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CHAPTER SIX

"'ROUND AN' AROUND AN' AROUND AN' AROUND!": Why Physics Turns Your Stomach

Circular Motion and Gravity

In the early days of the space program, we in the good ol' U.S.A put our astronauts on skyrockets, otherwise known as military missiles, and shot them off into space. We were forced to do this because the Soviet Union was launching cosmonauts quite regularly, and our civilian space program was pretty much a disaster. Now, what do you suppose a military missle is designed to do? It is, of course, supposed to get to its target as quickly as possible. This means the missile accelerates at a very high rate in order to attain a large speed in as short a time as possible. Newton's second law, from Chapter Two, says the net force acting on an object equals the object's mass times its acceleration,

(2.29) Fnet = ma.

So you see there has to be a large net force acting on an object of mass m in order for it to achieve a large acceleration. Because of the large accelerations involved, and to better relate acceleration to human sensation, jet fighter operations and the space program use "g"s to measure the acceleration and, through the proportionality between acceleration and force expressed by Eq. (2.29), the force experienced by the pilots and astronauts. One "g" is the acceleration of gravity at the Earth's surface (which, in this book, I take as 9.8 m/s2 or 32 ft/s2).

The force you experience as 1 g is the support force (due to, for example, the floor you are standing on) that holds you up against the force of gravity, that is, your weight. Should you be so unfortunate as to lose this 1 g support force, for example falling off the roof while adjusting the satellite dish, you will experience, briefly, the effects of 0 g: no support force. If you were lifting off the surface of the Earth in a rocket and experiencing 2 g's, you would feel as if you weighed twice as much as you normally do, just standing around in the kitchen waiting for dinner. To give you the sensation of 2 g's, you would need a support force, provided by your flight seat, that is twice your weight. Since you get 1 g by just hanging out with the usual 9.8 m/s2 acceleration of gravity that is always there, you must accelerate upward at 9.8 m/s2 to experience 2 g's. An excess, or net, force equal to your weight must therefore act on you to give you this upward acceleration. The net force acting on you is

Fnet = 2mg (upwards, due to your flight seat) - mg (downwards, your weight) = mg (net, upwards),

so that,

a = mg / m = g.

The 2mg force of the seat on your body gives you the sensation of "2 g's".

The facility of the "g" terminology is that it refers to the force you experience in multiples of your weight. We all regularly experience our own weight, a force of 1 g. If you were subjected to a force of 2 g's, as noted above, you would feel as if your weight had doubled. The large g's generated by the acceleration of a military missile may not be a problem for a nuclear warhead but was definitely a problem for the early Mercury and Gemini astronauts. (The first two Mercury Astronauts, Alan Shepard and Gus Grissom, had to endure forces up to 11 g's due to the use of the Redstone ballistic missile as a booster. This meant their flight seats exerted a force on them 11 times their actual weight! An astronaut weighing 150 pounds would feel as if he weighed 1650 pounds!! By Newton's second law, the astronauts experienced a net force 10 times their weight, accelerating them at the enormous rate of 98 meters per second every second. This is close to increasing your speed by one tenth of a kilometer per second every second or over 200 miles per hour every second! (I'd like to see somebody's 'Vette do that!) The subsequent Mercury flights used the Atlas booster, so that the astronauts on the following flights experienced forces of "only" 7.5-8 g's.) How did the astronauts prepare for such tremendous "g forces"? The medical teams spun them around in centrifuges, that's how.

(6.1) "Pat, I'd Like to Buy an ... Eeeeeeeee!!": Centripetal Acceleration

The Wheel of Fortune is technically pretty lame when compared to some of centrifuges used nowadays to train jet fighter pilots. Anyone who has ridden a merry-go-round or one of those carnival rides that swing you around in circles knows that going in a circle results in the experience of forces that grow stronger as the circular motion gets faster. Training facilities for jet pilots use this effect to simulate the extreme forces of jet fighter flight, and prospective pilots are often required to take what is sometimes called "cockpit physics", where they learn, among other things, the origin of the g forces they experience in training centrifuges.

To understand why a centrifuge can produce large forces on the object being swung around, you have to go back to the definition of acceleration: Acceleration is the time rate of change of velocity. This definition is expressed by the following equation,

(6.1) a = dv/dt.

