Solutions to Doconotes Quick Questions (+ Some New Ones)

Solutions to DocONotes Quick Questions (+ some new ones)

Section 1 Quick Question 1: a. What speed does a person at the equator move due to Earth’s rotation? Give your answer in mi/hr, km/hr, and m/s.

Solution. R⊕ = 6000 km Prot = 24 hr v = 2πR⊕ = 2π(6000 km) = 1600 km hr−1 equator Prot 24 hr −1 0.6 mi −1 vequator = 1600 km hr ( 1 km ) = 960 km hr −1 1000 m 1 hr −1 vequator = 1600 km hr ( 1 km )( 3600 s ) = 440 m s b. What is the speed of the Earth in its orbit around the Sun? Give your answer in AU/yr, km/s, mi/hr, and in terms of the fraction of the speed of light

vorb/c?

Solution. Porbit = 1 yr Rorbit = 1 au 2πRorbit 2π(1 au) −1 vorbit = = = 6.3 au yr Porbit 1 yr −1 1.5·108 km 1 yr −1 vorbit = 6.3 au yr ( 1 au )( 3.15·107 s ) = 30 km s −1 3600 s 0.6 mi −1 vorbit = 30 km s ( 1 hr )( 1 km ) = 65000 mi hr −1 9.3·107 1 yr −1 or vorbit = 6.3 au yr ( 1 au )( 24 hr(365) ) = 67000 mi hr 30 km s−1 −5 vorbit/c = 3·105 km s−1 = 1 · 10 c. The Sun is about 25, 000 ly from the center of the Milky Way, and takes about 200 million years to complete one “Galactic year”. What is the speed of Sun in its orbit around the Milky Way, in km/s. In ly/yr? In terms of the

fraction of the speed of light vorb/c?

Solution. Rorbitgal = 25 000 ly 6 Porb = 200 · 10 yr 2πRorbitgal 2π25 000 ly −4 −1 vorb = = 6 = 7.9 · 10 ly yr Porb 200·10 yr 7.9·10−4 ly yr−1 −4 vorb/c = 1 ly yr−1 = 7.9 · 10 −4 −1 9·1012 km 1 yr −1 vorb = 7.9 · 10 ly yr ( 1 ly )( 3.15·107 s ) = 230 km s – 2 –

Quick Question 2: The Sun has a radius of about 700, 000 km. a. How many solar radii in 1 au? In 1 ly?

Solution. R = 700, 000 km 1 au 150·106 km 1 au/R = 700,000 km = 700,000 km = 210 1 ly 9.45·1012 km 7 1 ly/R = 700,000 km = 700,000 km = 1.4 · 10

b. How many Earth radii RE in one solar radius R ?

Solution. R = 700, 000 km RE = 6000 km 700,000 km R /RE = 6000 km = 120 c. Solar neutrinos created in the sun’s core travel at very nearly speed of light but hardly interact with solar matter. How long does it take such core neutrinos to reach the solar surface? How long to reach us on Earth?

Solution. R = 700, 000 km c = 3 · 105 km s−1 700,000 km R /c = 3·105 km s−1 = 2.3 s 150·106 km au/c = 3·105 km s−1 = 500 s d. What then is the solar radius in light-seconds?

Solution. The solar radius is 2.3 light-seconds.

Quick Question 3: The Moon is about 240,000 miles from Earth. a. What is the Earth-Moon distance in km? In light-seconds? In Earth radii

RE? In solar radii R ?

5 Solution. DEM = 2.4 · 10 mi 5 1.61 km 5 DEM = 2.4 · 10 mi( 1 mi ) = 3.9 · 10 km 5 1 −1 DEM = 3.9 · 10 km( 3·105 km s−1 )( s ) = 1.3 ls RE = 6000 km 3.9·105 km DEM /RE = 6000 km = 65 R = 700, 000 km 3.9·105 km DEM /R = 700,000 km = 0.56 b. How many Earth-Moon distances in 1 AU? – 3 –

5 Solution. DEM = 3.9 · 10 km 150·106 km 1 au/DEM = 3.9·105 km = 380

Quick Question 4: -New Using the Universe’s phone number, calculate the relative sizes of the following pairs of objects: a. The Earth and Sun

Solution. The Sun is 100 times the size of Earth

b. Earth’s orbit and the Milky Way

Solution. The Milky Way is 1010 times the size of Earth’s orbit

c. An atomic nucleus and the Sun

Solution. The Sun is 1024 times the size of an atomic nucleus.

Quick Question 5: -New Using Figure 1.2, by what factor is a lifetime longer than the following time frames? a. A day

Solution. A lifetime is 104.5 times the length of a day.

b. Since the time of dinosaurs

Solution. A lifetime is 10−6 times the length of time since the dinosaurs.

c. Since the big bang

Solution. A lifetime is 10−8 times the length of time since the big bang. – 4 –

Section 2 Quick Question 1: A helium party balloon of diameter 20 cm ﬂoats 1 meter above your head. a. What is its angular diameter, in degrees and radians?

Solution. DBalloon = 0.2 m, dBalloon = 1 m α = DBalloon = 0.2 rad dBalloon ◦ 180 ◦ α = 0.2 rad( π rad ) = 11 b. What is its solid angle, in square degrees and steradians?

Solution. Since α is the angular diameter, we have: Ω = πα2/4 = π(0.2 rad)2/4 = 0.031 sterad ◦ 180 2 2 Ω = 0.031 sterad( π rad ) = 110 deg c. What fraction of the full sky does it cover?

Solution. Ω/Ωsky = 0.031 sterad/(4π sterad) = 0.0025 d. At what height h would its angular diameter equal that of the Moon and Sun?

Solution. π rad α = α = 0.5 deg( ◦ ) = 0.0087 rad moon sun 180 d = DBalloon = 0.2 = 23 m Balloon αmoon 0.0087

Quick Question 2: a. What angle α would the Earth-Sun separation subtend if viewed from a distance of d = 1 pc? Give your answer in both radian and arcsec.

Solution. 8 1 pc −6 dES = 1 au = 1.5 · 10 km( 3·1013 km ) = 5 · 10 pc −6 dES 5·10 pc −6 α = d = 1 pc = 5 · 10 rad −6 206265 arcsec α = 5 · 10 rad( 1 rad ) = 1 arcsec b. How about from a distance of d = 1 kpc?

Solution. −6 dES 5·10 pc −9 α = d = 1000 pc = 5 · 10 rad −9 206265 arcsec −3 α = 5 · 10 rad( 1 rad ) = 1 · 10 arcsec – 5 –

Quick Question 3: Over a period of several years, two stars appear to go around each other with a ﬁxed angular separation of 1 arcsec. a. What is the physical separation, in au, between the stars if they have a distance d = 10 pc from Earth?

Solution. s = αd = (1 arcsec)(10 pc) = 10 arcsec au arcsec−1 = 10 au

b. If they have a distance d = 100 pc?

Solution. s = αd = (1 arcsec)(100 pc) = 100 arcsec au arcsec−1 = 100 au

Quick Question 4: -New Friedrich Bessel was the ﬁrst to use a parallax to measure a star’s distance, for 61 Cygni, a binary system. Using the orbit of Earth, he found the parallax to be 0.3136 arcsec. Based on his measurement, what is the distance to 61 Cygni in parsecs and light-years?

Solution. Note that the parallax angle is only the radius of the Earth’s orbit, not the full orbit. α s/ au s/ au 2 au/ au arcsec = d/ pc =⇒ d/ pc = α/ arcsec = 2(0.3136 arcsec)/ arcsec = 3.189 =⇒ d = 3.189 pc 3.276 ly d = 3.189 pc( pc ) = 10.4 ly

Quick Question 5: -New Delta Cephei is a Cepheid Variable with a parallax measured to be 0.00377 arcsec. a. What is its solid angle in steradians?

Solution. 1 rad −8 α = 0.00377 arcsec( 2·105 arcsec ) = 1.89 · 10 rad Ω = πα2 = π(1.89 · 10−8 rad)2 = 1.12 · 10−15 sterad

b. What is its distance from us in parsecs?

arcsec arcsec Solution. d = α pc = 0.00377 arcsec pc = 265 pc

Quick Question 6: -New If an observer was placed on the Sun, what would be the angular diameter of the Earth? (In arcseconds and radians) – 6 –

Solution. RE = 6400 km 6 −5 α = 2 ∗ RE/ au = 2 ∗ 6400 km/(150 · 10 km) = 8.6 · 10 rad −5 2·105 arcsec α = 8.6 · 10 rad 1 rad = 17 arcsec – 7 –

Section 3 Quick Question 1: Recalling the relationship between an AU and a parsec from eqn. (2.6), use eqns. (3.8) and (3.9) to compute the apparent magnitude of the sun. What then is the sun’s distance modulus?

Solution.

By (3.9), M = 5−2.5 log(L/L ) = 5. From (3.8), we get m = M+5 log(d/10 pc) = 5 5 + 5 log(1 au/10 pc) = 5 + 5 log((1/2 · 10 pc)/10 pc) = −26.5 = m . m − M = 5 − 26.5 = −21.5 = m − M

Quick Question 2:

Suppose two stars have a luminosity ratio L2/L1 = 100. a. At what distance ratio d2/d1 would the stars have the same apparent brightness, F2 = F1?

L1 L2 L2 d2 2 d2 Solution. F1 = F2 =⇒ 2 = 2 =⇒ = ( ) =⇒ = 10 4πd1 4πd2 L1 d1 d1 b. For this distance ratio, what is the diﬀerence in their apparent magnitude, m2 − m1?

Solution. m2 − m1 = 2.5 log(F1/F2) = 2.5 log(1) = 0

c. What is the diﬀerence in their absolute magnitude, M2 − M1?

Solution. M2 − M1 = 2.5 log(L1/L2) = −5 d. What is the diﬀerence in their distance modulus?

Solution. (m2 − M2) − (m1 − M1) = M1 − M2 = 2.5

e. If the stars have a surface brightness ratio of I2/I1 = 100, what is the stellar radius ratio, R2/R1? -New Solution. 2 R L2/4πR2 L2 R1 2 R2 2 1 L2 2 I2/I1 = 100 = 1 = ( ) =⇒ ( ) = = 1 =⇒ = 1 L /4πR L1 R2 R1 100 L1 1 2 R1

Quick Question 3: 10 A white-dwarf-supernova with peak luminosity L ≈ 10 L is observed to have an apparent magnitude of m = +20 at this peak. a. What is its Absolute Magnitude M? – 8 –

10 Solution. M = 5 − 2.5 log( 10·10 L ) = −20 L b. What is its distance d (in pc and ly)?

Solution. d d d m − M = 5 log( 10 pc ) =⇒ 20 − −20 = 5 log( 10 pc ) =⇒ 8 = log( 10 pc ) =⇒ d 8 9 9 3.26 ly 9 10 pc = 1 · 10 =⇒ d = 1 · 10 pc d = 1 · 10 pc( 1 pc ) = 3.26 · 10 ly c. How long ago did this supernova explode (in Myr)?

