Openstax College Physics
Instructor Solutions Manual
Chapter 34
CHAPTER 34: FRONTIERS OF PHYSICS
34.1 COSMOLOGY AND PARTICLE PHYSICS
1.
Find the approximate mass of the luminous matter in the Milky Way galaxy, given it
11
has approximately 10 stars of average mass 1.5 times that of our Sun.
Solution The approximate mass of the luminous matter in the Milky Way galaxy can be found
by multiplying the number of stars times 1.5 times the mass of our Sun:
 
 


M  1011 1.5mS  1011 1.5 1.99 1030 kg  3 10 41 kg
2.
Find the approximate mass of the dark and luminous matter in the Milky Way galaxy.
11
Assume the luminous matter is due to approximately 10 stars of average mass 1.5
times that of our Sun, and take the dark matter to be 10 times as massive as the
luminous matter.
Solution
M  11 1011 1.5mS  1.11012 1.5 1.99 1030 kg  3 1042 kg
3.
(a) Estimate the mass of the luminous matter in the known universe, given there are
1011 galaxies, each containing 1011 stars of average mass 1.5 times that of our Sun.
(b) How many protons (the most abundant nuclide) are there in this mass? (c)
Estimate the total number of particles in the observable universe by multiplying the
9
answer to (b) by two, since there is an electron for each proton, and then by 10 , since
there are far more particles (such as photons and neutrinos) in space than in luminous
matter.
 




Solution (a) M L  1011 1011 1.5mS  1022 1.51.99 1030 kg  3 1052 kg
(b)
3  1052 kg
 1.8  1079  2  1079
27
1.67  10 kg
(c) 1.787  10 79 (2)10 9   3.6  1088  4  1088
4.
If a galaxy is 500 Mly away from us, how fast do we expect it to be moving and in
what direction?
Openstax College Physics
Instructor Solutions Manual
Chapter 34
Solution
v  H 0 d  20 km/s/Mly 500 Mly   1.0  10 4 km/s away from us.
5.
On average, how far away are galaxies that are moving away from us at 2.0% of the
speed of light?
Solution

v
0.020 3.00  105 km/s
v  H 0d  d 

 3.0  10 2 Mly  0.30 Gly
H0
20 km/s/Mly
6.
Our solar system orbits the center of the Milky Way galaxy. Assuming a circular orbit
30,000 ly in radius and an orbital speed of 250 km/s, how many years does it take for
one revolution? Note that this is approximate, assuming constant speed and circular
orbit, but it is representative of the time for our system and local stars to make one
revolution around the galaxy.
Solution




2r 2 3.00  10 4 ly 9.460  1012 km/ly

v
250 km/s
1y


 7.1327  1015 s 
 2.26  108 y
7 
 3.156  10 s 
T

7.


(a) What is the approximate velocity relative to us of a galaxy near the edge of the
known universe, some 10 Gly away? (b) What fraction of the speed of light is this?
Note that we have observed galaxies moving away from us at greater than 0.9c .
Solution (a) Using v  H 0 d , and the Hubble constant, we can calculate the approximate
velocity of the near edge of the known universe:


v  H 0 d  20 km/s  Mly  10  103 Mly  2.0  105 km/s
(b) To calculate the fraction of the speed of light, divide this velocity by the speed of
v 2.0 10 5 km/s 10 3 m/km 
light: 
 0.67, so that v  0.67c
c
3.00 108 m/s
8.
(a) Calculate the approximate age of the universe from the average value of the
Hubble constant, H 0  20 km/s  Mly . To do this, calculate the time it would take to
travel 1 Mly at a constant expansion rate of 20 km/s. (b) If deceleration is taken into
account, would the actual age of the universe be greater or less than that found here?
Explain.
Openstax College Physics
Solution
Instructor Solutions Manual