For you non-technical students, this equation, in English, states that the acceleration, a, (which is a vector; note the boldface) is the time rate of change, d/dt, of the velocity, v, (also a vector). Recall that a vector has both magnitude (how big it is) and direction (which way it is pointing). Consider a pilot traveling in a circle with constant speed in the gondola of a centrifuge. Although his speed, v = |v|, is not changing, nevertheless his motion, that is, his velocity, is changing because he is continually changing direction. By the definition of Eq. (6.1), he must be in a state of acceleration, despite his constant speed. Since velocity consists of both speed and direction of motion, if either one (or both) change for an object, the object must experience an acceleration. See Fig. 6.1, which is supposed to depict a pilot in the gondola of a centrifuge. (An artist I'm not.)

With reference to Fig. 6.1, note first of all that the speed of the pilot is constant in this example as he (or she) circles in the centrifuge. This is depicted in the figure by the "v"s, all of equal length (same speed), drawn tangent to the circle. By the first law of motion, the pilot (and the gondola in which he resides) would like to continue in a straight line at a constant speed due to his inertia (momentum). However, the centrifuge continually causes the gondola to change direction and must exert a force on the gondola to do so. This force produces acceleration expressed as the change in the direction of motion of the gondola.

Now direct your attention to the detail in the figure, where the velocities of the gondola at positions A and B. are depicted. Position B is just a bit farther along the circle than position A. When the gondola reaches position B, its velocity is directed at an angle θ ("theta") from what it had at position A. This is the same angle as that between the radii extending from the center of the circle to points A and B. (This is because both vA and vB are perpendicular to their respective radii, a relationship that requires the angle between vA and vB be equal to that between the two radii.) If you recall the tail-to-tip method of adding vectors, you will see that vA plus Δv equals vB, which means that Δv is the difference between vA and vB.

Now look more closely at the vector Δv = vB - vA, the change in the velocity of the gondola due to the change in the direction of motion. It is nearly perpendicular to both vA and vB, because θ is a small angle. That means Δv must point roughly in the same direction as the radius of the circle, since both velocity vectors are perpendicular to that direction. Now recall that the acceleration is the change of the velocity (per unit time). That means the acceleration of the gondola, and the pilot, must point roughly in the same direction as Δv, also toward the center of the circle.

However, Δv does not correspond to the direction of the instantaneous acceleration, since points A and B are separated by a "finite" (nonzero) distance along the circle, and thus the gondola passes through the two points at different instants. But if you move the points closer together, Δv would point more toward the circle's center, and the two points would be closer to being at the same instant. How do you find the direction instantaneous acceleration from this observation?

In calculus you talk about "limits", where you look at what a value approaches, even though the value can never actually reach that limit. In this case Δv points directly at the center "in the limit" where the two points merge. (Of course, since vA and vB are equal where the points merge, Δv goes to zero there. That's why you have to argue using the concept of the limit.) Since the acceleration points in the same direction as Δv, the acceleration also must point directly at the center of the circle in the limit. (The reason the acceleration doesn't go to zero is that it equals Δv divided by Δt, the time difference between when the gondola is at point A and point B. This quotient does not go to zero in the limit even though both Δv and Δt do. This quotient is, in fact, the instantaneous acceleration by definition.)

The above argument leads to the following conclusion. The instantaneous acceleration of a body moving at a constant speed in circular motion is directed toward the center of the circle of motion. This type of motion is called uniform circular motion, and the acceleration associated with this motion, a very important concept in physics, is called centripetal acceleration.

Techs-Mechs (Mechanics for Techies): Fascinating Mathematical Detail of Centripetal Acceleration
What follows is some mathemagical slight of hand that is not essential, unless you are a physics major or there is any possibility you will enter engineering graduate school. Otherwise, all you need to worry about is the result, which will be the formula for calculating the magnitude of the centripetal force, something pilots and designers of fighter jets are very interested in. Please refer to Fig. 6.2.

Note that two triangles, adapted from Fig. 6.1, are depicted here. One involves the change in position from point A to point B, and the other the change in the velocity between points A and B. Both triangles have two equal sides (same radius in the former and same speed in the latter) and the same angle θ. Hence the two triangles are "similar" (that is, they have the same shape, but not necessarily the same size). If the triangles are very slender, that is, θ is a very teeny angle, the side opposite θ, Δr in the first triangle, will be very nearly the same length as the arc, Δs, subtended by (sliced off the circle by) the angle θ. (I use the symbol, Δr, here, rather than the vector symbol Δr, because Δr is the magnitude of the change of the radius vector from point A to point B.) The angle θ in radians is equal to Δs/r for this triangle and, as a result of θ being a small angle, is very nearly equal to Δr/r. Since Δr/r must equal Δv/v (due to the triangles being similar), θ = Δs/r ≅ Δv/v. This result is called the small angle formula. Multiply both sides of the previous algebraic expression by v and get

(6.2) v(Δv / v) ≅ v(Δs / r)  →  Δv ≅ (v / r)Δs.