Solution. 3.26 Gyr

Quick Question 4: -New Reformulate Equation (3.3) to be in terms of the Sun’s ﬂux measured on Earth.

Solution. L = 4πd2F 2 L = 4π(1 au) F L 4πd2F 2 = 2 = d (F/F ) L 4π(1 au) F au 2 (L/L ) = d au(F/F )

Quick Question 5: -New −10 An observer measures the ﬂux of a star to be 10 F . It is independently measured to be 50 pc away.

a. Calculate the luminosity of this star in terms of L . (Hint: QQ#4 would make this easier)

Solution. 2·105 au 7 50 pc( 1 pc ) = 10 au 2 7 2 −10 4 4 (L/L ) = d au(F/F ) = (10 au) (10 F /F ) = 10 =⇒ L = 10 L b. Calculate the absolute magnitude of the star.

4 Solution. M = 5 − 2.5 log(L/L ) = 5 − 2.5 log(10 L /L ) = −5 c. Calculate the apparent magnitude of the star.

Solution. m = M + 5 log(d/10 pc) = −5 + 5 log(50 pc/10 pc) = −5 + 5 ∗ 0.7 = −1.5

(For simplicity of computation, you may take the absolute magnitude of the sun to be

M ≈ +5.) – 9 –

Section 4 Quick Question 1:

Two photons have wavelength ratio λ2/λ1 = 2. a. What is the ratio of their period P2/P1?

Solution. P2 = λ2/c = λ2 = 2 P1 λ1/c λ1

b. What is the ratio of their frequency ν2/ν1? 1 Solution. ν2 = c/λ2 = λ1 = ν1 c/λ1 λ2 2

c. What is the ratio of their energy E2/E1? 1 Solution. E2 = hν2 = ν2 = E1 hν1 ν1 2

Quick Question 2:

a. Estimate the temperature of stars with λmax =100, 300, 1000, and 3000 nm. (To simplify the numerics, you may take T ≈ 6000 K.) Solution. T = 6000 K 500 nm λmax 500 nm T100 = 6000 K 100 nm = 30000 K = T100 500 nm T300 = 6000 K 300 nm = 10000 K = T300 500 nm T1000 = 6000 K 1000 nm = 3000 K = T1000 500 nm T3000 = 6000 K 3000 nm = 1000 K = T3000

b. Conversely, estimate the peak wavelengths λmax of stars with T =2000, 10,000, and 60,000 K.

6000 K Solution. λmax = 500 nm T 6000 K λmax,2000 = 500 nm 2000 K = 1500 nm = λmax,2000 6000 K λmax,10000 = 500 nm 10000 K = 300 nm = λmax,10000 6000 K λmax,60000 = 500 nm 60000 K = 50 nm = λmax,60000 c. What parts of the EM spectrum (i.e. UV, visible, IR) do each of these lie in?

Solution. 1500 nm-IR 300 nm-UV 100 nm-UV – 10 –

Quick Question 3: a. Assuming the earth has an average temperature equal to that of typical spring day, i.e. 50◦F, compute the peak wavelength of Earth’s Black-Body radiation.

Solution. ◦ ◦ ◦ 5 ◦ ◦ T = 50 F = (50 F − 32 F) 9 C = 10 C = 283 K ≈ 300 K 6000 K λmax = 500 nm 300 K = 10000 nm = 10 µm b. What part of the EM spectrum does this lie in?

Solution. IR

Quick Question 4: -New a. What is the wave period at the peak wavelength emitted from the Sun?

λmax, 500 nm 500·10−9 m −15 Solution. P = c = 3·108 m s−1 = 3·108 m s−1 = 1.7 · 10 s = 1.7 fs b. What is the corresponding frequency?

−15 14 Solution. ν = 1/Podot = 1/(1.7 · 10 s) = 5.9 · 10 Hz = 590 THz c. What is the energy of a photon at this frequency?

−34 14 −19 Solution. Eγ, = hν = (6.6 · 10 J s)(5.9 · 10 Hz) = 3.9 · 10 J

Quick Question 5: -New

The Sun has a B-V index of 0.63, what is the ratio FB/FV measured on Earth? Solution.

B − V = mB − mV =⇒ 0.63 = 2.5 log(FV /FB) =⇒ FV /FB = 1.8 =⇒ FB/FV = 0.56 – 11 –

Section 5 Quick Question 1:

Compute the luminosity L (in units of the solar luminosity L ), absolute magnitude M, and peak wavelength λmax (in nm) for stars with (a) T = T ; R = 10R , (b) T = 10T ; R = R , and (c) T = 10T ; R = 10R . If these stars all have a parallax of p = 0.001 arcsec, compute their associated apparent magnitudes m.

2 4 Equations. L = L (R/R ) (T/T ) M = 5 − 2.5 log( L ) L 500 nm λmax = T/T d arcsec pc/α 1/0.001 m = M + 5 log( 10 pc ) = M + 5 log( 10 pc ) = M + 5 log( 10 ) = M + 10 a. 2 4 L = L (10R /R ) (T /T ) = 100L = La 100L M = 5 − 2.5 log( ) = 0 = Ma L 500 nm λmax = = 500 nm = λmax,a T /T m = 0 + 10 = 10 = ma b. 2 4 L = L (R /R ) (10T /T ) = 10000L = Lb 10000L M = 5 − 2.5 log( ) = −5 = Mb L 500 nm λmax = = 50 nm = λmax,b 10T /T m = −5 + 10 = 5 = mb c. 2 4 L = L (10R /R ) (10T /T ) = 1000000L = Lc 1000000L M = 5 − 2.5 log( ) = −10 = Mc L 500 nm λmax = = 50 nm = λmax 10T /T m = −10 + 10 = 0 = mc

Quick Question 2:

Suppose a star has a parallax p = 0.01 arcsec, peak wavelength λmax = 250 nm, and apparent magnitude m = +5 . About what is its: a. Distance d (in pc)?

Solution. d = arcsec pc/α = arcsec pc/0.01 arcsec = 100 pc – 12 –

b. Distance modulus m − M?

d 100 Solution. m − M = 5 log( 10 pc ) = 5 log( 10 pc ) = 5 c. Absolute magnitude M?

Solution. m − M = 5 =⇒ M = m − 5 = 0

d. Luminosity L (in L )? Solution. m = 5−2.5 log(L/L )+5 log(d/10 pc) = 5−2.5 log(L/L )+5 log(100 pc/10 pc) = 5 − 2.5 log(L/L ) + 5 = −2.5 log(L/L ) + 10 =⇒ −5 = −2.5 log(L/L ) =⇒ 100 = L/L =⇒ L = 100L

e. Surface temperature T (in T )? Solution. T = 500 nm T = 500 nm T = 2T λmax 250 nm

f. Radius R (in R )? Solution. 2 4 2 −4 2 L/L = (R/R ) (T/T ) =⇒ (R/R ) = L/L (T/T ) = (R/R ) = q −4 102 102 10 100L /L (2T /T ) = 24 =⇒ R/R = 24 = 22 = 2.5 =⇒ R =

2.5R g. Angular radius α (in radian and arcsec)?

Solution. 5 3.09·1013 km −10 α = R/d = (6.96 · 10 km)/(100 pc 1 pc ) = 2.25 · 10 rad −10 2·105 arcsec −5 α = 2.25 · 10 rad 1 rad = 4.5 · 10 arcsec h. Solid angle Ω (in steradian and arcsec2)?

Solution. Ω = π(2.25 · 10−10 rad)2 = 1.6 · 10−19 sterad Ω = π(4.5 · 10−5 arcsec)2 = 6.4 · 10−9 arcsec2

i. Surface brightness relative to that of the Sun B/B ?

4 σsbT 4 4 4 4 4 Solution. B = π =⇒ B/B = T /T = (2T ) /T = 2 = 16

Quick Question 3: Use Equation (5.5) to determine the radius of the Sun. – 13 –

q F (d) q 1400 W m−2 11 8 Solution. R = 4 d = −8 −2 −1 −4 4 (1.5·10 m) = 6.55 · 10 m σsbT (5.67·10 J m s K )(6000 K) – 14 –

6 Quick Question 1: On the H-R diagram, where do we ﬁnd stars that are: a. Hot and luminous? Upper left b. Cool and luminous? Upper right c. Cool and Dim? Lower right d. Hot and Dim? Lower left Which of these are known as: 1. White Dwarfs? d 2. Red Giants? b 3. Blue supergiants? a 4. Red dwarfs? c

Quick Question 2: -New 61 Cygni A has a temperature around 4500 K and an absolute magnitude of 7.5. a. What is its spectral type? K b. What is its luminosity class? V c. What band of stars does 61 Cygni fall into? Main Sequence

Quick Question 3: -New Beta Cassiopeiae, a star in the constellation Cassiopeia, has a B-V color index of 0.34 and an luminosity of 27L . a. What is its spectral type? F b. What is its luminosity class? III (IV is an appropriate guess) c. What band of stars does Beta Cassiopeiae fall into? Giant – 15 –

Section 7 Quick Question 1:

In CGS units, the sun has log g ≈ 4.44. Compute the log g for stars with:

2 2 GM GM∗/R∗ M∗ R∗ −1 Solve for log gast. g = 2 =⇒ g∗/g = 2 = ( 2 ) =⇒ log g∗ = R GM /R M R M∗ R∗ −2 M∗ R∗ log( ( ) ) + log g = 4.44 + log( ) − 2 log( ) M R M R

a. M = 10M and R = 10R

10M 10R Solution. log g∗ = 4.44 + log( ) − 2 log( ) = 3.44 M R

b. M = 1M and R = 100R

1M 100R Solution. log g∗ = 4.44 + log( ) − 2 log( ) = 0.44 M R

c. M = 1M and R = 0.01R

1M 0.01R Solution. log g∗ = 4.44 + log( ) − 2 log( ) = 8.44 M R

Quick Question 2:

The sun has an escape speed of Ve = 618 km/s. Compute the escape speed Ve of the stars in parts a-c of QQ1. q q 2GM −1 M/M Reformulate into solar units. Ve = =⇒ Ve = 618 km s R R/R a. q −1 10M /M −1 Solution. Ve = 618 km s = 618 km s 10R /R b. q −1 1M /M −1 Solution. Ve = 618 km s = 61.8 km s 100R /R c. q −1 1M /M −1 Solution. Ve = 618 km s = 6180 km s 0.01R /R

Quick Question 3:

The earth has an orbital speed of Ve = 2πau/yr = 30 km/s. Compute the orbital speed Vorb (in km/s) of a body at the following distances from the stars with the quoted masses: – 16 –

q q q GM GM/r M/M Solution. Vorb = =⇒ Vorb/Ve = = =⇒ Vorb = r GM /re r/re q −1 M/M 30 km s r/1 au

a. M = 10M and d = 10 au. q −1 10M /M −1 Solution. Vorb = 30 km s 10 au/1 au = 30 km s

b. M = 1M and d = 100 au. q −1 1M /M −1 Solution. Vorb = 30 km s 100 au/1 au = 3 km s

c. M = 1M and d = 0.01 au. q −1 1M /M −1 Solution. Vorb = 30 km s 0.01 au/1 au = 300 km s

Quick Question 4: -New

Compute the gravitational binding energy ratio, U/UE, , of the system for parts a-c of Quick Question 3, assuming the orbiting body has a mass of 10mE, 1mE, and 1mE respectively.

