. 22
( 28)


be carried out as long as we are willing to observe. When we talk about orbiting space probes, the
Because it is the closest, the Moon has been mass of the probe is much less than that of the
the subject of the most active space exploration. Sun or any planet. We therefore do not have to
Both the United States and Soviet Union launched consider the recoil of the Sun or any planet, mak-
a variety of unmanned probes to the Moon in the ing the problem a little easier.
late 1950s and 1960s. These started with crude
Example 22.2 Orbit and escape from the Earth
craft that (intentionally) crashed on the surface,
What is the speed for a circular orbit just above the
relaying brief series of images before impact. The
Earth™s atmosphere? How does this compare with
US Ranger series was an example of this. With
the escape velocity?
more sophisticated equipment, as well as more
powerful booster rockets to lift the more sophis-
ticated equipment, the probes were able to land,
For an object of mass m, in a circular orbit of radius
and send back data from the surface. The US
r, around a planet of mass M, the gravitational
Surveyor series carried out those missions.
force must provide the acceleration for circular
These were all a prelude for astronauts explor-
motion, so
ing the Moon. Between 1969 and 1971, there were
mvorb GMm
six manned lunar landings. Each carried out
extensive studies of the surface and also returned
samples to Earth for detailed laboratory analyses.
where vorb is the orbital speed. Solving for vorb gives
The Soviet Union also had an unmanned space-
GM 1>2
a b
craft land on the Moon, and then return to Earth
with rock and soil samples. All of these analyses
16.67 dyn cm2 s2 2 16 1027 g 2
are still continuing. 8 12
a b
The rest of the Solar System is so far away that
108 cm2
it has only been visited by unmanned probes,
105 cm s
which signal their data back to Earth. The Voyager 7.9
spacecraft had missions in which they flew close
7.9 km s
to planets, often a few planets in succession.
28 000 km hr
22.4 Traveling through the The escape speed is that with which we must
Solar System launch an object from the surface, such that it can
move infinitely far away with zero kinetic energy.
This means that we define potential energies to be
We have already seen that the distances between
zero when objects are infinitely far apart. The total
the planets are quite large, and that the Solar
energy of an object with the escape speed is zero. If
System is mostly empty space. However, by the
vesc is the escape speed, then
standards of the distances to the nearest stars,
a b mv2
the planets are quite close. They are close enough
that the distances can be traveled by interplane- r

Solving for vesc gives v
2GM 1>2
a b θ

Note that comparing vesc and vorb gives

12 vorb

Therefore, the escape speed for the Earth is 11 km/s.
In considering travel to other planets, we use
elliptical orbits. After all, an object whose orbit
touches both the Earth and Mars cannot be a cir-
cular orbit about the Sun. (We should note that
parabolic or hyperbolic orbits can be used for one- E
time flybys, such as Pioneer and Voyager.) When
we studied binary stars, we found that the veloc-
ity of a particle in an elliptical orbit of semi-
major axis a, when the object is a distance r, from
a mass M at the focus, is

GM c d
2 1
v2 (22.2)
r a 1

We can use this to find the total energy of the

a b mv2

a b GMm c d
1 2 1
r a r
Fig 22.14. Angular momentum and orbital eccentricity.
(a) If a rocket stops its powered ¬‚ight at a distance r from
Note that, if you are in an orbit of a given semi- the Earth while moving with a velocity v, its energy is deter-
major axis a, and fire a rocket such that your mined (as the sum of kinetic plus potential energies at that
point), independent of the direction of motion.This means
energy increases, you must go to a higher orbit,
that the semi-major axis of the ellipse is determined.
making your energy less negative. In the higher
However, the direction of motion determines the
orbit, your speed will actually be less than it was
eccentricity. (b) Ellipses with the same semi-major axis, but
in the lower orbit. Thus, firing the engine to
with different eccentricities.
“accelerate” the rocket has the interesting effect
of reducing its orbital speed.
Another consequence of equation (22.3) is
that the energy of an orbit depends only on the also determines the total angular momentum, L.
semi-major axis, and not on the eccentricity. If we Referring to Fig. 22.14, the angular momentum is
launch a rocket, once its fuel is used up its energy given by
is determined by its location and its speed. How,
L mvr sin
then, can we determine the eccentricity of the
resulting orbit? The direction of motion when The least eccentric orbit has the highest angu-
the fuel is used up determines the eccentricity. It lar momentum for a given semi-major axis. It

of the distance of the Earth from the Sun and the
Earth at
distance of Venus from the Sun. Therefore, the
semi-major axis is

11 2 2 1RE RV 2
us's Orb
Ven 0.86 AU
The eccentricity is defined by the separation
Venus at
between the foci, divided by 2a. The Sun is at one
focus, RV from perihelion. Therefore, the symmetry
of the orbit requires that other focus to be RV from
the aphelion. The distance between the foci is

