PHY 150 - Astronomy
Homework Assignment #5
November 8, 2007
1)
Why are the evolutionary tracks of high-mass stars different from those of low-mass stars?
For which kind of star is the evolution more rapid? Why?
In high-mass stars everything takes place more rapidly. Greater mass means greater gravity and
the protostar process is accelerated. Greater mass leads to greater core pressures and
temperatures, thus, a hotter more luminous star. The greater mass star consumes the available
hydrogen at a much higher rate, thus, the star spend less time on the main sequence. Greater
mass means that higher mass elements, such as carbon, can be burned. Finally, when the nuclear
fuels are exhausted the gravitational end is more spectacular.
2)
If a red star and a blue star both have the same radius and both are the same distance from
the Earth, which one looks brighter in the night sky? Explain why.
The blue star. The Stefan-Boltzmann Law states that the power radiated by an object of area A
and temperature T is P = σ AT4 . These two stars have the same area, but the blue star is hotter
than the red star. This means the blue star has a higher temperature and radiates more. Thus, the
blue star has a the greater absolute magnitude. If the two stars are the same distance from the
Earth then their apparent magnitudes are directly proportional to their absolute magnitudes.
3)
Sketch a Hertzsprung-Russell
diagram. Indicate the regions on your
diagram occupied by (a) mainsequence stars, (b) red giants, (c)
supergiants, (d) white dwarfs, and (e)
the sun.
4)
What is the mass-luminosity relation?
Does it apply to stars of all kinds?
The mass-luminosity relation is a
logarithmic relation between mass and
luminosity. It only applies to hydrogen
burning stars on the main sequence.
5)
In the spectrum of a particular star, the Balmer line H α has a wavelength of 656.41 nm.
The laboratory value for the wavelength of H α is 656.28 nm.
(A) Find the star’s radial velocity.
(B) Is this star approaching us or moving away? Explain.
(C) Find the wavelength at which you would expect to find H β is the spectrum of this star,
given that the laboratory wavelength of H β is 486.13 nm.
(D) Do your answers depend on the distance from the Sun to this star? Why or why not?
The Doppler Shift is given as
∆λ v
= , where a velocity away from the receiver is positive and
λ
c
gives rise to a red shift or increase in wavelength. In this case, )8 = 0.11 nm.
∆ λ 011
. nm
∆λ
A)
=
= 16
. x 10−4 and v =
c = 16
. x 10− 4 3 x 108 m s = 4.8 x 104 m s .
λ
656 nm
λ
B) The star is moving away.
C) ∆ λ
)(
(
β
λβ
=
)
v ∆λ α
∆λα
=
⇒ ∆λβ = λβ
= 48613
. nm 16
. x 10−4 = 0.08 nm,
c
λα
λα
(
)
thus, λ 'β = 48613
. + 0.08 = 486.21 nm.
D) No. Doppler Shift depends only on relative velocity, not distance.
6)
How much dimmer does the Sun appear from Neptune than from Earth? (Hint: the average
distance between a planet and the Sun equals the semimajor axis of the planet’s orbit.)
The brightness or luminosity of an object obeys the inverse square law; therefore,
 d2

 1  I
ISaturn = I Earth  2 Earth  = I Earth  2  = Earth = ( .0011) I Earth
 30 
900
 d Neptune 
7)
The star Procyon in Canis Minor (the small Dog) is a prominent star in the winter sky, with
an apparent brightness 1.3 x 10-11 that of the Sun. It is also one of the nearest stars, being
only 3.50 parsecs from Earth. What is the luminosity of Procyon? Express your answer as
a multiple of the Sun’s luminosity.
 1 AU 
 = 2 .0 6 x 1 0 5 AU ,
1 pc = 3 .0 9 x 1 0 16 m 
11 
 1.5 x 1 0 m
 2 .0 6 x1 0 5 AU 
 = 7 .2 1 x 1 0 5 AU
so dProcyon = 3 .5 pc 
pc


One can determine the luminosity of a star with respect to the sun by invoking the relationship
between luminosity, distance, and brightness, i.e. L = 4 π d 2 b . Taking the ratio between Procyon
and the Sun
L Pr ocyon
LSun
2
 d procyon  b procyon

=
b Sun
 d Sun 
2
 7.21 x105   13
. x10−11 
=
 
 = 6.8 or LPr ocyon = 6.8 LSun
1
1

 

8)
There is a good deal of evidence that our universe is no more than 15 billion years old (see
Chapter 28). Explain why no main-sequence stars of spectral class M have yet evolved into
red-giant stars.
There are a few ways of getting at this answer, mostly equivalent. I looked at Figure 17-17 and
identified a mid-class M star as one which has a surface temperature of 4000 K and a luminosity
of 10-2 Lu. I then looked at Figure 17-21, the mass-luminosity relationship and identified this star
as having a mass of 0.1 Mu. Following the reasoning given in lecture and demonstrated in Box
21-2 I identified the time on the main sequence as proportional to the hydrogen mass available
over the burning rate, i.e. t - M/L.
 L  M 
M
 1   1 0 −1 
 = 10 11 y
t≈
⇒ t m = t o  o   m  = (10 10 y) −2  


L
L
M
1
 m o 
10


Thus, a M class star might be expected to burn hydrogen for 100 billion years, much longer than
the supposed 15 billion years of the universe.
9)
What is the difference between Population I and Population II stars? In what sense can the
stars of one population be regarded as the “children” of the other population?
Population I stars are enriched in heavy elements (metals) compared to Population II stars.
Population I stars can be considered to be the “children” of the older Population II stars because it
was in the Population II stars, which have long since lived and died, that the heavy elements were
manufactured.
10) Do Starry Night excise 17-78.
(A)
Star
Apparent
Magnitude
Luminosity
(LSun)
Distance
(light years)
Fomalhaut
+1.17
18.9
25.1
Vega
+.03
61.9
25.3
Altair
+.76
86,000
525
Deneb
+1.25
320,000
3230
Rigel
+.18
700,000
773
Betelgeuse
+.45
41,000
427
Capella
+.08
180
42.2
Aldebaran
+.87
370
65.1
All of these stars are listed in Appendix 5. The most luminous is Rigel which is also the furthest
away.
(B) Only Vega, Deneb, and Capella are visible. The rest are below the horizon. These three
stars are close enough to the pole star that although they dip below the horizon at certain parts of
the year they do not disappear for long. As it happens, at midnight tonight and in six months time
they are above the horizon.