Download Tidal Locking and the Gravitational Interaction

Tidal Locking and the Gravitational Interaction
Of the Earth and Moon
Michael Koohafkan
The Evergreen State College
5/27/06
Abstract
The gravitational pull that the Earth exerts on the Moon is what keeps it locked in
orbit around us. The Moon exerts a gravitational pull on the Earth as well, the most
noticeable effect being the phenomenon of tides in the Earth’s oceans. However, there are
many other effects of the gravitational interactions between the Earth and Moon not
immediately obvious to us. Both the Moon and the Earth are distorted into egg-like
shapes, and the tidal friction that occurs as a result of these distortions causes changes in
the orbital speeds, rotational speeds, and even axial tilts of the Moon and Earth.
The purpose of my research is to examine these various effects, and particularly how they
contribute to the deceleration of the Earth’s rotational speed. My goal is to determine the
amount of time it would take for the Earth to decelerate in rotational speed to the point
where rotational period of the Earth matches the orbital period of the Moon—a
phenomenon known as ‘Tidal Locking’—and mathematically express the relationship
between the gravitational interaction of the Earth and Moon and the rotational
deceleration of the Earth.
You aim to find out how long the day would eventually be, and how long it will take for
the day to reach its maximum length.
Background Information
The Earth and Moon are bound together by their gravitational interactions. Since
the gravitational force between two objects is inversely proportional to the distance
between them, it follows that different areas of the Earth experience different magnitudes
of the gravitational interaction between the Earth and Moon, since different points on the
Earth’s surface vary in distance from the Moon. These differences in gravitational pull
result in tidal forces, and are responsible for both land and ocean tides on Earth. The
Moon’s gravitational pull distorts the Earth into an egg-like shape and causes a tidal
bulge, and a torque is exerted on the Earth. This torque slows the Earth’s rotation.
Gravitational force felt at different points on Earth are not
equal in magnitude, nor are they parallel in direction.
Earth distorted as a result of the Moon's gravitational pull.
The Different magnitudes of gravitational attraction felt by
different points of the Earth cause different forces at each
point (as described by Newton's Second Law). The Earth is
a rigid body, and hence distorts into an egg-like shape
instead of being pulled apart. Since these bulges are caused
by the Moon's gravitational pull, they will stay more or less
aligned with the Moon (as illustrated by the line connecting
the centers of the Earth and Moon). This situation shows
why any given area experiences two high tides and two low
tides each day. The bulges are areas at high tide, and the
adjacent regions are areas of low tide.
Since the Earth rotates on its axis far more quickly than the Moon orbits around
the Earth, there is a dragging effect as the Moon’s gravitational pull tries to keep the
Earth’s tidal bulges more or less aligned to an imaginary line drawn through the centers
of the Earth and Moon. This drag manifests itself as tidal friction. Mechanical energy is
converted to heat within the Earth and throughout the oceans, and the resulting reaction
force felt by the Moon—the force of equal and opposite magnitude as predicted by
Newton’s Third Law—manifests itself as an acceleration of the Moon’s velocity
tangential to its orbital path (as shown in S J Peale’s diagram) which causes a widening in
the Moon’s orbit.
Recall that only a small part of the force is tangential.
Image taken from ‘Tides’ by S J Peale
S J Peale’s diagram describing the
gravitational interaction between
the Earth and Moon. The Moon’s
gravitational pull distorts the Earth
into an egg-like shape, with two
distinctive bulges. The difference
in the Moon’s gravitational force
experienced by these bulges results
in a torque. As the Earth spins, the
moon essentially pulls these bulges
back towards it, which causes the
Earth’s spin to slow. The reaction
forces felt by the Moon cause it to
gain a tangential acceleration,
which widens its orbit around the
Earth.
Unresolved: How can tangential acceleration be consistent with an increasing orbit
radius?
Tidal Locking
Tidal locking occurs when a celestial object synchronizes either its rotational
period or its orbital period with a celestial object it is gravitationally bound to. A good
example of this phenomenon is our own Moon; the fact that we only ever see one face
from Earth is because the Earth’s gravitational pull has caused the Moon to synchronize
its orbital period with its rotational period. Because of this, the Moon no longer feels a
torque as it orbits the Earth; the tidal bulges of the Moon are always aligned with the
Earth.