's d 2a 2RV
ro b
Earth when
Probe Reaches
E arth's O rbit
Venus RE RV
Fig 22.15. Minimum energy orbit of a space probe to
The eccentricity is therefore
should be noted that 90 doesn™t give a circu- RE RV
lar orbit unless the speed is the right speed for a
circular orbit at that radius. However, 90
means that the point where the fuel is spent will
be the aphelion or perihelion of the orbit.
At launch, r RE so the launch speed is given by
The trick in launching a planetary probe is to
place it in an orbit that intersects the planet™s
GM a b
2 2
orbit when the planet is at the point of intersec- RE RE RV
tion. In choosing the direction of launch, we can
This gives v 27.3 km/s. Note that the Earth™s
take advantage of the Earth™s motion. There are
orbital speed is 30 km/s, so the probe must be
many different orbits that can be chosen.
going 2.7 km/s slower than the Earth. This means
However, the minimum energy orbit is the one
that we must launch the probe in the opposite
that has the Earth at aphelion and the planet at
direction from the Earth™s motion.
perihelion (if the planet is closer to the Sun) or
We can find the length of the trip by using Kepler™s
one that has the Earth at perihelion and the
third law. It tells us that the period of the orbit is
planet at aphelion (if the planet is farther from
given by
the Sun). The minimum energy orbit of probe to
Venus is shown in Fig. 22.15. P2 a3
Example 22.3 Minimum energy orbit to Venus For a 0.86 AU, we find P 0.80 yr. Therefore the
Find the semi-major axis and eccentricity of a mini-
trip to Venus will take half this time, or 0.40 yr
mum energy orbit from the Earth to Venus. Find
(approximately five months).
the necessary launch speed and the time for the
Once we know how long the trip will take, we
must launch when Venus is at the place in its
orbit to take it to the rendezvous place in the
travel time of the probe. The short period of
In this case the Earth is at aphelion and Venus is at
time when a launch is possible is called a launch
perihelion. The major axis of the orbit is the sum

is much less than the mass of the train, and
v that the collision is completely elastic. The train
is moving with speed u and the ball is moving
with speed v. As viewed from the train, the ball
is coming towards the train at a speed of u v.
In the elastic collision, the train is so massive
that it doesn™t recoil, so the ball simply reverses
(b) its velocity relative to the train. It is now mov-
ing away from the train at a speed u v. As
viewed from the ground, its speed is now v 2u.
v+u In the collision, the ball has taken a little
energy from the train (very small relative to the
total energy of the train), and its speed is now
greater than its initial speed by twice the speed
of the train.
v + 2u u
With a space probe, We cannot, of course,
cause a space probe to bounce directly off Jupiter.
Fig 22.16. Analogy for gravity assist. (a) A low mass ball
with speed v approaches a high mass train with speed u for a
head-on collision. (b) The same situation viewed from the
train.The train is at rest, and the ball has a speed v u.
(c) Still looking from the train, the elastic collision simply
reverses the direction of the ball, so it is now moving away Actual
from the train at a speed v u. (d) We revert to the view Trajectory
from the ground. If the ball is moving ahead of the train at a
speed v u, then the speed of the ball, relative to the
ground, must be this speed, plus the speed of the train u, Planet's Orbit
giving a speed v 2u. Planet

window. The size of the window depends on how
Trajectory that
far we are willing to deviate from the minimum would give
energy orbit. In particular, orbits with higher maximum boost
energy might be used since they have shorter
travel times. Since the launch requires the cor-
rect relative positions of the Earth and Venus,
the launch window opens once per synodic
period of Venus. This explains the spacing of
launches for a given planet.
There is a trick that can be used to minimize
the energy needed to visit a planet beyond
Jupiter. In this, we take advantage of an elastic Fig 22.17. Trajectories for gravity assist.The space probe
gravitational encounter between the space probe doesn™t have to collide with the planet; it only has to make a
and a massive planet. This is known as a gravity close gravitational encounter.The maximum boost would
assist. come from a trajectory that comes the closest to a head-on
To understand how this works, let™s look at collision. However, since the spacecraft is coming from well
inside the planet™s orbit, this is not practical.The actual tra-
an analogous situation, a ball bounced off the
jectory still gives almost half of the maximum boost in
front of an incoming train. This is illustrated in
Fig. 22.16. We assume that the mass of the ball

However, we can arrange the trajectory to have
the same effect (Fig. 22.17). We cannot generally
take advantage of a head-on collision, but the
speed of the probe can be increased by an amount
of the order of the speed of Jupiter. The technique
has been used for the Voyager 1 and 2 missions, as Jupiter at
shown in Fig. 22.18.
Earth at
Saturn at
(20 Aug 1977)
(9 July 1979)

(25 Aug 1981)

(24 Jan 1986)
Uranus at

Fig 22.18. Orbit of Voyager 2 showing the effect of
gravitational assist in traveling past each planet.

Chapter summary
Most of the objects in the Solar System orbit the (3) the square of the period of the orbit is simply
Sun close to the plane of the Earth™s orbit, the related to the cube of the semi-major axis.
ecliptic. There are nine planets, which fall into The Moon changes its position in the sky and
two distinct groupings. The four inner planets its appearance as it orbits the Earth. The phases
are small, with the Earth being the largest. The result from the fact that when we look at the
next four planets are larger, and have very differ- Moon we are seeing it by reflected sunlight. Since
ent compositions than the inner planets. The out- the Moon™s orbit is not in the plane of the ecliptic,
ermost planet, Pluto, is small, like the inner there is not a solar eclipse and a lunar eclipse each
planets. All but two of the planets have moons month. However, approximately twice per year
orbiting them. There is also a collection of arrangements favorable for eclipses take place. In a
smaller material in the Solar System. There are solar eclipse, the Moon™s shadow moves across a
the asteroids, found mostly between Jupiter and small band on the Earth. For observers in that
Mars. There are also comets, which are usually band, the Moon covers the face of the Sun. In a
faint, but occasionally brighten and develop a tail. lunar eclipse, the Moon passes through the Earth™s
The comets leave debris behind, which we see as shadow, and all observers on the night side of the
meteors. Earth can see the eclipse. Some sunlight passing
Since the Earth is orbiting the Sun, the through the Earth™s atmosphere gives the Moon a
motions of the planets, as viewed from Earth, faint red glow, even during totality.
appear complicated. However, they can be The Solar System is unique in that the objects
explained most easily in terms of a Sun-centered are so close that we can visit them with space
system. The orbits of the planets are ellipses, probes. (All other astronomical objects are too far
rather than simply being circles. The motions of away for this.) We have learned a lot from astro-
the planets are summarized in Kepler™s laws: nauts visiting the Moon, as well as from automated
(1) the planets move in elliptical orbits with the spacecraft. Robot spacecraft have visited most of
Sun at one focus; (2) a line from the Sun to a the other planets and have sent back images as
planet sweeps out equal areas in equal times; well as other data.

