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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 –


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 1800R, 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 ′′?

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 ′′. 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 ′′. How about from a distance of

d = 1 kpc?


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

arcsec= s/ au

d/pc=⇒ d/ pc = s/ au

α/ arcsec= 2 au/ au

2(0.3136 arcsec)/ arcsec= 3.189 =⇒ d =

3.189 pc

d = 3.189 pc(3.276 lypc

) = 10.4 ly


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( 1 rad2·105 arcsec

) = 1.89 · 10−8 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?

Solution. d = arcsecα

pc = arcsec0.00377 arcsec

pc = 265 pc


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

α = 8.6 · 10−5 rad2·105 arcsec1 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 getm = 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?

Solution. F1 = F2 =⇒ L1

4πd21= L2

4πd22=⇒ L2

L1= (d2

d1)2 =⇒ d2

d1= 10

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.

I2/I1 = 100 =L2/4πR2

2

L1/4πR12

= L2

L1(R1

R2)2 =⇒ (R2

R1)2 = 1

100L2

L1= 1 =⇒ R2

R1

= 1

Quick Question 3:

A white-dwarf-supernova with peak luminosity L ≈ 1010 L is observed to

have an apparent magnitude of m = +20 at this peak.

a. What is its Absolute Magnitude M?

– 6 –

Solution. M = 5− 2.5 log(10·1010LL

) = −20

b. What is its distance d (in pc and ly)?

Solution.

m −M = 5 log( d10 pc

) =⇒ 20 − −20 = 5 log( d10 pc

) =⇒ 8 = log( d10 pc

) =⇒d

10 pc= 1 · 108 =⇒ d = 1 · 109 pc d = 1 · 109 pc(3.26 ly

1 pc) = 3.26 · 109 ly

c. How long ago did this supernova explode (in Myr)?

Solution. 3.26 Gyr


Reformulate Equation (3.3) to be in terms of the Sun’s flux measured on

Earth.

Solution. L = 4πd2F

L = 4π(1 au)2FLL

= 4πd2F4π(1 au)2F

= d2au(F/F)

(L/L) = d2au(F/F)


An observer measures the flux of a star to be 10−10F. 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.

50 pc(2·105 au1 pc

) = 107 au

(L/L) = d2au(F/F) = (107 au)2(10−10F/F) = 104 =⇒ L = 104L

b. Calculate the absolute magnitude of the star.

Solution. M = 5− 2.5 log(L/L) = 5− 2.5 log(104L/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

P1= λ2/c

λ1/c= λ2

λ1= 2

b. What is the ratio of their frequency ν2/ν1?

Solution. ν2ν1

= c/λ2c/λ1

= λ1λ2

=1

2

c. What is the ratio of their energy E2/E1?

Solution. E2

E1= hν2

hν1= ν2

ν1=

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.)

Solution. T = 6000 K500 nmλmax

T100 = 6000 K500 nm100 nm

= 30000 K = T100

T300 = 6000 K500 nm300 nm

= 10000 K = T300

T1000 = 6000 K 500 nm1000 nm

= 3000 K = T1000

T3000 = 6000 K 500 nm3000 nm

= 1000 K = T3000

b. Conversely, estimate the peak wavelengths λmax of stars with T =2000,

10,000, and 60,000 K.

Solution. λmax = 500 nm6000 KT

λmax,2000 = 500 nm6000 K2000 K

= 1500 nm = λmax,2000

λmax,10000 = 500 nm 6000 K10000 K

= 300 nm = λmax,10000

λmax,60000 = 500 nm 6000 K60000 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 typi-

cal spring day, i.e. 50F, compute the peak wavelength of Earth’s Black-Body

radiation.

Solution.

T = 50 F = (50 F− 32 F)59C = 10C = 283 K ≈ 300 K

λmax = 500 nm6000 K300 K

= 10000 nm = 10µm

b. What part of the EM spectrum does this lie in?

Solution. IR


a. What is the wave period at the peak wavelength emitted from the Sun?