If the Moon is responsible for the deceleration of the Earth’s rotation, it follows
that given enough time the Earth’s rotation will slow to the point that it synchronizes with
the Moon’s orbital period. This is rather intuitive; The Earth’s rotation will continue to
slow until there is no longer a net torque acting on it. If the Earth rotates with the same
period as the Moon’s orbit, then the Earth will feel no torque due to the Moon; thus the
deceleration will cease to occur.
Did you generate this curve? If so, explain how. If not, ref?
Curve representing the lengthening of the Earth’s day
over time shown left, with the starting value
representing the current length of day. Length of day
increases as the angular velocity of the Earth decreases
over time. As the angular velocity of the Earth
decreases, the torque felt by the Earth also decreases,
thus lowering the rate of deceleration. Furthermore, the
decrease in the Moon’s gravitational effect on the Earth
due to the increase in orbital radius (from the tangential
acceleration caused by reaction forces) also contributes
to a decrease in the rate of deceleration. The horizontal
asymptote is expected due to the fact that the Earth’s
rotation is not expected to slow beyond the orbital
period of the Moon.
Same comment on each curve.
Curve representing the change in Earth’s rotational
velocity over time shown right. This curve represents
the same information as the Length of Day vs Time
curve, since the length of day is merely our
perception of the speed of the Earth’s spin (the length
of the day is inversely proportional to the rotational
velocity). The horizontal asymptote is expected since
a constant rotational velocity is expected to be
maintained once the torque caused by the Moon is no
longer
signifi
cant.
Curve representing the expected change in rate of rotational
deceleration of the Earth due to the gravitational effects of
the Moon shown left. Rate of deceleration will decrease as
the torque felt by the Earth decreases over time and as the
Moon’s gravitational effect on the Earth decreases (result of the increase in orbital radius due to the
tangential acceleration caused by reaction forces). Rotational deceleration is expected to approach zero due
to the fact that the Earth’s rotation is not expected to slow beyond the orbital period of the Moon.
Note that while this situation is similar to the one of our Moon, it is NOT the
same; our Moon has synchronized its orbital and rotational periods due to the
gravitational effects of the body it orbits—the Earth. The deceleration of the Earth’s
rotation is caused by the gravitational effects of its satellite, and thus the orbital period of
the Earth around the Sun does not come into play. A situation analogous to the Moon’s—
and one that I have not analyzed—would be the Earth becoming tidally locked with the
Sun. If the Earth were to become tidally locked with the Sun, then the Earth’s rotational
period would synchronize with its orbital period, and the tidal bulges caused by the Sun’s
gravitational pull would be aligned with the Sun (and thus no torque would be felt). It is
important to note, however, that the tidal bulges on the Earth caused by the sun are not
nearly as significant as those caused by the Moon, although they are measurable.
Technically anything gravitationally bound to another object can become tidally
locked. Tidal locking can occur in a variety of different situations: Planets can become
tidally locked with their satellites and vice versa, planets can become locked with the star
they orbit, and even two stars in a binary system can become tidally locked! Basically,
anything massive enough and close enough to another body to cause tidal forces and
exert a torque can force that body to become tidally locked—it’s just a matter of time.
Of course, the time that it hypothetically would take for a body to become tidally
locked with another can be a lot longer than those bodies are expected to be around. The
Earth will probably not be around long enough to become tidally locked with the Moon;
it might even take longer than our entire galaxy is expected to be around. In light of this
fact, trying to build a model to predict when tidal locking will occur might seem useless.
However, the trends are more important than the end results; analyzing the relationship
between the Earth and the Moon might have unexpected benefits, and may help answer
questions we have about the geologic history of the Earth (as discussed in the next
section).
Uses of Modeling Gravitational Interactions of Celestial Bodies Over Time
Developing mathematical models of gravitational interactions and tidal locking
may be quite valuable. A working model that can be used to extrapolate both backwards
and forwards in time and examine the state of the Earth-Moon system can help us
determine things such as relative age and how properties of celestial bodies in
gravitationally bound systems—such as rotational speeds, orbital speeds, and tidal
interactions—have changed over time. Many studies on tidal locking have been used in
efforts to explain the origin and age of the Moon, and these same studies may be able to
help us determine the ages of other planets in relation to their satellites. Furthermore, by
combining these studies with analyses of the geologic effects of the Moon’s gravitational
pull, we can begin to get a better picture of how the Moon has affected and changed the
Earth throughout time, as well as predict changes yet to occur.
It is important to note, however, that such studies are quite limited. Such studies
become impossible—at least with current technology—when we try to consider anything
larger than a two-body system. For this reason, the gravitational effects of the Sun are
ignored in such studies, and these models cannot be used when examining planets with
multiple satellites that gravitationally affect a planet to a significant degree.