*22.9. Mercury™s location in the sky makes it one
22.1. List the planets in order of increasing mass.
of the most difficult planets to observe. Why
Where do the most massive moons of each
is this?
planet fit into this? Where do the largest
22.10. The closest approach between Mars and Earth
moons of Jupiter, Saturn, Uranus and
occurs when they are at opposition. However,
Neptune fit into this? See Appendix D for
that distance is not always the same from
properties of the planets and satellites.
one opposition to the next. Why is that?
22.2. If you were going to construct a scale model
22.11. When Mars is at opposition (and close to the
of the Solar System, and the Earth was 1 m
Earth for good observing), would you expect
from the Sun in your model, how far are
it to be up at night or during the day?
the other planets from the Sun in that
*22.12. Suppose the Earth and Mars are at opposi-
tion. Estimate how long it will be before
22.3. Why is the distinction between inner and
they are at opposition again. Remember,
outer planets important beyond the simple
both are moving. You might want to draw a
question of where the planets are located?
diagram, and keep track of the motions of
22.4. How many moons in the Solar System are
both planets in fixed time intervals.
more massive than ours? See Appendix D for
22.13. Why is the frequency of launch windows for
properties of the planets and satellites.
spacecraft to a planet dependent on the syn-
22.5. Explain, in your own words, how retrograde
odic period of that planet?
motion is explained in a Sun-centered
22.14. Suppose that you lived on Mars. (a) Describe
how the Earth would move in your sky.
22.6. In the philosophy of science, we are always
(b) Would the Earth appear to go through
told to use the simplest model that explains
phases, like Venus?
all of the available data. Compare the helio-
22.15. At what point in its orbit does a planet move
centric and Earth-centered systems in terms
(a) fastest, (b) slowest?
of their simplicity.
22.16. Why is the location of Jupiter the place
22.7. Why was Galileo™s use of the telescope
where you would expect to find the most
important in supporting the heliocentric
massive planet?
view of the Solar System?
22.17. If we want to study Jupiter up close, we can
22.8. Why is there a limit on the maximum angle
send a spacecraft. Why is this not practical
that we can see between Venus and the Sun
for studying the nearby stars?
in the sky?

22.1. Use the tabulated sidereal periods of the 22.7. What is the escape velocity from our point
planets to find their synodic periods. in the Solar System?
22.2. What are the angles of greatest elongation 22.8. For a minimum energy orbit to Mars, find
for Mercury and Venus? the semi-major axis, the eccentricity, and
22.3. Fill in the trigonometric steps in finding the the time of flight. Draw a diagram showing
distance to a planet using Kepler™s method. the orbit, and the positions of the Earth and
22.4. How does the angular momentum per unit Mars at launch and landing.
mass vary with distance from the Sun? 22.9. Find the speed and distance from the center
22.5. Show how the inverse square law can be of the Earth for a satellite in synchronous
derived from Kepler™s third law for circular orbit around the Earth. (A synchronous orbit
orbits. is one that takes 24 hours.)
22.10. Express equation (22.2) in a form that gives v
22.6. Show that viewers on opposite sides of the
Earth see the Moon rotated through 2 . in terms of the escape velocity.

22.11. How does the kinetic energy of an object momentum, orbital plus rotational, of all
with the escape speed at a distance r com- the planets.
pare with that of an object in a circular 22.14. Estimate the material that can be cleared
orbit of radius r. out of the Solar System in 1 yr by a solar
wind with a mass loss of 1 M 106 yr and a
22.12. If a ball of mass m bounces elastically off the
front of a train of mass M, where M W m, speed of 200 km/s. Assume that the material
with the ball initially moving with a speed v to be cleared out is at the position of the
and the train initially moving at a speed u, Earth.
what fraction of the kinetic energy of the 22.15. Estimate the rate at which material can be
train is lost in the collision? swept out from the position of the Earth by
22.13. Find the angular momentum of the Sun, radiation pressure from the Sun.
and compare it with the angular

Computer problems

22.1. Assume that the Earth and Mars are both in cir- 22.4. Show that the moons of Jupiter and those of
cular orbits, and consider the part of their orbits Saturn obey Kepler™s third law.
where the Earth is overtaking Mars and passing 22.5. Find the masses of planets from satellite orbit data.
it. Draw a graph of the apparent position of Mars 22.6. Find escape velocities for all the planets and for
in our sky as a function of time, from the time the largest moon of each planet.
when the Earth is 15 behind Mars to the time 22.7. For minimum energy orbits to Mercury, Mars,
when it is 15 ahead. Saturn and Neptune find the semi-major axis, the
22.2. Given the sidereal periods of the planets, make a eccentricity, and the time of flight. Draw a dia-
table showing their synodic periods. gram showing the orbit, and the positions of the
22.3. Show that the planets obey Kepler™s third law. Earth and the planet at launch and landing.
Chapter 23

The Earth and the Moon

We start our discussion of the planets with the
one with which we are most familiar, the Earth.
In understanding the processes that are impor-
tant on the Earth, both now and in the past, we
are setting a framework for our understanding of
the other planets. Therefore, in this chapter we
will develop many of the ideas that should apply
to all planets, both in terms of what properties
are important and how we measure them.