The magnitude of the average centripetal acceleration, by definition, must be the magnitude of the velocity change, Δv, divided by the time, Δt, over which the velocity changed. Dividing both sides of Eq. (6.2) by Δt results in

(6.3) ac ≅ ac,av = Δv / Δt ≅ (v / r)(Δs / Δt),

where ac, the instantaneous centripetal acceleration, is approximately equal to the average centripetal acceleration if Δt is not too large. As θ gets teenier and teenier, this relationship gets closer and closer to an exact equation. The techies in the audience will recognize two derivatives here. There is the rate of change of the direction of the velocity with respect to time, dv/dt, and the rate of change of the distance of the gondola along the curve with respect to time, ds/dt. (dv/dt is the rate of change of the direction of the velocity, because Δv is the length of the vector that results from the velocity's change of direction.) ds/dt is the speed of the gondola (speed = the rate at which distance, ds, is covered with respect to time, dt) = v. Therefore, finally you get

(6.4) ac = (v / r)(ds/dt) = (v / r)v = v2 / r,

for the magnitude of the centripetal ("center-seeking") acceleration.

For you non-techies that skipped down to this point, what has been proved is that the magnitude of the centripetal acceleration is given by Eq. (6.4) above.

(6.4) ac = v2 / r,

The direction of the centripetal acceleration, as pointed out previously, is directed straight at the center of the circle of motion. For uniform circular motion, the speed remains constant, and therefore, according to Eq. (6.4), the magnitude of the acceleration remains constant. However, the direction of the acceleration must constantly change in order that it always point toward the center of the circle. See Fig. 6.3.

Also note that the acceleration increases as the square of the speed. Therefore, if the pilot doubles her speed in the centrifuge, the g's she experiences will increase four-fold.

Apply Eq. (6.4) to find out how fast a modern centrifuge would have to turn in order to produce the 11 g's experienced by Alan Shepard. The AMST centrifuge made in Germany, for example, has an eight-meter arm. Although it "only" produces 6 g's of centripetal acceleration, you can compute how fast it would be turning to get up to 11 g's. (This centrifuge is capable of higher g forces for non-centripetal acceleration, for example, forces involved in speeding up and slowing down. The centripetal force depends only on the speed through the expression v2/r, not the rate of change of the speed.)

The force acting to accelerate the pilot is proximally attributable to the force of the gondola seat on the pilot. This force is called the centripetal force, because it is the cause of the centripetal acceleration. The equation for the centripetal force is,

(6.5) Fc = mac = mv2 / r,

which is just Newton's second law of motion with the magnitude of the centripetal acceleration expressed explicitly.

You should be aware that there are many types of forces that can give rise to centripetal acceleration. Since a centripetal force is the cause of an acceleration (centripetal acceleration) it must be a net force. Therefore, the centripetal force may be a single force or the vector sum of a number of forces that combine to produce circular motion. The centripetal force we are currently talking about is simply the force the surface of the pilot's seat exerts on the pilot, and therefore is a "normal" force. (If the pilot is seated vertically, the force will be exerted mainly by the back of the seat and the leg supports.)

Here are some other examples. For a satellite in circular orbit above the Earth, the centripetal force is the force of gravity. For a rock being twirled in a sling, the force is, once again, a normal force, due to the contact between the rock and the leather pouch. This normal force is derived from the tension in the sling, which, in turn, is a result of the force exerted by the hand and arm of the thrower. The centripetal force involved in a child's swing is the sum of the tension in the rope, acting toward the point of rotation of the swing, and that component of the childs's weight that is directly opposite the tension. Hence, in this case, the centripetal force is the vector sum of two opposing forces.

An acceleration of 11 g's means the pilot must be experiencing a force 11 times his weight, or 11×mg. In the case of the centrifuge, this force would be a centripetal force, exerted by the seat. By Eq. (6.5) this means

11 mg = mv2 / r.