GMm GMm GM me Reformulate into solar units. U = − d =⇒ U/Ue, = − d / − 1 au = ( M )( m )( 1 au ) M me d a.

10M 10me 1 au Solution. U/Ue, = ( )( )( ) = 10 M me 10 au b.

1M 1me 1 au Solution. U/Ue, = ( )( )( ) = 0.01 M me 100 au c.

1M 1me 1 au Solution. U/Ue, = ( )( )( ) = 100 M me 0.01 au

Quick Question 5: -New Assuming a rocket, having a mass of 50, 000 kg loses negligible mass from fuel, how much energy is required to lift a rocket out of the following object’s gravitational pull? a. The Earth

GMm (6.67·10−11m3 kg−1 s−2)(5.97·1024 kg)(5·104 kg) 12 Solution. W = R = 6.4·106 m = 3.11 · 10 J – 17 –

b. The moon

GMm (6.67·10−11m3 kg−1 s−2)(7.34·1022 kg)(5·104 kg) 11 Solution. W = R = 1.7·106 m = 1.44 · 10 J c. The Sun

GMm (6.67·10−11m3 kg−1 s−2)(1.99·1030 kg)(5·104 kg) 15 Solution. W = R = 6.96·108 m = 9.54 · 10 J – 18 –

Section 8 Quick Question 1:

What are the luminosities (in L ) and the expected main sequence lifetimes (in Myr) of stars with masses:

a. 10 M ?

3 Solution. L/L ∝ (M/M ) M 2 tms = 10000 Myr( M ) M 2 tms = 10000 Myr( ) = 100 Myr = tms,a 10M 3 3 L/L = (10M /M ) =⇒ L = 10 L = La

b. 0.1 M ?

M 2 6 Solution. tms = 10000 Myr( ) = 10 Myr = tms,b 0.1M 3 −3 L/L = (0.1M /M ) =⇒ L = 10 L = Lb

c. 100 M ?

M 2 Solution. tms = 10000 Myr( ) = 1 Myr = tms,c 100M 3 6 L/L = (100M /M ) =⇒ L = 10 L = Lc

Quick Question 2: Suppose you observe a cluster with a main-sequence turnoﬀ point at a luminosity of 100L . What is the cluster’s age, in Myr. What about for a cluster with a turnoﬀ at a luminosity of 10, 000L ?

L −2/3 Solution. tcluster = 10000 Myr( ) L 100L −2/3 tcluster = 10000 Myr( ) = 464 Myr L 10000L −2/3 tcluster = 10000 Myr( ) = 22 Myr L

Quick Question 3: -New

61 Cygni A is a main sequence star and has a luminosity of 0.15L , a mass of 0.70M , and a radius of 0.67R . a. If 61 Cygni A was powered by chemical burning, what would be the timescale which it could maintain its luminosity?

2 Mc2 Mc2 M c Solution. tchem = =⇒ tchem/tchem, = / =⇒ tchem = L L L M/M tchem, L/L 0.70M /M tchem = 15, 000 yr = 70, 000 yr 0.15L /L – 19 –

b. Suppose instead, that 61 Cygni A was powered by gravitational contraction. What would be the expected lifetime of this process?

2 2 3 GM 2 3 GM 2 3 GM (M/M ) Solution. tKH = =⇒ tKH /tKH, = / = =⇒ 10 RL 10 RL 10 R L (R/R )(L/L ) 2 (M/M ) tKH = tKH, (R/R )(L/L ) 2 (0.70M /M ) tKH = 30 Myr = 146 Myr (0.67R /R )(0.15L /L ) c. As a main sequence star, 61 Cygni A is powered by nuclear fusion. What is its nuclear burning timescale?

M 2 Solution. tms = 10 Gyr( ) = 20 Gyr 0.7M

Quick Question 4: -New Based on Figure 8.1, what is the age of M55?

Solution. Lto ≈ 3L L 2/3 tcluster = 10 Gyr( ) = 4.8 Gyr 3L

Quick Question 5: -New

Suppose a newly born star is found to have a mass of 10M . Suppose that it has been found to be 5 · 106 pc away, is the star still on the main sequence in reference to its own position?

6 3.26 ly 7 Solution. 5 · 10 pc( 1 pc ) = 1.71 · 10 ly = 0.0171 Gly M 2 M 2 tms = 10 Gyr( ) = 10 Gyr( ) = 0.01M M 10M The star is no longer on the main sequence because its lifetime is shorter than the time it takes for its light to reach us. – 20 –

Section 9 Quick Question 1: A star with parallax p = 0.02 arcsec is observed over 10 years to have shifted by 2 arcsec from its proper motion. Compute the star’s tangential space velocity

Vt, in km/s.

−1 µ −1 2 arcsec/10 −1 Solution. Vt = 4.7 km s p = 4.7 km s 0.02 arcsec = 47 km s

Quick Question 2:

For the star in QQ#1, a line with rest wavelength λo = 600.00 nm is observed to be at a wavelength λ = 600.09 nm. a. Is the star moving toward us or away from us?

Solution. λobs > λo away b. What is the star’s Doppler shift z?

Solution. z = λobs−λo = 600.09 nm−600 nm = 1.5 · 10−4 λo 600 nm

c. What is the star’s radial velocity Vr, in km/s?

Vr −4 8 −1 −1 Solution. z = c =⇒ Vr = zc = 1.5 · 10 (3 · 10 m s ) = 45 km s

d. What is the star’s total space velocity Vtot, in km/s?

p 2 2 p −1 2 −1 2 −1 Solution. Vtot = Vr + Vt = (45km s ) + (47 km s ) = 65 km s

Quick Question 3: -New The transition Lyman α has a rest wavelength of 1216 nm. For the following observed wavelengths from stars, what is the star’s Doppler shift, z, what is the star’s radial velocity (in km s−1) and is the star moving towards or away from us?

Relevant Equations. z = λobs−λo = λobs−1216 nm , z = Vr =⇒ V = zc λo 1216 nm c r a. 1216.5 nm

1216.5 nm−1216 nm −4 Solution. z = 1216 nm = 4.1 · 10 −4 5 −1 −1 Vr = 4.1 · 10 (3 · 10 km s ) = 123 km s Away

b. 1215.5 nm – 21 –

1215.5 nm−1216 nm −4 Solution. z = 1216 nm = −4.1 · 10 −4 5 −1 −1 Vr = −4.1 · 10 (3 · 10 km s ) = −123 km s T owards

c. 1215.85 nm

1215.85 nm−1216 nm −4 Solution. z = 1216 nm = −1.2 · 10 −4 5 −1 −1 Vr = −1.2 · 10 (3 · 10 km s ) = −36 km s T owards

d. 1216.15 nm

1216.15 nm−1216 nm −4 Solution. z = 1216 nm = 1.2 · 10 −4 5 −1 −1 Vr = 1.2 · 10 (3 · 10 km s ) = 36 km s Away

Quick Question 4: -New A passenger jet takes oﬀ at an average of 250 km hr−1. If the jet is blue, roughly 400 nm, what wavelength of light is picked up by an observer at the takeoﬀ position?

−1 1000 m 1 hr −1 Solution. 250 km hr ( 1 km )( 3600 s ) = 69 m s ∆λ V ∆λ 69 m s−1 −7 λ = c =⇒ 400 nm = 3·108 m s−1 =⇒ ∆λ = 2.3 · 10 nm λ0 = λ + ∆λ = 400 nm + 2.3 · 10−7 nm ≈ 400 nm – 22 –

Section 10 Quick Question 1: Note that the net amount of stellar surface eclipsed is the same whether the smaller or bigger star is in front. So why then is one of the eclipses deeper than the other? What quantity determines which of the eclipses will be deeper?

Solution. Brightness or Intensity ∝ Flux ∝ T 4 The hotter star gives the deeper eclipse Temperature is the key quantity in determining which eclipse will be deeper.

Quick Question 2: Over a period of 10 years, two stars separated by an angle of 1 arcsec are observed to move through a full circle about a point midway between them on the sky. Suppose that over a single year, that midway point is observed itself to wobble by 0.2 arcsec due to the parallax from Earth’s own orbit. a. How many pc is this star system from earth?

arcsec Solution. d = 0.2 arcsec/2 pc = 10 pc b. What is the physical distance between the stars, in au.

d α 10 pc 1 arcsec Solution. s/ au = pc arcsec = pc arcsec = 10 =⇒ s = 10 au

c. In solar masses, what are the masses of each star, M1 and M2. Solution. 3 M1+M2 (s/ au) = 2 = 10 =⇒ M1 + M2 = 10M M (P/ yr) Since the center of mass is the midway point, M1 = M2 =⇒ M1 = 5M = M2

Quick Question 3: -New (corrected 11-Apr-20) Within a binary star system both stars have circular orbits. When one star has a maximum radial velocity of 60 km s−1 the other has a minimum radial velocity of −30 km s−1. These measurements of their velocities repeat every 20 days. What is the mass of each star in terms of solar masses?

V2 3 V1 2 V1 3 V2 2 Solution. M1 = ( ) Pyr(1 + ) M , M2 = ( ) Pyr(1 + ) M Ve V2 Ve V1 The maximum and minimum speeds occur when the star is heading directly towards, or away, from us and are thus tangential. The tangential velocity of a circular orbit is equivalent to its orbital velocity. – 23 –

−1 −1 V1 = 60 km s = 2Ve, |V2| = 30 km s = 1Ve M = ( Ve )3(20/365)(1 + 2Ve )2M = 0.49M = M 1 Ve 1Ve 1 M = ( 2Ve )3(20/365)(1 + 1Ve )2M = 0.99M = M 2 Ve 2Ve 2

Quick Question 4: -New A binary star system has been discovered that has an eclipse. From the

Doppler shifts of the two stars, it is known that the orbital speeds are V1 = −1 −1 20 km s and V2 = 10 km s . A stopwatch starts begins right before star 1 eclipses star 2. It reads roughly 5 hr when the star 1 fully eclipses star 2. The eclipse ends and the stopwatch is stopped at 20 hr. What are the star’s radii in terms of R ?

