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\begin{center}
{\bf Notes on Fluid Dynamics}
\end{center}
These notes are meant for my PHY132 lecture class, but all are free to use them
and I hope they help. The ideas are presented roughly in the order in
which they are taught in my class, and are designed to supplement
the text. I will be writing these notes as I teach the class, so
they will be constantly updated and modified.
Our initial topic this quarter will be fluids. We will first consider
fluids at rest and then generalize to fluids in motion. We will not
introduce any new fundamental "Laws of Physics", but rather apply
Newton's laws of mechanics to fluids.
\bigskip
\centerline{\bf Fluids at Rest}
By fluids we mean liquids and gasses. If a fluid is "at rest", and neutral,
there are only two macroscopic quantities needed to describe it: density
($\rho$) and pressure ($P$).\\
\noindent {\bf Density}
Density is a concept you probably learned in high school. It is simply
the amount of mass per volume. Given the symbol $\rho$, it is defined as
\begin{equation}
\rho \equiv {{mass} \over {volume}} = {m \over V}
\end{equation}
\noindent With this definition, one needs a finite amount of volume and
mass. One can also define the {\bf density at a point} in the fluid by
taking the limit as the volume (and mass) approach zero:
\begin{equation}
\rho \equiv \lim_{V \rightarrow 0} {m \over V}
\end{equation}
\noindent Since it will be useful to consider the properties of fluids at
different points in the fluid, the second definition is more useful than
the first.\\
\noindent {\bf Notes}:\\
\noindent 1. Density is a scalar. It has no direction.\\
\noindent 2. The density can change from point to point. However, for
most of our applications the density will be treated as
constant over the volume under consideration. One obvious example where
density changes with position is the density of air with altitude.\\
\noindent 3. The units of density are mass/volume. The most common
units are the metric units of kg/M$^3$ or g/cm$^3$.\\
\bigskip
\noindent {\bf Pressure}
You were also probably introduced to the concept of pressure in high school,
and remember that it has something to do with the force that a fluid exerts
on a surface. Suppose a small surface of area $A$ is placed in a fluid. The
molecules of the liquid or gas will bounce off the surface and exert a force
on it (the surface). The direction of the force on the "small" area will be
perpendicular to the surface. Here is the nice property of fluids: if we
double the size of the "small" surface then the force on the bigger surface
is twice as large. That is, the force on the surface is proportional to
its area (assuming the area is small enough). The ratio of the force to
area is thus independent of the area, and this ratio we call the pressure $P$:
\begin{equation}
P \equiv {{force} \over {area}} = {F \over A}
\end{equation}
\noindent where $F$ is {\it the force on} and $A$ is {\it the area of}
a small surface {\bf if it is placed in the fluid}. Note that there
does not need to be a surface in the fluid to define the pressure.
{\bf If} a surface were placed in the fluid, {\bf then} the force on the
surface would be $P$ times $A$.
You might be wondering if pressure is a scalar or a vector. Since it has
something to do with force, you might think it is a vector. However, the
direction of the force on the surface is always perpendicular to the
surface. So the direction of the force depends on the orientation of the
surface, and not on the ratio of $F$ to $A$. We can define an area vector as
follows: it's magnitude is equal to the area of the surface, and it's direction
is perpendicular to the surface. Then, the direction of the force, $\vec{F}$, is the
same as the direction of the area vector $\vec{A}$. Thus, we can write
\begin{equation}
\vec{F} = P \vec{A}
\end{equation}
\noindent and we see that {\bf pressure is a scalar}. $\vec{F}$ and $\vec{A}$
have directions, but not $P$.
The above definition requires a finite
sized area. Since it will be useful to consider the properties of fluids
at different points in the fluid, it is best to define pressure using the
above definition as the area approaches zero:
\begin{equation}
P \equiv \lim_{A \rightarrow 0} {F \over A}
\end{equation}
\noindent where $F$ is the {\it magnitude} of the force on and $A$ is the
{\it magnitude} of the area of a small surface {\bf if} it is placed in the fluid.
You will encounter this same sort of abstraction when you study electricity and
magnetism. For example, the electric field at a point can be defined as the force per charge
{\bf if} a charge is placed at the point. One can define quantities in empty
space this way. The price per weight of a quantity is a similar concept. Let the
price of bananas be 50 cents per pound. This value, 50 cents per pound, can exist
even if no one ever buys bananas. However, {\bf if} someone wants to buy bananas,
the bananas are weighed and the price is determined. If a surface is in a fluid,
the force on the surface equals the area times the pressure if the pressure is known.
The {\bf units} of pressure are force/area. The most common units are
$N/m^2$ in the metric system. One $N/m^2$ is called a Pascal (Pa). In the British
system, pounds/in$^2$ (PSI) is used. Pressure is also measured in mm of mercury, which
will be discussed later. Atmospheric pressure near sea level is around
$1.013 \times 10^5$ Pa, which in British units is $14.6$ PSI.
\bigskip
\noindent {\bf Variation of Pressure with Position}\\
You probably have experienced how pressure changes with depth while
swimming. As you dive deeper in water, you feel the water pressure
increase. Here we want to figure out exactly how the pressure changes
within a fluid.
If the fluid is at rest in a container that is "floating" in space, for example
in the space shuttle, then the
pressure is the same everywhere. In an inertial frame, objects at rest are
just floating, and the pressure will not change with position. For the pressure
to change with position, one needs to
be in an accelerating reference frame, or on a planet with a gravitational
field. Here, we will consider how the pressure changes in a fluid on earth.
Consider a vessel on the earth that contains a fluid at rest. For example a
swimming pool filled with water. To determine how the pressure changes with
depth in the fluid one can analyze a cylindrical volume of fluid (within the pool).
Let the cylinder have an area $A$ on its top and bottom, and have a height $h$, as
shown in figure 1. Let the axis of the cylinder be aligned with the direction
of gravity. Since the fluid is at rest, the net force on this volume of fluid
must be zero. The forces around the side of the cylinder balance out, so
we must have:
\begin{center}
(Force on bottom) - (Force on top) = weight of fluid in the cylinder
\end{center}
\noindent Since the force on the top and bottom is just the pressure times
the area, one obtains:
\begin{equation}
P_{bottom}A - P_{top}A = mg
\end{equation}
\noindent where $m$ is the mass of the fluid in the cylinder. Since
$m=\rho V = \rho A h$, we have
\begin{equation}
P_{bottom} A - P_{top} A = \rho A h g
\end{equation}
\noindent Canceling the area, we have
\begin{equation}
P_{bottom} - P_{top} = \rho g h
\end{equation}
\noindent Thus, the pressure on the bottom is larger than the top by an amount
$\rho g h$. As one dives deeper in a fluid (in the direction of gravity) the
pressure increases by an amount $\rho g$ times the change in depth.
