Download Homework Solutions Chapter 4

Homework Solutions
Chapter 4
Todd B. Krause
September 30, 2008
51. (a) If 2.5 · 1016 joules represents the energy of a major earthquake,
the energy of a 1-megaton bomb is smaller by a factor of
25 · 1015
2.5 · 1016 joule
=
= 5.
5 · 1015
5 · 1015 joule
A major earthquake releases as much energy as five 1-megaton
bombs.
(b) The annual U.S. energy consumption is about 1020 joules, and a
liter of oil yields about 1.2 · 107 joules. Thus, the amount of oil
needed to supply all the U.S. energy for a year would be
1020 joule
12
= 8 · 10 liter,
joule
1.2 · 107 liter
or about 8 trillion liters of oil (roughly 2 trillion gallons).
(c) We can compare the Sun’s annual energy output to that of the supernova by dividing: to be conservative, we use the lower number
from the 1044 – 1046 range for supernova energies:
supernova energy
1044 joule
= 34 = 1010 .
Sun’s annual energy output
10 joule
54. First, we find the U.S. energy consumption per minute by converting
the annual energy consumption into units of joules per minute:
1020
1
yr
1
day
1
hr
1.9 · 1014 joules
joules
·
·
·
≈
.
24 60 min
yr
365 days
hr
60
min
1
Next we divide this energy consumption per minute by the amount of
energy available through fusion of 1 liter of water (from table 4.1):
joules
1.9 · 1014 min
7·
joules
1013 liter
≈ 2.7
liters
.
min
In other words, it would take less than 3 liters of water per minute
— which is less than 1 gallon per minute — to meet all U.S. energy
needs through nuclear fusion. This is somewhat less than the rate at
which water flows from a typical kitchen faucet. So if we could simply
attach a nuclear fusion reactor to your kitchen faucet, we could stop
producing and importing oil, remove all the hydroelectric dams, shut
down all the coal-burning power plants, and still have energy to spare.
56. Newton’s version of Kepler’s third law has the form
p2 =
4π 2
a3 .
G(M1 + M2 )
Because the square of the period varies inversely with the sum of the
masses, the orbital period itself depends on the inverse square root of
the object masses:
s
4π 2
a3 .
p=
G(M1 + M2 )
Thus, if we have a star four times as massive
as the Sun, the the period
√
of a planet orbiting at 1 AU will be 1/ 4 = 1/2 that of the Earth, or
6 months.
57. (a) Using the Moon’s orbital period and distance and following the
method given in Mathematical Insight 4.3, we find the mass of
the earth to be about
MEarth =
4π 2 (aMoon )3
.
G(PMoon )2
Making sure that we use appropriate units, we find
m 3
· 1, 000 4π 2 384, 000 km
km
MEarth ≈ 2
3
s
m
hr
· 24 27.3
6.67 · 10−11 days
·
3,
600
2
day
hr
kg·s
24
= 6.0 · 10 kg.
2
(b) Using Io’s orbital period and distance and following the method
given in Mathematical Insight 4.3, we find the mass of Jupiter to
be about
m 3
· 1, 000 4π 2 422, 000 km
km
MJupiter ≈ 2
3
s
m
−11
42.5
hr
·
3,
600
6.67 · 10
2
hr
kg·s
= 1.9 · 1027 kg.
We find the same answer using Europa’s orbital properties; Kepler’s third law does not depend on the mass of either moon
because neither moon has a significant mass in comparison to the
mass of Jupiter.
(c) We again start with Newton’s version of Kepler’s third law:
p2 =
4π 2
a3 .
G(M1 + M2 )
We solve for the semimajor axis of the planet with a little algebra:
r
3 G(M1 + M2 )
p2 .
a=
4 π2
We convert the planet’s orbital period of 63 days into seconds:
·
63 days
60 s
60 hr
min
24 6
·
·
= 5.44 · 10 s.
1
1
hr
day
1
min
Planets are much less massive than their stars are, so we can
approximate M1 + M2 ≈ Mstar . From Appendix A, the mass of
the Sun is about 2 · 1030 kg. So we can calculate the semimajor
axis:
v
u
u
m3
(2 · 1030 kg)
6.67 · 10−11 kg·s
3
2
t
a=
(5.44 · 106 s)2
4π 2
= 4.64 · 1010 m.
Of course, this number would probably be more useful in astronomical units, so we should convert:
4.64 · 1010 m
·
1 AU
= 0.31 AU.
1.5 · 1011 m
The new planet is only 0.31 AU from its star, which is closer than
Mercury is to the Sun.
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59. In (a) and (c), the easiest way to find the escape velocities with the
given data is by comparison to the escape velocity from earth:
r
2
GMplanet
s
Rplanet
Mplanet REarth
vescape planet
= q
·
.
=
vescape Earth
MEarth Rplanet
2
GMEarth
REarth
Given that the escape velocity from Earth’s surface is about 11 km/s,
this formula becomes
s
Mplanet REarth
km
vescape planet = 11
·
·
.
s
MEarth Rplanet
(a) From the surface of Mars, the escape velocity is
r
km
MMars REarth
vescape = 11
·
·
s
M
RMars
r Earth
km
1
= 11
· 0.11 ·
s
0.53
km
= 5.0
.
s
(b) From the surface of Phobos, the escape velocity is
v u
u 2 · 6.67 · 10−11 m3 · 1.1 · 1016 kg
t
kg·s2
vescape =
12, 000 m
m
= 11
s
km
= 0.11
.
s
(c) From the surface of Jupiter, the escape velocity is
s
MJupiter REarth
km
vescape = 11
·
·
s
MEarth RJupiter
r
km
1
= 11
· 317.8 ·
s
11.2
km
= 58.6
.
s
4
(d) To find the escape velocity from the solar system, starting from
the Earth’s orbit, we use the mass of the Sun (since that is the
mass we are trying to escape) and the Earth’s distance from the
Sun of 1 AU, or 1.5 · 1011 m:
v u
u 2 · 6.67 · 10−11 m3 · 2.0 · 1030 kg
t
kg·s2
vescape =
1.5 · 1011 m
m
= 42, 200
s
km
= 42.2
.
s
(e) To find the escape velocity from the solar system starting from
Saturn’s orbit, we use the mass of the Sun (since that is the mass
we are trying to escape) and Saturn’s distance from the Sun of
1.4 · 1012 m:
v u
u 2 · 6.67 · 10−11 m3 · 2.0 · 1030 kg
t
kg·s2
vescape =
1.4 · 1012 m
m
= 13, 800
s
km
= 13.8
.
s
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