Solution. t1 = 0 s 4 3600 s 4 t2 = 5 · 10 hr( hr ) = 1.8 · 10 s 3600 s 5 t4 = 100 hr( hr ) = 3.6 ∗ ·10 s 4 −1 −1 R2 = (t2 − t1)(V1 + V2)/2 = (1.8 · 10 s − 0 s)(20 km s + 10 km s )/2 = 2.7 ∗ 5 1R 10 km( 6.96·105 km ) = 0.39R = R2 5 4 −1 −1 R1 = (t4 − t2)(V1 + V2)/2 = (3.6 ∗ ·10 s − 1.8 · 10 s)(20 km s + 10 km s )/2 = 6 1R 5.1 · 10 km( 6.96·105 km ) = 7.3R = R1

Quick Question 5: -New For main sequence stars, if you assume the scaling found in Figure 10.4, log(L/L ) = 0.1 + 3.1 log(MM ), is the exact scaling, what is the relative error in calculating the luminosity of a star from its mass when using L ∼ M 3 for the following masses?

a. 1M

3 3 3 Solution. Lapprox/L = (M/M ) =⇒ Lapprox = (M/M ) L = (M /M ) L = L log(L/L ) = 0.1+3.1 log(MM ) = 0.1+3.1 log(M M ) = 0.1 =⇒ L = 1.25L |Lapprox−L| |1−1.25| = L = 1.25 = 0.2 = 20%

b. 10M

3 Solution. Lapprox = (10M /M ) L = 1000L log(L/L ) = 0.1 + 3.1 log(10M M ) = 3.2 =⇒ L = 1600L |1000−1600| = 1600 = 0.38 = 38%

c. 0.1M – 24 –

3 Solution. Lapprox = (0.1M /M ) L = 0.001L log(L/L ) = 0.1 + 3.1 log(0.1M M ) = −3 =⇒ L = 0.001L |0.001−0.001| = 0.001 = 0 = 0%

d. 100M

3 6 Solution. Lapprox = (100M /M ) L = 10 L 6 log(L/L ) = 0.1 + 3.1 log(100M M ) = 6.3 =⇒ L = 2.0 · 10 L |106−2.0·106| = 2.0·106 = 0.5 = 50% – 25 –

Section 11 Quick Question 1:

A line with rest wavelength λo = 500 nm is rotational broadened to a full width of 0.5 nm. Compute the value of V sin i, in km/s.

Solution. ∆λfull 2V sin i ∆λfull 0.5 nm 5 −1 1 = =⇒ V sin i = c = (3 · 10 km s ) = 3 (3 · λo c 2λo 2·500 nm 2·10 5 −1 300 km s−1 −1 10 km s ) = 2 = 150 km s

Quick Question 2: -New If the star emitting the spectrum from QQ#1 is found to have a stellar rotation period of 30 days, what is the lower bound on its radius (in km and R )?

−1 Vrot sin iP 150 km s ∗30∗86400 s 7 Solution. Rmin = 2π = 2π = 6.2 · 10 km 7 1R Rmin = 6.2 · 10 km( 6.96·105 km ) = 89R – 26 –

Section 12 Quick Question 1 a. Suppose spherical dust grains have a radius r = 0.1 cm and individual 3 mass density ρg = 1 g/cm . What is their cross section σ, mass m, and associated opacity κ?

Solution. σ = πr2 = π(0.1 cm)2 = 0.03 cm2 = σ 4 3 −3 4 3 m = ρgV = ρg 3 πr = 1 g cm 3 π(0.1 cm) = 0.004 g = m σ 0.03 cm2 2 −1 κ = m = 0.004 g = 7.5 cm g = κ

−3 b. If the number density of these grains is nd = 1 cm , what is the mass density of dust ρd and the mean free path ` for light?

−3 −3 Solution. ρd = ndm = 1 cm (0.004 g) = 0.004 g cm = ρd 1 1 ` = = 2 −1 −3 = 30 cm κρd 7.5 cm g 0.004 g cm or 1 1 ` = = −3 2 = 30 cm = ` ndσ 1 cm 0.03 cm c. What is the optical depth at a physical depth 1 m into a planar layer of such dust absorbers?

R z dz0 z 100 cm Solution. τ = 0 ` = ` = 30 cm = 3

d. What fraction of impingent intensity Io makes it to this depth? Solution. I(t) = e−τ = e−3 = 0.05 Io

Quick Question 2:

Derive expressions for dinf /d in terms of both the absorption magnitude A and the optical depth τ.

q L q L Solution. dinf = , d = 4πFobs 4πFem √ p τ τ/2 dinf /d = Fem/Fobs) = e = e A = 1.08τ =⇒ τ = A/1.08 A/(2∗1.08) 0.46A dinf /d = e = e

Quick Question 3 -New

A star has AV = 15 magnitudes of extinction due to Mie scattering at 500 nm which is green light. – 27 –

Use the deﬁnitions of A and τ to solve for A in terms of AV . z z −1 A = 1.08τ = 1.08 ` = 1.08 1/(ρκ) = 1.08zρλ −1 1.08zρλx λx −1 λV Ax/AV = −1 = ( λ ) =⇒ Ax = AV λ 1.08zρλV V x

a. Calculate the fraction of intensity for green light that reaches Earth, I/Io.

−AV /1.08 −15/1.08 −7 Solution. IG/Io = e = e = 9.3 · 10 = IG/Io

b. What is AB, the magnitudes of extinction in blue light, 450 nm? Calculate the fraction of intensity for blue light that reaches Earth, I/Io.

500 nm Solution. AB = 15 450 nm = 16.7 = AB

−AB /1.08 −16.7/1.08 −7 IB/Io = e = e = 1.9 · 10 = IB/Io

c. What is AR, the magnitudes of extinction in red light, 750 nm? Calculate the fraction of intensity for red light that reaches Earth, I/Io.

500 nm Solution. AR = 15 750 nm = 10 = AR

−AR/1.08 −10/1.08 −5 IR/Io = e = e = 9.5 · 10 = IR/Io

d. What is AM , the magnitudes of extinction in mid-infrared, 5 µm? Calculate the fraction of intensity for mid-infrared light that reaches Earth, I/Io.

500 nm Solution. AM = 15 5 µm = 1.5 = AM −AM /1.08 −1.5/1.08 IM /Io = e = e = 0.25 = IM /Io

e. What is AIR, the magnitudes of extinction in far-infrared, 5 mm? Calculate the fraction of intensity for blue light that reaches Earth, I/Io.

500 nm Solution. AIR = 15 5 mm = 0.0015 = AIR −AIR/1.08 −0.0015/1.08 IIR/Io = e = e = 0.999 = IIR/Io

Quick Question 4 -New

Suppose a star is known to be 3.3 ly away and a luminosity of 100L . It is observed to have an apparent magnitude of 15. a. What is its absolute magnitude?

Solution. M = 5 − 2.5 log(L/L ) = 5 − 2.5 log(100L ) = 0 b. If the light did not have to travel through the ISM, what would be its apparent magnitude? – 28 –

Solution. 3.3 ly ≈ 1 pc m = M + 5 log(d/10 pc) = 0 + 5 log(1 pc/10 pc) = −5

c. What is the optical depth of the ISM?

Solution. mobs − m = 1.08τ =⇒ 15 − −5 = 1.08τ =⇒ τ = 18.5 d. What is the optical length of this material?

z z 1 pc 3·1013 km 12 Solution. τ = ` =⇒ ` = τ = 18.5 = 18.5 = 1.62 · 10 km

Quick Question 5 -New Suppose a star is 3.3 ly away. The optical length of the ISM is 30 cm. a. What is the optical depth?

Solution. 3.3 ly ≈ 3 · 1013 km z 3·1013 km 17 τ = ` = 30 cm = 10 b. What fraction of the intensity of the star reaches Earth?

−1017 Solution. I/Io = e – 29 –

Section 13 Quick Question 1: a. For a typical temperature on a spring day (∼ 50◦F), compute the scale height H for the earth (in km), and its ratio to earth’s radius, H/Re.

kT Solution. H = µg The main component of Earth’s atmosphere is N2 so µ = 28 mp = 28(1.67 · 10−24 g) = 4.7 · 10−23 g T = (F − 32)(5/9) + 273 K = (50 deg F − 32)(5 K/9◦ F) + 273 K = 283 K (1.38·10−16 erg K−1)(283 K) 5 H = (4.7·10−23 g)(980 cm2 s−1) = 8.5 · 10 cm = 8.5 km = H −3 H/Re = 8.5 km/6400 km = 1.3 · 10 b. Relative to values at sea level, compute the pressure and density at a typical height h = 300 km for an orbiting satellite.

ρ(h) P (h) −h/H −300 km/8.5 km −16 Solution. ρ = P = e = e = 4.7 · 10

Quick Question 2:

a. Compute the escape speed (in km/s) from stars with M = M and R = 2R ; with M = 2M and R = R . q q Vesc M/M M/M Solution. = =⇒ Vesc = Vesc, Vesc, R/R R/R q −1 M /M −1 Vesc = 620 km s = 440 km s = Vesc,1 2R /R q −1 2M /M −1 Vesc = 620 km s = 880 km s = Vesc,2 R /R b. For these stars, estimate the associated central temperature.

M/M Solution. Tint = 14 MK R/R M /M Tint = 14 MK = 7 MK = Tint,1 2R /R 2M /M Tint = 14 MK = 28 MK = Tint,1 R /R

Quick Question 3: -No Solution Found, Made One a. For a constant density stellar envelope, show that the mass within radius r is given by M(r) = M(R)(r/R)3, where R is the stellar radius. – 30 –

Solution.

ρ(r) = ρ(R) M(r) M(R) = V (r) V (R) M(r) M(R) = 4 3 4 3 3 πr 3 πR M(r) M(R) = r3 R3 r M(r) = M(R)( )3 R

b. For such a star, show by explicit integration of the hydrostatic equilibrium equation (13.8) that the core temperature is T (0) = 7 MK, and so half the value given by eqn. (13.9).

Pintµ Pintµ Solution. ρint = =⇒ Tint = kTint kρint dP 3 = − ρGM(r) = − ρGM(R)r = − ρGM(R)r =⇒ dP = − ρGM(R) rdr =⇒ dr r2 r2R3 R3 R3 R 0 ρGM(R) R R ρGM(R) R2 ρGM(R) dP = − 3 rdr =⇒ 0 − Pint = − 3 =⇒ Pint = Pint R 0 R 2 2R ρintGMµ GMµ Tint = = 2Rkρint 2Rk Which is half of eqn (13.9), thus Tint = 7 K

Quick Question 4: -New A main sequence star with a similar composition to the Sun has a temperature of 9000 K and is ten times as luminous as the Sun. What is the scale height as a fraction of the star’s radius?

Solution. L = ( T )4( R )2 =⇒ 10L = ( 1.5T )4( R )2 =⇒ R = 1.41 L T R L T R 3 p3 L/L = (M/M ) =⇒ M = 10L /L M = 2.15M H = 0.0005 T R/R = 0.0005 1.5T 1.41R /R = 0.00049 ≈ 0.0005 R T M/M T 2.15M /M – 31 –

Section 14 Quick Question 1: a. At what optical depth τ does the local temperature T in a stellar atmosphere equal the stellar eﬀective temperature Teff ?

