A glimpse of the universal glue...

In this section of the program we are going to focus on what
gravity does, not how it works. How it works is the subject of a
whole other course. What it does is of great interest in
predicting the future of mechanical systems since every
mechanical system we know about is subject to the effects of
gravity. In our examples so far we have arranged things so that
the effect of gravity was cancelled out by another force, for
example in the collision
situation the colliding objects were supported by a table of some
sort. Or as in the projectile
motion example we made a simplifying approximation that the
direction of the force of gravity was constant. Now we need to
look more closely at the nature of the gravitational force.
That requirement brings us back to Mr.
Newton. The story goes that he observed an apple falling from a
tree and at the same time noticed the moon, also up in sky but
not falling to Earth. The difference, he might have decided, was
that the moon had some sideways (tangential) velocity whereas the
apple did not.
The apple orbit, at least that part of it which was
observable, appears to be a straight line. We know that in the
case of simple harmonic motion a linear "orbit" is
possible. Just look at the combined
SHM display with the x and y parameters identical. It turns
out that a linear orbit is also possible with an inverse square
type central force. Unfortunately in the apple situation the
orbit was interrupted by a collision with the surface of the
planet. If it were possible for a hole to be drilled through the
entire earth, an apple or anything else falling into that hole
would oscillate back and forth along a diameter of the
planet.

One of Newton's major contributions to science was to come
up with one law of nature that could explain the behavior of
both the apple and the moon. By the way, he had to invent
calculus along the way to accomplish this feat. What he concluded
was that any two bodies are attracted to each other in a way
which is quantified by a simple law called the law of
"universal gravitation". That law says that the
attractive force between a pair of objects is proportional to the
product of their masses divided by the square of the distance
between their centers of mass. The force on the two objects is an
actionreaction pair in that they are equal and directed
oppositely.




The law of universal gravitation is written
f = G * m_{1} * m_{2} /
r^{2},
where G is the gravitational constant = 6.673e11 Newton meters
squared per kilogram squared. As well as can be determined this
value holds for any two masses, anywhere in the universe and has
been the same for all time. Such a truly universal number must
originate from the very foundations of the physical universe.
Perhaps later we can follow up on that notion.
We have a pretty good grip on the mass of an object at this
point but we should look in more detail at the implications of
the r^{2} in the law of universal gravitation. The law as
expressed is a scalar relationship, giving only the magnitude of
the force. It will be useful to have this expressed in vector
terms. Let r_{12} be the vector from the
particle of mass m_{1} to the particle of mass
m_{2}. Let r_{12} be the magnitude of
r_{12}. Then the force exerted on m_{2} by
m_{1}, call it f_{21}, is given by
f_{21} = G * m_{1} *
m_{2} / r_{21}^{3} *
r_{21},
where we have multiplied by the vector r_{21}
and divided by its magnitude r_{21} so as to include the
direction intelligence without changing the magnitude of the
quantities. Since r_{12} is
r_{21}, f_{12} =
f_{21} as required by the third law.

So what sort of motion can we expect from objects subject to
this universal law of gravitation? Let's take the vector form
of the law and expand it in terms of the 2 dimensional vector
components.
f_{21} = (G * m_{1}
* m_{2} / r_{12}^{3} *
(x_{12}* +
y_{12}*).
Pause a second to consider the
subscripting business. There is lots of chance for confusion in
this. In the expression above we have replaced the factor
r_{21} from the previous paragraph with
r_{12} and taken the negative sign that maneuver
produced and put it out in front of the G factor. We did this
because we chose to examine the effect on m_{2} of
m_{1}. That means that we will want m_{1} to be
at the origin of our reference frame and m_{2} to be out
there somewhere reacting to it. Therefore vector displacements
should be measured relative to m_{1}, that is in terms of
r_{12}. Also we have kept the subscripts on the
scalar components of the r_{12} vector. A scalar
has no directional information but it does have a sign which
determines the sense of the vector made up of components x* + y*. That is why we write
x_{12} and y_{12}, keeping the minus sign in
front of the G factor legitimate.
For convenience let's replace G * m_{1} *
m_{2}, with a constant k since for a given system they
are all constant. So
f_{21} = k /
r_{12}^{3} * (x_{12}* + y_{12}*).
Looking at the vector components separately,
f_{21x} = k /
r_{12}^{3} * x_{12},
and
f_{21y} = k /
r_{12}^{3} * y_{12}.
Except for that pesky r_{12}^{3}, this looks
like the force which resulted in SHM. Since r is always positive,
being the magnitude of a vector, r^{3} will also always
be positive. This means that the force is directed opposite to
the displacement so it is a restoring force, therefore we can
expect that the x and y components of the motion will be
oscillatory.
Obviously as r increases, the force gets weaker and weaker,
dropping to zero at infinite r. As r approaches zero the force
approaches infinity but this is not a problem since all real
masses have some size to them so that the particles will collide
before the distance between centers can get too small. In fact if
it were possible for objects to pass through one another, the
force drops off once the one object penetrates the other since
the mass outside the distance between the two centers can be
shown to make no contribution to the force. {Think about this one
a bit!}