Solution. P = λmax,c

= 500 nm3·108 m s−1 = 500·10−9 m

3·108 m s−1 = 1.7 · 10−15 s = 1.7 fs

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


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.

Equations. L = L(R/R)2(T/T)4

M = 5− 2.5 log( LL

)

λmax = 500 nmT/T

m = M + 5 log( d10 pc

) = M + 5 log( arcsec pc/α10 pc

) = M + 5 log(1/0.00110

) = M + 10

a.

L = L(10R/R)2(T/T)4 = 100L = La

M = 5− 2.5 log(100LL

) = 0 = Ma

λmax = 500 nmT/T

= 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(10000LL

) = −5 = Mb

λmax = 500 nm10T/T

= 50 nm = λmax,b

m = −5 + 10 = 5 = mb

c.

L = L(10R/R)2(10T/T)4 = 1000000L = Lc

M = 5− 2.5 log(1000000LL

) = −10 = Mc

λmax = 500 nm10T/T

= 50 nm = λmax

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( d10 pc

) = 5 log( 10010 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 nmλmax

T = 500 nm250 nm

T = 2T

f. Radius R (in R)?

Solution.

L/L = (R/R)2(T/T)4 =⇒ (R/R)2 = L/L(T/T)−4 = (R/R)2 =

100L/L(2T/T)−4 = 102

24=⇒ R/R =

√102

24= 10

22= 2.5 =⇒ R =

2.5R

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

Solution.

α = R/d = (6.96 · 105 km)/(100 pc3.09·1013 km1 pc

) = 2.25 · 10−10 rad

α = 2.25 · 10−10 rad2·105 arcsec1 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 = σsbT4

π=⇒ 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.

– 11 –

Solution. R =√

F (d)σsbT 4d =

√1400 W m−2

(5.67·10−8 J m−2 s−1 K−4)(6000 K)4(1.5·1011 m) = 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


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


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:

Solve for log gast. g = GMR2 =⇒ g∗/g = GM∗/R2

∗GM/R2

= M∗

M( R

2∗

R2

)−1 =⇒ log g∗ =

log(M∗M

( R∗R

)−2) + log g = 4.44 + log(M∗M

)− 2 log( R∗R

)

a. M = 10M and R = 10R

Solution. log g∗ = 4.44 + log(10MM

)− 2 log(10RR

) = 3.44

b. M = 1M and R = 100R


)− 2 log(100RR

) = 0.44

c. M = 1M and R = 0.01R


)− 2 log(0.01RR

) = 8.44

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.

Reformulate into solar units. Ve =√

2GMR

=⇒ Ve = 618 km s−1√

M/MR/R

a.

Solution. Ve = 618 km s−1√

10M/M10R/R

= 618 km s−1

b.


1M/M100R/R

= 61.8 km s−1

c.


1M/M0.01R/R

= 6180 km s−1

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 =√

GMr

=⇒ Vorb/Ve =√

GM/rGM/re

=√

M/Mr/re

=⇒ Vorb =

30 km s−1√

M/Mr/1 au

a. M = 10M and d = 10 au.

Solution. Vorb = 30 km s−1√

10M/M10 au/1 au

= 30 km s−1

b. M = 1M and d = 100 au.


1M/M100 au/1 au

= 3 km s−1

c. M = 1M and d = 0.01 au.


1M/M0.01 au/1 au

= 300 km s−1


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 = −GMmd

=⇒ U/Ue, = −GMmd/ − GMme

1 au=

( MM

)( mme

)(1 aud

)

a.

Solution. U/Ue, = (10MM

)(10me

me)( 1 au

10 au) = 10

b.


)(1me

me)( 1 au

100 au) = 0.01

c.


