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Solution Manual for Stars and Stellar Processes 1st Edition by Guidry

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Solution Manual for Stars and Stellar Processes 1st Edition by Guidry










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Uploaded on
February 5, 2025
Number of pages
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Written in
2024/2025
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1 Some Properties of Stars


1.1 Sirius is 8.6 ly = 2.6 pc away, corresponding to a parallax angle p = 1/d( pc) = 1/2.6 =
0.38′′ .
1.2 The absolute magnitude M and apparent magnitude m are related by Eq. (1.7),
 
d
m − M = 5 log ,
10 pc
so the absolute magnitude is
19.3 pc
   
d
M = m − 5 log = 9.7 − 5 log = +8.3.
10 pc 10 pc
The current apparent magnitude m1 and apparent magnitude in a million years m2 are
related by Eq. (1.3),

d2 2
  "  #
f1
m2 = m1 + 2.5 log = m1 + 2.5 log ,
f2 d1

since fluxes f vary as the inverse square of the distance d. Assuming a closest approach of
1 ly in a million years the apparent magnitude will be about 0.7. (Some estimates are that
Gliese 710 may pass within 0.2 ly of Earth, in which case the apparent magnitude would
decrease to −2.8.) The radial velocity may be estimated roughly as −18.9 km s−1 if the
star must cover 63 lightyears in about a million years. From the Simbad database [2] the
actual radial velocity is −13.8 km s−1 .
1.3 From Eq. (1.3) for an actual distance d1 and reference distance d2

d2 2
  "   #
f1
M = m + 2.5 log = m + 2.5 log ,
f2 d1

since fluxes f vary as the inverse square of the distance d. Taking d2 = 10 pc,

10 pc 2
"  #
M = m + 2.5 log
d1
10 pc
   
d1
= m + 5 log = m − 5 log ,
d1 10 pc
which is Eq. (1.7).
1.4 The total energy flux F emitted by a blackbody at temperature T can be expressed as
1
Z ∞
F = 2 π c2 h dλ .
0 λ 5 (ehc/λ kT − 1)
1

, 2 Some Properties of Stars



Using the substitutions
−hc
u = hc/λ kT λ = hc/ukT dλ = du,
u2 kT
the integral can be evaluated giving
4 Z ∞ u3

kT
2
F = 2π c h du
hc 0 eu − 1
2π c2hk4 T 4
= ζ (4)Γ(4)
h 4 c4
2π 5 k 4 4
= T ≡ σ T 4,
15h3c2
where the properties of the Riemann zeta function ζ (x)
1
Z ∞ x−1
u
ζ (x) ≡ du
Γ(x) 0 eu − 1
and the gamma function Γ(x) have been used, and σ is the Stefan–Boltzmann constant.
1.5 For parallax of 0.02′′ the distance is d = 1/0.02 = 50 pc, corresponding to a volume
4
V = π (50 pc)3 ∼ 5.2 × 105 pc3 .
3
Multiplying by an average density of 0.08 stars pc−3 then gives about 4.2 × 104 stars in that
volume. If instead one can measure parallax to 0.001′′ the accessible distance increases to
1000 pc, the accessible volume to ∼ 4.2 × 109 pc3 , and multiplying that by the average
density of stars gives about 3.4 × 108 stars accessible to parallax distance measurements.
Repeating the same analysis for GAIA, parallax distances for 2.7 × 1015 stars could be
measured in principle with a parallax resolution of 5 × 10−6 arcsec.
1.6 The distance in parsecs is given by d = 1/p, where p is the parallax angle in seconds
of arc. Then d = 1/0.286 = 3.5 pc = 11.4 ly, since there are 3.26 lightyears in a parsec.
1.7 From the following diagram,

m1 v2
F1 F2

v1 r1 r2
CM m2


circular motion about the center of mass requires that
m1 v21 Gm1 m2 m2 v22 Gm1 m2
|F
F 1| = = |F
F 2| = = .
r1 (r1 + r2 )2 r2 (r1 + r2 )2
The angular velocity is ω = |ω | = v/r and the period is P = 2π /ω = 2π r/v, from which
v2 = 4π 2r2 /P2 . Insert this into the above equation for |F
F 1 | and solve for P2 ,
4π 2 r1
P2 = (r1 + r2 )2 .
G m2
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