Well, the entire image is a 90" square, but V838 doesn't extend all the way to the edges, so both I and the key guesstimated that it was about 70" in diameter.astro124 wrote:AlphaTauri,
Why did you say for problem 2b, it was 70 arcseconds instead 90 arcseconds (the image says 90).
And yes, you could easily answer for distance in LY or pc if you had the linear diameter in LY/pc as well, but I used meters because it was the most convenient unit (since I calculated the expansion time in seconds, and I know the speed of light in m/s).Also, wouldn't there be an easier way of just using LY or parsecs to put your answer in (I'm not talking about converting once you get your solution at the very end, but rather converting at the very beginning).
I think you meant 1d? It's a LRT (Luminosity-Radius-Temperature) problem; the equation isAlso, for the problem 2d , do you know why the instructor squared the radius and temperature?
Okay, so the period of a Cepheid varstar is related to its luminosity (more luminous Cepheids have longer periods). The exact luminosity of a Cepheid can be calculated by the Period-Luminosity Relationship. Once you know the luminosity of the star, you can work out its absolute magnitude - or, some forms of the P-L Relationship directly spit out absolute magnitude. Once you know absolute magnitude, you can take the star's apparent magnitude from your observations and use the distance modulus to work out how far it must be.Finally, how do use Cepheids to calculate distance.
Hope that helps! Feel free to ask if you have any more questions :]