The "m"s cancel and r = 8 m. Therefore, solving for v,

v = [(r)(11)g]1/2 = [(8 m)(11)(9.8 m/s2)]1/2 = 29.4 m/s (= 66 mph).

This result may not be all that informative to you. You might get a better feel for the situation by knowing how many revolutions this is per minute or per second. In physics we call this quantity angular velocity. Finding an equation to compute its magnitude is not too difficult. First, use the fact that the speed of the gondola, if constant, is the circular distance traveled, s, divided by the time it took to travel that distance, t. That is, v = s / t. Now use the fact that the number of revolutions that are made over some time interval is equal to the total distance traveled by the gondola divided by the distance traveled in one revolution, that is, the circumference. The circumference is equal to 2πr, so the number of revolutions is N = s / (2πr). Turning this around, s = 2πrN. Plugging this expression for s in v = s / t and solving for N / t (the number of revolutions per unit time) gives

(6.6) N / t = v / (2πr).

If, in this equation, v is in m/s and r is in m, then N / t will have the units of revolutions per second. Apply this to the AMST centrifuge.

N / t = (29.4 m/s) / [2π(8 m)] = 0.584 rev/s,

which is more than one revolution every two seconds. If you don't think this is very fast, recall that the gondola is out on an 8-m arm, that is, nearly 9 yards from the center of rotation. Also recall that the centrifuge does not attain this level of g force.

Revolutions per second are quite meaningful both to physicists and ordinary, benighted, people. More meaningful in calculations, however, are the units of radians per second, because the radian is the natural unit of angular measurement. Both the second and the radian are SI units. Since there are 2π radians in one revolution, the angular velocity, from now on called ω ("omega"), is found from Eq. (6.6) as

(6.7) ω [= no. of radians / s = (2π radians / revolution)(no. of revolutions / s)] = (2π)(N / t) = (2π)(v / 2πr) = v / r.

(2 π radians per revolution times the number of revolutions, N, equals the number of radians.) This equation only applies to the angular velocity given in radians per second. If you want to convert to degrees per second, for example, you need to multiply the right-hand side of Eq. (6.7) by 180° / π.

(6.2) "You Got To Learn How To Fall Before You Learn To Fly": Orbital Motion

One of the goals of the Mercury Program was to put a man in orbit. The first U. S. astronaut to do this was, of course, John Glenn. Mr. Glenn was above the Earth going around in circles (actually about two of them), so certainly he was experiencing g forces as if he were in some gigantic centrifuge, right? Naturally, I wouldn't even ask a question in this way unless the answer were, "Wrong!!" Well, he did experience a g force, a force of zero g. How can this be happening if he is in circular motion? Isn't his body experiencing a centripetal acceleration? "Yes," I reply, appearing distracted. Then he must have a centripetal force acting on him, right? "(Yawn) Yes, again." But a centripetal force has to be a net force, and if a net force is acting on Mr. Glenn, how come you say he is experiencing zero g? "Because it is and he is." (I slouch in my chair, put on my very best bored expression, and stare out the window.) OK, you pompous physics professor! Why don't you just explain this matter and quit this "I know something you don't know" BS? "If you insist," I mutter with disgust, slowly rousing myself to an upright position.

Many people think that astronauts are weightless in orbit (experiencing zero g) because there is no gravity up there. However, it is the Earth's gravity that is keeping them in orbit. If somehow a skyscraper could be erected to, say, the operating altitude of the space shuttle, the people on the top floor would not be weightless. They would weigh less than people on the surface of the Earth (about 8% less, depending on the exact height of the building) but not zero. You have to recall the idea of apparent weight, which is the weight registered by a scale. If you are floating around in the shuttle, your apparent weight is zero. This is what is meant by zero g, loosely referred to as weightlessness. This does not mean, however, that no force is acting on you. You still have a weight, but that weight is not being supported. Imagine the following scenario.

You have built the world's tallest structure, which extends beyond the troposphere, beyond the stratosphere, beyond the mesosphere, and into the reaches of near-Earth space. You decide to go to the top of your structure and put yourself in orbit. So you get on the roof and jump off. Unfortunately, all that happens to you is you descend at an ever increasing speed into the atmosphere and burn up like a meteor. Fortunately, you have been genetically engineered with a gene from Crash Bandicoot such that you are instantly regenerated for another go.