Since we can let the area $A$ go to zero, the equation applies to a {\it point} at the
top and bottom of the cylinder. We can
write the above equation in a nicer way as follows:\\
\noindent 1. Take the y-axis to point opposite to the direction of gravity.
That is, let $\vec{g} = -g \hat{j}$, where $\hat{j}$ is the unit vector in the
+y direction.\\
\noindent 2. Label the bottom point "1" and the top point "2".\\
\noindent With these conventions, $P_{bottom}$ can be called $P_1$ or
$P(y_1)$. The pressure at the top $P_{top}$ is then $P_2$ or $P(y_2)$.
The height difference $h$ is just $h=y_2 - y_1$. Plugging into the above
equation gives:
\begin{equation}
P_1 - P_2 = \rho g (y_2 - y_1)
\end{equation}
\noindent rearrainging the terms, we have:
\begin{equation}
P_1 + \rho g y_1 = P_2 + \rho g y_2
\end{equation}
\noindent which can also be written as
\begin{equation}
P(y_1) + \rho g y_1 = P(y_2) + \rho g y_2
\end{equation}
This is an interesting result. It states that if one moves around
in a fluid (which is at rest) the sum of the pressure plus $\rho g y$
remains constant. That is
\begin{equation}
P(y) + \rho g y = constant
\end{equation}
\noindent for a fluid at rest. If there is no gravity, then the
pressure is the same everywhere. The pressure only changes in the
direction of gravity! A useful distance to remember for water is
the change in height to cause a change in pressure of one atmosphere.
This can be found by solving the equation $\rho g h = 1.013 \times 10^5$Pa,
or $h = 1.013 \times 10^5 / (\rho g) = 1.013 \times 10^5 / 1000/9.8 \approx
10.3$m. For every $10.3$ meters (or around 33 ft) of depth in water
the pressure increases by one atmosphere.
We will do a number of examples in class, where in some cases we will
calculate the net force on a surface or object. If you hold a piece
of paper, that has an area of 1 in$^2$, in the air, the force on
one side is $14 \times 2 = 28$ pounds!. Why doesn't the paper move?
It is because there is also a force of $28$ pounds on the other side.
Only if there is a difference in pressure will there be a net force
on an object. This should be clear from the examples we do in lecture.
If the density does change with position, one can generalize the above formula
to:
\begin{equation}
P_1 - P_2 = \int_{y_1}^{y_2} \rho g dy
\end{equation}
\bigskip
\noindent {\bf Archemedes Principle}\\
The {\bf bouyant force} on an object in a fluid is the net upward force
the object feels due to the fluid. If history is correct, Archemedes
was the first one to figure out what the bouyant force on an object
is. Archemedes principle is a statement (without an equation) of the
bouyant force on an object {\it in a fluid that is at rest}:\\
{\bf The bouyant force on an object is equal to the weight of the
fluid that the object displaces.}\\
Although Archemedes Principle can be derived from Eqs. 8-12 above, I am
not sure if Archemedes did it that way. His reasoning might have been
as follows:
The bouyant force is the sum of the forces the fluid exerts
on all the surfaces of the object. Thus, the bouyant force can only depend
on the size (and perhaps shape) of the object. Suppose that the object in the fluid is replaced by the
fluid itself. Then the bouyant force on the "replaced" fluid must
equal the weight of the fluid, since it is at rest. If you put the
object back, the bouyant force is the same as on the "replaced" fluid, since
it only depends on the size of the object. Thus, the bouyant force must
equal the weight of the fluid that is displaced by the object.\\
If this is confusing, then one can derive Archemedes Principle using Eq. 8
above. Consider a cylinder placed in a fluid with its axis in the direction
of the gravitational field. Let the top and bottom area be $A$, and the height
$h$. Since the forces on the sides cancel out, the net force, or net bouyant
force, is equal to the upward force on the bottom minus the downward force
on the top:
\begin{equation}
F_{bouyant} = F_{bottom} - F_{top}
\end{equation}
\noindent Using $F = PA$ we have
\begin{eqnarray*}
F_{bouyant} & = & P_{bottom} A - P_{top} A\\
& = & A(P_{bottom} - P_{top})\\
& = & A (\rho_{fluid} g h)\\
& = & \rho_{fluid} g (Ah)\\
& = & \rho_{fluid} g V\\
& = & m_{fluid \: displaced} g
\end{eqnarray*}
\noindent The last line is recognized as the weight of the fluid displaced. Note that
the $\rho$ on the right side is the density of the {\bf fluid} and has nothing to do
with the object in the fluid. Thus, {\bf the bouyant force only depends on the
volume of the object} and {\bf the density of the fluid}. It does not depend
on the density or weight of the object itself. Although
the formula was derived using a cylinder, using vector calculus one can show that
this result is true for an object of any shape. We will do a number of examples
using Archemedes principle in lecture.
\bigskip
\centerline{\bf Fluids in Motion}
In this introductory class, we will limit our treatment to moving fluids
whose density doesn't change and ones that are at steady state. There
are two main relationships that we will derive and apply. The first one
is called the "continuity equation", and the second one Bernouli's equation.
By steady state, we mean that the pressure and velocity do not change in
time in the fluid, although they may change with position.
For fluids at rest, we only needed to consider two quantites,
density and pressure. If the fluid is flowing (or moving) we need one
more quantity, the velocity of the fluid. Last quarter we introduced
the concept of velocity as the velocity of a particle. What do we mean
by the velocity in a fluid. The velocity in a fluid is the velocity
of a "small" volume of the fluid. More specifically, it is the
velocity of a volume of the fluid as the volume approaches zero.
In general the velocity in a fluid can change from one point to another,
so we can speak of the velocity {\it at a point} in the fluid. At
every point in the fluid, we can ascribe a velocity vector, which
is the velocity of a small volume of the fluid at that point.
The velocity in a fluid is an example of a "vector field". If there is
a vector assigned to every point in space, this collection of vectors is
called a vector field. You will see vector fields in PHY133. The
electric field and magnetic field are both vector fields. They are vectors
that can be defined at every point in space. In this class, we will
not deal with the properties of vector fields, but next quarter we will.