Chapter 34


1 Mly  10 6 ly/Mly 9.46  1012 km/ly  4.73  1017 s
d
1 Mly


v 20 km/s
20 km/s
1y


10
 4.73  1017 s 
  1.5  10 y  15 billion years
7
 3.156  10 s 
(a) t 


(b) Younger, since if it was moving faster in the past it would take less time to travel
the distance.
9.
Assuming a circular orbit for the Sun about the center of the Milky Way galaxy,
calculate its orbital speed using the following information: The mass of the galaxy is
11
equivalent to a single mass 1.5  10 times that of the Sun (or 3  10 41 kg ), located
30,000 ly away.
Solution
mv 2 GMm


r
r2
 GM 
v

 r 
10.
12



 6.67  10 11 N  m 2 / kg 2 3  10 41 kg 
5

  2.7  10 m/s
15
30,000 ly  9.46  10 m/ly




(a) What is the approximate force of gravity on a 70-kg person due to the Andromeda
13
galaxy, assuming its total mass is 10 that of our Sun and acts like a single mass 2
Mly away? (b) What is the ratio of this force to the person’s weight? Note that
Andromeda is the closest large galaxy.
Solution
(a) FG 

 

GMm 6.67  10 11 N  m 2 / kg 2 1013 1.99  10 30 kg 70 kg 

2
r2
2  10 6 ly 9.46  1015 m/ly



 2.6  10 10 N  3  10 10 N
(b) w  mg  70 kg 9.8 m/s 2   6.86  10 2 N, so that
11.
FG
 3.8  10 13  4  10 13
w
Andromeda galaxy is the closest large galaxy and is visible to the naked eye. Estimate
12
its brightness relative to the Sun, assuming it has luminosity 10 times that of the Sun
and lies 2 Mly away.
Solution The relative brightness of a star is going to be proportional to the ratio of surface
areas times the luminosity, so:
Openstax College Physics
Instructor Solutions Manual
Chapter 34
2
 rSun 
4πr 2
 .
Relative Brightness  luminosity 
 1012 
2
4πRAndromeda
 RAndromeda 
 
11
From Appendix C, we know the average distance to the sun is 1.496  10 m , and we
are told the average distance to Andromeda, so:
10 1.496 10 m

2 10 ly 9.46 10 m/ly 
12
Relative Brightness
11
6
15
2
2
 6 10 11 .
Note: this is an overestimate since some of the light from Andromeda is blocked by its
own gas and dust clouds.
12.
(a) A particle and its antiparticle are at rest relative to an observer and annihilate
(completely destroying both masses), creating two  rays of equal energy. What is
the characteristic  -ray energy you would look for if searching for evidence of protonantiproton annihilation? (The fact that such radiation is rarely observed is evidence
that there is very little antimatter in the universe.) (b) How does this compare with the
0.511-MeV energy associated with electron-positron annihilation?
Solution (a) Each proton is converted to a  ray, so to speak. So each  ray has an energy
m p c 2  938.27 MeV.
(b)
938.27 MeV
 1.84 10 3 larger than electron-positron annihilation.
0.511 MeV
13.
The average particle energy needed to observe unification of forces is estimated to be
1019 GeV . (a) What is the rest mass in kilograms of a particle that has a rest mass of
1019 GeV/ c 2 ? (b) How many times the mass of a hydrogen atom is this?
Solution
 1.67  10 27 kg 
  1.78  10 8 kg  2  10 8 kg
(a) 1019 GeV/c 2 
2 
0
.
9383
GeV/c


(using proton masses from Table 33.2 as a convenient conversion factor)
10


GeV/c 2 10 3 MeV/GeV
1.78  10 -8 kg
19

1

10
or
 1  1019
1.007825 u  931.5 MeV/c 2 u
1.67  10 - 27
19
(b)