23.1 History of the Earth
23.1.1 Early history
The main steps in the history of the Earth are
shown in Fig. 23.2. Somehow, the Earth accreted
from the material in the original solar nebula.
We will discuss more about the solar nebula in
Fig 23.1. In this photo of the Earth from space we can
Chapter 27. Enough material collected together
really appreciate that it is just another planet, one that has a
so that its own gravitational pull was able to keep
moon.This image was taken on 16 Dec 1992, by the Galilelo
most of the material from escaping. As particles
spacecraft from a distance of 6.2 million km. [NASA]
fell towards a central core, they were moving
closer together, so their gravitational potential
energy decreased. This means that their kinetic sank into the center, while lighter elements, such
energy increased. This kinetic energy was then as aluminum, silicon, sodium and potassium,
floated to the surface. This process is called differ-
available to heat the forming planet. In addition,
entiation. The iron and nickel form the current
heat was provided by the radioactive decay of
potassium, thorium and uranium. Such decays core of the Earth. So much heat was trapped by
led to heating, because the energetic particles “ the Earth that its core is still molten. The tho-
alpha, beta, gamma “ were absorbed by the sur- rium and uranium were squeezed out of the core,
rounding rock. The relatively massive alpha par- and carried along in crystals to the surface. These
ticles were particularly effective in this heating. radioactive materials provide heating close to the
The heating resulted in a liquid, or molten, surface.
interior. Since materials are free to move in a liq- The molten core is responsible for the Earth™s
uid, heavier elements, such as iron and nickel, magnetic field. The rapid rotation of the Earth

Fig 23.3. Cooled volcanic rock is an important
component of the Earth™s surface. [USGS]

potassium), and sedimentary rocks, that have been
deposited gradually. Both types of rocks can be
altered by high pressures, and are then called meta-
morphic. The lower part of the crust is composed of
Crust gabbro and basalts. These are dark rocks containing
silicates of sodium, magnesium and iron. (Silicates

Mantle are compounds in which silicon and oxygen are

added to the other atoms in various combina-

tions.) Gabbros have coarser crystals than do
basalts. Larger crystals result from slower cooling.
An important clue to the Earth™s interior

comes from volcanic activity. This activity results

from the heating of the crust, producing the

molten material. Volcanic activity is very impor-
Fig 23.2. Diagram showing steps in the formation and
tant in mountain building, as shown in Fig. 23.4.
development of the Earth.
It also provides a means of transporting certain
materials from the interior to the surface. In par-
ticular, we think that this is the source of carbon
leads to a dynamo process. In this process, a small
dioxide (CO2 ), methane (CH4) and water (H2O), as
magnetic field plus convection creates electrical
well as sulfur-containing gases in our atmos-
currents flowing through the fluid. These cur-
phere. Visitors to active volcanoes on Earth, such
rents produce a larger magnetic field, and so on.
as Kilauea on Hawaii, shown in Fig. 23.4, often
This process takes energy from the Earth™s rota-
note a strong sulfur smell.
tion. The Earth™s magnetic field is not fixed. The
As the water vapor was ejected into the atmos-
magnetic pole wanders irregularly. In addition,
phere, the temperature was low enough for it to
geological records indicate that the direction of
condense. Other gases dissolved in the water, and
the magnetic field has reversed every few hun-
combined with calcium and magnesium leached
dred thousand years. The actual reversals take
from surface rocks. This had the important effect
tens of thousands of years, and there is a period
of removing most of the carbon dioxide from the
when the field is very weak. The causes of the
Earth™s atmosphere.
reversals are not well understood.
The upper part of the Earth™s crust is composed
23.1.2 Radioactive dating
of several different types of rocks. There are igneous
Much of what we know about the age of the
rocks, such as granite, which are formed in vol-
Earth™s crust comes from radioactive dating. In this
canos (and are enriched in silicon, aluminum and

Fig 23.4. Images of an active
volcano, Kilauea, in Hawaii.This is
an image from space, to give you
an idea of the quality images we
can make from orbit around
other planets. (a) An overhead
view from radar studies on the
shuttle Endeavor.This is an inter-
ferometric image. (b) A three-
dimensional perspective (in false
color) reconstructed from the
radar images. [NASA]


which is the time for the number in the
sample to fall to half of its original value,
1 t> 1>2
N1t 2 N0 a b (23.1b)

Comparing these two expression (see Problem
23.2), tells us that
1/2 e

(b) The behavior of N(t) vs. t is shown in Fig. 23.5.
Half-lives of many nuclei are measured in the
technique, we are studying the products of laboratory (see Problem 23.4). Therefore, if we
know N0 and N(t), we can solve for t, the time that
radioactive decay. In radioactive dating we take
has elapsed since the sample had N0 nuclei of the
advantage of the fact that if we start with some
number N0 of nuclei of a radioactive isotope, the particular isotope. This is the essence of radioac-
number left after a certain time t is given by tive dating. It is important to choose an isotope
N1t 2
whose half-life is comparable to the time period
N0 e (23.1a)

you are trying to measure. If the time period is
The quantity e is the time for the number in much longer than the half-life, there will be too
the sample to fall to 1/e of its original value. We can few left to measure. If the time period is much
also write this expression in terms of the half-life, less than the half-life, very few decays will have