In summary, there are {\bf three quantites} we will consider in a fluid:
density, pressure, and velocity. Each of these are defined at a point
in the fluid, and can vary from point to point.\\
\noindent {\bf Continuity Equation}\\
Consider a fluid that is flowing through a pipe. The pipe has a cross
sectional area that is not constant. Let the area on the left end
of the pipe be $A_1$ and the area on the right end be $A_2$. Let the
velocity of the fluid entering the pipe from the left be labeled
$v_1$ and the velocity of the fluid leaving the pipe from the right be
$v_2$. See figure 2.
The question we want to consider is how are $v_1$, $A_1$, $v_2$ and
$A_2$ related to each other. If we assume that the density of the
fluid is the same throughout the pipe, i.e. that the fluid is
incompressible, then there is a relationship amoung the variables:
\noindent In a time $\Delta t$ the {\it volume} of fluid entering from the left end is
$A_1 v_1 \Delta t$, and the {\it volume} of fluid leaving from the right end
is $A_2 v_2 \Delta t$. If the fluid is incompressible, then the volume of
fluid leaving in a time $\Delta t$ equals the volume entering in a time
$\Delta t$:\\
\centerline{Volume entering = Volume leaving}
\begin{equation}
A_1 v_1 \Delta t = A_2 v_2 \Delta t
\end{equation}
\noindent Canceling the $\Delta t$,
\begin{equation}
A_1 v_1 = A_2 v_2
\end{equation}
\noindent If $A_2$ is smaller than $A_1$, then $v_2$ must be larger
than $v_1$ so the amount of water coming out equals the amount
going in.
\bigskip
\noindent {\bf Bernouli's Equation}\\
Let's first state Bernouli's equation, then derive it from the laws of mechanics.
The equation deals with how the pressure and velocity change with position
in a fluid that is flowing (or at rest):
\begin{equation}
P + {\rho \over 2} v^2 + \rho g y = constant
\end{equation}
\noindent where $P$ is the pressure at, $\rho$ is the density at, $v$ is
the speed at, and $y$ is the height of a point in a fluid. If one
multiplies the equation by $V$, the
$V \rho v^2/2$ term is similar to the "kinetic energy" of a volume $V$
of the fluid. The term $V \rho g y$ is similar to gravitational
potential energy, and $PV$ has units of force $\times$ distance or
work. Thus, the physics of the work-energy theorum is somehow contained
in the equation, and we will use it to derive Bernouli's equation.
Consider a small volume of fluid within the bigger fluid. Let the
small volume be a cylindrical shape with the axis of symmetry in
the direction of its velocity. Let the area of the cylinder be $A$,
the height $d$, the mass $\Delta m$, and the volume $\Delta V$. Suppose the volume is flowing to the
right as shown in figure 3. We label the point to the left of the small volume as $1$,
and the point to the right labeled as $2$. At point number 1, the fluid will
have a density, pressure and velocity vector. At point 2, the fluid will also have
a density, pressure and velocity, which might be different than at point 1. Note
that if the volume is moving to the right, $P_1$ will be greater than $P_2$.
We can apply the work-energy theorum as the left side of the volume of fluid moves
from point 1 to 2. From the work-energy theorum:
\begin{center}
Net Work = Change in K.E.\\
(Work done by gravity) + (Work done by the Pressure) = Change in K.E.
\end{center}
\noindent The work done by gravity is $\Delta m g (y_1 - y_2)$, where $y_1$ ($y_2$) is
the "y" coordinate at point 1 (2). To find the work done by the pressure,
we use work equals force times distance. Since force is pressure times
area, the {\it net average force} on the small volume as the left side moves from $1 \rightarrow 2$
is $(P_1 A - P_2 A)$. This force acts through a distance $d$, so
\begin{equation}
(P_1 A - P_2 A)d + \Delta m g(y_1 - y_2) = {{\Delta m} \over 2} v_2^2 - {{\Delta m} \over 2} v_1^2
\end{equation}
\noindent Factoring out the $A$ and dividing by $\Delta V = A d$ we have:
\begin{equation}
P_1 - P_2 + \rho g (y_1 - y_2) = {\rho \over 2} v_2^2 - {\rho \over 2} v_1^2
\end{equation}
\noindent where $\rho = \Delta m / \Delta V$. Rearrainging the terms gives
us a more useful relationship:
\begin{equation}
P_1 + \rho g y_1 + {\rho \over 2} v_1^2 = P_2 + \rho g y_2 + {\rho \over 2} v_2^2
\end{equation}
\noindent As the volume moves a small distance $d$ through the fluid, the combination
of Pressure plus $\rho g y$ plus $\rho v^2 /2$ doesn't change:
\begin{equation}
P + \rho g y + {\rho \over 2} v^2 = constant
\end{equation}
\noindent We can follow the small volume as it moves through the fluid, with
the result being that $P + \rho g y + \rho v^2 /2 =$ constant throughout
the whole fluid!
This is a remarkable result that there is such a simple relationship between
these variables for all points in a fluid, whether it is flowing or not. As one moves
around within a fluid $P + \rho g y + \rho v^2 /2$ remains the same. WOW!\\
The work done by the pressure difference derived above is somewhat non-rigorous. Why did we use the
average force equal to $(P_1 - P_2)A$? A more detailed approach is the following.
Let the initial position be labeled as "$1$" and the final position as "$2$".
The net work done by the pressure difference is:
\begin{equation}
W_{net} = \int_1^2 F_{net} \; dx = \int_1^2 A(P(x) - P(x+d)) \; dx
\end{equation}
\noindent where $P(x)$ is the pressure at the location $x$. $P(x)-P(x+d)$ is
the pressure difference between the left and right sides of the cylinder
for any value of $x$. If
$d$ is small enough, then $P(x) - P(x+d) \approx - d (dP)/(dx)$ from the definition
of the derivitive. With this approximation, we have
\begin{eqnarray*}
W_{net} & = & \int_1^2 A(P(x) - P(x+d)) \; dx \\
& \approx & A \int_1^2 (-d) {{dP} \over {dx}} dx \\
& \approx & -Ad (P(2) - P(1)) \\
& \approx & V (P_1 - P_2) \\
\end{eqnarray*}
\noindent Using the average force gives the same result. Note that in the
discussion above position "$2$" is not necessarily at the other side of
the cylinder.