)(1me

me)( 1 au

0.01 au) = 100


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

Solution. W = GMmR

= (6.67·10−11m3 kg−1 s−2)(5.97·1024 kg)(5·104 kg)6.4·106 m

= 3.11 · 1012 J

– 15 –

b. The moon

Solution. W = GMmR

= (6.67·10−11m3 kg−1 s−2)(7.34·1022 kg)(5·104 kg)1.7·106 m

= 1.44 · 1011 J

c. The Sun

Solution. W = GMmR

= (6.67·10−11m3 kg−1 s−2)(1.99·1030 kg)(5·104 kg)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. 10M?

Solution. L/L ∝ (M/M)3

tms = 10000 Myr(MM

)2

tms = 10000 Myr( M10M

)2 = 100 Myr = tms,a

L/L = (10M/M)3 =⇒ L = 103L = La

b. 0.1M?

Solution. tms = 10000 Myr( M0.1M

)2 = 106 Myr = tms,b

L/L = (0.1M/M)3 =⇒ L = 10−3L = Lb

c. 100M?

Solution. tms = 10000 Myr( M100M

)2 = 1 Myr = tms,c

L/L = (100M/M)3 =⇒ L = 106L = Lc

Quick Question 2:

Suppose you observe a cluster with a main-sequence turnoff point at a lumi-

nosity 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

tcluster = 10000 Myr(100LL

)−2/3 = 464 Myr

tcluster = 10000 Myr(10000LL

)−2/3 = 22 Myr


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?

Solution. tchem = εMc2

L=⇒ tchem/tchem, = εMc2

L/εMc2

L=⇒ tchem =

tchem,M/ML/L

tchem = 15, 000 yr0.70M/M0.15L/L

= 70, 000 yr

– 17 –

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

tion. What would be the expected lifetime of this process?

Solution. tKH = 310GM2

RL=⇒ tKH/tKH, = 3

10GM2

RL/ 3

10

GM2

RL= (M/M)2

(R/R)(L/L)=⇒

tKH = tKH,(M/M)2

(R/R)(L/L)

tKH = 30 Myr (0.70M/M)2

(0.67R/R)(0.15L/L)= 146 Myr

c. As a main sequence star, 61 Cygni A is powered by nuclear fusion. What

is its nuclear burning timescale?

Solution. tms = 10 Gyr( M0.7M

)2 = 20 Gyr


Based on Figure 8.1, what is the age of M55?

Solution. Lto ≈ 3Ltcluster = 10 Gyr( L

3L)2/3 = 4.8 Gyr


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?

Solution. 5 · 106 pc(3.26 ly1 pc

) = 1.71 · 107 ly = 0.0171 Gly

tms = 10 Gyr(MM

)2 = 10 Gyr( M10M

)2 = 0.01MThe 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.

Solution. Vt = 4.7 km s−1 µp

= 4.7 km s−1 2 arcsec/100.02 arcsec

= 47 km s−1

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−λoλo

= 600.09 nm−600 nm600 nm

= 1.5 · 10−4

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

Solution. z = Vrc

=⇒ 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?

Solution. Vtot =√V 2r + V 2

t =√

(45km s−1)2 + (47 km s−1)2 = 65 km s−1


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−λoλo

= λobs−1216 nm1216 nm

, z = Vrc

=⇒ Vr = zc

a. 1216.5 nm

Solution. z = 1216.5 nm−1216 nm1216 nm

= 4.1 · 10−4

Vr = 4.1 · 10−4(3 · 105 km s−1) = 123 km s−1

Away

b. 1215.5 nm

– 19 –


= −4.1 · 10−4

Vr = −4.1 · 10−4(3 · 105 km s−1) = −123 km s−1

Towards

c. 1215.85 nm


= −1.2 · 10−4

Vr = −1.2 · 10−4(3 · 105 km s−1) = −36 km s−1

Towards

d. 1216.15 nm


= 1.2 · 10−4

Vr = 1.2 · 10−4(3 · 105 km s−1) = 36 km s−1

Away


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?