1 doesn™t stay bound to the rocks. We can collect
the argon and study the relative abundances of
various isotopes.
0.8 In considering the history of the Earth, we
must remember that the atmosphere has eroded
Fraction Remaining

the surface, altering its characteristics. (As we
0.6 will see below, the motions of the continents also
alter the surface.) The oldest rocks that we see are
dated at 3.7 billion years. (The oldest fossil cells
are dated at 3.4 billion years.) We think that the
surface underwent significant alteration 2.2 to
2.8 billion years ago. This was a period of volcanic
mountain building with the Earth having a very
thin crust. The crust was probably broken into
smaller platelets.
The age of the Earth, even as an approxima-
1 2 3 4 5
tion, is an important reference. It gives us an idea
t / Lifetime
of what the appropriate time scales are for
Fig 23.5. Radioactive decay.The vertical axis is the number
understanding the Solar System. For example, if
of remaining nuclei N, divided by the original number N0.
the Earth formed as a by-product of the forma-
tion of the Sun, then the Sun must be at least as
old as the Earth. This tells us that in understand-
ing the Sun and other stars, we must be able to
taken place. We believe that the Earth™s crust is
billions of years old, so the alpha decay of 238U explain how they can shine for billions of years.
On the other end of the time scale, we can see
(uranium-238), with a half-life of 4.6 billion years,
that, as far back as ancient history seems to us,
is ideal for studying its age.
some 5000 years, it is but a blink on the time
A practical problem is that we often don™t
scales of the Earth™s history.
know how much of a given radioactive material
At that time, the atmosphere was mostly
the Earth started out with. This means that we
water, carbon dioxide, carbon monoxide and
must employ indirect methods, involving knowl-
nitrogen. It is thought that any ammonia and
edge of the end products of the decays. For exam-
ple, the alpha decay of 238U is followed by a series methane could not have lasted very long. There
was little oxygen, since oxygen is a product of
of alpha decays, eventually leading to the stable
isotope 206Pb (lead-206). If we can assume that all plant life. Since there was no oxygen, there was
of the 206Pb on the Earth came from such decays, no ozone (O3) layer to shield the Earth from the
we could use the amount of 206Pb as an indicator solar ultraviolet radiation. We think that this
of the original amount of 238U. ultraviolet radiation must have stimulated the
chemical reactions to make the simplest organic
However, the Earth may have been formed
with some 206Pb, so we have to use even more compounds. It has been shown in the laboratory
that such reactions are greatly enhanced in the
involved techniques to correct for that effect.
presence of ultraviolet radiation. We will talk
These techniques often involve measuring the rel-
more about the early evolution of the Earth™s
ative abundances of certain isotopes, such as
Pb and 206Pb. Such measurements require a atmosphere in Chapter 27.
means of separating the isotopes. This separation
23.1.3 Plate tectonics
is much easier for gases than for solids. For this
The layer below the thin crust is kept heated by
reason, we actually use a different age tracer. For
example, 40K (potassium-40) beta decays with a radioactive decay. The amount of heat is not suf-
ficient to melt the material completely, but it
half-life of 1.3 billion years. The decay product
is 40Ar (argon-40). Since argon is a noble gas, it keeps it from being completely solid. It has the

180° 90° 90° 180°

45° 45°

45° 45°

90° 90°

Fig 23.6. The main tectonic plates are outlined by areas of
geological activity. Earthquakes and volcanos are marked by
When one plate is forced under another, the
red dots. Arrows indicate the direction of motion of the
resulting upward pressure can build great moun-
plates. [NASA]
tain ranges, such as the Himalayas, as shown in
Fig. 23.8. These plate boundaries also show a high
frequency of volcanoes and earthquakes. The vol-
consistency of plastic. This means that if you
canoes result from material being pushed
press on it, the material doesn™t deform instantly,
upward. The earthquakes result from the fact
as a liquid would. However, under a steady pres-
sure, it will flow slowly. The solid layer above the
plastic region is called the lithosphere. The Earth™s
lithosphere is broken into plates. The name is
meant to suggest that they are much larger in
extent along the surface than they are thick. The
plates float on top of the plastic layer.
As they float, the plates move slowly. Since
they carry the continents with them as they
move, we refer to this motion as continental drift.
The general term for any process involved in the
movement or deformation of planetary surfaces
is tectonics, so continental drift is also called plate
tectonics. Fig. 23.6 shows the Earth™s main tectonic
plates. Their motion is being driven by material
being forced up from below into some narrow
gaps between the plates. One such region, shown
in Fig. 23.7, is called the mid-Atlantic ridge.
Throughout the ridge, fresh material is appear-
ing on the sea floor, as the plates move away from
Fig 23.7. A computer generated image of what the mid-
the ridge.
Atlantic ridge would look like if there were no water in the
The regions where the plates meet are charac-
ocean. [USGS]
terized by a high level of geological activity.

Fig 23.8. (a) Space radar image
of Mt Everest, in the Himalayan
mountain range. (b) Optical
image. Both (a) and (b) were
taken from the shuttle Endeavor
(10 Oct 1997). Images show an
area 70 38 km across. (c) Part
of the San Andreas fault line in
California. [NASA]



which this slippage takes place is called a fault
line. One famous fault line “ the San Andreas
fault in California “ is shown in Fig. 23.8(c).

