STUDENT SOLUTIONS MANUAL for Stars and Stellar Processes 1st Edition by Mike Guidry

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Student Solutions Manual: Stars and Stellar Processes Mike Guidry This document gives the solutions for select problems (those marked with ***) at the ends of chapters for the first edition of Stars and Stellar Processes

by Mike Guidry (Cambridge University Press, 2019). Unless otherwise indicated, literature references, equation numbers, figure references, table references, and section numbers refer to the print version of that book.

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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′′ .

1.4 The total energy flux F emitted by a blackbody at temperature T can be expressed as F = 2 π c2 h

Z ∞ 0

Using the substitutions u = hc/λ kT

1

λ 5 (ehc/λ kT − 1)

λ = hc/ukT

dλ .

dλ =

−hc du, u2 kT

the integral can be evaluated giving

kT F = 2π c h hc 2

2π c2hk4 T 4

4 Z ∞ 0

u3 du eu − 1

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)

ζ (x) ≡

1 Γ(x)

Z ∞ 0

ux−1 du eu − 1

and the gamma function Γ(x) have been used, and σ is the Stefan–Boltzmann constant.

1.7 From the following diagram, m1

F1

v1

r1

CM

F2

v2

r2

m2

circular motion about the center of mass requires that F 1| = |F

m1 v21 Gm1 m2 = r1 (r1 + r2 )2

F 2| = |F

m2 v22 Gm1 m2 . = r2 (r1 + r2 )2

The angular velocity is ω = |ω | = v/r and the period is P = 2π /ω = 2π r/v, from which F 1 | and solve for P2 , v2 = 4π 2 r2 /P2 . Insert this into the above equation for |F P2 =

r1 4π 2 (r1 + r2 )2 .

G m2

1

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Some Properties of Stars

2

Impose the center of mass condition r1 = m2 (r1 + r2 )/(m1 + m2 ) and rearrange to give m1 + m2 =

4π 2 (r1 + r2 )3 G P2

for the sum of the masses.

1.8 In Kepler’s third law the factor 4π 2 /G is 4π 2 = 5.9152 × 108 gs2 cm−3 = 1.0M⊙ yr2 AU−3 .

G Thus, if time is measured in years, masses in solar masses, and distances in AU, the factor 4π 2 /G = 1 and from Kepler’s law m1 + m2 =

(r1 + r2 )3 .

P2

This reduces to P2 = r3 if m1 >> m2 = 1M⊙ and r1 + r2 ≃ r2 ≡ r for circular orbits.

1.13 From the Planck law (1.10) the intensity is Bλ (T ) =

2hc2 1 . λ 5 ehc/λ kT − 1

Find the wavelength corresponding to the maximum by setting the derivative equal to zero: ! hc/λ kT dB 2hc2 −5(ehc/λ kT − 1) + λhc kT e = 6 = 0. dλ λ (ehc/λ kT − 1)2 Letting x ≡ hc/λ kT , this is equivalent to xex /(ex − 1) − 5 = 0. This equation doesn’t have a solution in terms of elementary functions; it may be solved numerically (for example, Maple, Mathematica, or Newton–Raphson iteration) to give x ∼ 4.965114. Thus

0.2898 K cm 2.898 × 106 K nm hc hc λ = λmax = = = 0.201405 = xkT kT T T for the wavelength corresponding to the maximum of the Planck distribution (Wien law).

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2

The Hertzsprung–Russell Diagram 2.3 For absolute bolometric magnitude M = −5, the ratio of luminosities is L = 10(M⊙ −M)/2.5 , L⊙

from which L = 7870L⊙ , using the absolute bolometric magnitude M⊙ = 4.74 for the Sun. Solving L = 4πσ R2T 4 for the radius R gives R = R⊙

T⊙ T

2 s

L = 329, L⊙

where T is the effective surface temperature and where the luminosity computed above and an effective solar surface temperature T⊙ = 5780 K have been used. From the known radius of the Sun, this gives a radius of about 1.5 AU (approximately the radius of the orbit of Mars) and an average density of about 3.9 × 10−7 g cm−3 . The large size and low density and surface temperature indicate that this star is a red supergiant.

2.9 This problem is adapted from a discussion in Carrol and Ostlie [68]. The Balmer absorption line is produced by excitation of neutral hydrogen with the electron initially in the first excited state. The Ca H and K lines are produced by excitation of singly-ionized Ca initially in the ground state. Thus, the number of neutral hydrogens with electrons in the first excited state must be compared with the number of Ca+ ions in their ground state.

In Example 2.3 it was estimated that in the solar surface H + /H ≃ 10−4 , so the hydrogen may be assumed to be all neutral. The amount of this neutral hydrogen in the first excited electronic state is given by the Boltzmann formula, g2 2(2)2 −(−3.4−(−13.6))kT n2 e = 5.5 × 10−9, = e−(E2 −E1 )/kT = n1 g1 2(1)2 where an excitation energy of 3.4 eV for the first electronic state and an ionization energy of 13.6 eV have been used, and we have assumed T = 5800 K. Thus, the fraction of the hydrogen capable of producing the Balmer absorption series is

n2 n2 n2 nI ≃ = = 5.5 × 10−9. nT n1 + n2 nT n1 Now let’s do a similar analysis for the Ca ions. From Eq. (2.4) the ratio of singly-ionized to neutral calcium in the solar surface is given by , log

n+ u+ 5 5040 = log + log T − Ei (eV) − logPe − 0.179 = 2.66, n0 u0 2 T

3

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