4 3 4 T 4 3 Solution. T = Teff (τ + 2/3) =⇒ 4 = (τ + 2/3) 4 Teff 4 2 T = T =⇒ 1 = 3 (τ + 2/3) =⇒ 4 = τ + 2/3 =⇒ τ = eff 4 3 3

b. At about what optical depth τ does the local temperature T = 10Teff ?

4 3 4 Solution. T = 10Teff =⇒ 10 = 4 (τ + 2/3) =⇒ τ = 1.3 · 10 c. What is the temperature at the top of the atmosphere (hint: Its where τ = 0). -New

4 3 4 3 4 1 4 −1/4 Solution. T = 4 Teff (τ + 2/3) = 4 Teff (0 + 2/3) = 2 Teff =⇒ T = 2 Teff = 0.84Teff

Quick Question 2: a. Near the Sun’s surface where the temperature is at the eﬀective temperature T = T∗ = Teﬀ ≈ 5800 K, compute the scale height H (in km).

−24 Solution. µ = 0.6mp ≈ 10 kT 1.38·10−16 erg K(5800 K) 7 H = µg = 10−24 g(2.7·104 cm s−1) = 3.0 · 10 cm = 300 km Note: 90km is the answer in the solutions, but I could not recreate that number from the solution. Please look this over. I used this value for H in the next part.

b. Using the fact that the mean-free-path ` ≈ H near this surface, compute the mass density ρ (in g/cm3) assuming the opacity is equal to the electron 2 scattering value given in §C.1, i.e. κe = 0.34 cm /g.

1 1 1 −8 −3 Solution. ` = κρ = H =⇒ ρ = κH = (0.34 cm2 g−1)(3.0·107 cm) = 9.8 · 10 g cm

Quick Question 3: -New a. What is the optical depth to a temperature equal to the interior temperature of the Sun? – 32 –

4 3 4 6 4 3 4 Solution. Tint, = 4 Teff, (τint, + 2/3) =⇒ (1.4 · 10 K) = 4 (6000 K) (τint, + 9 2/3) =⇒ τint, + 2/3 = 3.95 · 10 ≈ τint b. What is the optical depth to a temperature equal to the interior of a star? The equation should be in solar units.

Solution. 3 3 (T 4 = T 4 (τ + 2/3))/(T 4 = T 4 (τ + 2/3)) (1) int 4 eff int int, 4 eff, int, 4 4 4 4 (Tint = Teff τint)/(Tint, = Teff, τint, ) (2)

Tint, 4 Teff, 4 τint = τint, ( ) ( ) (3) Tint Teff – 33 –

Section 15 Quick Question 1: -New

A proto-star of mass M = M , Luminosity L = L and temperature T = 4000 K has recently evolved out of its fully convective phase and started to develop a radiative core. a. What is the fully convective phase called? Solution: Hayashi Track b. Assume its starting the horizontal-track Kelvin-Helmholtz contraction. It will evolve to have R = R and T = T = 6000 K, what is its current radius?

Solution. Horizontal-track Kelvin-Helmholtz contraction means Lf /Li ≈ 1 4 2 4 2 4 2 4 2 4 2 L = σsbT 4πR =⇒ 1 = Lf /Li = σsbTf 4πRf /σsbTi 4πRi = Tf Rf /Ti Ri =⇒ q q 4 2 4 2 2 Tf 4 2 6000 K 4 T R = T R =⇒ Ri = R ( ) = R ( ) = 2.3R f f i i f Ti 4000 K

c. As a proto-stellar cloud, it had a luminosity L = 10L , what was its radius at this point?

Solution. During the Hayashi Track, Tf /Ti ≈ 1 4 2 4 2 4 2 4 2 4 2 2 L = σsbT 4πR =⇒ Lf /Li = σsbTf 4πRf /σsbTi 4πRi = Tf Rf /Ti Ri = (Rf /Ri) =⇒ q q 2 Li 2 10L Ri = R = (2.3R ) = 7.3R f Lf 1L – 34 –

Section 16 Quick Question 1: -New Calculate the radius of main sequence stars with the following masses:

a. 10M

M 0.7 10M 0.7 Solution. R = ( ) R = ( ) R = 5R M M

b. 0.1M

0.1M 0.7 Solution. R = ( ) R = 0.2R M

c. 100M

100M 0.7 Solution. R = ( ) R = 25R M

d. 0.5M

0.5M 0.7 Solution. R = ( ) R = 0.6R M

e. 50M

50M 0.7 Solution. R = ( ) R = 15R M

f. 25M

25M 0.7 Solution. R = ( ) R = 9.5R M

Quick Question 2: -New Based on the lower and upper limits for main sequence stars calculate the approximate range in orders of magnitude for the following stellar properties: a. Mass:

Solution. Mmin = 0.1M , Mmax = 195M ≈ 100M Mmag,range = log(Mmax/Mmin) = log(100M /0.1M ) = 3 b. Radius:

0.7 0.7 Solution. Rmag,range ∼ Mmag,range = 3 = 2.2 ≈ 2 c. Luminosity:

3 3 Solution. Lmag,range ∼ Mmag,range = 3 = 27 d. Temperature: – 35 –

8 1/8 1/8 Solution. Lmag,range ∼ Tmag,range =⇒ Tmag,range ∼ Lmag,range = 27 = 1.5 e. Age:

M 2 −4 Solution. tms,min = 10 Gyr( ) = 2.6 · 10 Gyr = 0.26 Myr ≈ 0.1 Myr 195M M 2 tms,max = 10 Gyr( ) = 1000 Gyr 0.1M 7 trange = 1000 Gyr/0.1 Myr = 10 tmag,range = 7 – 36 –

Section 17 Quick Question 1: a. Because of general relativistic eﬀects, it turns out the lowest stable orbit around a black hole is at radius of 3Rbh. What is the luminosity eﬃciency for accreting to this radius?

Solution. = Rbh = Rbh = 1/12 4Ra 4·3Rbh ˙ b. What is the accretion luminosity, in L , for a mass accretion rate Ma = −6 10 M /yr to this radius?

˙ 2 1 −6 −1 2 6 Solution. L = Mac = 12 · 10 M yr c = 1.2 · 10 L

c. Challenge problem: For a black hole with mass Mbh = 3M , use the Stefan-Bolzmann law to derive the radiative ﬂux that would balance the local gravitational heating at this radius r = 3Rbh, and then solve for the local blackbody temperature T (r = 3Rbh). Express this ﬁrst as a ratio to sun’s surface temperature T , and then also in Kelvin. -Note: No solution yet

Quick Question 2: -New Assume a black hole is a sphere. a. Derive an equation for the density of a black hole.

Mbh Solution. Rbh = 3 km M ρ = Mbh = 3Mbh = 3Mbh = M ( Mbh )−2 bh 3 Mbh 3 3 Vbh 4πRbh 4π(3 km ) 36π km M M 33 M 1 km 3 2.0·10 g 16 −3 3 ( 5 ) ( ) = 1.8 · 10 g cm 36π km 10 cm M

16 −3 Mbh −2 ρbh = 1.8 · 10 g cm ( ) M b. How large must a black hole be to have the same density as water?

Solution.

16 −3 Mbh −2 ρbh = 1.8 · 10 g cm ( ) M 1 g cm−3 = 1.8 · 1016 g cm−3( Mbh )−2 M M ( bh )2 = 1.8 · 1016 M M bh = 1.3 · 108 M 8 Mbh = 1.3 · 10 M – 37 –

Quick Question 3: -New Calculate the radii of white dwarfs with the following masses:

a. 1.4M

M 1/3 M 1/3 Solution. Rwd = 0.01R ( ) = 0.01R ( ) = 0.0089R M 1.4M

b. 1M

M 1/3 Solution. Rwd = 0.01R ( ) = 0.01R 1M

c. 0.5M

M 1/3 Solution. Rwd = 0.01R ( ) = 0.013R 0.5M

d. 0.25M

M 1/3 Solution. Rwd = 0.01R ( ) = 0.016R 0.25M

e. 0.1M

M 1/3 Solution. Rwd = 0.01R ( ) = 0.022R 0.1M

Quick Question 4: -New Calculate the radii for a neutron star with the following masses:

a. 1.2M

M 1/3 M 1/3 Solution. Rns = 10 km( ) = 10 km( ) = 9.4 km M 1.2M

b. 1.5M

M 1/3 Solution. Rns = 10 km( ) = 8.7 km 1.5M

c. 2M

M 1/3 Solution. Rns = 10 km( ) = 7.9 km 1.5M

d. 2.5M

M 1/3 Solution. Rns = 10 km( ) = 7.4 km 2.5M

e. 3M

M 1/3 Solution. Rns = 10 km( ) = 6.9 km 1.5M – 38 –

Quick Question 5: -New Calculate the radii of a black hole with the following masses:

a. 0.5M

M 0.5M Solution. Rbh = 3 km = 3 km = 1.5 km M M

b. 1M

1M Solution. Rbh = 3 km = 3 km M

c. 10M

10M Solution. Rbh = 3 km = 30 km M

d. 50M

10M Solution. Rbh = 3 km = 150 km M

e. 100M in km and R Solution. 100M Rbh = 3 km = 300 km M R −4 300 km( 6.96·105 km ) = 4.3 · 10 R

4 f. 10 M in km and R Solution. 4 10 M 4 Rbh = 3 km = 3 · 10 km M 4 R −2 3 · 10 km( 6.96·105 km ) = 4.3 · 10 R

7 g. 10 M in km and R Solution. 7 10 M 7 Rbh = 3 km = 3 · 10 km M 7 R 3 · 10 km( 6.96·105 km ) = 43R

Quick Question 6: -New

If the Schwarzschild radius is R , what is the mass of this black hole?

M M 5 M Solution. Rbh = 3 km =⇒ R = 3 km =⇒ 6.96 · 10 km = 3 km =⇒ M M M 5 M = 2.32 · 10 M – 39 –

Section 18 Quick Question 1:

a. Recalling that the equilibrium blackbody temperature of the earth is Te ≈ p (T /2) 2R /au ≈ 280 K, show that the corresponding temperature at a dis- p tance d from the sun is given by T = Te au/d. Solution.

Ein = Eout

L Acrosssection,e = Le SAdistancefromsun σT 4 4πR2 (πR2) = σT 44πR2 4πd2 T 4 R2 = T 4 4d2 r R T = T d r 2R T = T 4d r T 2R T = 2 d √ r1 T = T au e d rau T = T e d

b. Compute the temperature for d =1 pc.

q au p au Solution. T = Te 1 pc = 280 K 2·105 au = 0.626 K c. Compute the temperature at a distance d = 1 pc from a hot star with

T∗ = 10 T ≈ 60,000 K.

Solution. Thotstar = 10(Tsun) = 10(0.626 K) = 6.26 K

Quick Question 2: a. For a GMC with molecular Hydrogen density n = 100 cm−3, compute the associated mass density ρ. – 40 –

−24 −3 −24 −3 Solution. ρ = mH2 n = 2mpn = 2(1.67 · 10 g)100 cm = 3.34 · 10 g cm

b. For UV light with λ = 100 nm and dust with size a = 0.1µm and solid 3 −3 density ρd = 1g/cm and the solar abundance mass fraction Xd = 2 × 10 , use the geometric cross section opacity derived in the text to compute the mean free path ` (in pc) for this GMC.