Solution. 250 km hr−1(1000 m1 km

)( 1 hr3600 s

) = 69 m s−1

∆λλ

= Vc

=⇒ ∆λ400 nm

= 69 m s−1

3·108 m s−1 =⇒ ∆λ = 2.3 · 10−7 nm

λ′ = 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?

Solution. d = arcsec0.2 arcsec/2

pc = 10 pc

b. What is the physical distance between the stars, in au.

Solution. s/ au = dpc

αarcsec

= 10 pcpc

1 arcsecarcsec

= 10 =⇒ s = 10 au

c. In solar masses, what are the masses of each star, M1 and M2.

Solution.M1+M2

M= (s/ au)3

(P/ yr)2= 10 =⇒ M1 +M2 = 10M

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 = (V2Ve

)3Pyr(1 + V1V2

)2M, M2 = (V1Ve

)3Pyr(1 + V2V1

)2MThe 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 = 1VeM1 = (Ve

Ve)3(20/365)(1 + 2Ve

1Ve)2M = 0.49M = M1

M2 = (2VeVe

)3(20/365)(1 + 1Ve2Ve

)2M = 0.99M = M2


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

t2 = 5 · 104 hr(3600 shr

) = 1.8 · 104 s

t4 = 100 hr(3600 shr

) = 3.6 ∗ ·105 s

R2 = (t2 − t1)(V1 + V2)/2 = (1.8 · 104 s − 0 s)(20 km s−1 + 10 km s−1)/2 = 2.7 ∗105 km( 1R

6.96·105 km) = 0.39R = R2

R1 = (t4 − t2)(V1 + V2)/2 = (3.6 ∗ ·105 s− 1.8 · 104 s)(20 km s−1 + 10 km s−1)/2 =

5.1 · 106 km( 1R6.96·105 km

) = 7.3R = R1


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 ∼M3 for the

following masses?

a. 1M

Solution. Lapprox/L = (M/M)3 =⇒ Lapprox = (M/M)3L = (M/M)3L =

Llog(L/L) = 0.1+3.1 log(MM) = 0.1+3.1 log(MM) = 0.1 =⇒ L = 1.25Lε = |Lapprox−L|

L= |1−1.25|

1.25= 0.2 = 20%

b. 10M

Solution. Lapprox = (10M/M)3L = 1000Llog(L/L) = 0.1 + 3.1 log(10MM) = 3.2 =⇒ L = 1600Lε = |1000−1600|

1600= 0.38 = 38%

c. 0.1M

– 22 –

Solution. Lapprox = (0.1M/M)3L = 0.001Llog(L/L) = 0.1 + 3.1 log(0.1MM) = −3 =⇒ L = 0.001Lε = |0.001−0.001|

0.001= 0 = 0%

d. 100M

Solution. Lapprox = (100M/M)3L = 106Llog(L/L) = 0.1 + 3.1 log(100MM) = 6.3 =⇒ L = 2.0 · 106Lε = |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.∆λfullλo

= 2V sin ic

=⇒ V sin i =∆λfull

2λoc = 0.5 nm

2·500 nm(3 · 105 km s−1) = 1

2·103(3 ·

105 km s−1) = 300 km s−1

2= 150 km s−1


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)?

Solution. Rmin = Vrot sin iP2π

= 150 km s−1∗30∗86400 s2π

= 6.2 · 107 km

Rmin = 6.2 · 107 km( 1R6.96·105 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.

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.

dinf =√

L4πFobs

, d =√

L4πFem

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 = fλ1. 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.

– 26 –

Chapter 13

Quick Question 1:

What is the limiting magnitude for a naked eye viewing through a 5m tele-

scope?

Solution. m ≈ 7.5 + 5 log(D/cm) = 7.5 + 5 log(500cm/cm) = 21

Quick Question 2:

Estimate the angular resolution of a 5 m optical telescope in space.

Solution. Optical ⇒ λ ≈ 0.5µm, so α = 0.25 arcsec λ/µmD/m

= 0.25(0.5/5) =

0.025 arcsec

Quick Question 3:

What is the diameter D of a single telescope with the same collecting area of

the 4 × 8 m-diameter VLT telescopes in Chile?

