Firstly i worked out the scale factor of the universe
R(t)/R(t0) = 1/1+z = 1/1+11.1 = 1/12.1 = 12.1^3 = 1/1772
The distance between the galaxies were 12.1 times less than today and the volume was 1772 times smaller than today.
Then I think the average density in the universe at that time is...
Here, i have provided the description of the mean molecular weight from my course notes as well as my workings.
I'm not sure about my values for X and Y. It states that X=1 if all of the hydrogen is ionised which i think the 1 represents 100% hydrogen but the core is also made up...
Distance is d=1/0.07 = 14.3 parsecs
The Doppler shift of one star is, Δλ = 512 - 512.04 = -0.04
So the radical of the velocity of the star is = (-0.04/512) x (3.00 x 10^5 km/s) = 23.4km/s which is the same for both stars because they have the same mass.
This is as far as I've got.
Homework Statement
I need to work out the errors of my corrected apparent magnitudes.
The Colour Excess is 0.36 with uncertainty = 0.01
The star was observed in two wavebands.
r-band = 2.285
g-band = 3.303
The 2.285 and 3.303 are the ratios R_V for the SDSS (Sloan digital sky servery) r and...
I've did my calculations again and got the absolute magnitude uncertainty of 0.21.
I did 0.4343 x 500/6300
∆log(d) = 0.0345
∆5log(d) = 0.0345 × 5 = 0.1725
∆M = sqaureroot of (0.033030)^2 + (0.1725)^2
∆M = 0.20553 rounded up to 0.21
Homework Statement
I have a star that has an apparent magnitude of 13.73 with uncertainty of 0.03303
It's distance Modulus is 13.9967 so it's absolute magnitude is -0.26
The distance is 6300 parsecsHomework Equations
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The uncertainty on log10(d) is given by
Δ(log10)≈0.4343 Δd/d
ΔQ) =...