Pertinent Aspects of the Earth-Moon System
As in any scientific model, certain aspects of a gravitationally bound system must
be taken into account to maintain accuracy while studying the phenomenon of tidal
locking, while others can be safely ignored. The mass of the objects within the system is
of course vital to determine the magnitude of the gravitational force each object exerts.
By the same token, the distance between objects is also necessary. Rotational speeds and
the rates of change are what allow us to model such changes over time. Orbital speeds of
objects can either be ignored or taken into account depending on the situation; in
examining the tidal locking of Earth with the Moon, the orbital period of Earth can be
ignored (since the Moon does not affect the Earth’s orbit around the Sun) while the
orbital period of the Moon must be taken into account (since that period is what the Earth
is expected to synchronize its rotation with).
There are other second-order factors that may be taken into account. Factors such
as orbital eccentricity, axial tilt, and tilts in orbital planes can be taken into account to
gain a greater degree of accuracy. However, adding these factors can make a model
impossibly complicated. A balance between simplicity and accuracy is probably best, and
new models can always be designed to analyze specific factors as needed.
One factor in particular is ignored in the study of the Earth-Moon system,
although its effects are from negligible; the gravitational effects of the Sun. The reason
for this is that trying to analyze the gravitational interactions of a system containing more
than two bodies become near impossible. Because of this, the Sun’s effects are always
ignored, even though we lose a great deal of accuracy by doing so.
In this study, I have only taken into account the masses and rotational periods of
the Earth and Moon, the orbital period of the Moon and the torque it exerts on the Earth,
and the distance between the two bodies.
A Linear Model of Tidal Locking
Surprisingly enough, it is possible to measure the current rate of Earth’s rotational
deceleration. By setting up reflectors on the Moon’s surface, scientists nowadays use
lasers to measure the distance between the Earth and Moon simply by measuring the time
it takes for the laser pulse to make a roundtrip journey between the Earth and Moon.
Using this same method, scientists have determined that the Earth’s day varies by a few
thousandths of a second each year. Although slightly different values have been given in
different texts, I have decided to use S J Peale’s rate of 0.0016 seconds/century in this
calculation.
Another discovery made using these methods is integral to our calculations.
According to J.E. Faller and J.O. Dickey, the distance between the Earth and the Moon is
increasing by 3.7 centimeters each century. This increase is caused by the tangential
acceleration of the Moon as a result of reaction forces as mentioned earlier.
Of course, this number only applies at this time. As the Moon drifts away and the
gravitational pull it exerts on the Earth decreases, the reaction force will also decrease.
This will, in turn, will decrease the rate at which the Moon drifts away from the Earth. At
this time, the change in the Moon’s rate of drift will not be considered, and the drift rate
will be held constant at 3.7 cm a year.
Manipulating the linear model to reflect a changing rate of rotational deceleration
over time is a relatively simple exercise. Using S J Peale’s rate of rotational deceleration
(0.0016 seconds per century) and the current length of day, we first construct a simple
linear equation D(t), the length of the Earth’s day as a function of time:
•
•
•
Let D0 be the current length of the Earth day, 86,400 seconds
Let t represent time, in years.
Let K0 be the current constant rate of change of Earth’s day. The value 0.0016
seconds per century will be converted to reflect the rate of change in a year, so the
adjusted value is 1.6x10-5 seconds per year. Let K represent the rate of change at
time t.
Thus, the deceleration of the Earth’s rotation can be modeled by
D(t) = D0 + Kt
In this model, we are assuming that the deceleration of the Earth’s rotation is due
to a torque. We know that the torque is related to the gravitational force, and we know
that the gravitational force is inversely related to the square of the distance. Thus, K is
inversely related to the square of the distance, or r2. From this, we can postulate that K
must be equivalent to some factor X over r2.
The change in distance r can also be modeled by a simple linear model. J.E. Faller
and J.O. Dickey claim that laser measurements of the distance between the Earth and
Moon show that the Moon is drifting away from Earth at a rate of 3.7 centimeters each
year. I have chosen to maintain that this orbital drift is constant—at least for now. Thus,
the change in distance between the Earth and Moon can be modeled with a linear
equation r(t), the distance between the Earth and Moon as a function of time:
• Let r0 be the current distance between the Earth and Moon, 384x106 meters, and r
be the distance between the Earth and Moon at time t.
• Let C be the rate of increase in the distance r, 0.004 m/year.
