Ch. 45
Page 1
CHAPTER 45 – Astrophysics and Cosmology
Note:
A factor that appears in the analysis of energies is
e
2
/4p
Å
0
= (1.60
×
10
–19
C)
2
/4p(8.85
×
10
–12
C
2
/N
·
m
2
) = 2.30
×
10
–28
J
·
m = 1.44 MeV
·
fm.
1.
Using the definition of the parsec, we find the equivalent distance in m:
D
=
d
/
φ
;
1 pc = (1.50
×
10
11
m)/[(1.000
″
)/(3600
″
/°)(180°/p rad)] = 3.094
×
10
16
m.
From the definition of the lightyear, we have
1 pc = (3.094
×
10
16
m)/(3.00
×
10
8
m/s)(3.16
×
10
7
s/yr) = 3.26 ly.
2.
We find the distance from
D
= 1/
φ
= [1/(0.38
″
)](3.26 ly/pc) =
8.6 ly
.
3.
We find the distance from
D
= 1/
φ
= [1/(0.00019°)(3600
″
/°)](3.26 ly/pc) =
4.8 ly
.
4.
(
a
) We find the parallax angle from
D
= 1/
φ
;
36 pc = 1/
φ
, which gives
φ
=
0.028
″
.
(
b
) We convert this to degrees:
φ
= (0.028
″
)/(3600
″
/°) =
(7.7
×
10
–6
)°.
5.
We find the parallax angle from
D
= 1/
φ
;
(55 ly)/(3.26 ly/pc) = 1/
φ
, which gives
φ
=
0.059
″
.
The distance in parsecs is
D
= (55 ly)/(3.26 ly/pc) =
17 pc
.
6.
We find the distance in lightyears:
D
= (35 pc)(3.26 ly/pc) = 114 ly.
Thus the light takes
114 yr
to reach us.
7.
The star farther away will subtend a smaller angle, so the parallax angle will be
less
.
From
D
= 1/
φ
, we see that
D
1
/
D
2
= 2 =
φ
2
/
φ
1
,
or
φ
1
/
φ
2
=
!
.
8.
(
a
) The apparent brightness of the Sun is
¬
= 1.3
×
10
3
W/m
2
.
(
b
) The
absolute luminosity of the Sun is
L
=
¬
A
= (1.3
×
10
3
W/m
2
)4p(1.5
×
10
11
m)
2
=
3.7
×
10
26
W
.
9.
The apparent brightness is determined by the absolute luminosity and the distance:
L
=
¬
A
=
¬
4p
D
2
.
If we form the ratio for the apparent brightness at the Earth and at Jupiter, we get
¬
E
/
¬
J
= (
D
J
/
D
E
)
2
;
(1.3
×
10
3
W/m
2
)/
¬
J
= (5.2)
2
, which gives
¬
J
=
48 W/m
2
.
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Ch. 45
Page 2
10. The diameter of our Galaxy is 100,000 ly so the angle subtended by our Galaxy is
φ
Galaxy
=
d
1
/
D
1
= (100,000 ly)/(2
×
10
6
ly) = 0.05 rad =
2.9°.
The angle subtended by the Moon at the Earth is
φ
Moon
=
d
2
/
D
2
= (3.48
×
10
6
m)/(3.84
×
10
8
m) = 9.06
×
10
–3
rad =
0.52°.
Thus we have
φ
Galaxy
˜ 6
φ
Moon
.
11. If we assume negligible mass change, as a red giant, the density of the Sun will be
ρ
=
M
/
V
=
M
/
)
p
r
3
;
= (2
×
10
30
kg)/
)
p(1.5
×
10
11
m)
3
=
1.4
×
10
–4
kg/m
3
.
12. The angle subtended at the Earth by the Sun as a white dwarf will be
φ
=
d
/
D
= (3.48
×
10
6
m)/(1.5
×
10
11
m) = 2.3
×
10
–5
rad =
4.8
″
.
13. The density of the white dwarf is
ρ
dwarf
=
M
/
V
=
M
/
)
p
r
3
;
= (2.0
×
10
30
kg)/
)
p(6.38
×
10
6
m)
3
=
1.8
×
10
9
kg/m
3
.
The ratio of densities will be the ratio of masses:
ρ
dwarf
/
ρ
Earth
=
M
dwarf
/
M
Earth
= (2.0
×
10
30
kg)/(5.98
×
10
24
kg) =
3.3
×
10
5
.
14. The density of the neutron star is
ρ
ns
=
M
ns
/
V
ns
=
M
ns
/
)
p
r
ns
3
;
= (1.5)(2
×
10
30
kg)/
)
p(11
×
10
3
m)
3
=
5.4
×
10
17
kg/m
3
.
From the result for Problem 13 we see that
ρ
ns
= (3
×
10
8
)
ρ
dwarf
.
For the density of nuclear matter we calculate the density of a proton:
ρ
p
=
m
p
/
)
p
r
p
3
= (1.67
×
10
–27
kg)/
)
p (1.2
×
10
–15
m)
3
= 2.3
×
10
17
kg/m
3
.
Thus we see that
ρ
ns
˜
ρ
nuclear matter
.
15. For the reaction
He
2
4
+ He
2
4
→
Be
4
8
+
γ
, we determine the
Q
value:
Q
= [
M
(
4
He) +
M
(
4
He) –
M
(
8
Be)]
c
2
= [2(4.002603 u) – (8.005305 u)]
c
2
(931.5 MeV/u
c
2
) = – 0.0922 MeV =
– 92.2 keV
.
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 Spring '09
 Fielding
 Physics, Energy, Potential Energy, kg, Hubble, Schwarzschild Radius

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