23.2 Temperature of a planet

The temperature of the Earth is determined by a
balance between the energy absorbed from the
Sun and the energy given off by the planet. For
planets like the Earth, heat from the inside does
not have much effect on the surface temperature.
The planetary temperature for which these bal-
ance is called the equilibrium temperature of the
planet. The actual energy transport might be
complicated by the presence of an atmosphere,
but we will first calculate the equilibrium tem-
that slippage of the plates past each other is not
perature, ignoring atmospheric effects. In the
smooth. For long periods the plates might not
absence of an atmosphere, the calculation is
move as the pressure increases. Eventually, the
essentially the same as for that of an interstellar
pressure becomes too great, and there is a sudden
grain (see Chapter 14).
movement along the boundary. The line along

We start by calculating the energy received albedo and emissivity inside the integral. For
per second. The energy per second given off by example, remembering the luminosity of the
the Sun is its luminosity, L . (We could leave this Sun,
in terms of the luminosity, or rewrite the lumi-
4 R2 B 1T 2 d
nosity as 4 R2 T 4 .) At the distance d of the L
planet from the Sun, the luminosity is spread 0

out over a surface area of 4 d2. This means that where B is the Planck function. The power
the luminosity per surface area is L 4 d2. As absorbed by the Earth is then
seen from the Sun, the projected area of the
R2 R2
c d 11 a 2 B 1T 2 d
2 q
planet is RP . The planet therefore intercepts a P
Pabs (23.5)
power equal to this area multiplied by the power 0
per surface area. Finally, not all of the sunlight
Similarly, the power radiated is
is absorbed by the planet. A fraction a, the
4 R2 e B 1TP 2 d
albedo, is reflected. The amount absorbed is q
Prad (23.6)
equal to (1 a) multiplied by the amount that 0
actually strikes. In this calculation, we are
Equating these gives
assuming that the albedo is the same at all wave-
lengths. Therefore the power absorbed by the R2
4 e B 1TP 2 d c d 11 a 2 B 1T 2 d
q q
planet is d2
0 0
L 11 a 2 R2 This equation cannot be solved directly for TP
Pabs (23.2)
4d even if we know a and e . But since we know T ,
we can evaluate the right-hand side of the equa-
We now look at the power radiated. We
tion. We then try different values of TP on the
assume that the planet rotates fast enough that
left-hand side, probably using a computer to eval-
there is no great difference between day and
uate the integral, until we find a value of TP
night temperatures so we can treat the tempera-
which makes the left-hand side equal to the
ture of the planet as being the same everywhere.
right-hand side. To be even more rigorous, we
(This is a good approximation for the Earth but
should also account for the fact that the temper-
not for the Moon.) The power radiated per unit
ature is not constant across the surface of a
surface area is e TP , where e is the emissivity. The
emissivity can range from zero to one, and is one
We can take advantage of the fact that most of
for a perfect blackbody. Multiplying this by the
the Sun™s energy is in the visible and most of the
planet™s surface area 4 RP, we obtain the total
energy given off by the Earth is in the infrared.
power radiated:
We can assume that there is a constant albedo,
4 R2 e T 4
Prad (23.3) aV, in the visible, and a constant emissivity, eIR, in
p p
the infrared. Since aV and eIR are constant over
If we equate the power absorbed with the
the region of each integral where B is signifi-
power radiated, we solve for the equilibrium tem-
cant, we can factor them out of the integral. The
perature of the planet, giving
result is similar to equation (23.4):
L 11 a2 1>4
c d 11 aV 2
Tp (23.4) 1>4
c d
16 d2 e TP (23.8)
16 d2 eIR
This calculation doesn™t account for the fact
On the Earth, the albedos are different for the
that the albedo and emissivity vary with wave-
oceans and for land. They are also different for
length. We must integrate the energy received
cloud cover. When we take these into account,
over the spectral energy distribution of the Sun,
the equilibrium temperature is 246 K. However,
and integrate the energy radiated over the spec-
this is still not the temperature we measure at
tral energy distribution of the planet. In each
the ground. We have not yet considered the
case, we incorporate a frequency dependent

results shown in Fig. 23.9(a). Here we present a
graph, in which we show the equilibrium tem-
700 perature at different distances, and note the loca-
Equilibrium Temperature

tions of the planets. Notice how this temperature
decreases with increasing distance.
23.3 The atmosphere

The Earth™s atmosphere provides us with a multi-
tude of phenomena whose complexity may make
them seem beyond understanding. However, we
can apply many of the basic physical ideas that
100 we have already discussed for stars, such as hydro-
static equilibrium and energy transport, to form
a reasonable understanding of those phenomena.
0 5 10 15 20 25 30
With the aid of supercomputers, these ideas have
Distance from Sun (AU)
been applied to the atmosphere in some detail.
Also, the concepts that we develop in studying
the Earth™s atmosphere will be directly applicable
to studying the atmospheres of other planets. In
fact, one of the checks on computer models for
the Earth™s atmosphere is to see if they can pre-
dict the properties of the other atmospheres in
the Solar System. In this section we look at some
of the concepts common to all planetary atmos-
pheres, and see how they apply to the Earth™s
Though the Earth is a sphere, and the atmos-
phere is a spherical shell around it, the atmos-
phere is very thin (a few hundred kilometers)
compared with the radius of the Earth (6380 km).
This means that, if we stand on the ground, the
effects of the curvature in the atmosphere are
very small. This is shown in Fig. 23.10(a). This
tells us that we can treat the atmosphere like a
thin layer, and only worry about how things
change as we go to higher altitudes. In studying
Fig 23.9. (a) Diagram showing graph of the equilibrium the Earth™s atmosphere, we want to understand
temperature vs. distance from the Sun. (b) Temperature varia-
how the pressure changes with altitude (the pres-
tions across the surface of the Earth, as shown in near infrared
sure distribution), how the temperature changes
images from space.This image is at a wavelength of 1 m from
with altitude (temperature distribution), what
the Galileo spacecraft at a distance of 2.1 million km. Lighter
the composition is, and how it changes with alti-
areas are giving off more infrared emission. [(b) NASA]
tude, and how energy is transferred through the
important effects of radiative transfer in the In studying the atmosphere, it is convenient to
atmosphere. divide it into layers, according to what conditions
We can also do this calculation for planets at are prevalent. These layers are shown in part (b)
any distance from the Sun, and we obtain the of Fig. 23.10. Each of the layers has a different

must have an equation of state. We can treat plan-
etary atmospheres as ideal gases, so the equation
of state is simply
1 >m 2kT
P (23.9)