 
Openstax College Physics
14.
Solution
Instructor Solutions Manual
Chapter 34
The peak intensity of the CMBR occurs at a wavelength of 1.1 mm. (a) What is the
9
energy in eV of a 1.1-mm photon? (b) There are approximately 10 photons for each
9
massive particle in deep space. Calculate the energy of 10 such photons. (c) If the
average massive particle in space has a mass half that of a proton, what energy
would be created by converting its mass to energy? (d) Does this imply that space is
“matter dominated”? Explain briefly.
(a) E  hv 
hc


4.14  10
21


MeV  s 3.00  10 8 m/s
 1.1  10 -3 eV
1.1  10 -3 m
(b) 1.129  10 -3 eV 10 9   1 10 6 eV  1 MeV
(c) E  mc 2 


1
938.27  10 6 eV/ c 2 c 2  4.69  10 8 eV  5  10 2 MeV
2
(d) Yes, much more energy is associated with the mass of massive particles (about
500 times as much).
15.
(a) What Hubble constant corresponds to an approximate age of the universe of 1010
y? To get an approximate value, assume the expansion rate is constant and calculate
the speed at which two galaxies must move apart to be separated by 1 Mly (present
10
average galactic separation) in a time of 10 y. (b) Similarly, what Hubble constant
10
corresponds to a universe approximately 2  10 -y old?
Solution (a) Since the Hubble constant has units of km s  Mly , we can calculate its value by
considering the age of the universe and the average galactic separation. If the
10
10
universe is 10 years old, then it will take 10 years for the galaxies to travel 1
Mly. Now, to determine the value for the Hubble constant, we just need to
determine the average velocity of the galaxies from the equation d  v  t :
d
1 Mly 1  10 6 ly 9.46  1012 km
1y
, so that v  10 


 30 km/s.
10
t
1 ly
10 y
10 y
3.156  10 7 s
30 km/s
Thus, H 0 
 30 km/s  Mly
1 Mly
v
10
(b) Now, the time is 210 years, so the velocity becomes:
v
1 Mly
1 106 ly 9.46  1012 km
1y



 15 km/s.
10
10
2  10 y 2  10 y
1 ly
3.156  107 s
Openstax College Physics
Instructor Solutions Manual
Thus, the Hubble constant would be approximately H 0 
16.
Solution
15 km/s
 15 km/s  Mly
1 Mly
Show that the velocity of a star orbiting its galaxy in a circular orbit is inversely
proportional to the square root of its orbital radius, assuming the mass of the stars
inside its orbit acts like a single mass at the center of the galaxy. You may use an
equation from a previous chapter to support your conclusion, but you must justify its
use and define all terms used.
Setting the centripetal force equal to the gravitational force gives: F 
Solving the equation for the velocity gives: v 
17.
Chapter 34
mv 2 GMm
.

r
r2
GM
.
r
The core of a star collapses during a supernova, forming a neutron star. Angular
momentum of the core is conserved, and so the neutron star spins rapidly. If the initial
core radius is 5.0  10 5 km and it collapses to 10.0 km, find the neutron star’s angular
velocity in revolutions per second, given the core’s angular velocity was originally 1
revolution per 30.0 days.
Solution The angular momentum L  Iω is conserved. Therefore,
I f ωf  I i ωi  ωf 
Ii
ωi
If
The momentum of inertia I is proportional to r 2 , and the angular velocity in degrees
per second ω is proportional to the angular frequency in revolutions per second f .
 5.0  10 5 km 
r2

Therefore, f f  i 2 f i  
1
rf
 1.00  10 km 
2
 1 rev  1 day 


 964 rev/s
4 
 30.0 days  8.64  10 s 
18.
Using data from the previous problem, find the increase in rotational kinetic energy,
given the core’s mass is 1.3 times that of our Sun. Where does this increase in kinetic
energy come from?
Solution
KE 
1 2
2
I , I  Mr 2
2
5
Openstax College Physics
KEi 
Instructor Solutions Manual
1
1
2
Mr 2i2  Mr 2 2f 
5
5

