Great Videos. Helps me study for Science olympiad astronomy
@arindammukhopadhyay43593 жыл бұрын
Think of Brightness suffix B ( Blue Light Filter) i.e. b suffix B and Brightness suffix V ( Visible Light Filter) i.e. b suffix V. In the Apparent Magnitude Scale increase in Negative Number means increase in Brightness. While using B Filter, if in the Telescope with Photo meter more Blue Light enters for this particular wavelength then this will give large Negative m compared to observing the same object with Visible Light Filter. So V - B result will be Negative. From this we can conclude the object is emitting more Blue Light compared to Visible Light. More Blue Light means object is emitting lesser wavelength photons in comparision to higher wavelength photons. As frequency is inversely proportional to wavelength the the Core of the object is so hot that it is emitting high frequency photons which contains much energy than Visible Light photons. Thus, also the Surface Temperature of the object is high relative to a object which emits more Visible Light Photons. Here, as an example Sirius is taken. As compared to our Sun Sirius emits more Blue Lights. So for Sirius the B - V is a small number. For Sun, the B - V value is greater than B - V value of Sirius. As Sirius emits more Blue Light in comparision to our Sun the Core and also the Surface Temperature of Sirius is higher than Sun. In this context I am sure that when these High Energy Photons collide with the Hydrogen Atoms in the Photosphere of Sirius they got absorbed and from the above concept it can be concluded that in the Absorption Spectra of Sirius the H-Gamma and H-Delta lines are much prominent in comparision with our Sun's H-Gamma and H-Delta lines. (Layman's Explanation)
@Physics-Learner-fs8xd8 ай бұрын
michel does this require a formula to find the temperature of a star or this is you explaining it by words?
@MichelvanBiezen8 ай бұрын
The temperature of a star can be determined by measuring the peak of the blackbody radiation curver of the star. This peak is found by observing the intensity of the light of the star as seen through several filters. Once this peak wavelenght if found the temperature in Kelving is: T = (0.0029) / wavelength of the peak
@MichelvanBiezen8 ай бұрын
You can find the temperature of a star using Wien's law T = 0.0029 / wavelength of the peak of its radiation.
@arnaskazakevicius89556 жыл бұрын
Hello! I am facing a problem and I am wondering if you could help me. I need to find how much nova's radius would increase if it's brightness gets 10 times bigger. Hope you can help me :)
@MichelvanBiezen6 жыл бұрын
If the brightness only depends on the size and not on the surface temperature, then since luminosity is proportional to the Area, then the radius would be sqrt (10) larger.
@arnaskazakevicius89556 жыл бұрын
@@MichelvanBiezen Thank you :) really helped!
@transtemporal_artist3 жыл бұрын
We can determine the temperature of a star using weins law by first taking observation using the filters then sketching the curve of a black body and finding the peak wavelength right ???? Very helpful videos ☺️
@MichelvanBiezen3 жыл бұрын
That is correct. Also based on the intensity of the light through several filters and then looking those number up on a table. (we have a video on that as well)
@transtemporal_artist3 жыл бұрын
@@MichelvanBiezen yeah i have watched all the videos .... One in which you used a table with B-v index with tempratures ..... But we would require to refer to it when calculating and i suppose for exam purposes that must not be suitable. Anyways thank you so much for responding 🙏☺️❤️
@0219-n1e5 жыл бұрын
I think it is better to say negative numbers instead of small number.