\bigskip
\noindent {\bf Final Ideas:}\\
Bernouli's equation states that if one moves around in the fluid, points of fast velocity
are points of low pressure, and points of lower speed have higher pressure. This does make
"sense", since to obtain a large velocity places of larger pressure somewhere else are needed to "push"
the fluid to these higher speeds where the pressure is lower. Note, that if the fluid is
at rest, $v$ is zero everywhere, and Bernouli's equation reduces to the equation for a fluid
at rest: $P + \rho g y =$ constant. If one is in a "weightless" environment where $g=0$, the equation
is also valid: $P + \rho v^2 /2 =$ constant. One cannot understate the importance of
Bernouli's equation to applications in society. It is the "physics" behind the
invention and development of the airplane. We will do interesting examples and demo's in lecture
which demonstrate Bernouli's equation.
In deriving Bernouli's equation we did not introduce any new fundamental principle of
physics, but rather applied Newton's laws of mechanics to fluid motion. Bernouli
presented his equation in 1738. Newton lived from 1642-1727. You might wonder why
it took so long to apply Newton's laws to fluids, or why didn't Newton himself
come up with Bernouli's equation? I can only guess that since Newton's laws of
mechanics dealt with particles, it was not clear if they could be
applied to fluids. It took the genius of Bernouli to extend Newton's ideas to
continuous media. In the next section we will discuss a truely fundamental
principle of physics. We will ponder over the nature of gravity, and learn one of the most interesting ideas
of physics ever discovered: Newton's Law of Universal Gravity.
\begin{figure}
\includegraphics[width=14cm]{fig1321a.png}
\end{figure}
\newpage
\begin{center}
{\bf Universal Gravitation}
\end{center}
\bigskip
The discovery of universal gravitation was a result of our curiousity about
the stars and planets. The history of the scientific development of
our understanding about the universe warrants a course in itself. The interested
student might consider taking our upper division G.E. elective Phy303:
The Universe in 10 weeks. Here I will briefly mention two important
scientists that gave Newton the necessary background to formulate his famous
theory of universal gravitation: the experimentalist Tycho Brahe (1546-1601)
and the theoretician Johannes Kepler (1571-1630).\\
\noindent {\bf Tycho Brahe}:\\
Brahe was born in Denmark. A predicted solar eclipse that he observed fueled
his interest in astronomy. He devoted his life to making extremely accurate
measurements of the positions of the planets, because he knew such data was
necessary for a correct understanding of their movement. He was funded by
King Frederick II of Denmark, and was given an island to do his measurements.
Since the telescope had not yet been invented, he built large sextants for
accurate measurements. He took data for over $20$ years, and was able to obtain an accuracy of $1/60$ of a
degree in his angular measurements. When Brahe died in Prague, he entrusted his data
to his collegue Johannes Kepler.\\
\noindent {\bf Johannes Kepler}\\
Kepler was very talented in mathematics, and was interested in discovering
mathematical relationships among the planets. He believed in the
Copernicus model of the solar system and devoted his efforts to
understanding the motion of the planets. In analyzing Brahe's data for the orbit of
Mars, Kepler came across a problem: Mars' orbit did not appear to be
circular. He had $40$ good data points for Mars, and no matter what
circular shape he tried Brahe's data was off by $8/60$ of a degree.
Brahe was no longer alive, but Kepler knew that the experimental data was
accurate to $1/60$ of a degree. Trying different shapes, Kepler
discovered that an eliptical shape fit the Mars data well. After
nearly $30$ years of analyzing Brahe's data, Kepler discovered
three quantitative relationships for the planets. These are known
as "Kepler's Laws":\\
\noindent {\bf Kepler's Laws}\\
\noindent I. {\bf Each planet moves in an elliptical orbit with the
sun at one focus.} Who ordered an ellipse? It was believed that the
circle was the perfect shape, and the planets should travel in
circles about the sun. The discovery that the planets travel in
elliptical orbits was revolutionary, but the data supported it.
Kepler was correct.\\
\noindent II. {\bf The vector drawn from the sun to a planet
sweeps out equal areas in equal times.} This means that when
a planet is furthest from the sun it moves slower than when it
is closer. We shall see that this is because the force of
gravity is a central force and angular momentum is conserved.\\
\noindent III. {\bf The square of the period of a planet is proportional
to the semi-major axis cubed.} This is probably better expressed
mathematically. Let $T$ be the period of the planet. The period is the
time it takes the planet to orbit the sun. It is the planet's year.
Let $a$ be the semi-major axis. If the orbit were to be circular,
this is the radius of the orbit. Then,
\begin{equation}
T^2 \propto a^3
\end{equation}
Kepler's laws allowed astronomers to predict the positions of the planets much more
accurately then before. Before Kepler, it was assumed that the planets
moved in circular orbits (the perfect shape), and corrections had to be
made from these predictions. Kepler's work also emphasized the importance
of believing the data, even though it is contrary to your pre-conceived
ideas. It took $20$ years of accurate
data taking and $20$ more years of data analysis to discover these three
properties of planetary motion. It was a tremendous accomplishment.
The reason why these "laws" were true were not understood by
Kepler and other scientists at the time. It took the genius of
Issac Newton (1642-1727) to discover that Kepler's 3 laws are the
result of only one law of nature: Universal Gravitation.\\
\noindent{\bf Issac Newton}
As the story goes, Newton was sitting under an apple tree and could
see the moon in the sky. All of a sudden an apple fell from the tree.
He then realized that the same type of force (gravity) that was attracting the apple to
the earth was attracting the moon to the earth as well. Never before
has anyone thought that the earth's gravitational force could extend to
the moon and beyond. Newton attacked the problem of gravity quantitatively.
He considered first the force between two small objects.
Suppose there are two "small" objects or particles. Label one of them "1"
and the other "2" having masses $m_1$ and $m_2$ respectively. Let them
be separated by a distance $r$. Newton supposed that there exists a
force of gravity attracting them to each other. From his third law,
the force on "1" due to "2" is equal in magnitude (but opposite in
direction) to the force on "2" due to "1": $\vec{F}_{12} = - \vec{F}_{21}$.
What could the magnitude of this force, $|\vec{F}_{12}| \equiv F_{gravity}$ depend on?
Since $w=mg$ on earth, the gravitational force is proportional to the objects mass.
Since the forces are equal in magnitude, the source of the force must be proportional
to the sources mass. Thus, it is a good guess that the force of gravity is proportional
to the product of the two particle's masses:
\begin{equation}
F_{gravity} \propto m_1 m_2
\end{equation}
\noindent How should the magnitude of the force depend on the distance between the
particles $r$? One would think that the larger $r$ is, the smaller the force.
