At what temperature is the light in the B filter equal to the light in the R filter (i.e. the peak of the curves are the same height)? What is the lowest temperature at which U>B>V>R?
Which star is brighter in the B image? Which is brighter in the R image?
Send a screen shot of the 4 profiles to Chris Martin and post your analysis of which star is the hotter, A or B, on the blog.
At 5720K, the light passing through R filter is same as the light passes through V filter.
ReplyDeleteAt 9410K, light passing through the filters will be ordered as U›B›V›R.
In R image Star B is brighter and in B image Star A is brighter.
Which one is hotter?
Star A will be hotter than Star B. As the light passing through R filter increases, star has lower temperatures. In the same manner, as the light passing through B filter, star has higher temperatures. According to my analysis, I found that Star A is brighter when B filter is used which means that it is hotter than star B. And when R filter is used Star B is observed brighter which means that it is cooler than Star A.
The temperature at which B = R is approximately 5750K. The lowest temp at which U>B>V>R is 7620K.
ReplyDeleteIn B image,Star A is brigter and Star B is brighter in R image. From the profile(blue filter) , Star A and B has a peak at 11261 counts and 8989 counts respectively. In the red filter, Star A and Star B has a peak at 20356 counts and 61433 counts respectively.
Computing for the ratio of blue to red: star A has 0.6 and Star B has 0.1.
Therefore, Star A is hotter than Star B
The R filter will have the same peak as B filter at 5720K and the lowest temperature at which U>B>V>R is approximately equal to 7610K.
ReplyDeleteusing the DN values: star A is brighter than star B in blue filter and star B is brighter than star A in red filter.
Using the profile analysis: In the blue filter: star A has a peak at 10224 and star B has a peak at 8170
In the red filter: star A has a peak at 17401 and star B has apeak at 57930. Taking the ratio of the blue to red peak values. I found out that star A is hotter than star B
The temperature at which B=R is approximately 5750K. The lowest temp at which U>B>V>R is approximately 7620K.
ReplyDeleteStar A is brighter than Star B in blue filter. Star B is brighter than Star in A in red filter.
In blue filter, Star peak is at 9958.69 counts and star B peak is at 8631.48 counts.
In the R filter,Star A peak is at 18375.25 and star B peak is at 59801.61 counts.
Taking the counts ratio of the blue to red filter, we can get the relative temperature of the stars. Star A has 0.5 and Star B has 0.1. From this, I can say that Star A is hotter than Star B.
The B/R ratio of star A is 0.6 and B/R ratio of star B is 0.2. From these ratios, I can say that star A is hotter than star B
ReplyDeleteAt 5730 K, the peaks of blue and red are approximately equal. 7630 K is the minimum temperature at which U>B>V>R.
ReplyDeleteIn the B image, Star A appears to be a little brighter. In the R image star B appears to be significantly brighter
From these images, it can be determined that Star A is hotter than Star B. Star A is giving off significantly more blue light than Star B. Previous experimentation with the blackbody simulator showed that as the temperature increased, the peak frequency of light emitted shifted towards shorter wavelengths. From the profiles, it can be seen that the ratio of blue light to red light emitted is higher for Star A than it is for Star B.
My blue to red ratios for each star are as follows:
ReplyDeleteStar A
B/R
10800/20873 = 0.52
Star B
B/R
8268/ 53440 = 0.15
It appears that B is equal to R at T=5730 K and U>B>V>R is at T=7630 K.
ReplyDeleteBy just looking on the image, star A is a little bit brighter thank star B in the blue filter. But in the red filter I can easily say that star B is brighter than Star A
I tried to have a different way in finding out which star is hotter. Using the values from the profile analysis and the color index formula, I got these measurements:
Using the Blue Filter: Star A peak at 11432 cts and Star B peak at 9363 cts
Using the Red Filter: star A peak at 20727 cts and Star B peak at 56000 cts
using the index formula: star A is 0.6 (6350K)
and star B is 1.9 (3710K)
using the B/R ratio: star A is 0.6 and star B is 0.2
Both of the these two methods show that Star A is hotter than star B
B equals R at the temperature equal to 5710K and U>B>V>R at T= 7610K
ReplyDeleteStar A is brighter than Star B in Blue filter
Star B is brighter than Star A in Red filter.
In the blue filter, star A’s highest peak is 11443.26 and star B’s highest peak is at 9189.91
In the red filter, star A’s highest peak is 18943.67 and star B’s highest peak is at 59500.75
To get the relative temperature of the stars, I computed for the ratio of B/R: star A is 0.6 and star B is 0.2
Therefore I conclude that Star A is hotter than Star B
This comment has been removed by the author.
ReplyDeleteThe temperature at which B equals R is 5740K. The lowest temperature for the four filters where U>B>V>R is at 9540K where U:2.79 > B:2.77 >V:2.76 > R:2.75.
ReplyDeleteI observed that in R filter, Star B is brighter than star A and in the B filter, star A is sligtly brighter than Star B.
In the blue filter: Star A peak is at 11320.16 counts and Star B peak is at 6703.72 counts
In the red filter: star A pea is at 16659.36 counts and star B peak is at 60126.65 counts
computing for the ratio of B to R, Star A has 0.7 B/R ratio and star B has 0.1 B/R ratio. I can say then that Star A is hotter than Star B.
OOPs. I submitted my reply for this assignment in the previous assignment.
ReplyDeleteB = R at 5720 K
ReplyDeleteThe lowest temperature at which U>B>V>R is 7670 K
In the “b” image the “A” star is brighter, in the “r” image the “B” star is brighter
From the image analysis the b/r ratio is higher for the A star than the B star. The A stars relative “blueness” is greater than that of the B star. Since Wein’s law says that bluer stars are hotter then it only makes sense that a star with a greater b/r ratio is hotter, thus the A star is hotter in this case.
This is the data measured using ImajeJ for M26 using r and b filtered images for brightness:
ReplyDeleteStar A B
red filter 23,113 61,187
blue filter 12,634 9,519
b/r ratio 0.547 0.156
My answers to the previous questions were not the same as some posted. I will redo my work.
The temperature at which the B = R values is 5730 K and the temperature at which U>B>V>R is 7630 K.
ReplyDeleteIn the ref filter, Star B is brighter than Star A and in the blue filter, star A is a little bit brighter than Star B.
From the profile graphs (blue filter):
Star A has a peak of 11332 counts and star B has a peak of 9354 counts
From the profile graphs (red filter):
Star A has a peak of 16465 counts and star B has a peak of 59761 counts. Taking the ratio of b/r . Star A has 0.688 and star B has 0.157
Star A is hotter than Star B