Solution.√

4× 64 m= 16 m

Quick Question 4:

What is the angular resolution of the Hubble Space Telescope at a wavelength

of 200 nm?

Solution. HSTD = 2.4m;⇒ α = 0.25 arcsec λ/µmD/m

= 0.25(0.2/2.4) = 0.021 arcsec

Quick Question 5: Two infrared sources in the Orion nebula, which has a

distance d = 500 pc, are separated by just s = 0.1 pc. What diameter telescope

would be needed to resolve them at a wavelength λ = 100µm?

Solution. α = 0.1pc/(500pc) = 2× 10−4 radian ⇒D = λ/α = 100µm/(2× 10−4) = 2 m.

– 27 –

Chapter 14

Quick Questions

1. For solar wind mass loss rate M ≈ 10−14M/yr and speed v = 450 km/s, what is the

associated kinetic energy luminosity Lsw (in both L and W) of the solar wind?

Solution. Lsw = Mv2/2

= 10−14M/yr × (2× 1030)kg/M yr/(3.15× 107s) (4.5× 105)2

= 6.4× 1019W = 1.6× 10−7L .

2. The Sun’s surface temperature is about T = 5800 K, but that of a sunspot is about

Tspot = 3800 K. What then is the ratio of the surface flux of a sunspot compared to

the rest of the Sun, Fspot/Fsun.

Solution. Fspot/F sun = (Tspot/T)4 = (3800K/5800K)4 = 0.18 .

3. Derivation – no quick solution provided.



– 28 –

Chapter 15

Quick Questions

1. If we double the Sun’s radius, what happens to its surface gravity?

Solution. g/g = (M/M)/(R/R)2 ⇒ g/g = 1/22 = 0.25 .

2. If we double the Sun’s mass, what happens to its surface gravity?

Solution. g/g = (M/M)/(R/R)2 ⇒ g/g = 2/1 = 2 .

3. If we double the Sun’s surface gravity, what happens to its surface scale height H?

Solution. H/H = (T/T)/(g/g) ⇒ H/H = 1/2 = 0.5 .

4. If we double the Sun’s surface temperature, what happens to its surface scale height

H?

Solution. H/H = (T/T)/(g/g) ⇒ H/H = 2/1 = 2 .

5. If we double the Sun’s radius, what happens to its core temperature?

Solution. T/T = (M/M)/(R/R) ⇒ T/T = 1/2 = 0.5 .

6. If we double the Sun’s mass, what happens to its core temperature?

Solution. T/T = (M/M)/(R/R) ⇒ T/T = 2/1 = 2 .

– 29 –

Chapter 16

Quick Questions

1. If we flip a coin repeatedly, about how many flips will it take, on average, before the

number of heads is 100 more than the number of tails?

Solution. n =√N ⇒ N = n2 = 1002 = 10, 000 .

2. If a star has twice the mass and twice the radius of the Sun, about what is the optical

depth from its center to surface?

Solution. τ ≈ κρR with ρ ∼M/R3. Assuming same opacity, κ:

τ/τ = (M/M)/(R/R)2 = 2/22 = 0.5 , with τ ≈ 1011.

3. If we double the Sun’s mass, about how long would it take light to diffuse from the

center to the surface?

Solution. tdiff = τR/c ∼M/R, so:

tdiff/t = (M/M)/(R/R) = 2/1 = 2 , with t ≈ 7000 yr.

4. If we double the Sun’s radius, about how long would it take light to diffuse from the

center to the surface?

Solution. tdiff = τR/c ∼M/R, so:

tdiff/t = (M/M)/(R/R) = 1/2 = 0.5 , with t ≈ 7000 yr.

5. At about what optical depth τ is the temperature of a star 100 times its effective

temperature Teff?