• Let t be time, in years.
Thus, the change in the Earth-Moon distance can be modeled by
r(t) = r0 +Ct
Combining these equations together, we can create a model that expresses a changing rate
over time. At t = 0, we know that K = K0 and r = r0. Thus
K = K0 = x / r0
By manipulating this relationship, we get
x = K0(r0)2
Thus, at time t,
K = K0(r0)2 / [r(t)]2
Or
K = K0(r0)2 / [r0 + Ct]2
Plugging this formula for K into the original equation, we get
D(t) = D0 + (K0(r0)2 / [r0 + Ct]2)t
Analyzing the New Model
The graph above shows the change in the Earth’s length of day over time as
predicted by the modified linear equation. Unfortunately, we can see by the graph that
this method is invalid. When extrapolating forward through time, the values simply do
not coincide with predictions. At first, one might think that this simply implies that the
prediction is wrong; however, examining the graph shows that the trend does not, in fact,
make any logical sense. The trend displayed by this model shows a sharp increase in the
length of day, and then a severe drop. This does not make any sense; since we are looking
at the Earth-Moon system as a closed two body system without any outside interference,
the only changes in the Earth’s rotational speed should be those caused by the Moon—
and the Moon will never cause a decrease in the length of day. The only two possibilities
in this case are for the Earth’s rotation to slow to the point where it synchronizes with the
Moon’s orbital period, or for the deceleration of the Earth’s rotation to cease as the Moon
drifts away from the Earth. We see neither of these eventualities in the model, and thus
we must look for another way to model this system.
Analyzing Angular Momentum and the Earth-Moon System
Another to way to look at this problem is to consider the angular momentum of
the Earth-Moon system. Since we are considering the Earth-Moon system as a closed
system, the angular momentum of the system will be conserved—it will remain constant.
With this in mind, we can assume a direct tradeoff between the change in the Earth’s spin
angular momentum (as a result of the rotational deceleration) and the changes in the
Moon’s Orbital Angular Momentum (as a result of the reaction forces) and Spin Angular
Momentum.
Let’s stop to consider why we would assume changes in the spin angular
momentum of the Moon. Take a moment to think about why tidal locking occurs. When a
body is tidally locked to another, it is in a low-energy situation. The process of tidal
locking occurs because that is the most balanced and static situation a system can be in.
Because of this, ‘unlocking’ a body requires a huge amount of work, and a body’s orbital
angular momentum can be expected to compensate for a change in spin angular
momentum (and vice versa) to maintain the tidal lock.
Thus, if the Moon’s orbital period became shorter as a result of the reaction forces
caused by the Earth’s gravitational pull, we would expect the Moon’s rotational period to
shorten as well to maintain its tidal lock with the Earth. Is this what has happened,
historically? Or hasn’t the rotational period gradually lengthened to match the steadier
orbital period?
The diagram to the left shows the
Earth-Moon system in terms of angular
momentum. (Note that r is from center to
center.) The total angular momentum of
the Earth-Moon system is the sum of all
angular momenta of the Earth and
Moon—the sum of the Earth’s spin
angular momentum (LSE) and orbital
angular momentum (LOE) and the Moon’s
spin angular momentum (LSM) and orbital
angular momentum (LOM):
Ltot = LSE + LOE + LSM + LOM = Constant
Since the Earth’s orbital angular
momentum is not affected when
examining the Earth and Moon as a closed system, it will remain constant. Thus Ltot –
LOE = constant = LSE + LSM + LOM. (Ref: Zita, personal communication, May 2006)
Modeling the Changing Angular Momenta of the Earth and Moon
Spin angular momentum is defined as the product of a body’s Moment of Inertia
and its spin angular velocity, or LS = IωS (where ωS = 2π/T, T being the period of the
rotating body) (cf Giancoli Ch.__ for basic definitions of angular properties). Thus, the
Earth’s spin angular momentum at time t is modeled by the equation LSE = IEωSE(t), or
IE2π/TSE, where TSE(t) is the change in Earth’s rotational period as a function of time. The
only aspect of Earth’s spin angular momentum that changes significantly is the rotational
period (and thus the angular velocity); the Earth’s moment of inertia, the product of its
mass and the square of its radius remains constant throughout all calculations in this
model. (While events such as the undersea earthquake that led to the catastrophic
Indonesian tsunami of Dec.2004 have been observed to suddenly change Earth’s moment
of inertia and therefore its spin, such effects are generally small? It would be interesting
to compare the size of that change to those calculated below…) The current value of
Earth’s spin angular momentum, LSE0, can be easily calculated, as shown below in the
appendix.
