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.
s/ au
au
au/ au
α
= d/
=⇒ d/ pc = α/s/arcsec
= 2(0.31362arcsec)/
= 3.189 =⇒ d =
arcsec
pc
arcsec
3.189 pc
ly
d = 3.189 pc( 3.276
) = 10.4 ly
pc
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.
α = 0.00377 arcsec( 2·1015rad
) = 1.89 · 10−8 rad
arcsec
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
Solution. d = arcsec
pc = 0.00377
pc = 265 pc
α
arcsec
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
α = 2 ∗ RE / au = 2 ∗ 6400 km/(150 · 106 km) = 8.6 · 10−5 rad
5 arcsec
α = 8.6 · 10−5 rad 2·101 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 log(1 au/10 pc) = 5 + 5 log((1/2 · 105 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 ?
L2
L2
L1
d2 2
Solution. F1 = F2 =⇒ 4πd
=⇒ dd21 = 10
2 = 4πd2 =⇒ L = ( d )
1
1
1
2
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.
L /4πR2
R1 2
R2 2
1 L2
I2 /I1 = 100 = L12 /4πR21 = LL12 ( R
) =⇒ ( R
) = 100
= 1 =⇒
L1
2
1
2
Quick Question 3:
A white-dwarf-supernova with peak luminosity L ≈ 1010 L
have an apparent magnitude of m = +20 at this peak.
a. What is its Absolute Magnitude M ?
R2
=1
R1
is observed to
–6–
10
Solution. M = 5 − 2.5 log( 10·10L L ) = −20
b. What is its distance d (in pc and ly)?
Solution.
m − M = 5 log( 10dpc ) =⇒ 20 − −20 = 5 log( 10dpc ) =⇒ 8 = log( 10dpc ) =⇒
d
= 1 · 108
10 pc
ly
=⇒ d = 1 · 109 pc d = 1 · 109 pc( 3.26
) = 3.26 · 109 ly
1 pc
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 = 4πd2 F
L = 4π(1 au)2 F
2F
L
= 4π(14πd
= d2au (F/F )
L
au)2 F
(L/L ) = d2au (F/F )
Quick Question 5: -New
An observer measures the flux of a star to be 10−10 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.
5 au
50 pc( 2·10
) = 107 au
1 pc
(L/L ) = d2au (F/F ) = (107 au)2 (10−10 F /F ) = 104 =⇒ L = 104 L
b. Calculate the absolute magnitude of the star.
Solution. M = 5 − 2.5 log(L/L ) = 5 − 2.5 log(104 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. PP12 = λλ21 /c
= λλ12 = 2
/c
b. What is the ratio of their frequency ν2 /ν1 ?
2
= λλ12 =
Solution. νν12 = c/λ
c/λ1
1
2
c. What is the ratio of their energy E2 /E1 ?
2
2
Solution. E
= hν
= νν12 =
E1
hν1
1
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.)
nm
Solution. T = 6000 K 500
λmax
500 nm
T100 = 6000 K 100
= 30000 K = T100
nm
500 nm
T300 = 6000 K 300
= 10000 K = T300
nm
500 nm
T1000 = 6000 K 1000 nm = 3000 K = T1000
500 nm
T3000 = 6000 K 3000
= 1000 K = T3000
nm
b. Conversely, estimate the peak wavelengths λmax of stars with T =2000,
10,000, and 60,000 K.
K
Solution. λmax = 500 nm 6000
T
K
λmax,2000 = 500 nm 6000
= 1500 nm = λmax,2000
2000 K
6000 K
= 300 nm = λmax,10000
λmax,10000 = 500 nm 10000
K
6000 K
λmax,60000 = 500 nm 60000
= 50 nm = λmax,60000
K
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.
T = 50◦ F = (50◦ F − 32◦ F) 59 ◦ C = 10◦ C = 283 K ≈ 300 K
K
λmax = 500 nm 6000
= 10000 nm = 10 µm
300 K
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?
−9
500 nm
500·10 m
−15
= 3·10
s = 1.7 fs
Solution. P = λmax,
8 m s−1 = 3·108 m s−1 = 1.7 · 10
c
b. What is the corresponding frequency?
Solution. ν = 1/Podot = 1/(1.7 · 10−15 s) = 5.9 · 1014 Hz = 590 THz
c. What is the energy of a photon at this frequency?
Solution. Eγ, = hν = (6.6 · 10−34 J s)(5.9 · 1014 Hz) = 3.9 · 10−19 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
FB /FV = 0.56
=⇒ 0.63 = 2.5 log(FV /FB ) =⇒ FV /FB = 1.8 =⇒
–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.