In this expression, m is the average mass per
particle. For the Earth™s atmosphere, this is
approximately 29 times the mass of the proton,
reflecting the fact that the atmosphere is mostly
N2 (molecular weight 28) and O2 (molecular weight
32). For a surface temperature of 300 K, the den-
sity at the surface is 1.1 10 3 g/cm3.
The pressure at the bottom of the atmosphere,
near the surface of the Earth, is called one atmos-
phere or one bar. It is quite large, approximately
105 N for every square meter (106 dyn/cm2 or
15 lb/in2). Remember the weight of a typical per-
son is about 750 N, so it is like having over 100
people stand on every square meter. We don™t
normally see the effects of this pressure because
in most situations it tends to cancel. For example,
for a wall, the air on opposite sides is pushing
with equal and opposite forces, resulting in a net
Mesosphere 90 km force on the wall of zero. We see the effects of the
pressure if we remove it from one side, by using
Ozone Layer an air pump, for example.
50 km
23.3.1 Pressure distribution
Tropopause The vertical distribution of pressure and den-
18 km
Troposphere sity is governed by the condition of hydrostatic
equilibrium, just as in stars. The weight of each
layer is supported by the pressure difference
(b) between the bottom and the top of that layer.
For stars, we treated the layers as spherical
Fig 23.10. (a) Photograph showing the Earth and its
shells. We could do that for the Earth, also.
atmosphere from space; note how thin the atmosphere is.
However, the Earth™s atmosphere is so thin that
(b) Diagram showing the layers in the atmosphere.
[(a) NASA] we can treat it as a plane parallel layer (see
Problem 23.8).
The equation of hydrostatic equilibrium then
temperature distribution, as we will discuss in becomes
Section 23.3.2. The bottom layer, to which we are
mostly confined, is called the troposphere. It is g (23.10)
only 14 km thick. At the top of the troposphere is
the thin tropopause. Above that is the stratosphere, We now have two equations, the equation of
in which some high altitude aircraft fly. At the state and the equation of hydrostatic equilib-
top of the stratosphere is the ozone layer (which rium. However, we have three unknowns, T, P and
will be discussed below). . When we look at the temperature distribution,
To relate the basic variables that describe a discussed later in this section, we find that, espe-
gas “ temperature, density and pressure “ we cially in the lower atmosphere, the temperature

doesn™t deviate by large amounts from its value at 35
the ground, T0. We therefore make the approxi-
mation that the temperature is constant.
Knowing T, we can use equation (23.9) to substi-
tute for the density in equation (23.10), so that the 3H
only remaining variable is pressure. This gives 25

a bP

Altitude (km)
dz 20
To integrate this, we want all of the P dependence
on one side and all of the z dependence on the 15
other side. This gives

a b dz
dP 10 H
P kT0

We now integrate this, with the limits on pres- 5
sure being P0, the surface pressure, and P, the
pressure at the altitude of interest, and the limits
on altitude being zero and z. That is, 1
0 0.2 0.4 0.6 0.8 1
Pressure (ATM)
a b dz¿
mg Z
P dP¿
P¿ kT0 0
Fig 23.11. Pressure vs. altitude in the Earth™s atmosphere.

In plotting quantities about the atmosphere, we usually plot
Integrating and substituting the limits gives altitude on the vertical axis, even though it is the independ-
ent variable.The scale height is indicated by H.
ln a b a bz
P0 kT0

where we have used the fact that ln(P/P0) ln(P) Example 23.1 Scale heights
ln(P0). If we raise e to the left-hand side, it Compute the scale height for the Earth™s atmos-
should equal e raised to the right-hand side. phere as well as that for an atmosphere of pure
Remembering that eln x x, this gives oxygen (O2) and pure hydrogen (H2).

1mg>kT0 2z
For a gas of molecular mass Amp (where mp is the
The quantity kT0/mg has dimensions of length, proton mass), the scale height is

1300 K 2 11.38 erg>K 2
and is the distance over which the pressure falls 16
to 1/e of its original value. We call this quantity
11.67 g 2 1980 cm>s2A
H 24
the scale height, H, where
107 cm>A
a b
H (23.12)
102 km>A
mg 2.53

For A 29, H 8.7 km. For oxygen, A 32, so
In terms of the scale height, the pressure distri-
H 7.9 km; for hydrogen, A 2, so H 125 km.
bution becomes
The drop off shown in Fig. 23.11 explains sev-
P P0 e (23.13)
eral things. If we go up to an altitude of 2 km, the
The variation of pressure with altitude is shown atmospheric pressure is already down to 80% of
in Fig. 23.11. what it is at sea level. This explains why it is