You ponder what went wrong and decide that perhaps you should take a running jump off the roof. At first it appears as if this might work, but, sadly, you still end up plummeting downward, this time at some distance from the building, and again flame out. On your third try you drive off the roof in a sports car. You go farther but end up as a spectacular fireball. (By this time word has gotten around that there is this awesome meteor shower, and people have gathered from far and wide around the base of the building, looking up expectantly.)

You are now down to your last reincarnation. You go over what you have done so far. "When I just dropped off the building, I went straight down. When I ran off the building, I made some lateral progress but still ended up curving toward the Earth. Finally, when I drove my sports car off the building, I made even more lateral progress, but I nevertheless could not avoid plummeting toward the ground." Finally, it hits you. "If I were to leave the roof at a fast enough speed, my curved path would parallel the curvature of the Earth. Then, I would never get any closer to the surface. I would be in orbit!"

You strap a booster rocket on your back and confidently blast off into the final frontier. With just the right amount of thrust, maintained for just the right amount of time, you achieve what you dreamed of: a curved path that mirrors that of the Earth below. You are falling, just like the other times when your path was curved too steeply and you became a spectacle. But now, you never reach the surface of the Earth, because you maintain a constant altitude above the Earth's surface. You are in orbit! You are exhilirated! You are ecstatic! You are, after one complete orbit, about to hit the other side of your building!

(Yes I know!. And, of course, you are not really going fast enough falling off the top of your building to make a fireball.)

Fig. 6.4 was originally devised by Isaac Newton to explain how orbit is achieved, except he used a boring old cannon shooting a cannonball. An ordinary cannon, fired horizontally, would shoot a cannonball that would strike the ground at some distance from where it was fired, because its curved path takes it into the Earth. Newton reasoned that if you were to fire the cannonball at faster and faster speeds, eventually the curvature of the cannonball's path would match that of the Earth below. Assuming a perfectly spherical Earth (and no air to slow the cannonball down), the cannonball should be able to circle the Earth. It would have achieved orbit. It would also be traveling in uniform circular motion, since its speed would remain constant: no thrust to speed it up, no air to slow it down.

There is no essential difference between the cannonball that is shot at a relatively small speed and one that is shot with sufficient speed to put it in orbit, except one hits the ground and the other doesn't. Both are falling. However, the orbiting cannonball falls just as fast as the Earth's surface curves away, so it never strikes the ground. When something falls, it is unsupported and its apparent weight (scale weight) is zero. This is why the astronauts are "weightless". Actually, they do have weight, but, since it is unsupported (they are falling), their apparent weight is zero, and we say they are weightless, or experiencing zero g: zero support force.

In Chapter Two the weight of an object was defined as the force of gravity on that object: mathematically, W = mg, where g is the acceleration of gravity. The acceleration of gravity is about 9.8 m/s2 at mean sea level. At a near-Earth orbit altitude of 200 miles (about 320 km), the acceleration of gravity is close to 8.9 m/s2. Therefore, if you weigh 150 lb (about 670 N) at the surface of the Earth, you would weigh 136 lb on the shuttle at that altitude. You would feel weightless, however, because your weight is not supported. If you were suddenly placed in the shuttle and unable to see out a porthole, you couldn't tell whether you were in orbit or simply falling toward Earth. The sensation of unsupported weight, or weightlessness, would be the same either way.

(6.3) One Turn Deserves Another: How to Take Your Turn Using Physics

Leaning in the Right Direction: How to Turn on a Bicycle

Remember when you first learned how to ride a bicycle? You may have done something like this when you first tried to make a turn. What could be more natural than to simply turn the handlebars so that the wheel points in the direction you want to go? This maneuver probably resulted in you flying over the handlebars in obedience to the first law of motion. After that, the law of gravity directed your attention to the approaching pavement. Likely, you, poor child, ran to your mommy with a major owie. Later you learned that you had to lean into the turn in order to successfully change your direction and maintain your balance. But, wait a minute. Why would leaning be necessary? Isn't it true that if you get on your bicycle when it is stationary and then lean over you will topple? Why don't you topple when you lean into a turn while in motion? The answer lies in the same considerations that underlie the operation of a centrifuge.

What we learned earlier in this chapter was that you need a force, a centripetal force, to cause you to move in a circle. Merely turning the wheel of your bike does not create a centripetal force. However, if you lean into the turn while turning the wheel, you do execute a circular arc. There must be a centripetal force involved, but what is it? To get to the answer, study Fig. 6.5, which shows the forces acting on you as you make your turn. (Again, please hold the comments on the art work.)

