Recent content by astronomystudent

3.) m = -.54137808 Here are the other questions: 4.) If the Hubble Space Telescope measure a parallax angle of 0.0217 arcsec for a given star, how far away from us is it in parsecs and light-years? 5.) If a star has a mass seven times greater than our Sun, what is the luminosity of the...

No, sorry I typed it wrong. It's supposed to be: 1705. Thanks for catching that. Here are the rest of my answers, are the others correct? 1.) distance - 11.17 parsecs 36.41 light -years 2.) m = 1.39 3.) M = 22.86 4.) distance - 46.08 parsecs 150.302 light-years 5.) 2401 times...

1. Given that Arcturus has an apparent magnitude of -0.06 and an Absolute magnitude of -0.30, calculate its distance in parsec in light-years. Answer = 11.17 parsecs & 36.4 light years 2. Given that Sirius has an apparent magnitude of -1.47 and a distance of 2.67 parsecs, what is the...

5.) Here is the equation I found to possibly work this problem that compares the two star/sun: F2/F1 = 100 ^(m1-m2)/5

5.) I don't understand the brightness idea and how to figure that out. I see that my past answer doesn't make sense. But I don't know what else to do. Should I multiply the 50 by the brightness of the sun. I checked my book and didn't find an answer for brightness of the sun or a formula...

ANSWERS: 4.) When I looked up a table on this website (listed below) it told me that it would be spectral class: A and that it would have an apparent magnitude of 2.0 - 3.0. http://outreach.atnf.csiro.au/education/senior/astrophysics/spectral_class.html#specclasstable 5.) I still have no...

Could someone please help me finish these problems?

I turned these questions in but I am still interested in figuring out how to solve them.

These problems are due in an hour and I am still have no answer as to whether I am done with problem three or what I am to do now once I have completed Wien's Law on problem 4. I am working on problem 5.

Answers: On the previous page you haven't seen those answers. But I think I did number 5 wrong maybe it should be: 5.) m = 5 M = 4.83 5 - 4.83 = 1.7 x10^-1 (1.7 x10^-1)/5 = 3.4 x 10^-2 10^3.4 x 10^-2 = 1.08 x 10^0 distance = 1.08 x 10^0 parsecs i am still stuck on what would it be...

Am I done with problem 3 or do I still need to look for an equation to figure out how bright it appears from Earth? I am confused on that point, sorry.

ANSWERS 2.) Different equation so the units would equal yrs. I found it in my notes and it also says that M^4 = L so I guess I will use that thus: L = (11)^4 L = 1.4641 x 10^4 W L = (0.09)^4 L = 6.561 x 10^-5 W Lifetime Equation Lifetime = [1/mass^4] (1 x 10^10 yrs) L = [1/1.4641 x...

ANSWERS: 2.) The units would be (years), I think. And I checked my my on the lifetime of the star with the mass of 0.09. I still got the 4.115226337 which in scientific notation would be 4.12 x 10^2. 3.) 5log(d/10pc)+M=m 5log(6.66x10^1/10pc)+3 = m m= 7.12 x 10^0 Does little "m" solve...

ANSWERS: 2.) L = M^3.5 L= (11)^3.5 L = 4414.427596 L = 4.41 x 10^3 L = M^3.5 L = (0.09)^3.5 L = 2.187x10^-4 I think this is the equation for what it's lifetime would be on the main sequence. T = M/L T = 11/4.41 x 10^3 T = .0024918293 T = M/L T = (0.09)/2.187 x 10^04 T =...

ANSWERS: 1.) L=R^2 * T^4 L = (10)^2 (7/10)^4 L = 100 (.2401) L = 24.01 The Luminosity is 24.01 times greater than the Sun's. 2.) L = M^3.5 L= (11)^3.5 L = 4414.427586 The star's luminosity is about 4.14 x10^3 times that of the Sun's. L = M^3.5 L = (0.09)^3.5 L = 2.187 x 10^-4...