Equations. L = L (R/R )2 (T /T )4
M = 5 − 2.5 log( LL )
nm
λmax = 500
T /T
pc/α
m = M + 5 log( 10dpc ) = M + 5 log( arcsec
) = M + 5 log( 1/0.001
) = M + 10
10 pc
10
a.
L = L (10R /R )2 (T /T )4 = 100L = La
M = 5 − 2.5 log( 100L
) = 0 = Ma
L
λmax = T500/Tnm = 500 nm = λmax,a
m = 0 + 10 = 10 = ma
b.
L = L (R /R )2 (10T /T )4 = 10000L = Lb
M = 5 − 2.5 log( 10000L
) = −5 = Mb
L
500 nm
= 50 nm = λmax,b
λmax = 10T
/T
m = −5 + 10 = 5 = mb
c.
L = L (10R /R )2 (10T /T )4 = 1000000L = Lc
M = 5 − 2.5 log( 1000000L
) = −10 = Mc
L
500 nm
λmax = 10T
= 50 nm = λmax
/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 ?
Solution. m − M = 5 log( 10dpc ) = 5 log( 10100pc ) = 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 )?
nm
nm
Solution. T = 500
T = 500
T = 2T
λmax
250 nm
f. Radius R (in R )?
Solution.
L/L = (R/R )2 (T /T )4
=⇒
2
100L /L (2T /T )−4 = 10
24
=⇒
(R/R )2 = q
L/L (T /T )−4 = (R/R )2 =
R/R
=
102
24
= 10
= 2.5
22
2.5R
g. Angular radius α (in radian and arcsec)?
Solution.
13 km
α = R/d = (6.96 · 105 km)/(100 pc 3.09·10
) = 2.25 · 10−10 rad
1 pc
5
arcsec
α = 2.25 · 10−10 rad 2·101 rad
= 4.5 · 10−5 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 ?
Solution. B = σsbπT
4
=⇒ B/B = T 4 /T 4 = (2T )4 /T 4 = 24 = 16
Quick Question 3:
Use Equation (5.5) to determine the radius of the Sun.
=⇒
R =
– 11 –
Solution. R =
q
F (d)
d=
σsb T 4
q
1400 W m−2
(1.5·1011 m) =
(5.67·10−8 J m−2 s−1 K−4 )(6000 K)4
6.55 · 108 m
– 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∗ /R∗
M∗ R∗ −1
Solve for log gast . g = GM
= M
=⇒ log g∗ =
=⇒ g∗ /g = GM
( R2 )
R2
/R2
M∗ R∗ −2
M∗
log( M
( R ) ) + log g = 4.44 + log( M
) − 2 log( RR∗ )
a. M = 10M and R = 10R
Solution. log g∗ = 4.44 + log( 10M
) − 2 log( 10R
) = 3.44
M
R
b. M = 1M and R = 100R
) − 2 log( 100R
) = 0.44
Solution. log g∗ = 4.44 + log( 1M
M
R
c. M = 1M and R = 0.01R
) − 2 log( 0.01R
) = 8.44
Solution. log g∗ = 4.44 + log( 1M
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
M/M
−1
Reformulate into solar units. Ve = 2GM
=⇒
V
=
618
km
s
e
R
R/R
a.
−1
q
10M /M
10R /R
= 618 km s−1
Solution. Ve = 618 km s−1
q
1M /M
100R /R
= 61.8 km s−1
q
1M /M
0.01R /R
= 6180 km s−1
Solution. Ve = 618 km s
b.
c.
−1
Solution. Ve = 618 km s
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 –
Solution. Vorb =
q
M/M
−1
30 km s
r/1 au
q
GM
r
=⇒
Vorb /Ve =
q
GM/r
GM /re
=
q
M/M
r/re
=⇒
Vorb =
a. M = 10M and d = 10 au.
q
10M /M
−1
Solution. Vorb = 30 km s
= 30 km s−1
10 au/1 au
b. M = 1M and d = 100 au.
q
1M /M
−1
= 3 km s−1
Solution. Vorb = 30 km s
100 au/1 au
c. M = 1M and d = 0.01 au.
q
1M /M
−1
Solution. Vorb = 30 km s
= 300 km s−1
0.01 au/1 au
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.
Reformulate into solar units. U = − GMd m =⇒ U/Ue, = − GMd m / − GM1 aume =
( MM )( mme )( 1 dau )
a.
1 au
e
)( 10
Solution. U/Ue, = ( 10M
)( 10m
) = 10
M
me
au
b.
1 au
e
)( 1m
)( 100
Solution. U/Ue, = ( 1M
) = 0.01
M
me
au
c.