−3 −3 −3 2 −1 Xd 0.1 µm 1 g cm 2 −1 2·10 0.1 µm 1 g cm Solution. κd = 150 cm g −3 = 150 cm g −3 −3 = 2·10 a ρd 2·10 0.1 µm 1 g cm 150 cm2 g−1 1 1 19 1 au 1 pc ` = = 2 −1 −24 −3 = 2.00 · 10 cm · 13 · 5 = κdρd (150 cm g )(3.34·10 g cm ) 1.49·10 cm 2.06·10 au 6.52 pc

c. For a GMC of diameter D = 30 pc, compute the optical depth τ and reduction fraction Fobs/F for a star behind the cloud that emits such UV light.

D 30 pc Solution. τUV = ` = 6.52 pc = 4.60 = τUV Fobs Fobs = e−τ = e−4.60 = 0.0101 = F F d. Use this to compute the associated extinction magnetic for this UV light,

AUV .

Solution. AUV = 1.08τUV = 1.08(4.60) = 4.97

e. Assuming a reddening exponent β = 1, now compute the extinction AV for visible light with λ = 500 nm, and the extinction ANIR for near IR light with λ = 2µm.

100 nm Solution. AV = AUV 500 nm = 0.92 = AV 100 nm AIR = AUV 2000 nm = 0.23 = AIR

Quick Question 3: -New The ”pillars of creation” are cold clouds. a. What is an approximate temperature range for the clouds? 10 − 100 K b. If the clouds are at the higher end of that range, what would be their approximate number density? 10 cm−3 – 41 –

Section 19 Quick Question 1: Recall the elementary physics result that an object falling under gravitational acceleration g drops a distance s = gt2/2 in time t. Fixing the gravity at a 2 constant value g = GM/R , use this simple relation to solve for the time tg(R) to fall through a stellar radius (i.e., by setting s = R). Compare this with the free fall time in (19.7) by evaluating the ratio tg(R)/tff .

r 3 2 q 3 √ R Solution. s = gt2/2 =⇒ R = ( GM ) tg =⇒ t = 2R = 2 = t g R2 2 g GM GM g P 1 q 4π2R3 π q R3 tff = = = 4 √4 GM 2 GM tg/tff = 2/(π/2) = 0.90 = tg/tff

Quick Question 2: -New Find the normalization constant, K, for (19.9) in terms of the total number of stars, Ntot, for the three power model. Solution. dN = Km−α dm Z Ntot R ∞ −α dN = 0 Km dm 0 R 0.08 −0.3 R 0.5 −1.3 R ∞ −2.35 Ntot = 0 Km dm + 0.08 Km dm + 0.5 Km dm m0.7 0.08 1 0.5 1 ∞ Ntot = K[ 0.7 ]0 + K[ −0.3m0.3 ]0.08 + K[ −1.35m1.35 ]0.5

Ntot = 0.244K + 3.01K + 1.89K

Ntot = 5.14K

K = 0.195Ntot

Quick Question 3: -New Integrate (19.9) using the three power form for the fraction of stars in the following mass regions (You need to ﬁnd the normalization constant, K, ﬁrst). a. Within 1% of the mass of the sun.

R 1.01 0.195N m−2.35dm Solution. f(0.99 < m < 1.01) = 0.99 tot = 0.0039 Ntot – 42 –

b. Within 10% of the mass of the sun.

R 1.1 0.195N m−2.35dm Solution. f(0.9 < m < 1.1) = 0.9 tot = 0.040 Ntot c. Red dwarf stars.

R 0.5 0.195N m−1.3dm Solution. f(red dwarf) = f(0.08 < m < 0.5) = 0.08 tot = 0.59 Ntot

d. Within 10% of the mass of Betelgeuse, ∼ 12M . Solution. R 13 0.195N m−2.35dm f(0.9∗12 < m < 1.1∗12) = f(11 < m < 13) = 11 tot = 0.0011 Ntot

e. Within 10% of the mass of Proxima Centauri, ∼ 0.12M . Solution. R 0.13 0.195N m−1.3dm f(0.9 ∗ 0.12 < m < 1.1 ∗ 0.12) = f(0.11 < m < 0.13) = 0.11 tot = Ntot 0.062

Quick Question 4: -New Using the Saltpeter model, calculate the fraction of stars within the mass regions from Quick Question 3. a.

Solution. R 1.00 0.57N m−0dm+R 1.01 0.57N m−2.35dm f(0.99 < m < 1.01) = 0.99 tot 1.00 tot = 0.011 Ntot b.

R 1.0 0.57N m−0dm+R 1.1 0.57N m−2.35dm Solution. f(0.9 < m < 1.0) = 0.9 tot 1.0 tot = 0.11 Ntot c.

R 0.5 0.57N m−0dm Solution. f(red dwarf) = f(0.08 < m < 0.5) = 0.08 tot = 0.24 Ntot d.

Solution. R 13 0.57N m−2.35dm f(0.9 ∗ 12 < m < 1.1 ∗ 12) = f(11 < m < 13) = 11 tot = 0.0033 Ntot e.

Solution. R 0.13 0.57N m−0dm f(0.9 ∗ 0.12 < m < 1.1 ∗ 0.12) = f(0.11 < m < 0.13) = 0.11 tot = Ntot 0.011 – 43 –

Quick Question 5: -New Warm clouds rarely form into star clusters from gravitational contraction.

For a warm ISM cloud composed of hydrogen (not H2), with a temperature of 5000 K and number density of 0.1 cm−3 calculate the following properties: a. Jean’s Radius

Solution. R = 9.6 pc( T )1/2 mp = 9.6 pc( 5000 )1/2 mp = 2100 pc J n µ 0.1 mp b. Jean’s Mass

Solution. M = 92M T 3/2 ( mp )2 = 92M 50003/2 ( mp )2 = 1.0 · 108M J n1/2 µ 0.11/2 mp c. Contraction time

51 Myr q 2mp 51 Myr q 2mp Solution. tff = √ = √ = 230 Myr n µ 0.1 mp – 44 –

Section 20 Quick Question 1: a. Generalize the equilibrium temperature equation (20.1) to the case that a body has a ﬁnite albedo, a > 0, i.e. that it reﬂects a fraction a of incoming light, instead of absorbing it.

Solution. 4 2 L 2 σT 4πR 2 4 2 2 2 Ein = (1 − a) 4πd2 (πr ) = (1 − a) 4πd2 (πr ) = (1 − a)σT πR r /d 4 2 Eout = σT 4πr

Ein = Eout (4) 4 2 2 2 4 2 (1 − a)σT πR r /d = σT 4πr (5) 4 2 2 4 (1 − a)T R /4d = T (6)

q 1/2 (1−a) R T = T 2d (7)

q (1−a)1/2(0.0047 au) T = 6000 K 2d (8) q (1−a)au T = 290 K d (9)

b. Derive Earth’s equilibrium temperature given its mean albedo a = 0.3.

q (1−a)au q (1−0.3)au Solution. T = 290 K d = 290 K 1 au = 243 K

Quick Question 2: -New Ceres, a dwarf planet, has a semi-major axis of 2.77 au. a. What is the equilibrium temperature at Ceres?

p au p au Solution. T = 290 K d = 290 K 2.77 au = 174 K b. Accounting for Ceres’ albedo of 0.090, what is its equilibrium temperature (using Quick Question 1)?

p au p au Solution. T = 290 K (1 − a) d = 290 K (1 − 0.090) 2.77 au = 166 K

Quick Question 3: -New Find the equilibrium temperature of the following points in the solar system: a. The orbit of Mercury - 0.38 au

p au p au Solution. T = 290 K d = 290 K 0.38 au = 470 K – 45 –

b. The outer edge of the asteroid belt - 3.2 au

p au Solution. T = 290 K 3.2 au = 160 K c. The orbit of Jupiter - 5.2 au

p au Solution. T = 290 K 5.2 au = 130 K d. The orbit of Saturn - 9.6 au

p au Solution. T = 290 K 9.6 au = 94 K e. The orbit of Uranus - 19 au

p au Solution. T = 290 K 19 au = 67 K f. The orbit of Neptune - 30 au

p au Solution. T = 290 K 30 au = 53 K g. The orbit of Pluto in the Kuiper Belt - 39 au

p au Solution. T = 290 K 39 au = 46 K

Quick Question 4: -New Jupiter has a mean radius of 70, 000 km and a semi-major axis of 5.2 au.

a. In terms of Earth radii, RE = 6, 400 km, what is the radius of Jupiter?

4 1RE Solution. RJ = 7 · 10 km( 6400 km ) = 10.9RE

b. In terms of Sun radii, R ?

4 1R Solution. RJ = 7 · 10 km( 6.96·105 km ) = 0.10R c. What is Jupiter’s angular size from Earth? (In radians and arcseconds)

1.5·108 km 8 Solution. 5.2 au( 1 au ) = 7.8 · 10 km α = 2R/d = 2(7 · 104 km)/7.8 · 108 km = 1.8 · 10−4 rad −4 2·105 arcsec α = 1.8 · 10 rad( 1 rad ) = 36 arcsec d. In steradians, what is Jupiter’s solid angle in Earth’s sky?

Solution. Ω = πα2 = π(1.8 · 10−4 rad)2 = 1.0 · 10−7 sterad – 46 –

Quick Question 5: -New At 0.1 bar the mean temperature in Jupiter’s atmosphere is 110 K. Jupiter has a mass of 1.9 · 1027 kg and a composition of 90% Hydrogen and 10% Helium. At 1 bar, what is the scale height as a fraction of its radius and in absolute terms?

Solution. H = 0.0005 T/T R/R R µ/µ M/M 5 30 T = 6000 K, µodot = 0.6mp, R = 6.96 · 10 km, M = 1.99 · 10 kg mp mp µ = 2X+3Y/4 = 2(0.9)+3(0.1)/4 = 0.53mp

H 110 K/6000 K 70000 km/6.96·105 km H = 0.0005 27 30 = 0.0011 = R 0.53mp/0.6mp 1.9·10 kg/1.99·10 kg R H = 0.0011R = 0.0011(70000 km) = 77 km = H

Quick Question 6: -New

As shown in §16, the upper mass limit for brown dwarf stars is about 0.08M . Their small mass means that their masses are often quoted in Jupiter units. 27 What, in units of MJ = 1.9 · 10 kg, is the upper limit on brown dwarfs?

30 1.99·10 kg 1MJ Solution. 0.08M ( )( 27 ) = 84MJ 1M 1.9·10 kg

Quick Question 7: -New With a mass of 1.9 · 1027 kg and a radius of 70, 000 km, Jupiter is the largest planet in our solar system.

a. In terms of gE = 9.8 m/ s, what is the acceleration of gravity on the edge of its atmosphere?