 0.2 1.31.99  0

 2  3.858 10 s   7.60 10
kg 10 10 m  2  964.5 s   1.90 10 J
 0.2 1.3 1.99  030 kg 5.0 108 m
KEf
Chapter 34
30
3
2
2
7
2
-1 2
-1 2
2
35
J
45
Increase  1.9 10 45 J
The increase comes from decreased gravitational energy.
19.
Solution
Distances to the nearest stars (up to 500 ly away) can be measured by a technique
called parallax, as shown in Figure 34.15. What are the angles θ1 and θ 2 relative to
the plane of the Earth’s orbit for a star 4.0 ly directly above the Sun?
tan 1 


 4.0 ly  9.46  1015 m/ly 
4 ly
, so that 1  tan 1 
  89.999773   2
rS
1.5  1011 m


A better way is to calculate complementary angle  in the triangle (because we have
only two significant figures).
1.5 1011 m
 
using small angle approximat ion
4.0 ly  9.46 1015 m


4
 3.964 10 rad  2.3 10 degrees (using conversion 1 rad  57.30)
6
1   2  90    89.999773
20.
Solution
(a) Use the Heisenberg uncertainty principle to calculate the uncertainty in energy for
a corresponding time interval of 10 43 s . (b) Compare this energy with the 1019 GeV
unification-of-forces energy and discuss why they are similar.
(a) ΔE 
(b)
h
4.14  10 24 GeV  s

 3.29  1018 GeV  3  1018 GeV
 43
4t
4 10 s


ΔE 3  1018 GeV

 0.3
E
1019 GeV
Expect symmetry-breaking of the three forces to occur shortly after the
separation of gravity from the unification force (near the Planck time interval).
The uncertainty in time then becomes greater. Hence the energy available
becomes less than the needed unification energy.
Openstax College Physics
Instructor Solutions Manual
Chapter 34
34.2 GENERAL RELATIVITY AND QUANTUM GRAVITY
22.
What is the Schwarzschild radius of a black hole that has a mass eight times that of
our Sun? Note that stars must be more massive than the Sun to form black holes as a
result of a supernova.
Solution
2GM 2 6.67  10 11 N  m 2 / kg 2 ( 8 ) 1.99  10 30 kg
RS  2 
 2.36  10 4 m  23.6 km
8
2
c
3.00  10 m/s
23.
Black holes with masses smaller than those formed in supernovas may have been
created in the Big Bang. Calculate that has a mass equal to the Earth’s.
Solution
24.
Solution

RS 
 







2GM 2 6.67 10 11 N  m 2 / kg 2 5.98 10 24 kg

 8.86 10 3 m  8.86 mm
2
2
8
c
3.00 10 m/s

Supermassive black holes are thought to exist at the center of many galaxies. (a)
9
What is the radius of such an object if it has a mass of 10 Suns? (b) What is this
radius in light years?
(a) RS 

 


2GM 2 6.67  10 11 N  m 2 / kg 2 10 9 1.99  10 30 kg

 2.95  1012 m
2
8
c2
3.00  10 m/s

1 ly


4
(b) 2.950  1012 m 
  3.12  10 ly
15
 9.46  10 m 
34.3 SUPERSTRINGS
26.
Solution
The characteristic length of entities in Superstring theory is approximately 10 35 m .
(a) Find the energy in GeV of a photon of this wavelength. (b) Compare this with the
average particle energy of 1019 GeV needed for unification of forces.
4.14  10
21

MeV  s 3.00  10 8 m/s

1  10 -35 m
 1.242  10 23 MeV  1  10 20 GeV
(a) E  hv 
hc


Openstax College Physics
(b)
Instructor Solutions Manual
Chapter 34
1  10 20 GeV
 10
1019 GeV
34.4 DARK MATTER AND CLOSURE
27.
If the dark matter in the Milky Way were composed entirely of MACHOs (evidence
shows it is not), approximately how many would there have to be? Assume the
average mass of a MACHO is 1/1000 that of the Sun, and that dark matter has a mass
11
10 times that of the luminous Milky Way galaxy with its 10 stars of average mass
1.5 times the Sun’s mass.
M DM  (10)(1011 )1.5mS ;
Solution