1 au
e
Solution. U/Ue, = ( 1M
)( 1m
)( 0.01
) = 100
M
me
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
−11 m3 kg−1 s−2 )(5.97·1024 kg)(5·104 kg)
m
Solution. W = GM
= (6.67·10
R
6.4·106 m
= 3.11 · 1012 J
– 15 –
b. The moon
−11 m3 kg−1 s−2 )(7.34·1022 kg)(5·104 kg)
m
Solution. W = GM
= (6.67·10
R
1.7·106 m
= 1.44 · 1011 J
c. The Sun
−11 m3 kg−1 s−2 )(1.99·1030 kg)(5·104 kg)
m
Solution. W = GM
= (6.67·10
R
6.96·108 m
= 9.54 · 1015 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 ?
Solution. L/L ∝ (M/M )3
tms = 10000 Myr( MM )2
M
tms = 10000 Myr( 10M
)2 = 100 Myr = tms,a
L/L = (10M /M )3 =⇒ L = 103 L = La
b. 0.1 M ?
M
)2 = 106 Myr = tms,b
Solution. tms = 10000 Myr( 0.1M
L/L = (0.1M /M )3 =⇒ L = 10−3 L = Lb
c. 100 M ?
M
)2 = 1 Myr = tms,c
Solution. tms = 10000 Myr( 100M
L/L = (100M /M )3 =⇒ L = 106 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 ?
Solution. tcluster = 10000 Myr( LL )−2/3
)−2/3 = 464 Myr
tcluster = 10000 Myr( 100L
L
tcluster = 10000 Myr( 10000L
)−2/3 = 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
Solution. tchem = MLc
tchem,
=⇒
tchem /tchem,
M/M
L/L
/M
tchem = 15, 000 yr 0.70M
= 70, 000 yr
0.15L /L
2
= MLc / ML c
2
=⇒
tchem =
– 17 –
b. Suppose instead, that 61 Cygni A was powered by gravitational contraction. What would be the expected lifetime of this process?
2
3 GM
Solution. tKH = 10
RL
2
GM 2
2
(M/M )
3 GM
3
=⇒ tKH /tKH, = 10
/ 10
= (R/R
=⇒
RL
R L
)(L/L )
2
(M/M )
tKH = tKH, (R/R
)(L/L )
2
/M )
tKH = 30 Myr (0.67R(0.70M
= 146 Myr
/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
Solution. tms = 10 Gyr( 0.7M
)2 = 20 Gyr
Quick Question 4: -New
Based on Figure 8.1, what is the age of M55?
Solution. Lto ≈ 3L
L
)2/3 = 4.8 Gyr
tcluster = 10 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?
ly
) = 1.71 · 107 ly = 0.0171 Gly
Solution. 5 · 106 pc( 3.26
1 pc
M
tms = 10 Gyr( MM )2 = 10 Gyr( 10M
)2 = 0.01M
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.
arcsec/10
Solution. Vt = 4.7 km s−1 µp = 4.7 km s−1 20.02
= 47 km s−1
arcsec
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?
nm−600 nm
o
= 600.09600
= 1.5 · 10−4
Solution. z = λobsλ−λ
nm
o
c. What is the star’s radial velocity Vr , in km/s?
Solution. z = Vcr =⇒ Vr = zc = 1.5 · 10−4 (3 · 108 m s−1 ) = 45 km s−1
d. What is the star’s total space velocity Vtot , in km/s?
p
p
Solution. Vtot = Vr2 + Vt2 = (45km s−1 )2 + (47 km s−1 )2 = 65 km s−1
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?
−1216 nm
o
Relevant Equations. z = λobsλ−λ
= λobs1216
, z = Vcr =⇒ Vr = zc
nm
o
a. 1216.5 nm
nm−1216 nm
Solution. z = 1216.51216
= 4.1 · 10−4
nm
Vr = 4.1 · 10−4 (3 · 105 km s−1 ) = 123 km s−1
Away
b. 1215.5 nm
– 19 –
nm−1216 nm
Solution. z = 1215.51216
= −4.1 · 10−4
nm
Vr = −4.1 · 10−4 (3 · 105 km s−1 ) = −123 km s−1
T owards
c. 1215.85 nm
nm−1216 nm
Solution. z = 1215.851216
= −1.2 · 10−4
nm
Vr = −1.2 · 10−4 (3 · 105 km s−1 ) = −36 km s−1
T owards
d. 1216.15 nm
nm−1216 nm
= 1.2 · 10−4
Solution. z = 1216.151216
nm
Vr = 1.2 · 10−4 (3 · 105 km s−1 ) = 36 km s−1
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?
m
1 hr
Solution. 250 km hr−1 ( 1000
)( 3600
) = 69 m s−1
1 km
s
∆λ
69 m s−1
−7
= Vc =⇒ 400∆λnm = 3·10
nm
8 m s−1 =⇒ ∆λ = 2.3 · 10
λ
0
−7
λ = λ + ∆λ = 400 nm + 2.3 · 10 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?