Although your weight and the normal force are not directly involved in making you turn, recall that frictional force can not exist without two surfaces being pressed together, which is the province of the normal force. The weight of you plus your bike is transmitted, via internal stress, to the bottom of your tires, which press down on the pavement. By Newton's third law of motion, the pavement presses up on the tires. This pressing together of the tires and pavement through the normal forces makes the frictional force possible.

Around You Go in the Wild Blue Yonder: Turning an Airplane

I'll bet you weren't aware that an airplane turns pretty much like a bicycle. Once again, the most obvious way to turn, turning the rudder, won't get the job done any more than merely turning your bicycle's front wheel. If you just turn the plane's rudder, the aircraft will yaw (what you would call "fishtail" in an automobile) but won't turn effectively.

You need a centripetal force to turn the plane just like you need to turn your bicycle. Recall that the centripetal force needed to turn the bike was supplied by the force of static friction acting on the wheels. You had to lean the bike into the turn in order to invoke these forces. Similarly, you have to bank the aircraft to perform a turn. But there is no surface to supply friction, so what is the force that plays the role of the centripetal force for the plane? Fig. 6.6 supplies the answer.

The aircraft is kept in the air by the lift on the wings, L. When the plane makes a turn, it must bank into the turn so that the lift has a horizontal component, Lsinθ. It is this horizontal component that is responsible for the centripetal acceleration, so this is the centripetal force. Meanwhile, for a perfect turn, you want to maintain altitude. However, note that the vertical component of the lift is less than it was in level flight. That is, in level flight a balance between lift and weight would be L = mg. When the plane is banked, however, the force countering the weight is Lcosθ, which is smaller than L (cosθ < 1 if θ is nonzero). You must increase the lift in order to maintain altitude so that Lcosθ = mg. This can be accomplished by pulling back on the controls to increase your "angle of attack", which points the nose of the aircraft higher than it was before.

Let's see what we can learn about the g forces on an aircraft versus banking angle, θ. If the aircraft is in vertical equilibrium, then the vertical component of the lift must balance the weight, that is,

(6.8) Lcosθ = mg.

In the horizontal direction, perpendicular to the velocity of the plane, the horizontal component of the lift, Lsinθ, is responsible for the centripetal acceleration, ac.

(6.9) Lsinθ = mac = mv2 / r.

Dividing Eq. (6.9) by (6.8) you get

(6.10) (Lsinθ) / (Lcosθ) = (mac) / (mg) → tanθ = ac / g

The centripetal acceleration can therefore be expressed as

(6.11) ac = gtanθ.

As an example of how you might use Eq. (6.11), imagine you are turning your aircraft through a fairly steep angle, say 45°. Your centripetal acceleration will then be

ac = g tan(45°) = g.

You might think this means you experience a g-force of 1; however, this result does not take into account explicitly the force of your seat on you. This "support" force is the one that determines how many g's you experience. You need another free-body diagram (FBD), like the one for the plane in Fig. 6.6 to find what this force is (Fig. 6.7).

Taking a look at this figure, you see that the net force acting on you is the centripetal force of 1 g. This force is the vector sum of your weight,mg, plus the support force supplied by your seat, N. Since your weight is, by definition, 1 g, you need a force of about 1.41 g acting at a 45° angle to the horizontal added to this to get the 1 g horizontal centripetal force. (Apply the Pythagorean theorem:

[(1 g)2 + (1 g)2]1/2 = 1.41 g.)

Hence, you experience 1.41 g's, or a 41% increase in your weight, when making this turn.

Let's put some algebra to this and come up with a general result. The support force for the aircraft in Fig. 6.6 is the lift L. This same diagram applies to the pilot, except the support force is the normal force N rather than the lift. If you replace L with N, Eqs. (6.8)-(6.10) apply to the pilot, and therefore so does Eq. (6.11). Substituting the expression for ac in Eq. (6.11) for ac in Eq. (6.9), replacing L with N, you get

(6.12) N = mg / cosθ.

In terms of "g force", this is

(6.13) N / mg = 1 / cosθ.

