Recent content by Thomas Smith

I thought that in regards to the number of hydrogen atoms you needed to include it

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...

Thank you for your reply, what units is my answer in? mH ?

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...

Ah sorry, i missed out the inclination. It says the orbital plane is inclined to the plane of the sky by 90 degrees

Well considering the separation of 0.04nm every 26 years, I believe the full orbit period is 52 years.

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.

I'm unsure about the brackets in the equation. Is this correct? ## \sqrt{((a/b)^2 + (x/y)^2)}## Or should it be ## \sqrt{(a/b)^2 + (x/y)^2}##

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

∆d is 500 parsec sorry

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 [/B] The uncertainty on log10(d) is given by Δ(log10)≈0.4343 Δd/d ΔQ) =...