Of course, the y component of force
behaves the same way along its dimension as the x does along its.
So we find that the motion component along the x and y axes are
oscillatory, but certainly not SHM since the magnitude of the
force is not directly proportional to the displacement but to the
reciprocal of displacement squared. To see how a simple harmonic
oscillator's position vs. time graph compares to one of an
object subject to an inverse square type central force, run the
Gravity and SHM display. In the
inverse square law part of this display, we assume that the
moving object can pass through the object creating the central
force so that a linear orbit is possible, allowing a direct
comparison with the linear SHM.
Notice that the inverse square central force produces a
waveform that is generally broader near the peaks and steeper at
the crossing of the axis than that produced by the directly
proportional force. The mathematical form of the spring like
force waveform is just the cosine function. The mathematical form
of the inverse square force waveform is an infinite series of
sine and cosine terms.
One way to model the motion of a particle subject to a central
force like gravity might be to treat the x and y components as
separate oscillators as we did in converting SHM to circular
motion. This idea has two drawbacks. First, since the oscillatory
motion is not simple harmonic, initial conditions affect not only
the phase angle but the frequency. Second, the model is very
sensitive to the magnitude of the time increment, D t and becomes unstable if D t is not small enough. Besides these
technical difficulties, our everyday experience with gravity
makes it feel like a force so we will go back to Newton's
second law to model gravitational attraction.

First we will look at a strictly made up situation. We want to
predict the future of a something the mass of a hamburger in
orbit around something the mass of a sports car in a universe
where the gravitational constant is 1.0. This is essentially the
situation set up in the next display called Orbital Vibrations.
Here we will see a relatively light object interacting with an
object of more mass but still movable. The display is a three
dimensional one but looking at the (x,y) plane edgewise along the
x axis.
We will consider our objects in this
model to be spherical so the force of gravity is spherically
symmetric. Otherwise the math is too hard for me. With a
spherically symmetric force, the orbits will lie in a plane
determined by the initial position and velocities. I chose the
initial conditions so the (x,y) plane contains the orbit. Due to
this symmetry property of the gravitational field, later we will
often use two dimensional displays for twobody interactions.
The size of the objects on the display is somewhat related to
mass but the smaller object is kept big enough so you can see it.
Notice that motion of our quarter pounder sort of fits the
description of vibration, which we discussed in the previous
section. It dithers around in the vicinity of the heavier mass.
The blue mass by the way appears transparent to the green one in
this display. Notice that the oscillations while symmetrical in
space are not symmetrical in time. The green mass goes faster
when moving to the left than when to the right. Run the Orbital Vibrations display.



Unlike the uniform circular motion we observed in combining
simple harmonic vibrations, the orbital speed in a gravitational
field changes noticeably in different parts of the orbit.
Remember the conservation of angular momentum? Angular momentum,
L, was defined as rXp where r
was the position vector of the object and p was the
linear momentum vector. But p is the mass times the
velocity v so
L = m * rXv.
The vector v is just D
r / D t so
L = m * rX D r / D
t,
and rX D r is
proportional to the area of the triangle swept out by the vector
r in the time D t since the
magnitude of the vector rX D
r is r times D r times the
sine of the angle between them. Since the angular momentum is
conserved then the area swept out by the position vector as the
object goes around in its orbit must be the same for every D t, requiring the object to go faster when
r is shorter.