But is the dependence $1/r$, $1/r^2$, $1/r^3$ or $1/r^x$ where $x$ is some non-
integer? From the available data at the time, Newton was able to determine that
$F \propto 1/r^2$. We will discuss how he was able to do this later. Putting the
two dependences together gives:
\begin{equation}
F_{gravity} \propto {{m_1 m_2} \over r^2}
\end{equation}
\noindent The proportionality can be changed to an equality by adding a
proportionality constant:
\begin{equation}
F_{gravity} = G {{m_1 m_2} \over r^2}
\end{equation}
\noindent with the direction of the force being along the line joining the
particles and is attactive.
The constant $G$ is called the universal gravitational constant. This is different
than "little" $g$ which is the acceleration near the surface of a planet. $g$ can
change from place to place, but $G$ is universal: it is the same everywhere in the
universe. It was first accurately measured by Cavandish in 1798 to be
$6.67 \times 10^{-11} \; Nm^2/Kg^2$. \\
\noindent Comments on Newton's universal law of gravity:\\
\noindent 1. The above equation pertains only to "point" particles. That is
objects whose size is much smaller than the separation of the objects.\\
\noindent 2. The gravitational force $F_{gravity}$ is always attractive. Both objects
attract each other.\\
\noindent 3. No matter how large $r$ is, there is still some gravitational attractive
force between the particles. This means that you are attracted to every piece of matter
in the entire universe. You are attracted to every star in the universe, a moth in
Tibet, the student sitting next to you, etc. Of course, the farther away the objects
are, the weaker is the magnitude of the force.\\
\noindent 4. Since $G$ is small (in metric units), one of the particles needs to have
a large mass for the force to be a significant number of Newtons. For example, the
force between two $1$ kg objects that are separated by a distance of $1$ meter is
only $6.67 \times 10^{-11}$ Newtons. Don't worry about the force from the student
sitting next to you, it is pretty small.\\
\noindent 5. In the above equation (Newton's law of gravitation), mass plays a different role than
in the force-motion equation $\vec{F}_{net} = m \vec{a}$. In the law of gravitation, {\bf mass
is the source} of the gravitational force. In the force-motion law, {\bf mass is a measure of
inertia} or the difficulty to change the objects velocity. It turns out that the
"gravitational mass" is equal to the "inertial mass". This equivalence motivated Einstein to
formulate his law of gravity.\\
\noindent 6. The force of gravity is {\bf always attractive}. In the case of the electrostatic
force, charges can attract or repel. This is not the case with gravity. As far as we
know, all matter (regular or anti-matter) attract each other.\\
Since force is a vector, it would be nice to express Newton's law in vector form. We
can do this by defining a unit vector that points along the line joining the two
particles. Let $\hat{r}_{12}$ be defined as a vector of length {\it one unit} that
points from particle "1" to particle "2". Similarly, $\hat{r}_{21}$ points from particle "2"
to particle "1" and is of length one. Then the force {\bf on particle "2"} from
particle "1" is:
\begin{equation}
\vec{F}_{21} = G {{m_1 m_2} \over r^2} (- \hat{r}_{12})
\end{equation}
\noindent or
\begin{equation}
\vec{F}_{21} = -G {{m_1 m_2} \over r^2} \hat{r}_{12}
\end{equation}
\noindent The minus sign in the above equation signifies that
the force on particle "2" is towards particle "1", i.e. attractive.
Note that the notation in the book of $\vec{F}_{ij}$ is opposite to
the definition used here.
The equation can also be written in terms of {\bf the force on
particle "1"} due to particle "2":
\begin{equation}
\vec{F}_{12} = G {{m_1 m_2} \over r^2} \hat{r}_{12}
\end{equation}
\noindent {\bf Principle of Superposition}\\
Suppose we have three "point" objects, labeled "1", "2", and "3", with respective
masses $m_1$, $m_2$, and $m_3$. What is the force on object "1" {\bf due} to the other
two objects? Your first thought might be to simply add $\vec{F}_{12}$ to $\vec{F}_{13}$
using the laws of vector addition. This turns out to be correct, but must be verified
by experiment, and is. This principle, which is a law of nature, is called the
{\bf principle of superposition}: the net gravitational force on an object is the
vector sum of the gravitational forces on the object from all the other objects in the system.
\begin{equation}
\vec{F}_1(Net) = G {{m_1 m_2} \over {r_{12}^2}} \hat{r}_{12} + G {{m_1 m_3} \over {r_{13}^2}}
\hat{r}_{13}
\end{equation}
If there are $N$ "point" particles in the system, the net force on particle "1" is the vector
sum of the forces due to (from) the other $N-1$ particles:
\begin{equation}
\vec{F}_1(Net) = \sum_{j=2}^N G {{m_1 m_j} \over {r_{1j}}} \hat{r}_{1j}
\end{equation}
\noindent where $r_{ij}$ is the distance between particle $i$ and particle $j$. The force
on the other particles are determined in a similar way.
We will do a number of examples in class involving Newton's law of the gravitational
force between point particles and the principle of superposition.\\
\noindent {\bf Justifying the inverse square law for gravitation}\\
How did Newton know the gravitational force decreased as $1/r^2$? One way was
to calculate the acceleration of the moon as it orbits the earth and compare it
with the acceleration of an object near the surface of the earth. Since the moon
is moving in approximately circular motion, its acceleration is
$a_{moon} = v^2 /r = (2 \pi r/T)^2 /r = (4 \pi^2 r/T^2$. Using $r = 3.84 \times
10^8$m, and $T = 28$ days, one gets $a_{moon} = 2.72 \times 10^{-3}$ m/s$^2$.
The acceleration near the surface of the earth is $9.8$ m/s$^2$. The moon is
around $60$ earth radii away, and since $9.8/60^2 = 2.72 \times 10^{-3}$ we
see that the gravitational force on the moon is $1/60^2$ that of an object on
the surface of the earth, verifying the inverse square decrease of the gravitational
force.
Another observation that supports the inverse square nature of the force is
Kepler's third law which states that $T^2 \propto a^3$. Using $F=ma$:
\begin{equation}
F_{grav} = G {{m_1 m_2} \over r^2} = m_1 {v^2 \over r}
\end{equation}
\noindent Since $v = 2 \pi r / T$ we obtain
\begin{equation}
G {{m_1 m_2} \over r^2} = m_1 {{(2 \pi r / T)^2} \over r}
\end{equation}
\noindent reducing to
\begin{equation}
T^2 = {{4 \pi} \over {G m_2}} r^3
\end{equation}
\noindent Thus, the data of the planet's motion $T^2 \propto r^3$ is supported by
the inverse square law decrease of the gravitational force.