The Moon’s orbital angular momentum, LOM, is represented by a different
equation. Orbital angular momentum, LO, is the product of a body’s mass, velocity, and
distance between it and its axis of rotation. For the Moon, this would be the product of
the Moon’s mass (constant), its velocity (changing increasing? decreasing?) and the
distance between it and the Earth (also changing increasing? decreasing?, as modeled by
our previous equation #?): LOM = MMVOM(t)r(t), where VOM(t) and r(t) are both functions
of time. The current value of the Moon’s orbital momentum, LOM0, is also easily
calculable, as shown in the appendix.
The Moon’s spin angular momentum, LSM, can be modeled in the same way as
the Earth’s. Once again, the moment of inertia will not? change, so the only variable
factor in the Moon’s spin angular momentum is its angular velocity. Thus, LSM =
IMωSM(t) and LSM0 = IMωSM0. However, we have another way of expressing ωSM, a way
that will allow us to match it with LOM. Compare for a moment the rotational angular
velocity (ωS) and orbital angular velocity (ωO) of the Moon. Since the Moon is tidally
locked with the Earth, ωSM = ωOM. And ωOM = VOM/r, the tangential velocity of the
Moon divided by the radius of the Moon’s orbit. Since we know how VOM and r are
changing over time, LSM = IMVOM(t)/r(t) increases or decreases?. The current value of
LSM, LSM0, is also easily calculated, as shown in the appendix.
We have seen that the angular momentum of the system can be expressed by Ltot
= LOE + LSE + LOM + LSM, and Ltot is a constant. Furthermore, since LOE will remain
constant, Ltot – LOE will be a constant and equal to LSE + LOM + LSM.
Since Ltot is a constant over time, we know that Ltot now = Ltot later. Thus, (LSE +
LSM + LOM) now = (LSE + LSM + LOM) later. Using basic algebra, we can show that LSE =
Ltot – (LSM + LOM) later. Substituting the previous equations for the angular momenta of
the Earth and Moon, we get the following equation:
IωS = Ltot – [IMVOM(t)/r(t) + MMVOM(t)r(t)]
Unfortunately, an expression for the change in the Moon’s orbital velocity as a
function of time has yet to be determined; this is the last missing piece, and without it we
cannot determine if this method of modeling the Earth-Moon system is valid.
You have two expressions for the Moon’s changing orbital speed – one from the
tangential force and once from the increasing distance. Are these consistent? If not,
which do you believe, and why?
Final Points
Angular Momentum is most likely the simplest way to describe the Earth-Moon
system, and also gives a great deal of freedom. Since this method allows us to substitute
different equations in for each individual aspect of a bodies angular momentum,
refinements and changes are easy to implement within the model. For example, the
current model of the Moon’s orbital drift is represented as a linear function; however, this
is not the case, since the rate of orbital drift will depend on the change in gravitational
force between the Earth and Moon. However, once this change is modeled accurately, we
can simply substitute that new equation in for r(t) without changing other aspects of the
model. Furthermore, it should be a relatively simple affair to add in factors such as
changes in axial tilt and orbital plane tilts—although these changes may not be easy to
model themselves. On the flipside, we could simplify the model even more and assume
linear changes in the various aspects of angular momentum, and the general form of this
model will remain the same; we can make the model as simple or as accurate as we want
using this method.
The final piece of the equation, VOM(t), may be harder to calculate than previously
thought. In fact, determining changes in the reaction force may prove impossible to
model. However, I am confident that some proportionality exists between the general
gravitational force and the tidal force that the Moon exerts on the Moon. With this in
mind, it may be possible to construct an expression for VOM without dealing directly with
the changing tidal forces.
