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.

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, f_21x = -k / r₁₂³ * x₁₂, and f_21y = -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.

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.

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 * v_m² - 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?