GM (6.67·10−11 m3 kg−1 s−2)(1.9·1027 kg) Solution. g = R2 = (7·107 m)2 = 26 m/ s 1gE g = 26 m/ s( 9.8 m/ s ) = 2.7gE

b. What is the acceleration of gravity in terms of g ≈ 27gE?

g Solution. g = 2.7gE( ) = 0.1g 27gE

Quick Question 8: -New The asteroid belt, composed of ﬁve million asteroids with an average diameter of 1 km, spans the region of 2.1 au to 3.4 au. Assume the asteroid belt is a disk, and at the outer reaches, the inclination angle of orbits reach a maximum of 10◦. How many times, on average, would a spacecraft have to traverse the asteroid belt before hitting an asteroid? -Could be an exercise – 47 –

1 Solution. `free−path = nσ σ = πr2 = π(0.5 km)2 = 0.79 km2 h = 2(3.4 au tan(10◦)) = 1.2 au 2 2 2 2 3 150∗106 km 3 V = πh(R2 − R1) = π1.2((3.4 au) − (2.1 au) ) = 22 au ( 1 au ) = 7.43 · 1025 au3 n = N/V = (5 · 106)/(7.43 · 1025 au3) = 6.7 · 10−20 km−3 1 19 `free−path = (6.7·10−20 km−3)(0.79 km2) = 1.9 · 10 km 150∗106 km 8 z = 3.4 au − 2.1 au = 1.3 au( 1 au ) = 1.95 · 10 km 19 8 11 Ntraverse = `free−path/z = 1.9 · 10 km/(1.95 · 10 km) ≈ 10 – 48 –

Section 21 Quick Question 1: If we approximate the outer and inner limits of a habitable zone as ranging from where the equilibrium temperature is in the range 0 C < T < 50 C, derive expressions, analogous to eqns. (21.2) and (21.3), for the inner and outer values for the orbital period Pi and Po, and for the orbital distances di and do. Solution. Starting with equation (21.2), P T M T∗ M 1/6 yr 1/3 o ∗ 3 1/2 273 K = To = 290 K( )( ) ( ) =⇒ = 1.2( ) ( ) T M∗ Po yr T M∗ P T M T∗ M 1/6 yr 1/3 i ∗ 3 1/2 323 K = Ti = 290 K( )( ) ( ) =⇒ = 0.90( ) ( ) T M∗ Pi yr T M∗ T T∗ q au ∗ 2 273 K = Ti = 290 K( ) =⇒ di/au = 1.1( ) T di T T T∗ p au ∗ 2 323 K = To = 290 K( ) =⇒ do/au = 0.81( ) T do T

Quick Question 2: -New

For a red dwarf with T∗/T = M∗/M = 1/2, use the results from Quick Question 1 to approximate the limits of the habitable zone.

Solution. Pi T∗ 3 M 1/2 3 −1/2 = 1.2( ) ( ) = 1.2(0.5) (0.5) = 0.21 =⇒ Pi = 77 dy yr T M∗ T∗ 2 2 do/au = 1.1( ) = 1.1(0.5) = 0.28 =⇒ do = 0.28 au T Po T∗ 3 M 1/2 3 −1/2 = 0.90( ) ( ) = 0.90(0.5) (0.5) = 0.16 =⇒ Po = 58 dy yr T M∗ T∗ 2 2 di/au = 0.81( ) = 0.81(0.5) = 0.20 =⇒ di = 0.20 au T

Quick Question 3: -New Rigel is a supergiant star that is one of the brightest stars in the sky. It has a surface temperature of 2.0T and a mass of 21M . What would be this star’s habitable zone?

Solution. Po T∗ 3 M 1/2 3 −1/2 = 1.2( ) ( ) = 1.2(2.0) (21) = 2.09 =⇒ Po = 2.09 yr yr T M∗ T∗ 2 2 do/au = 1.1( ) = 1.1(2.0) = 4.4 =⇒ do = 4.4 au T Pi T∗ 3 M 1/2 3 −1/2 = 0.90( ) ( ) = 0.90(2.0) (21) = 1.6 =⇒ Pi = 1.6 yr yr T M∗ T∗ 2 2 di/au = 0.81( ) = 0.81(0.5) = 3.2 =⇒ di = 3.2 au T – 49 –

Quick Question 4: -New A star of mass 3/Msun wobbles with a periodicity of 1.5 yr. Calculate the planet’s mass in terms of ME if the observed star wobbles with an orbital velocity of: a. 1 cm/ s

M∗/M 1/3 Mp Mp V∗ M∗/M −1/3 Solution. V∗ = 212 km/ s( ) =⇒ = 5 ( ) = P/day M M 2.12·10 cm/ s P/day 1 cm/ s 3M /M −1/3 −5 2.12·105 cm/ s ( 1.5∗356 dy/day ) = 2.7 · 10 −5 −5 30 1ME Mp = 2.7 · 10 M = 2.7 · 10 (1.99 · 10 kg)( 5.98·1024 kg ) = 8.98ME b. 5 cm/ s

Mp 5 cm/ s 3M /M −1/3 −4 Solution. = 5 ( ) = 1.3 · 10 M 2.12·10 cm/ s 1.5∗356 dy/day −4 −4 30 1ME Mp = 1.3 · 10 M = 1.3 · 10 (1.99 · 10 kg)( 5.98·1024 kg ) = 43ME c. 0.5 cm/ s

Mp 0.5 cm/ s 3M /M −1/3 −5 Solution. = 5 ( ) = 1.3 · 10 M 2.12·10 cm/ s 1.5∗356 dy/day −5 −5 30 1ME Mp = 1.3 · 10 M = 1.3 · 10 (1.99 · 10 kg)( 5.98·1024 kg ) = 4.3ME – 50 –

Section 22 Quick Question 1: At the distance d ≈ 8 kpc of the Galactic Center, the sun turns out to have an orbital speed Vo ≈ 220 km/s. How long is one “galactic year”, i.e., the sun’s orbital period (in Myr) around the galaxy?

3·1016 km 17 Solution. 8 kpc · ( 1 kpc ) = 2.4 · 10 km 2πR 2π(2.4·1017 km) 15 1 dy 1 yr 8 P = = −1 = 6.9·10 s·( )·( ) = 2.2·10 yr = 220 Myr Vo 220 km s 86400 s 365 dy

Quick Question 2: What is the energy E, in eV, of the hyperﬁne, spin-ﬂip transition that gives rise to the 21-cm emission line of neutral Hydrogen.

Solution. E = 1.2 eV µm/λ = 1.2 eV µm/21 cm = 5.7 · 10−6 eV

Quick Question 3: At the distance d ≈ 8 kpc of the Galactic Center: a. What is the physical size s (in AU) of an angle α = 1 arcsec?

α d 1 arcsec 8 kpc Solution. s/au = ( arcsec )( pc ) = ( arcsec )( pc ) =⇒ s = 8000 au

b. What is the angular radius αc (in arcsec ) of the central parsec cluster? s d 206265 au 8000 pc αc/arcsec = ( au )/( pc ) = ( au )/( pc ) = 25.8 =⇒ αc = 25.8 arcsec

Quick Question 4: -New What is the Schwarzschild Radius for the super massive black hole at our galactic center in km, au and solar radius?

Solution. 6 Mbh 4·10 M 7 Rbh = 3 km = 3 km = 1.2 · 10 km M M 7 1 au 1.2 · 10 km( 1.5·108 km ) = 0.08 au 7 1R 1.2 · 10 km( 7.0·105 km ) = 17R – 51 –

Section 23 Quick Question 1: -New Using equation (23.3), ﬁnd the Hubble time for the following estimates of the Hubble constant: a. Hubble’s original estimate of 500(km/s)/Mpc,

Solution. t ≈ 10 Gyr 100( km/ s)/ Mpc = 10 Gyr 100( km/ s)/ Mpc = 2 Gyr H Ho 500( km/ s)/ Mpc b. Current approximation of 69(km/s)/Mpc.

Solution. t ≈ 10 Gyr 100( km/ s)/ Mpc = 10 Gyr 100( km/ s)/ Mpc = 14.5 Gyr H Ho 69( km/ s)/ Mpc – 52 –

Section 24 Quick Question 1:

For accretion down to a radius that is a factor Racc/Rs times the Schwarzschild 2 radius of black hole, compute the energy eﬃciency factor = Eg/mc for the 2 gravitational energy gain Eb as a fraction of the rest mass energy mc of the accreted mass. Conﬁrm that = 0.1 for Racc/Rs = 5.

2 Solution. Rs = 2GM/c Eg GMm/Racc GM Rs = 2 = 2 = 2 = mc mc Raccc 2Racc 1 = 2·5 = 0.1

Quick Question 2: Suppose a quasar shows absorption from a Lyman-alpha cloud at an observed wavelength λobs = 183 nm. a. What is the redshift z for this cloud. -Is the λLα right? I used what was in the solution, but it seems like λLα ≈ 122 nm instead

Solution. z = λobs−λLα = 183 nm−91.5 nm = 1 λLα 91.5 nm

b. What is its inferred recession speed vr?

5 −1 5 −1 Solution. vr = zc = 1(3 · 10 km s ) = 3 · 10 km s

c. For a Hubble constant Ho = 67 (km/s)/Mpc, what is its distance?

3·105 km s−1 Solution. d = vr/Ho = 67 (km/s)/Mpc = 4500 Mpc

Quick Question 3: -New The ﬁrst quasar discovered was aﬀectionately named 3C48. a. It has a redshift of 0.367, what was the observed wavelength of the Lyman α transition (122 nm)?

λobs−λLα Solution. z = =⇒ λobs = λLα(z + 1) = 122 nm(0.367 + 1) = 167 nm λLα

b. What is the speed at which 3C48 moves away from us assuming its speed has been constant?

Solution. V = zc = 0.3673 · 105 km s−1 = 1.10 · 105 km s−1

c. Assuming that speed has been constant over cosmological time, how far away is 3C48 (in Mpc and ly)? – 53 –

5 −1 −1 Solution. V = Hod =⇒ d = V/Ho = (1.10 · 10 km s )/(70 km s /Mpc) = 1570 Mpc 9 3.26 ly 9 1570 Mpc = 1.57 · 10 pc( 1 pc ) = 5.12 · 10 ly = 5.12 Gly d. 3C48 has an apparent magnitude of 16.2, what is the absolute magnitude of 3C48?

Solution. M = m − 5 log(d/ pc) = 16.2 − 5 log(1570 Mpc/ pc) = −29.8

e. How much luminosity is generated from accretion?

Solution. M = 4.8 − 2.5 log(L/L ) =⇒ −29.8 = 4.8 − 2.5 log(L/L ) =⇒ 13 13.8 = log(L/L ) =⇒ 6.31 · 10 L

−1 f. Assuming it has an accretion rate of 1M yr , what is the conversion eﬃciency? Does this seem like its physically possible?