The inverse square law shows up in different situations in physics: light intensity, sound intensity,
and the electostatic force also decrease as $1/r^2$. Is there a deep
reason for the inverse square law? For light and sound intensity there is. It
has to do with geometry. The surface area of a sphere
equals $4 \pi r^2$, so if a quantity is spread out evenly on the surface of an
expanding sphere, its intensity decreases as $1/r^2$. Sound intensity and
light intensity follow this principle. Force is not an intensity, so
one cannot really use this argument. However, advanced theories of gravity and
electrodynamics give the same geometric result when speeds are slow enough.
Time factors out of these equations and the remaining three dimensional space geometry
produces a $1/r^2$ decrease of the gravitational and electrostatic force.
\bigskip
\noindent {\bf Graviational force for extended objects}\\
What is the gravitational force between a "point" particle and a larger or
extended object. By an extended object we mean an object that has a finite
size: a solid object. Do we need any new physics to find
the graviational force in this case. No, we just need divide up the solid
object into small pieces, and use the superposition principle and Newton's
gravitational force equation for point particles. Let's explain the
method using two examples.\\
\noindent {\it The force between a ring and a point particle}\\
Suppose we have a ring of mass $M$ and radius $a$. A point particle of
mass $m$ is placed on the axis of the ring a distance $x$ from the
center (See the figure at the end of this section). What gravitational
force does the point particle feel due to the ring?
\begin{figure}
\includegraphics[width=14cm]{fig1321b.png}
\end{figure}
First we divide up the ring into small "point" pieces of mass $\Delta M$.
Next, we find the magnitude of the force, $|\Delta \vec{F}|$, between the point particle and the small piece of
the ring:
\begin{equation}
|\Delta \vec{F}| = G {{m \Delta M} \over {x^2 + a^2}}
\end{equation}
\noindent The direction of the force $\Delta \vec{F}$ points from the point particle to
the edge of the ring where $\Delta M$ is. Now we need to sum up the forces due to all
the little pieces of the ring. This involves an integration: $\int \Delta \vec{F}$.
However, for the ring this is simple. Upon integrating around the ring, the only component
that survives is the one along the axis: $|\Delta \vec{F}| cos \theta$ or
\begin{equation}
\Delta F_x = G {{m \Delta M} \over {x^2 + a^2}} cos(\theta)
\end{equation}
\noindent towards the ring. Since $x$, $a$ and $\theta$ are the same for all the pieces of the
ring, the integral is simple:
\begin{equation}
F_x = G {{m} \over {x^2 + a^2}} cos(\theta) \int \Delta M
\end{equation}
\noindent The integral over $\Delta M$ is just $M$, so
\begin{equation}
F_x = G {{mM} \over {x^2 + a^2}} cos(\theta)
\end{equation}
\noindent towards the ring. Since $cos\theta = x/\sqrt{x^2 + a^2}$, we have
\begin{equation}
F_x = G {{mMx} \over {(x^2 + a^2)^{3/2}}}
\end{equation}
\noindent with the direction towards the ring. \\
\noindent {\it The force between a point and a long thin rod}\\
Suppose we have a long thin rod of mass $M$ and length $l$. A point
particle of mass $m$ is placed on the axis of the rod a distance
$d$ from one end. See the figure at the end of the section.
What gravitational force does the point particle feel due to the
rod?
First we divide up the rod into small pieces. Lets divide it up
into $N$ equal pieces. Each piece will have a mass $\Delta M$ and
a length $\Delta x$. Consider the force on the point particle
due to a small piece of rod a distance $x$ from the end.
The magnitude of this force, $\Delta F_x$, from Newton's law of gravitation
is:
\begin{equation}
\Delta F_x = G {{m \Delta M} \over {(d+x)^2}}
\end{equation}
\noindent Since we will integrate over $x$, we need to express the
mass of the small piece $\Delta M$ in terms of $\Delta x$. Since
the whole rod has mass $M$ and length $l$, the mass of the small
piece is $\Delta M = M (\Delta x / l)$. Thus, we have
\begin{equation}
\Delta F_x = G {{m M \Delta x} \over {l (d+x)^2}}
\end{equation}
\noindent The final step is to add up the contribution from all
the little pieces of the rod as $\Delta x \rightarrow 0$. This
leads to the integral expression:
\begin{equation}
F_x = \int_0^l G {{m M \; dx} \over {l (d+x)^2}}
\end{equation}
\noindent with the result being:
\begin{equation}
F_x = G {{Mm} \over {d(d+l)}}
\end{equation}
\noindent with the force directed towards the rod. The above
equation deserves some comments:\\
\noindent 1. It was a real theoretical breakthrough of Newton to
come up with a method of finding the force between two "complicated"
extended objects. The superposition principle is the key to the
simplification: the forces from many sources add like vectors.
One needs only to understand the force between two "point" particles,
then integrate over the solid objects to find the total force.
It turns out that the force between two point particles is simple:
$\sim m_1 m_2/r^2$. This same method will be used for the electrostatic
and magnetic forces next quarter. It was a triumph to break down a
complicated problem into two (or more) simple problems.\\
\noindent 2. One can check the result of the particle and rod in the
limiting case of $d>>l$. As $d$ becomes much greater than $l$, the
term $(d+l)$ is very close to $d$. Thus the limit of $F_x$ as
$d \rightarrow \infty$ is $GMm/d^2$ which is the force between two
point objects as expected.\\
\noindent 3. Note that the force between the particle and the rod is not
equal to $G M m/(d+l/2)^2$:
\begin{equation}
F_x \neq G {{Mm} \over {(d + l/2)^2}}
\end{equation}
\noindent That is, {\bf in general the gravitational force is not
the same as if all of the objects mass were located at its center of mass}.
There is one exception to this: the force between an object with spherical symmetry
and a point particle outside the object. We consider this case next.\\
\noindent {\it The force between a point particle and a spherical shell}\\
Suppose we have a thin spherical shell of mass $M$ and radius $R$. A point
particle of mass $m$ is placed outside the shell and distance $d$ from the
center of the shell. What gravitational force
does the point particle experience due to the shell?
