mirror of
https://asciireactor.com/otho/as-500.git
synced 2024-11-22 18:55:05 +00:00
79 lines
2.3 KiB
Plaintext
79 lines
2.3 KiB
Plaintext
|
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.
|