Fundamentals of Astrophysics: Quick Question Solutions

Chapter 1

Quick Questions

1. Discuss how we can estimate the temperature of warm or hot object, e.g. a stove, without touching it.

2. Discuss the ways we estimate distances and sizes in our everyday world.

3. Discuss what sets our perception limits on the smallest intervals of time. How might this differ in creatures of different size, e.g. a fly vs. a human?

4. For a typical car highway speed of 100 km/hr, about how long does it take to travel from coast to coast? How does this compare to how long it would it take drive to the moon? To the Sun? To alpha centauri?

5. Why might astronomical observations be useful in measuring the speed of light?

6. How does the speed of sound on Earth compare to the speed of light?

7. About how old is the oldest living thing on Earth? What about the oldest animal?

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. – 2 –

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. – 3 –

Chapter 2

Quick Questions

1. Betelgeuse has a diameter of about 1800 R , and a distance of d ≈ 220 pc. What is its angular diameter in milli-arcsec?

2. What is the distance (in pc)to a star with a parallax of 0.1 00?

3. The average separation between human eyes is s ≈60 mm. Derive then a general formula for the distance d (in m) to an object with visual parallax angle α.

4. If we lived on Mars instead of Earth, what would be the length of a parsec (in km).

5. Over a period of several years, two stars appear to go around each other with a ﬁxed angular separation of 1 00. What is the physical separation, in au, between the stars if they have a distance d = 10 pc from Earth?

6. 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 00. How about from a distance of d = 1 kpc?

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 – 4 –

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)

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 – 5 –

Chapter 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? – 6 –

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.) – 7 –

Chapter 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 – 8 –

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 – 9 –

Chapter 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 – 10 –

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. – 11 –

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) – 12 –

Chapter 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 – 13 –

Chapter 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: – 14 –

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 – 15 –

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 – 16 –

Chapter 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 – 17 –

b. Suppose instead, that 61 Cygni A was powered by gravitational contrac- tion. 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. – 18 –

Chapter 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 – 19 –

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 – 20 –

Chapter 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. – 21 –

−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 – 22 –

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% – 23 –

Chapter 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 rota- tion 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 – 24 –

Chapter 12 Quick Question 1: “Opacity” of people vs. electrons a. Seen standing up, about what is the cross section (in cm2) of a person with height 1.8 m and width 0.5 m?

Solution. σ = h × w = (180 cm) × (50 cm) = 9000 cm2 = σ

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 cm2/60, 000g = 0.15 cm2/g = κ

c. How does this compare with the (Thomson) electron scattering opacity κe for a fully ionized gas with solar abundances?

Solution. 2 κe = 0.34 cm /g ≈ 2.1 × κperson

Quick Question 2:

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

Solution. q L q L 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:

What is the electron scattering opacity κe for a fully ionized medium with the mass fractions of Hydrogen, Helium, and metals given by X = 0, Y = 0.98 and Z = 0.02.

Solution. 2 2 κe = (1 + X)0.2 cm /g = 0.2 cm /g. – 25 –

Quick Question 4: What is the extinction magnitude A for a star behind an interstellar cloud with optical thickness τ = 4?

Solution. A = 2.5 log eτ = 1.086 τ = 4.34.

Quick Question 5: Suppose dust absorption has a reddening exponent β = 1. Write a formula

for the ratio in extinction magnitude A1/A2 between that at a wavelength λ1 and that at λ2 = fλ1. Evaluate this for f = 2, 5, 10, 100.

Solution. −β A1/A2 = (1.086τ1/1.086τ2) = τ1/τ2 = λ2/λ1 = f, since τ ∼ κ ∼ λ ∼ 1/λ. So for f = 2, 5, 10, 100, ratio is A1/A2 = 2, 5, 10, 100.