What is a cepheid's distance if its brightness is 4 x 10^-12 watts/m^2?

Question

its period is 5 days

Answer 1

Hmm....well, since Cepheids are variables it is important to know which part of its 5 day cycle that brightness measurement was taken during. This is rather important since a cepheids brightness can vary by a factor of 4, so off the bat your estimate could be a factor of 2 off (inverse square law, right?)


Assume for now that is the peak brightness. First convert this energy flux value to the apparent magnitude. (You should have an equation for this someplace if this problem was assigned to you, or you can look it up online or extrapolate from this plot: http://physics.lakeheadu.ca/courses/Astro/2330/Propweb/apmagdef.gif)

Once you get that go to your handy "cepheid period to absolute magnitude" conversion chart (there is a plot here http://www.astro.lsa.umich.edu/Course/MMSS/Interactive/Ex1.4/ , but again you should have something like this at your disposal already; otherwise, extrapolate from the plot near the bottom of that page).

Now you use that old chestnut of the magnitude distance relationship,

d=10^((m-M)/5+1)

where m is the apparent, and M is absolute magnitude.

Answer 2

-8x +10 = -4(x+3) + 6 -8x +10 = -4x - 12 + 6 (distributed the damaging 4) -8x + 10= -4x - 6 (mixed -12 and 6) -8x + 4 = -4x (subtracted 6 from the two side) 4 = 4x (further 8x to the two side) a million=x (finally, only divide by utilising 4)