difficult to breathe, even at these “modest” moun- enters the atmosphere. The ultimate source of
tain altitudes. As an example, Denver is at an alti- energy for the atmosphere is the Sun. However, it
tude of 1.6 km, meaning that its pressure is only is not the direct source of heat over most of the
83% that at sea level. It is also interesting to com- atmosphere.
pare pressure changes with altitude with those Most of the Sun™s visible energy passes through
associated with weather changes on the Earth. directly to the ground, and is not absorbed by the
Severe storms are usually associated with regions atmosphere. This is true even if there are clouds.
of low pressure. However, even the most severe The clouds tend to scatter (rather than absorb) the
storms only have a drop in pressure to about 90% of visible radiation. This scattered light can either be
normal pressure. More typically, the weather directed back into space, and have its energy lost,
changes at sea level produce pressure changes of or it can bounce around in the clouds, and even-
only a few percent. So these changes are much tually reach the ground. This explains why it is
smaller than the changes that one encounters by still light on a cloudy day, but not as light as on a
climbing mountains. This is why we often use pres- clear day. The visible solar radiation reaches the
sure meters as altimeters, that is, devices that tell ground, and some is reflected back (mostly from
us how high above sea level we are. Another impor- the oceans), and the rest is absorbed.
tant constituent in the Earth™s atmosphere is water The heated ground then gives off radiation
vapor. Equation (23.13) is not a good description of characteristic of its temperature, so the radiation
its distribution. This is because the atmospheric from the ground is mostly in the infrared.
temperature is close to that at which water con- Infrared radiation from the ground is then
denses. The water vapor may have a normal distri- trapped in the lower atmosphere. Thus, the
bution at low altitudes. However, it may be almost ground is the immediate source of energy for the
totally absent at the cooler, higher altitudes. lower atmosphere, explaining why the tempera-
ture in the lower atmosphere decreases as one
23.3.2 Temperature distribution moves farther from the ground. As the ground
The temperature distribution with altitude is heats the air just above it, that air expands and
shown in Fig. 23.12. Notice that it is more com- rises. This convection is another means of energy
plicated than the pressure distribution. The com- transport from the ground to the lower atmos-
plexity in the temperature distribution reflects phere. (The above description is very simplified.
the variety of mechanisms by which energy In certain situations, “temperature inversions”
are present in which warmer air is on top of
cooler air, and there is little convection. With lit-
tle convection, pollution can build up.)
Altitude (km)

The Sun™s radiation that is absorbed goes into
heating the ground. This process is shown in
Fig. 23.13. If there were no atmosphere, the
Mesosphere ground would heat up to the equilibrium tem-
50 perature that we discussed above. However, the
calculation of that equilibrium temperature was
done on the assumption that all of the energy
radiated by the Earth escaped into space. Since
the Earth is at a temperature in the range
200 300 400
250“300 K, it is much cooler than the Sun, and its
blackbody spectrum peaks at longer wavelengths.
Fig 23.12. Temperature vs. altitude in the Earth™s So, while the Sun gives off most of its energy in
atmosphere. Layers are divided according to important the visible part of the spectrum, the Earth gives
energy balance mechanisms, so temperature behavior differs off most of its energy in the infrared part. Many of
from layer to layer.
the molecules in the lower atmosphere, especially

Fig 23.13. (a) Diagram showing the greenhouse effect.
(b) Global (meaning averaged over all measuring stations)
tropospheric deviations from monthly average temperatures,
Infrared is from 1979 to 2000. [(b) NASA]
Visible light
absorbed and
passes through
heats air
and heats ground




Degrees C





79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 00
Calendar Year

the water vapor, carbon monoxide and carbon were no glass. (In real greenhouses, the blocking
dioxide, are very efficient at absorbing that of the wind is also important.) This effect also
infrared radiation. Therefore, instead of being works in your house. Sunlight can pass through
radiated into space, some of the energy given off the windows, heating the interior, which pro-
by the Earth is trapped in the lower atmosphere. duces infrared radiation, which is trapped by the
This results in the surface of the Earth being hot- windows. This explains how you can have useful
ter than if there were no atmosphere. solar heating, even in the winter.
This effect, in which the visible light from the On the Earth, the greenhouse effect is modest.
Sun heats the ground, and the infrared radiation It raises the temperature by about 25 K. We will see
from the ground heats the air above the ground, in the next chapter that the presence of large
is called the greenhouse effect. The name results amounts of carbon dioxide on Venus has produced
because this effect is similar to the way that an extreme greenhouse effect on that planet. This
greenhouses work. In a greenhouse, the window leads us to worry that a similar thing could hap-
replaces the Earth™s atmosphere. The window lets pen on Earth, if we build up the concentrations of
the visible radiation through, heating the gases that trap the infrared radiation near the
ground. The ground then gives off infrared radia- ground. That is why atmospheric scientists are
tion, which is trapped by the windows, and the concerned over the by-products of human activity,
air inside the greenhouse is hotter than if there from fires, to automobile and factory exhausts,

Each time a photon is absorbed and re-emitted,
its direction is changed in a random way. We say
the photon does a random walk. If the walk is not
T random, all of the steps are in the same direction.
If there are N steps of length L, the distance from
the original point is NL. However, in a random
walk, a lot of time is spent backtracking. It can be
Te shown (see Problem 23.10) that for a random walk
Earth in one dimension (steps back and forth along a
line), the distance from the origin after N steps of
Fig 23.14. Radiative transport in the troposphere.We are length L is N1/2L, and for a three-dimensional walk
considering material a height z above the ground, and the
the distance is (N/3)1/2L. (In each case, the total dis-
top of the troposphere is at a height zt.
tance traveled by the photons is NL. In the ran-
dom walk, some of this is lost in the doubling
producing gases that will enhance the greenhouse
effect. If there is an increase in these gases, then it
Using this, the number of steps required for a
is possible that the Earth™s temperature will go
photon to do a random walk of length zt z is
into steady increase. This phenomenon is called
N 3[(zt
global warming. The problem with measuring the
effects of global warming is that the increase in
where L is the photon mean free path. The length of
any one year is small, and fluctuations due to vari-
each step is L, so the time for this number of steps is
ations in weather and climate cycles are much
larger. However, atmospheric scientists are search- t NL>c

13>Lc 2 1zt z22
ing for steady trends (Fig. 23.13b).
The lower part of the atmosphere, heated by


. 22
( 28)