR diagram. This diagram is a plot of luminosity (absolute magnitude) against the colour of the stars ranging from the high-temperature blue-white stars on the left side of the diagram to the low temperature red stars on the right side.
From SIMBAD Query:
HD331246 (B3V star)
B=9.00 V=9.00
Apparent magnitude: 9
Absolute Magnitude from HR diagram = 0.2
Distance= 575 parsec