M MACHO  1.00  10 3 mS , so that
N
(10)(1011 )1.5mS
M DM

 1.5  1015
3
M MACHO
1.00  10 mS


28.
The critical mass density needed to just halt the expansion of the universe is
2
3
approximately 10 26 kg / m 3 .(a) Convert this to eV / c  m . (b) Find the number of
neutrinos per cubic meter needed to close the universe if their average mass is
7 eV / c 2 and they have negligible kinetic energies.
Solution
 938.27  10 6 eV/c 2 
  5.62  10 9 eV/c 2 /m 3  6  10 9 eV/c 2  m 3
(a) 10 26 kg/m 3 
 27
1
.
67

10
kg


(b)
29.
Solution
5.62  10 9 eV/c 2
 8  10 8
2
7 eV/c
Assume the average density of the universe is 0.1 of the critical density needed for
closure. What is the average number of protons per cubic meter, assuming the
universe is composed mostly of hydrogen?
N


ρc ( 0.10 ) 10 26 kg/m 3 ( 0.10 )

 0.6 m 3
 27
mp
1.67  10 kg
Openstax College Physics
Instructor Solutions Manual
Chapter 34
30.
To get an idea of how empty deep space is on the average, perform the following
calculations: (a) Find the volume our Sun would occupy if it had an average density
equal to the critical density of 10 26 kg / m 3 thought necessary to halt the expansion
of the universe. (b) Find the radius of a sphere of this volume in light years. (c) What
would this radius be if the density were that of luminous matter, which is
approximately 5% that of the critical density? (d) Compare the radius found in part
(c) with the 4-ly average separation of stars in the arms of the Milky Way.
Solution
M
M
1.99  10 30 kg
V 

 1.99  10 56  2  10 56 m 3
(a) ρc 
 26
3
V
ρc 1  10 kg/m
(b) V  4 r 3 
3


13
 3 1.99  10 56 m 3 

 
4


1 ly
 3.62  1018 m 
 383 ly  4  10 2 ly
9.46  1015 m
 3V 
r 

 4 
 1 
(c) 

 0.05 
(d)
13
13
383 ly   1040 ly  1  10 3 ly
1040
 260 times larger than stellar separation
4
34.6 HIGH-TEMPERATURE SUPERCONDUCTORS
31.
A section of superconducting wire carries a current of 100 A and requires 1.00 L of
liquid nitrogen per hour to keep it below its critical temperature. For it to be
economically advantageous to use a superconducting wire, the cost of cooling the
wire must be less than the cost of energy lost to heat in the wire. Assume that the cost
of liquid nitrogen is $0.30 per liter, and that electric energy costs $0.10 per kW·h.
What is the resistance of a normal wire that costs as much in wasted electric energy
as the cost of liquid nitrogen for the superconductor?
Solution
 1 kWh 
Power  P  I 2 R , Energy  E  Pt  I 2 Rt (in joules), E( in kW  h )  I 2 Rt 

6
 3.6  10 J 
It costs $0.30 to cool the wire for 1 hour. So, we want to know what resistance in the
wire would cost $0.30 to run the current for 1 hour (3600 s).
Openstax College Physics
Instructor Solutions Manual
Chapter 34
 $0.10 
 1 kW  h  $0.10 
2
Dollar Cost  E (in kWh )
  $0.30  $0.30  I Rt 


6
 kW  h 
 3.6  10 J  kW  h 


$0.30 3.6  10 6 J
$0.30 3.6  10 6 J
R

 0.30 
$0.10 I 2 t 
$0.10100 A 2 3600 s 

 



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