Now let's take a look at an example
with more realistic values. We will show in the Satellite Orbit display, the Earth with its
true radius and mass, and give you an opportunity to place a
satellite in orbit at any altitude and with a velocity you set.
Then you can observe the orbit produced by your choices. The
distances will be in kilometers and the velocities will be in
meters per second in this display.
This Satellite Orbit display is sort of fun but to do serious
research into the effect of gravity you need to use the
Physics_T
program offered by M. Casco. In
that program you can set the positions, masses and velocities of
up to three objects and track their trajectories or plot the
variation in coordinates with time. Also numerical results may be
tabulated to any required precision. In Physics_T the model setup
may be saved and reloaded at a later time. That kind of power in
mathematical modeling is just not available with Java applets of
the sort displayable over the world wide web.



Next we will look at the work and energy aspects of gravity.
In the section on work and kinetic energy we introduced the
notion of work done against the
force of gravity and in the section on potential energy and
fields we described a scalar
potential field due to gravity. In that case we were talking
about the approximation where the force of gravity was constant
in space. Now we know more about how gravity depends on position
so let's work out what the scalar potential field would be
like in the vicinity of a large object, like the Earth.
Remember that we can only
assign a specific potential energy to each point in space if the
forces involved are
conservative in nature. To demonstrate that the force of
gravity is conservative consider two points out in space and a
small test mass moving from one to the other along some arbitrary
path. Any point on that path may be reached from any other point
by a combination of motion along a circular arc centered on the
origin, and motion along a radius from the origin. Look at the Conservative Central Force display for an
illustration.
The arbitrary path we have could be approximated by dividing it
as finely as we wish into arcs and radial movement. Now consider
the work done against a central force with its center at the
origin. Remember that the work done against the force is the
negative of the work done by the force,
D w = f
· D r.
Along the circular arc parts of the path, the dot product is zero
since the motion is perpendicular to the force. Along the radial
parts of the path the dot product is just f*
D r since the motion is parallel to the force. In the
radial direction the work done against the force is positive when
the direction of motion is outward from the center and negative
when the direction of motion is inward. So, whatever the path,
the net work done depends only on the difference in distance from
the force center to the path end points. This is because the net
radial displacement is the only component contributing to the
work. So we have demonstrated that all central forces are
conservative. Therefore gravity must be a conservative force.

Well we got off onto that tangent because we wanted to establish
that a potential energy field is created in space by the
universal gravitational attraction. So coming back to that point,
let's consider a two body system where one of the bodies is
huge, like the Earth with mass M, and the other is small enough
that its own contribution to the potential energy field is
negligible, like a 1kg test mass with mass symbolized by m.
Remember that the change in gravitational potential energy is the
negative of the work done by the gravitational force in going
from some initial system configuration to a final one.
D pe =  D w = f * D r.
Notice that we can use scalar quantities now because we know that
since gravity is a central force the pe depends only on the
distance r, the magnitude of the radius vector, and the magnitude
of the force f. The directional information is now conveyed by
the minus sign. Now dividing by D r we
get
D pe / D r = f.
The rate of change of the potential energy with respect to D r is the negative of the gravitational
force.
D pe / D r = G*M*m / r^{2} =
G*M*m*r^{2}.
By now you probably know what is coming next. When we have an
expression for the rate of change of a quantity and want to know
how the quantity itself depends on the independent variable, we
take the antiderivative. That consists of increasing the
exponent of the independent variable by 1 and dividing by the new
exponent. That gives us
pe(r) = (G*M*m*r^{1}) / (1) =
G*M*m*(1/r).
Pay close attention to the accumulation of
minus sign in the discussion above. We got one because the change
in pe is the negative of the work done by gravity. We got one
because the gravitational force is directed opposite to the
radius vector. And we got one because the antiderivative
required dividing by minus one.
There is an implicit assumption in taking the potential energy
to be the negative of the work done by gravity and that is that
the zero of potential energy is at infinite displacement. That
makes intuitive sense since the force goes to zero at that
distance. Remember that we can make an arbitrary choice where we
want potential energy zero to be since we always deal in changes.
Customarily the for gravitational potential energy, it is in fact
chosen to be at infinity. Run the
1/r Potential
display to see a plot of potential vs. distance.
Notice that the gravitational potential energy goes to
infinity as the distance to the center goes to zero. This makes
physicists nervous. We do not like our natural laws to have
singularities (blow up) unless it happens way out at infinity
where it is not going to bother anyone. For ordinary sized
objects, we found a way to wiggle out of this difficulty by
noticing that objects will collide before the distance between
their centers goes to zero and a different force law applies in a
collision. Suppose however we have a planet sized object with a
hole drilled through its center. The center of mass of that
object will lie in the middle of this drilled passage and thus be
accessible to another object dropped into the hole. Under this
scenario, the centers of mass of the drilled object and the
dropped object could coincide.