This problem is solved the same way as the last two. One divides up the
shell into rings, and adds up the force due to each ring. I show the
solution to this problem at the end of the lecture notes, since it is
easiest to solve for the graviational potential energy between a
point particle and a thin spherical shell. You might think that the
result would be complicated, but as we will show it is amazingly simple:
\begin{equation}
F = G {{Mm} \over d^2}
\end{equation}
\noindent where $F$ is the magnitude of the gravitational force, and
the direction of the force on the particle is towards the
center of the shell. Thus, for the thin shell, it is as if all the mass were
located at the center of the shell. This is only true if the particle is
{\bf outside the shell}. If the particle is {\bf inside the shell}, the net force on
the particle is zero! We note that this simple result is only true in the
special case of the inverse square law force, which is the case for the gravitational and
electrostatic forces.
The above result for the thin shell enables us to find the gravitational force
between a point particle and any spherically symmetric object, since a spherically
symmetric object can be divided up into thin shells. We will only consider one
example in this class: the gravitational force between a point particle and
a solid sphere of uniform density.\\
\noindent {\it The force between a point particle and a solid sphere}\\
A solid sphere can be divided up into thin shells. If particle of mass $m$ is
located outside the sphere, the magnitude of the gravitational force it feels due to the sphere
is just
\begin{equation}
F = G {{Mm} \over r^2}
\end{equation}
\noindent where M is the mass of the sphere, and $r$ is the distance the particle is
from the center of the sphere. Note that the radius of the sphere does not enter
into the formula (as long as the point particle is outside the sphere).
We can use this nice result to determine the acceleration due to gravity at the
surface of a planet. Let $M$ be the mass of a planet, and $R$ its radius. Let a
particle, of mass $m$, be located a distance $h$ above the surface of the planet. The
magnitude of the gravitational force between the point object and the
planet is
\begin{equation}
F = G {{Mm} \over {(R+h)^2}}
\end{equation}
\noindent If $h <<< R$ the force is approximately
\begin{equation}
F \approx G {{Mm} \over {R^2}}
\end{equation}
\noindent Since the weight of the object is the force due to gravity, $mg$,
we have:
\begin{equation}
mg \approx G {{Mm} \over {R^2}}
\end{equation}
\noindent Actually, the $m$ on the left is the inertial mass, and the $m$ on
the right is the gravitational mass. Since they are equivalent they cancel
and all objects have the same acceleration $g$:
\begin{equation}
g \approx G {{M} \over {R^2}}
\end{equation}
\noindent where $M$ is the mass and $R$ is the radius of the planet. It is easy
to measure $g$, with the result on earth being around $9.8$ m/s$^2$. The radius
of the earth was known from ancient times, and is $R \approx 6.37 \times 10^6$m.
When Cavandish accurately measured $G$, he was also measuring the mass of the
earth! By observing the motion of moons and planets, astonomers can determine
their masses using Newton's law of gravitation.\\
\centerline{\bf Gravitational Potential Energy}
\bigskip
For our final topic on universal gravitation, we want to derive an expression
for the potential energy for the gravitational force. Last quarter (Phy131)
we derived the expression: $P.E. = mgy$. This was the case if the force of
gravity is constant, which is only valid if one is close to the surface of
a planet. Here we will handle the situation in general, and let the particles
be far apart.
Let's start, as physicists often do, with the simplest case: two small "point"
particles. For example, two small marbles. Let one have a mass $m_1$ and
the other a mass of $m_2$. Let particle $m_2$ be fixed in space, and not
be able to move. We want to calculate the work done by the force of
gravity if the particle $m_1$ is moved from one position to another. Remember
that work is the component of force in the direction of path.
Suppose $m_1$ is moved in a circular path with constant radius, say $r_i$. Since
the force of gravity is outward from mass $m_2$, this path is perpendicular to
the direction of the force. Thus, the force of gravity does no work in this case.
The force of gravity only does work on $m_1$ if $m_1$ is moved away from
$m_2$. Let's calculate the work that the force of gravity does on $m_1$ when
it is moved {\bf radially} away from $m_2$. Suppose it ($m_1$) starts at
a distance $r_i$ and is moved to a distance $r_f$, where $r_i < r_f$. At
any position, the force of gravity on $m_1$ is:
\begin{equation}
|\vec{F}_{12}| = G {{m_1 m_2} \over r^2}
\end{equation}
\noindent and is directed towards $m_2$. Work, $W$, is force times distance.
In this case, the force changes with position, so we must integrate
$W = \int \vec{F} \cdot d\vec{r}$. In our case, $\vec{F}$ is exactly opposite
to $\Delta \vec{r}$, so we have:
\begin{equation}
W_{r_i \rightarrow r_f} = \int_{r_i}^{r_f} -G {{m_1 m_2} \over r^2} dr
\end{equation}
\noindent where the minus sign is because $\vec{F}$ points opposite to the
direction that $m_1$ is moved. Evaluation of the integral gives:
\begin{equation}
W_{r_i \rightarrow r_f} = G {{m_1 m_2} \over {r_f}} - G {{m_1 m_2} \over {r_i}}
\end{equation}
\noindent Note that since $r_i < r_f$, gravity does negative work in this case.
If $m_1$ were to go from $r_f \rightarrow r_i$ (further away to closer) then
gravity would do positive work on $m_1$. Let's determine a possible form for
the potential energy by using the work-energy theorum.
The work-energy theorum states that the net work done equals the change in K.E.
Let the particle of mass $m_1$ have an initial speed of $v_i$ directed away from
$m_2$, and let the speed when it reaches $r_f$ be $v_f$. Then from the work-energy
theorum:
\begin{equation}
G {{m_1 m_2} \over {r_f}} - G {{m_1 m_2} \over {r_i}} =
{{m_1 v_f^2} \over 2} - {{m_1 v_i^2} \over 2}
\end{equation}
\noindent rearrainging terms, we have
\begin{equation}
{{m_1 v_i^2} \over 2} - G {{m_1 m_2} \over {r_i}} =
{{m_1 v_f^2} \over 2} - G {{m_1 m_2} \over r_f}
\end{equation}
Note that the left side of the above equation only contains $r_i$ and $v_i$,
and that the right side only contains $r_f$ and $v_f$. Since $r_f$ is
arbitrary, we see that as particle $m_1$ moves under the influence of the
force of gravity from $m_2$ that
\begin{equation}
{{m_1 v^2} \over 2} - G {{m_1 m_2} \over {r}} = constant
\end{equation}
\noindent where $r$ is the distance that particle "1" is away from
particle "2". The first term on the left is recognized as the
kinetic energy of particle "1". The second term only depends on
position and has units of energy, and is thus interpreted as the
potential energy function, $U(r)$. The above statement is one of
mechanical energy conservation, and the potential energy function
for the gravitational force can be taken as
\begin{equation}
U(r) = -G {{m_1 m_2} \over r}
\end{equation}
\noindent Some comments on $U(r)$:\\
\noindent 1. $U(r)$ is the potential energy function for two point particles. If we
have more than 2 particles, then we have use the above formula and sum over all the
pairs of particles. If the objects are solid and have finite size, then one needs
to integrate over the volume (or volumes) of the objects. We will only do simple cases
in this class.\\
\noindent 2. The potential energy function is not absolute. One can always add a
constant value to $U(r)$, since only the difference $U(r_f) - U(r_i)$ is important.
