Homework Day 8, Sept 21
1) NGC 7027 has a distance of about 920 pc and diameter of about 14
seconds of arc

a. How does its diameter compare with its thickness (from previous
homework)?

    Diameter D = θ(radians) × Distance d
               = 6.7787×10⁻⁵ radians × 920 pc
               = 0.0624 pc.

    This is very comparable to the result from HW7 (0.617). Well within 10%.


b. Its shell is expanding at about 20km/s. How old is it? (how long
ago was the shell ejected?)

    age t = (0.0624 pc) / (20 km s⁻¹ × 3.24×10⁻¹⁷ pc / 10⁻³ km) 
            × (3.17×10⁻⁸ years / 1.00 s)
          = 4.50×10⁷ years.


2) The Orion nebula is about 400 pc distant and has an apparent diameter of about 60 minutes of arc.

a. What is its physical diameter?

    D = 6.98 pc.

b. The gas density is about 10 4 cm -3 and its temperature is about 10 4 K. The turnover frequency is obvious on Wilson’s SED plot. How thick is the H + layer along our line of sight?

    turnover ν = 500 MHz.

    At turnover, τ = 1.

    Gaunt factor g = 10.6 + 1.9 log(T) - 1.26 log(Z ν)
                   = 7.24.

    κ′ₛ = 0.178 Z² g ν⁻² T^(-3/2) nₑ nₕ
        =  5.15×10⁻¹⁶ cm⁻¹.

    Thickness l = τ / κ′ₛ.



3) The Perseus cluster of galaxies is about 73.6 Mpc distant and has an angular diameter of 860 arc minutes. The total mass in stars is about 2×10 13 M sun and the virial mass is about 10 15 M sun . It is one of the strongest x-ray sources in the sky, with L x = 4x10 45 erg/s. X-rays come from 5 million Kelvin brems.

a. Use total luminosity to find the emission measure.

    L = 4π j V.

    4π j = (From AGN3) 1.42×10⁻²⁷ Z² √T g[ff] n².

    I will adopt an average g[ff] = 1.3.

    Emission measure n² V = L / (4π 1.42×10⁻²⁷ Z² √T g[ff])
                          = 7.71×10⁶⁷ cm⁻³.




b. Use diameter to find V

    Spherical V = 4/3 π r³
                = 4/3 π (2.84×10²⁵ cm)³
                = 9.61×10⁷⁶ cm³.

c. Solve for the density

    n² = 7.71×10⁶⁷ cm⁻³ / 9.61×10⁷⁶ cm³
       = 8.03×10⁻¹⁰.

    n = √n² = 2.83×10⁻⁵ cm⁻³.

d. What is the total mass in hot gas?

    M = μ n V

    μₕ = 1.67×10⁻²⁷ kg × 2.83×10⁻⁵ cm⁻³ × 9.61×10⁷⁶ cm³
       = 2.29×10¹⁵ Mₛₒₗ.