Appendix: List of Equations and Values




Linear model: D(t) = r0 + K0t
Moon’s orbital drift model: r = r0 + Ct
Modified linear model: D(t) = (t) = D0 + (K0(r0)2 / [r0 + Ct]2)t
Angular momentum models:
o Ltot = LSE + LOE + LSM + LOM
o Constant = LSE + LOM + LSM
o LSE = Ltot – [LSM(t) + LOM(t)]
 LSE = IEωSE(t) = IE2π/TSE(t)
 LSM = IMVOM(t)/r(t)
 LOM = MMVOM(t)r(t)
Values of constants, variables, etc.:



ME = 5.97x1024 kg
(Radius of Earth) RE = 6.38x106 m
D0 = TSE0 = 24 hrs = 86400 s








MM = 7.35x1022 kg
RM = 1.74x106 m
TOM = 28.5 days = 2.4624x106 s
ωSM = ωOM = 2π/TOM = 2.552x10-6 s-1
r0 = 384x106 m
C = .0037 m/yr
K0 = 1.6x10-5 s/yr
VOM0 = 979.8 m/s




LSE0 = IEωSE = (2/5)MERE2ωSE (known formula) = 7.069x1033 kg . m2 / s
LSM0 = IMωSM = (2/5)MMRM2ωSM = 2.271x1029 kg . m2 / s
LOM0 = MMVOM0r0 = 2.765x1034 kg . m2 / s
Ltot = LSE0 + LSM0 + LOM0 = 3.472x1034 kg . m2 / s
Bibliography
S J Peale, Tides. Encyclopedia of Astronomy & Astrophysics. © IOP Publishing Ltd 2006. ISBN
0333750888
Zita, “Will the Earth tidally lock with the Moon? If so, when?” personal communication, 16 May 2006
Giancoli, Physics for scientists and engineers, ed._, Prentice Hall 200_
Summary
This article describes the phenomenon of tides from a scientific standpoint. Peale describes how
the tides are caused by the gravitational interaction between the Earth and the Moon, and offers scientific
equations to mathematically describe the interacting forces and how the mass and distance between two
bodies are related to the gravitational pull they exert on each other. The author later describes the physical
warping of a planet or satellite’s shape that occurs due to this gravitational force—described as a ‘tidal
bulge’—and finally applies the phenomenon of tides to other planets and their satellites in our solar system.
Assessment
This article provided a large amount of background information to the phenomenon of tides and
tidal locking. The equations outlined in the article were invaluable in determining the gravitational
interaction between the Earth and Moon. The author also brought up an important point: the orbital
eccentricity of a mass and its axial tilt effect the movement of the tidal bulge and the change in axial
rotation caused by tidal forces.
D D McCarthy. Polar Motion and Earth Rotation. Reviews of Geophysics and Space Physics, vol 17, no 6,
Sep 1979. pg 1397-1403.
Summary
This article focuses on the changes in axial tilt that the Earth undergoes due to changes in the
Earth’s shape due to gravitational interactions with the Sun and Moon. The article covers what can cause
changes in the Earth’s orientation (such as tidal forces) and expounds on the differences between periodic
changes and irregular changes in the Earth’s acceleration and axial orientation. Methods of observing these
changes are also discussed.
Assessment
This article provides some background information on the changes in the Earth’s rotation and how
this applies to the conservation of angular momentum, but does not go into any great mathematical detail.
This reference was useful in providing basic information to be used in a presentation, but did not really
assist in actually calculating tidal effects and changes in the Earth’s rotation. The portion that describes the
methods used to calculate these changes in the Earth’s rotation, however, was invaluable both in describing
how this data is collected and in guiding my searches for the actual data.
Douglas C Giancoli. Physics for Scientists and Engineers 3rd e. © Prentice Hall 2000
Summary
This college textbook covers a wide range of topics and fundamental concepts in physics, and is
integral to a calculus-based physics course. This book includes sections on angular momentum and
gravitational force.
Assessment
This textbook was the most comprehensive and useful source of information on how to
mathematically model many aspects of tidal locking, including the known formulas for angular momentum.
All of the information provided was general, and did not apply specifically to tidal locking; thus the
information had to be put into the proper context for it to be used.
Roger A Freedman, William J Kaufmann III. Universe 7th e. © W.H. Freeman and Company 2005.
Summary
This textbook covers a wide variety of astronomy topics and concepts, and is integral to a noncalculus-based college astronomy course. The text provides a variety of information on individual bodies in
our solar system, including a large amount of data on the Earth and Moon.
Assessment
This text provided countless facts and useful data on the Earth and Moon, as well as how they
interact. The information in this book was vital to forming an accurate understanding of the Earth-Moon
system.
J E Faller, J O Dickey. Precisely Measuring the Distance to the Moon. Earth In Space, vol. 3, Dec 1990.
pg 6,7
Summary
This article describes the methods that scientists use to calculate distances between the Earth and
Moon, as well as detailing how the equipment used was set up. A few facts and values for changes in the
Earth-Moon system were also mentioned as the result of such research.
Special Thanks to Dr. E J Zita, for her unending support throughout this research presentation.