What is the potential energy of our
system when one of the objects gets inside the other? Without
spending a lot of time on this I will give what is called a
plausibility argument that the gravitational force exerted on a
mass inside a spherical shell by the shell itself is zero.
Consider a situation where the mass is at the center of the shell
centered on the origin of our reference frame. The force exerted
by every little D m of mass in the
shell would be exactly cancelled by the same amount of mass on
the opposite side of the shell, so in the case of a centered
object the gravitational force from the spherical shell is
clearly zero. For the next part of the discussion refer to the
Spherical Shell display.

Now consider a solid spherical object or uniform density, r, that either is penetrable by the test
mass or has a drilled hole in it through which the test mass may
pass. The mass M of this object is its density times its volume,
M = 4/3 * p *
R^{3} * r.
When the test mass is at distance r from the center of the
spherical object and r is outside the radius R of the object, the
inverse square law for force applies so that
f = G*M*m / r^{2} = G * 4/3
* p * R^{3} * r * m / r^{2}.
When the test mass is inside the radius of the large object, all
the large object's mass which lies outside a sphere of radius
r has no net effect on the test mass so the effective mass of the
spherical object is reduced from 4/3*
p * r *R^{3} to 4/3*
p * r
*r^{3}. So
f = G * 4/3 *
p * r^{3} / r^{2} *
r * m.
Inside the object the force on the test mass is proportional to
the displacement, just like a simple harmonic oscillator. The Penetrable Solid display illustrates this
effect. In this display a test mass is released form its location
on the positive x axis and oscillates back and forth under the
influence of gravity. Plots of the force on the test mass and the
potential energy of the system as functions of test mass position
are included.
While we are dealing with
this little makebelieve twobody system, we should also look at
the kinetic energy and total energy of the system as the red
object goes along its path. Remember that kinetic energy is 1/2 *
m v^{2}. If both objects were moving in the reference
frame the kinetic energy of the system would be the sum of that
expression for each object. In our system only one object is
moving. The total energy then is
E = 1/2 * m * v_{m}^{2} 
G*M*m / r.
Notice that E may be positive, negative or zero depending on the
velocity v_{m}. The
Gravitational Energy display shows kinetic energy and total
energy in addition to the force and potential energy shown above.
Also in this display the moving object is made invisible so that
is does not damage the plots.



The potential energy field goes to zero at infinity and produces a
force that attracts the moving object to the origin of the
reference frame. The kinetic energy is a measure of the moving
object's ability to overcome this attractive force. As the
object becomes more distant from the origin, the potential energy
increases (becomes less negative) and the kinetic energy
decreases. If we run out of kinetic energy before the potential
energy gets up to zero, the object is held captive by the
gravitational force and will return to its starting point,
producing a closed orbit. If there is more than enough kinetic
energy to compensate for the negative potential energy, the
object will not be held captive by the gravitational force so the
orbit will be open. Look at the
Open Orbit
display to see this.
There are numerous examples of gravitational effects posed as
problems in the physics texts. To be able to set them up as
mathematical models and see the results you need the M. Casco
Associates
Physics_T program of which we
have spoken earlier.
In the next section of this course we will move on to the
study of mechanical waves.
Are there any questions?

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