What is the visual luminosity of the star in solar units?

Assume a normal interstellar extinction law with R(v) = 3.1 and M(v,sun) = 4.83mag. A star with E(B-V) = 3mag and a distance of 1.7kpc has an apparent magnitude of m(v) = 16.4mag.



Find the visual luminosity of the star in solar units.What is the visual luminosity of the star in solar units?R(v) = 3.1 = A(V)/E(B-V)
E(B-V) = A(V)/3.1 or
A(V) = 3.1(E(B-V)
E(B -V) = A(B) - A(V) = (10^0.4)^3 =10^1.2 = 15.8
A(V)/3.1 = 15.8
A(V) = 49 = (10^0.4)^m
0.4m = log 49
m = 4.22 magnitudes fainter due to extinction
So the star would appear 4.22 magnitudes brighter than 16.4 or 12.18 apparent magnitude if not for extinction
12.18 = M -5 + 5* log 1,700
M = 17.18 -5* log 1,700 = 1.03
The star has an absolute magnitude of 1.03
The star is (10^0.4)^(4.83-1.03) = 33.1 times the Sun's brightnessWhat is the visual luminosity of the star in solar units?This looks like undergrad course work, which I will not answer on principle. However I will point you in the right direction.



You know the distance and apparent magnitude so you can find the unextincted apparent magnitude using the R(v) above, which BTW seems far too high. Now once you have this figure use the it and the distance to find the absolute magnitude. It's then a straight forward calculation to change from absolute mag to luminosity.



You must have the equations in your course work, if not Google it or look it up in a text book (I suggest Palla).