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-

SOLUTION

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

2

mvorb GMm

six manned lunar landings. Each carried out

r2

r

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

vorb

r

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

10

16.4

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

GMm

1

the planets are quite close. They are close enough

0

esc

that the distances can be traveled by interplane- r

2

440 PART VI THE SOLAR SYSTEM

Solving for vesc gives v

2GM 1>2

a b θ

vesc

r

Note that comparing vesc and vorb gives

12 vorb

vesc

Therefore, the escape speed for the Earth is 11 km/s.

r

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

(a)

we studied binary stars, we found that the veloc-

ity of a particle in an elliptical orbit of semi-

3

major axis a, when the object is a distance r, from

a mass M at the focus, is

2

GM c d

2 1

v2 (22.2)

r a 1

We can use this to find the total energy of the

a

orbit,

a b mv2

GMm

1

E

r

2

a b GMm c d

GMm

1 2 1

(22.3)

r a r

2

(b)

GMm

Fig 22.14. Angular momentum and orbital eccentricity.

2a

(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

22 OVERVIEW OF THE SOLAR SYSTEM 441

of the distance of the Earth from the Sun and the

Earth at

distance of Venus from the Sun. Therefore, the

Launch

semi-major axis is

11 2 2 1RE RV 2

a

us's Orb

Ven 0.86 AU

it

The eccentricity is defined by the separation

Venus at

between the foci, divided by 2a. The Sun is at one

Launch

Sun

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

Probe

therefore

it

Reaches

rb

Venus

O

's d 2a 2RV

e

ro b

P

Earth when

RE RV 2RV

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

Venus.

RE RV

e

2a

should be noted that 90 doesn™t give a circu- RE RV

lar orbit unless the speed is the right speed for a

RE RV

circular orbit at that radius. However, 90

0.16

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

v2

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

trip.

must launch when Venus is at the place in its

orbit to take it to the rendezvous place in the

SOLUTION

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

442 PART VI THE SOLAR SYSTEM

is much less than the mass of the train, and

u

v that the collision is completely elastic. The train

is moving with speed u and the ball is moving

(a)

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

v+u

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

(c)

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.

(d)

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,

Spacecraft

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

speed.

Fig. 22.16. We assume that the mass of the ball

22 OVERVIEW OF THE SOLAR SYSTEM 443

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

Launch

shown in Fig. 22.18.

Earth at

Saturn at

Launch

Launch

(20 Aug 1977)

(9 July 1979)

(25 Aug 1981)

(24 Jan 1986)

Uranus at

Launch

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.

444 PART VI THE SOLAR SYSTEM

Questions

*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-

model?

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

system.

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?

Problems

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 OVERVIEW OF THE SOLAR SYSTEM 445

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

448 PART VI THE SOLAR SYSTEM

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

O

Mantle are compounds in which silicon and oxygen are

ce

an

added to the other atoms in various combina-

s

tions.) Gabbros have coarser crystals than do

Core

basalts. Larger crystals result from slower cooling.

An important clue to the Earth™s interior

C

on

comes from volcanic activity. This activity results

tin

en

from the heating of the crust, producing the

t

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

23 THE EARTH AND THE MOON 449

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]

(a)

1/2,

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)

2

Comparing these two expression (see Problem

23.2), tells us that

(0.693)

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

t>

N0 e (23.1a)

e

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

450 PART VI THE SOLAR SYSTEM

1 doesn™t stay bound to the rocks. We can collect

the argon and study the relative abundances of

0.9

various isotopes.

0.8 In considering the history of the Earth, we

must remember that the atmosphere has eroded

Fraction Remaining

0.7

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

0.5

dated at 3.7 billion years. (The oldest fossil cells

0.4

are dated at 3.4 billion years.) We think that the

surface underwent significant alteration 2.2 to

0.3

2.8 billion years ago. This was a period of volcanic

0.2

mountain building with the Earth having a very

thin crust. The crust was probably broken into

0.1

smaller platelets.

0

The age of the Earth, even as an approxima-

1 2 3 4 5

0

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

204

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

23 THE EARTH AND THE MOON 451

0°

180° 90° 90° 180°

45° 45°

0°

0°

45° 45°

90° 90°

0°

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.

452 PART VI THE SOLAR SYSTEM

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]

(a)

(b)

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

(c)

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

23 THE EARTH AND THE MOON 453

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

q

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)

d2

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)

P

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

(23.7)

planet is d2

0 0

L 11 a 2 R2 This equation cannot be solved directly for TP

P

Pabs (23.2)

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

4

surface area is e TP , where e is the emissivity. The

planet.

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

2

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

L

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

454 PART VI THE SOLAR SYSTEM

results shown in Fig. 23.9(a). Here we present a

800

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

600

decreases with increasing distance.

500

23.3 The atmosphere

400

The Earth™s atmosphere provides us with a multi-

300

tude of phenomena whose complexity may make

them seem beyond understanding. However, we

200

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

0

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.

(a)

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

atmosphere.

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

(b)

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

atmosphere.

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

23 THE EARTH AND THE MOON 455

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-

(a)

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

Ionosphere

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

Stratosphere

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

Earth

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

dP

mostly confined, is called the troposphere. It is g (23.10)

dz

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

456 PART VI THE SOLAR SYSTEM

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.

30

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

mg

dP

Altitude (km)

kT0

dz 20

2H

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

mg

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

0

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¿

(23.11)

P¿ kT0 0

Fig 23.11. Pressure vs. altitude in the Earth™s atmosphere.

P0

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

mg

P

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

P

e SOLUTION

P0

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

10

to 1/e of its original value. We call this quantity

11.67 g 2 1980 cm>s2A

H 24

10

the scale height, H, where

107 cm>A

2.53

a b

kT0

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-

z>H

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

23 THE EARTH AND THE MOON 457

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

Thermosphere

100

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

Stratosphere

done on the assumption that all of the energy

radiated by the Earth escaped into space. Since

Troposphere

the Earth is at a temperature in the range

200 300 400

250“300 K, it is much cooler than the Sun, and its

T(K)

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

458 PART VI THE SOLAR SYSTEM

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

Troposphere

Earth

(a)

0.8

0.6

0.4

Degrees C

0.2

0

’0.2

’0.4

’0.6

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

Calendar Year

(b)

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,

23 THE EARTH AND THE MOON 459

Each time a photon is absorbed and re-emitted,

its direction is changed in a random way. We say

zt

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

z

walk, a lot of time is spent backtracking. It can be

Troposphere

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

back.)

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

z)/L]2

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