Solution. T 4 = (3/4)T 4eff(τ + 2/3), so:

τ = (4/3)(T/Teff)4 − 2/3 ≈ (4/3)(100)4 = 1.33× 108 .

– 30 –

Chapter 17

Quick Questions

1. Estimate the luminosity L (in L) of a radiative stellar envelope of mass M = 30M?

Solution. L/L = (M/M)3 = 303 = 27, 000 .

2. What is the mass M (in M) of a radiative stellar envelope with luminosity L =

64, 000L?

Solution. M/M = (L/L)1/3 = (64000)1/3 = 40 .

– 31 –

Chapter 18

Quick Questions

1. About what are the thermal de Broglie wavelengths of electrons and protons in the

core of the Sun? Which of the above is closer to the size of the atomic nucleus?

Solution. λ = h/√

2mkT ; using Tc ≈ 1.5× 107K, we find:

λe = 3.4× 10−11m; λp = 8.0× 10−13m ;

λp closer in size, but still about 800 times nucleus, for which s ≈ 10−15m = 1fm.

2. What is the maximum mass Mmax for a star to have evolved off the main sequence in

the 14 Gyr age of the universe?

Solution. tms ≈ 10Gyr(M/M)−2 ⇒M/M =√

10Gyr/tms =√

1.4 = 1.18 .

3. About what is the main-sequence lifetime (Gyr) of star with mass just above the Brown

dwarf limit?

Solution. tms ≈ 10Gyr(M/M)−2 = 10Gyr 1/(0.08)2 = 1560Gyr .

– 32 –

Chapter 19

Quick Questions

1. Note that pressure can be written as P = Nvp for speed v and momentum p = mv

in a medium with N atoms of mass m. Show that this recovers the ideal-gas form for

pressure by identifying the thermal speed scaling with temperature, kT/2 = mv2th/2.

Solution. P = Nvp = Nmv2 = NkT = ρkT/m.

2. Derive a general formula for the escape speed vesc for a white dwarf of mass M . What

is the value (in km/s) for M = M and how does this compare to the Sun’s escape

speed?

Solution. vesc = 620 km/s√

(M/M)/(R/R). Using Rwd ≈ 0.01R(M/Mwd)1/3, we

have:

vesc ≈ 6200 km/s (M/M)2/3 . For sun, vesc, = 620 km/s.

3. What is the maximum escape speed (in km/s) of a white dwarf?

Solution. For Chandrasekhar mass Mch = 1.4M,

vesc,max ≈ 6200 km/s (1.4)2/3 = 7340 km/s .

4. Derive a general formula for log g of a white dwarf star of mass M , and compare this

to log g.

Solution. g/g = (M/M)/(R/R)2 = 104 (Mwd/M)5/3, with log g = 4.4 gives:

log gwd = 8.4 + (5/3) log(Mwd/M) .

– 33 –

Chapter 20

Quick Questions

1. Derive a general formula for the escape speed vesc for a neutron star of mass M . What

is the value (in km/s) for M = 2M and how does this compare to the Sun’s escape

speed, and to the speed of light?

Solution. vesc =√

2GM/R. Using Rns ≈ 10 km/s (M/Mns)1/3, we have:

vesc ≈ 1.63× 105 km/s (Mns/M)2/3 , vs. vesc, = 620 km/s and c = 3×1010 km/s.

2. What is the maximum escape speed (in km/s) of a neutron star?

Solution. Mns,max = 2.1M, so

vesc ≈ 1.63× 105 km/s (2.1)2/3 = 6.67× 105 km/s .

3. Derive a general formula for log g of a neutron star of mass M , and compare this to

log g, and also to log gwd.

Solution. g = GM/R2 = 1.3× 1014 cm/s2 (Mns/M)5/3.

log gns = 14.1 + (5/3) log(Mns/M) vs.

log gwd = 8.4 + (5/3) log(Mwd/M) and log g = 4.4.

fundamentals of astrophysics: quick question solutionsowocki/phys333/qqsolns-ch1-20.pdffundamentals...

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