This preview shows pages 1–3. Sign up to view the full content.
Serway/Moses/Moyer
Modern Physics, 3
rd
edition
Chapter 16 Answers and Solutions
3.
(a) We can estimate the distance with Hubble’s law:
R
=
v
H
=
(0.55)(3.00
×
10
8
m / s)
23
×
10

3
m / (s
⋅
lightyear)
=
7.2
×
10
9
lightyears
(b)
t
=
R
v
=
7.2
×
10
9
ly
0.55
c
=
7.2
×
10
9
c
(1 y)
0.55
c
≈
13
billion years
4.
(a) Let
H
(
t
)
=
b
, a constant. From equation 16.14
Ý
a
(
t
)
a
(
t
)
=
b
, which may be integrated to give
da
a
∫
=
bdt
+
C
∫
, where
C
is another constant.
Integrating gives
ln
a
=
bt
+
C
or
a
=
Ae
bt
where
A
=
e
C
(b) Equation 16.21 for an Einsteinde Sitter universe is
a
=
Ct
2/3
Thus,
H
=
Ý
a
a
=
2
3
t

1
5. Using Equation 16.22 gives
a
(
t
e
)
=
a
(
t
0
)
6
so the size is 1/6 the present
linear size.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentUsing
Z
= 5 in Equation 16.23 yields
t
e
=
1/6
(29
3
/2
t
0
≈
0.07
t
0
or about 7% of
the present age.
6.
Answers: (a) 590.2 nm, (b) 599.1 nm, (c) 688.6 nm
Solutions:
v
=
HR
,
H
=
(2.3
×
10

2
m / s) / ly
(a)
v
(2
×
10
6
ly)
=
4.6
×
10
4
m/s
′
λ =λ
1
+
v
/
c
1

v
/
c
=
(590 nm)(1.00031)
=
590.2 nm
(b)
v
(2
×
10
8
=
4.6
×
10
6
′
λ =
(590 nm)
1 + 0.01533
1

0.01533
=
599.1 nm
(c)
v
(2
×
10
9
=
4.6
×
10
7
′
(590 nm)
1 + 0.1533
1

0.1533
=
688.6 nm
7.
Answers: (a) 0.383
c
(b) 5.0
×
10
9
ly
Solutions:
(a) Solving Equation 16.1 for
v
gives
v
=
′
λ
2
λ
2
′
λ
2
+λ
2
c
=
650
2

434
2
650
2
+
434
2
c
=
0.383
c
(b)
v
=
HR
R
=
v
/
H
=
(0.383)(3
×
10
8
m / s) / [(2.3
×
10

2
m / s) / ly]
=
5.0
×
10
9
ly.
8.
Solution:
In our frame of reference, Hubble’s law is exemplified by
v
1
=
HR
1
and
v
2
=
HR
2
. From these we may form the equations

v
1
=
HR
1
and
v
2

v
1
=
H
(
R
2

R
1
)
. These equations express Hubble’s law as seen by the
observer in the first galaxy cluster, as she looks at us to find

v
1
=
H
(

R
1
)
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '09
 Staff

Click to edit the document details