−1 12 M˙ acc 13 12 1M yr Solution. L = 1.4·10 L −1 =⇒ 6.31·10 L = 1.4·10 L −1 =⇒ 0.1 M yr 0.1 M yr = 4.51

g. Whats the minimum accretion rate that would lead to the conversion eﬃciency to be 100%?

12 M˙ acc 13 12 1 M˙ acc Solution. L = 1.4·10 L −1 =⇒ 6.31·10 L = 1.4·10 L −1 =⇒ 0.1 M yr 0.1 M yr ˙ −1 Macc = 4.51M yr i. Now, assume the eﬃciency is 10%, what is the accretion rate?

12 M˙ acc 13 12 0.1 M˙ acc Solution. L = 1.4·10 L −1 =⇒ 6.31·10 L = 1.4·10 L −1 =⇒ 0.1 M yr 0.1 M yr ˙ −1 Macc = 45.1M yr – 54 –

Section 25 Quick Question 1: -New The average density based on our local region is ∼ 7·10−24 g cm−3 (from §18). -This problem could easily turn into questions of magnitudes as these are rough estimates. I ﬁgured removing is quicker than adding. a. The Milky Way has a radius of 15 kpc and is 0.3 kpc thick. It has a mass 12 10 M . What is the average density of the Milky Way? How does it compare to our local density?

Solution. 2 2 3 3.09·1021 cm 3 66 3 V = πR H = π(15 kpc) (0.3 kpc) = 212 kpc ( 1 kpc ) = 6.25 · 10 cm 12 2·1033 g 45 M = 10 M ( ) = 2 · 10 g 1M 45 66 3 −22 −3 ρMW = M/V = 2 · 10 g/6.25 · 10 cm = 3.2 · 10 g cm The average density of the Milky Way is about a factor of 100 greater than our local density.

15 b. Our local supercluster, Virgo, has a mass of 1.48 · 10 M and a diameter of 33 Mpc. Assuming it is a sphere, what is its average density and how does it compare to our local density and that of the Milky Way? -New

4 3 4 3 5 3.09·1024 cm 3 78 3 Solution. V = 3 πr = 3 π(33 Mpc) = 1.51·10 Mpc( 1 Mpc ) = 4.46·10 cm 15 2·1033 g 48 48 78 3 M = 1.48·10 M ( ) = 2.9·10 g ρLS = M/V = 2.9·10 g/4.46·10 cm = 1M 6.50 · 10−31 g cm−3 It is a factor 107 smaller than that of our local region and 109 smaller than of the Milky Way. – 55 –

Section 26 Quick Question 1: a. What is the age (in Gyr) of an “empty” universe with constant expansion and Hubble constant Ho = 67 (km/s)/Mpc?

Solution. R(t) = 1 + Hot R(−tage) = 0 =⇒ 0 = 1 + Ho(−tage) =⇒ tage = 1/Ho = 10 Gyr/ho = 10 Gyr/0.7 = 14.3 Gyr

b. What is the age (in Gyr) of a “critical” universe (Ωm = 1) and Hubble constant Ho = 67 (km/s)/Mpc?

3 2/3 Solution. R(t) = (1 + 2 Hot) 3 2/3 3 R(−tage) = 0 =⇒ 0 = (1 + 2 Ho(−tage)) =⇒ 0 = 1 − 2 Ho(tage) =⇒ H t = 2 =⇒ t = 2/3 = (2/3)10 Gyr ≈ (2/3)10 Gyr = 10 Gyr o age 3 age Ho ho 2/3

Quick Question 2:

Calculate the Hubble time, tH , and Hubble distance, dH , for universes with the following Hubble constants: a. 500( km/ s)/ Mpc

Solution. 1 Mpc π·107 s −10 −1 500( km/ s)/ Mpc( 3·1019 km )( yr ) = 5.24 · 10 yr −10 −1 tH = 1/Ho = 1/(5.24 · 10 yr ) = 1.91 Gyr = tH 5 dH = c/Ho = (3 · 10 km/ s)/500( km/ s)/ Mpc = 600 Mpc = dH b. 67( km/ s)/ Mpc

Solution. 1 Mpc π·107 s −11 −1 67( km/ s)/ Mpc( 3·1019 km )( yr ) = 7.02 · 10 yr −11 −1 tH = 1/Ho = 1/(7.02 · 10 yr ) = 14.2 Gyr = tH 5 dH = c/Ho = (3 · 10 km/ s)/67( km/ s)/ Mpc = 4480 Mpc = dH c. 69( km/ s)/ Mpc

Solution. 1 Mpc π·107 s −11 −1 69( km/ s)/ Mpc( 3·1019 km )( yr ) = 7.23 · 10 yr −11 −1 tH = 1/Ho = 1/(7.23 · 10 yr ) = 13.8 Gyr = tH 5 dH = c/Ho = (3 · 10 km/ s)/67( km/ s)/ Mpc = 4350 Mpc = dH – 56 –

d. 70( km/ s)/ Mpc

Solution. 1 Mpc π·107 s −11 −1 70( km/ s)/ Mpc( 3·1019 km )( yr ) = 7.33 · 10 yr −11 −1 tH = 1/Ho = 1/(7.33 · 10 yr ) = 13.6 Gyr = tH 5 dH = c/Ho = (3 · 10 km/ s)/67( km/ s)/ Mpc = 4290 Mpc = dH e. 74( km/ s)/ Mpc

Solution. 1 Mpc π·107 s −11 −1 74( km/ s)/ Mpc( 3·1019 km )( yr ) = 7.75 · 10 yr −11 −1 tH = 1/Ho = 1/(7.75 · 10 yr ) = 12.9 Gyr = tH 5 dH = c/Ho = (3 · 10 km/ s)/74( km/ s)/ Mpc = 4050 Mpc = dH – 57 –

Section 27 Quick Question 1: -New Most estimates of the Hubble constant lie within 67 and 74 ( km/ s)/ Mpc. Assuming an object is independently known to have a distance of 1 Gly, what would be the range of Doppler shifts measured for this object? (This distance is short enough to be approximately linear)

Solution. Ho V = zc & V = Hod =⇒ zc = Hod =⇒ z = c d 306 Mpc 1 Gly( 1 Gly ) = 306 Mpc 67( km/ s)/ Mpc zlower = 3·105 km s−1 (306 Mpc) = 0.068 74( km/ s)/ Mpc zupper = 3·105 km s−1 (306 Mpc) = 0.075 0.068 ≤ z ≤ 0.075 – 58 –

Section 28 Quick Question 1: -New Using the Planck analysis, what will be the Hubble constant in the distant future?

Solution.

1 − Ωm = ΩΛ = 0.682 Ω = Ω + Ω = 0.317 m √b dm √ ˙ R∞ = 1 − Ωm = 1 − 0.682 = 0.564 ˙ H∞ = Ho ∗ R∞ = 68.15 ( km/ s)/ Mpc ∗ 0.5639 = 38.43 ( km/ s)/ Mpc – 59 –

Section 29 Quick Question 1: -New Use dimensional analysis to derive an expression for the length associated with combining Newton’s gravitation constant G, Planck’s constant h, and the speed of light c. This represents the “Planck” length scale at which gravity and quantum physics meld together.

Solution. G = 6.67 · 10−11 m3 kg−1 s−2 h = 6.63 · 10−34 kg m2 s−3 c = 3 · 108 m s−1 Gh = 4.42 · 10−44 m5 s−3 The only way to get rid of kg c3 = 2.70 · 1025 m3 s−3 To get rid of s q √ Gh −69 2 −35 c3 = 1.64 · 10 m = 4.05 · 10 m – 60 –

Section 30 – 61 –

Appendix A Quick Questions 1:

a. Compute the wavelengths (in nm) for Paschen-α λ43 and the Paschen limit λ∞3. Solution. λ = 91.2 nm mn 1 − 1 n2 m2 λ = 91.2 nm = 1880 nm = λ 43 1 − 1 43 32 42 λ = 91.2 nm = 820 nm − λ ∞3 1 − 1 ∞3 32 ∞2

b. What are the associated changes in energy (in eV), ∆E43 and ∆E∞3.

1 1 Solution. ∆Emn = 13.6 eV( n2 − m2 ) 1 1 ∆E43 = 13.6 eV( 32 − 42 ) = 0.66 eV = ∆E43 1 1 ∆E∞3 = 13.6 eV( 32 − ∞2 ) = 1.5 eV = ∆E∞3 – 62 –

Appendix B Quick Question 1:

The n = 2 level of Hydrogen has g2 = 8 states, while the ground level has just g1 = 2 states. Using the energy diﬀerence ∆E21 from the Bohr atom, compute the Boltzmann equilibrium number ratio n2/n1 of electrons in these levels for a temperature T = 100, 000 K. -Not conﬁdent in my solution

n2 g2 2 −∆E1/kT Solution. = ( 3 )e n1 g1 neΛ g2 = 8 = 4 g1 2 1 1 ∆Emn = 13.6 eV( n2 − m2 ) 1 1 ∆E1 = ∆E∞1 = 13.6 eV( n2 − ∞2 ) = 13.6 eV -Pretty sure its the ionization energy and not the transition energy that is meant ∆E1 13.6 eV kT = (8.62·10−5 eV K−1)(105 K) = 1.58 −15 Λ = √ h = √ 4.14·10 eV s = c · 7.87 · 10−19 s = 2.36 · 2πmekT 2π·5.11·105 eV/c2(8.62·10−5 eV K−1)(105 K) 10−10 m −11 3 30 3 ne = N/V = 1/(4π · (5.29 · 10 m) /3) = 1.62 · 10 m -Guess, 1 electron per volume of hydrogen. Probably way overestimating 2 2 3 = 30 3 −10 3 = 0.047 -This probably isn’t what is intended? neΛ (1.62·10 m )(2.36·10 m) 2 10 3 = 10 -Based on assuming its a stellar atmosphere and taking the value neΛ from text. If this is meant, then the question should clarify. If this, then the calculation of Λ is not needed either. I use this to calculate the final answer. Another alternative that could be desired would be to use PV = nkT to find the volume of one mole of gas and calculate ne of one mole of Hydrogen gas, but I didn’t try this one as there is no pressure so I was figuring that it isn’t what was meant. n2 = 4 · 1010 · e−1.58 = 8.24 · 109 ≈ 1010 n1 – 63 –

Appendix C Quick Question 1: a. Seen standing up, what is the cross section (in cm2) of a person with height 1.8 m and width 0.5 m?

Solution. σ = hw = 180 cm · 50 cm = 9000 cm2

b. If this person has a mass of 60 kg, what is his/her “opacity” κ = σ/m, in cm2/g?

Solution. κ = σ/m = 9000 cm2/60000 g = 0.15 cm2/g

c. How does this compare with the typical opacity of stellar material?

Solution. It’s about half the opacity of stellar material. – 64 –

Appendix D Quick Question 1: -New The Sun has a surface temperature of 5800 K, what is its surface ﬂux?

4 −8 −2 −4 4 7 −2 Solution. F = σsbTeff = 5.67 · 10 W m K (5800 K) = 6.42 · 10 W m