Gravity
 The work of a curious fellow

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 action-reaction pair in that they are equal and directed oppositely.
 The law of universal gravitation is written f = G * m1 * m2 / r2, where G is the gravitational constant = 6.673e-11 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 r2 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 r12 be the vector from the particle of mass m1 to the particle of mass m2. Let r12 be the magnitude of r12. Then the force exerted on m2 by m1, call it f21, is given by f21 = G * m1 * m2 / r213 * r21, where we have multiplied by the vector r21 and divided by its magnitude r21 so as to include the direction intelligence without changing the magnitude of the quantities. Since r12 is -r21, f12 = -f21 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. f21 = -(G * m1 * m2 / r123 * (x12* + y12*). 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 r21 from the previous paragraph with r12 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 m2 of m1. That means that we will want m1 to be at the origin of our reference frame and m2 to be out there somewhere reacting to it. Therefore vector displacements should be measured relative to m1, that is in terms of r12. Also we have kept the subscripts on the scalar components of the r12 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 x12 and y12, keeping the minus sign in front of the G factor legitimate. For convenience let's replace G * m1 * m2, with a constant k since for a given system they are all constant. So f21 = -k / r123 * (x12* + y12*). Looking at the vector components separately, f21x = -k / r123 * x12, and f21y = -k / r123 * y12. Except for that pesky r123, this looks like the force which resulted in SHM. Since r is always positive, being the magnitude of a vector, r3 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 two-body 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 / r2 = 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 anti-derivative. 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 anti-derivative 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 * R3 * 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 / r2 = G * 4/3 * p * R3 * r * m / r2. 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 *R3 to 4/3* p * r *r3. So f = G * 4/3 * p * r3 / r2 * 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 make-believe two-body 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 v2. 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 * vm2 - G*M*m / r. Notice that E may be positive, negative or zero depending on the velocity vm. 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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