Solution. d = 0.2arcsec
pc = 10 pc
arcsec/2
b. What is the physical distance between the stars, in au.
d
α
arcsec
Solution. s/ au = pc
= 10pcpc 1arcsec
= 10 =⇒ s = 10 au
arcsec
c. In solar masses, what are the masses of each star, M1 and M2 .
Solution.
(s/ au)3
M1 +M2
= (P/
= 10 =⇒ M1 + M2 = 10M
M
yr)2
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?
Solution. M1 = ( VV2e )3 Pyr (1 + VV12 )2 M , M2 = ( VV1e )3 Pyr (1 + VV21 )2 M
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 –
V1 = 60 km s−1 = 2Ve , |V2 | = 30 km s−1 = 1Ve
e 2
M1 = ( VVee )3 (20/365)(1 + 2V
) M = 0.49M = M1
1Ve
e 3
e 2
M2 = ( 2V
) (20/365)(1 + 1V
) M = 0.99M = M2
Ve
2Ve
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 =
20 km s−1 and V2 = 10 km s−1 . 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
s
t2 = 5 · 104 hr( 3600
) = 1.8 · 104 s
hr
s
t4 = 100 hr( 3600
) = 3.6 ∗ ·105 s
hr
R2 = (t2 − t1 )(V1 + V2 )/2 = (1.8 · 104 s − 0 s)(20 km s−1 + 10 km s−1 )/2 = 2.7 ∗
1R
= R2
105 km( 6.96·10
5 km ) = 0.39R
R1 = (t4 − t2 )(V1 + V2 )/2 = (3.6 ∗ ·105 s − 1.8 · 104 s)(20 km s−1 + 10 km s−1 )/2 =
1R
5.1 · 106 km( 6.96·10
= R1
5 km ) = 7.3R
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(M M ), 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
Solution. Lapprox /L = (M/M )3 =⇒ Lapprox = (M/M )3 L = (M /M )3 L =
L
log(L/L ) = 0.1+3.1 log(M M ) = 0.1+3.1 log(M M ) = 0.1 =⇒ L = 1.25L
−L|
= |1−1.25|
= 0.2 = 20%
= |Lapprox
L
1.25
b. 10M
Solution. Lapprox = (10M /M )3 L = 1000L
log(L/L ) = 0.1 + 3.1 log(10M M ) = 3.2 =⇒ L = 1600L
= |1000−1600|
= 0.38 = 38%
1600
c. 0.1M
– 22 –
Solution. Lapprox = (0.1M /M )3 L = 0.001L
log(L/L ) = 0.1 + 3.1 log(0.1M M ) = −3 =⇒ L = 0.001L
= |0.001−0.001|
= 0 = 0%
0.001
d. 100M
Solution. Lapprox = (100M /M )3 L = 106 L
log(L/L ) = 0.1 + 3.1 log(100M M ) = 6.3 =⇒ L = 2.0 · 106 L
6 −2.0·106 |
= |10 2.0·10
= 0.5 = 50%
6
– 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.
∆λf ull
= 2V csin i
λo
=⇒
s
105 km s−1 ) = 300 km
2
−1
V sin i =
= 150 km s
∆λf ull
c
2λo
−1
0.5 nm
1
= 2·500
(3 · 105 km s−1 ) = 2·10
3 (3 ·
nm
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
sin iP
= 150 km s 2π∗30∗86400 s = 6.2 · 107 km
Solution. Rmin = Vrot2π
1R
Rmin = 6.2 · 107 km( 6.96·10
5 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
cm /g?
2
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.
κe = 0.34 cm2 /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
q
dinf = 4πFLobs , d = 4πFLem
p
√
dinf /d = Fem /Fobs ) = eτ = eτ /2
A = 1.08τ =⇒ τ = A/1.08
dinf /d = eA/(2∗1.08) = e0.46A
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.
κe = (1 + X)0.2 cm2 /g = 0.2 cm2 /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 = f λ1 . Evaluate this for f = 2, 5, 10, 100.
Solution.
A1 /A2 = (1.086τ1 /1.086τ2 ) = τ1 /τ2 = λ2 /λ1 = f , since τ ∼ κ ∼ λ−β ∼ 1/λ. So for
f = 2, 5, 10, 100, ratio is A1 /A2 = 2, 5, 10, 100.