For the form $U(r) = -G m_1 m_2 / r$ we have chosen as a reference zero potential
the point(s) at $r = \infty$: $U(\infty) \equiv 0$.\\
\noindent 3. If $m_1$ is "let go" at a distance $r_i$ away from $m_2$, it will
move towards it, since the force of gravity is attractive. Thus, points closer
to $m_2$ will be at "lower" potential energy. Since $U(\infty) = 0$, values
of $r<\infty$ will have negative potential energy. This is the reason for the
minus sign in the expression. It is basically because the gravitational force
is attractive.\\
\noindent 4. $U(r)$ is a {\bf scalar}. There is no direction to potential
energy. There is a direction to force.\\
\noindent 5. Notice that $U(r) \propto 1/r$, whereas the force $F \propto 1/r^2$.
In phy131 we discussed how force and potential are related, one is the derivative
of the other. In our case, $F_r = - dU/dr$. Check it out, it works.\\
\bigskip
\noindent {\it The gravitational potential energy of a ring and a point particle}\\
As an example of the graviational potential energy of a small "point"
object and an extended object the gravitational potential energy between
a thin ring of mass $M$ and radius $R$ and a small object of mass $m$
located on the axis of the ring a distance $x$ from it's center.
To solve the problem, we divide the ring up into small little pieces as
we did in finding the force between a "point" object and a ring. Let
the small piece of ring contain an amount of mass equal to $\Delta M$.
The gravitational potential energy, $\Delta U$, between the "point" object of mass
$m$ and the small piece of the ring is
\begin{figure}
\includegraphics[width=14cm]{fig1321c.png}
\end{figure}
\begin{equation}
\Delta U = - {{Gm (\Delta M)} \over {\sqrt{x^2+R^2}}}
\end{equation}
\noindent The factor $\sqrt{x^2+R^2}$ is the distance from $m$ to the ring
segment (see the figure). You might wonder if the sin or cos of an angle
should enter in the formula as it did with the force between a "point"
object on the axis of a thin ring. Potential energy is a scalar. There is
no direction to $\Delta U$. Only the distance between $m$ and $\Delta M$
matters, a nice feature of potential energy.
Now we can sum up the $\Delta U$ contributions from every piece of the ring.
\begin{eqnarray*}
U & = & \sum \Delta U \\
& = & -\sum {{Gm (\Delta M)} \over {\sqrt{x^2+R^2}}}\\
& = & - {{Gm} \over {\sqrt{x^2+R^2}}} \sum (\Delta M) \\
U & = & - {{GmM} \over {\sqrt{x^2+R^2}}}
\end{eqnarray*}
\noindent Since the distance from the "point" object to every piece of
the ring is the same, $\sqrt{x^2+R^2}$, this term can be brought out of
the sum. The sum consists of adding up the mass of each piece of the
thin ring, which adds up to $M$. The result is nice and simple.
Note that this expression uses as a reference for zero potential energy the
configuration for the ring and object to be infinitely far apart. That is,
$U=0$ when $x=\infty$.
From this expression for $U(x,0,0)$, one can obtain the force between the "point"
mass and the thin ring using $F_x=-\partial U/\partial x$.
\begin{equation}
F_x = -{{\partial U} \over {\partial x}} = -G {{mMx} \over {(x^2 + a^2)^{3/2}}}
\end{equation}
\noindent in agreement with the result we obtained before for the force between a "point"
object and a thin ring.\\
\bigskip
\noindent {\it The gravitational potential energy of a thin spherical shell and a point particle}\\
We are ready to solve a classic problem: the gravitational potential energy
between a small "point" object of mass $m$ and a thin spherical shell of
mass $M$. Let the radius of the thin shell be $R$, and let the small object
be located a distance $r>R$ from the center of the shell.
We divide the spherical shell up into many thin rings. Consider a ring that is
at an angle $\theta$ from the axis, and subtends an angle $\Delta \theta$ as
shown in the figure. The amount of mass $\Delta M$ in the ring is equal to the total
mass $M$ of the shell times the ratio of the rings area to that of the shell:
\begin{equation}
\Delta M = M {{2 \pi R sin(\theta) R (\Delta \theta)} \over {4 \pi R^2}}=
M {{sin(\theta) (\Delta \theta)} \over {2}}
\end{equation}
\noindent The distance from a point on the ring and the object $m$ is just
$d=\sqrt{r^2+R^2-2rRcos(\theta)}$ using the law of cosines. Thus, the gravitational
potential energy $\Delta U$ between the "point" object and the ring is
\begin{equation}
\Delta U = -Gm {{M \; sin(\theta) (\Delta \theta)/2} \over {\sqrt{r^2+R^2-2rRcos(\theta)}}}
\end{equation}
\noindent All the thin ring segments are added up by integrating over $\theta$ from
$\theta = 0$ to $\theta = \pi$:
\begin{equation}
U = -GmM \int_0^\pi {{sin(\theta) \; d\theta} \over {2\sqrt{r^2+R^2-2rRcos(\theta)}}}
\end{equation}
\noindent The integral is solvable with the substitution $y=r^2+R^2-2rRcos(\theta)$.
With this substitution we have $dy=2rRsin(\theta) d\theta$, and the integral
becomes
\begin{equation}
U = -{{GMm} \over {4rR}} \int_{(r-R)^2}^{(r+R)^2} {{dy} \over \sqrt{y}}
\end{equation}
\noindent Solving the integral yields:
\begin{eqnarray*}
U & = & -{{GMm} \over {2rR}} \sqrt{y} |_{(r-R)^2}^{(r+R)^2} \\
& = & -{{GMm} \over {2rR}} [(r+R) - (r-R)] \\
U & = & -{{GMm} \over {r}} \\
\end{eqnarray*}
\noindent for $r>R$. WOW!! The gravitational potential (and also force) between a
spherically symmetric object of mass $M$ and a small "point" object (outside
the spherically symmetric mass) is the same as if all the mass $M$ were
located at the center of the spherical object.
If $r