You can see that, for the previous example (θ = 45 °) the g force on the pilot is 1 / cos45° = 1.41 g's, the same result as before. As the banking angle approaches 90°, cosθ approaches zero, and the g force gets "large without bound", that is, it approaches infinity. In a real aircraft this would not be the case, because my analysis has assumed that the vertical component of the lift balances the weight for any banking angle. At large banking angles this would require physically impossible magnitudes for the lift. Therefore, at large banking angles the aircraft will invariably lose altitude as it loses vertical lift sufficient to balance its weight. High performance aircraft, however, can generate enough thrust that, with their noses pointed above the horizontal, the thrust can make up for what the lift can't accomplish.

Take It to the Bank: How an Automobile Turns

My final example in the fine art of turning has to do with America's favorite machine, the automobile. Recall that you had to lean into the curve on your bicycle in order to make a turn. Well, it is a little hard to lean into a curve in your car. In fact, as you well know, you don't lean into a curve in an automobile. You simply turn the wheel, something that does not work for the bicycle or the airplane. So, what's the diff? Actually, there are strong similarities. Friction is almost always involved, just as for the bicycle and, similar to the airplane, a banked curve is much more amenable to turning than a horizontal one. Nevertheless, there is the fact that you can turn on a level surface by just turning the wheel. Fig. 6.8 illustrates the situation.

The difference between the bicycle and the automobile is the fact that the automobile has four tires arranged in a rectangular pattern. When the front tires are turned, the car tries to go straight, according to the first law of motion, but the friction between the tires and the road attempts to change the car's direction. (The front tires are no longer pointing ahead and would slide over the road were it not for the friction between the tires and the road acting to prevent a skid.) Because the car doesn't lean into its turn, and the center of mass tries to continue in a straight line, the friction on the tires acting toward the center of the turn produces an effort to rotate the car, called a torque, indicated by the curved arrow in Fig. 6.8. Torque, the "turning force", will be covered in a future chapter.

This torque, if great enough, would cause the car to roll over in the direction away from the center of the turn. (If you are not driving a Ford Explorer with Firestone tires, however, your vehicle will probably just lean toward the outside of the curve, increasing the normal force on the tires on the outside of the curve.) This is indicated in Fig. 6.8 by the arrows for the normal forces on the outer tires (No) being longer than those arrows on the inner tires (Ni). These details aside, there is nevertheless a net force, Fi + Fo = the sum of the frictional forces on the inner and outer tires, acting perpendicular to the direction of the motion of the car. This friction constitutes a centripetal force that causes the car to turn.

The car does not topple over like the bike or yaw like the airplane for the following reasons. Unlike the bike, the car experiences a counter torque to that tending to roll it over due to the normal forces on the outer tires. The airplane does not have the friction between the tires and the road to help prevent yaw (known as skidding for an automobile).

A close examination of Fig. 6.8 will indicate that it is the normal forces on the outside tires that prevent the car from rolling by producing a torque that counters that produced by friction. (The curved arrow only indicates the direction of the latter torque.) In the event the act of turning is severe, two different possibilities arise, neither of which is wholly desirable. The more benign possibility is that the forces of static friction will reach their maximum value and the car will begin to skid. A more dangerous eventuality is that the clockwise torque (Fig. 6.8) due to the friction will win over the counterclockwise torque due to the outside normal forces. In this case the car will roll. To avoid these unpleasantries, engineers design banks for high-speed turns to use the normal forces to help create the centripetal force of the turn. This is just what happens for turning airplanes. Fig. 6.9 shows the forces on a car in a banked turn with a banking angle θ.

Note from the free-body diagram that there are three forces involved: the weight, mg; the normal force, N; and static friction, Fs. Since I am treating the car as if it were a point, I show the forces acting on the car's center of mass. (This "pointilist" approach ignores any torques acting on the car.) Also ignored are any forces due to the motion of the car through the air (which should be minor at moderate speeds). Since the car is turning at the moment pictured, the frictional forces (a force on each tire, adding up to the total frictional force, Fs) are in the direction shown, keeping the car from sliding. However, it is possible to design the banking angle, θ, such that the frictional forces are nearly zero for a car turning at a certain speed. More about this later.

Often, when the forces acting on an object on an inclined plane are analyzed, you resolve the components of the force vectors parallel and perpendicular to the plane. For example, take a block sliding down an inclined plane and picking up speed. The acceleration is parallel to the plane; therefore the net force is also parallel to the plane. In order to use the second law of motion (Fnet = ma), you would like to resolve the forces parallel to the plane to get the (non-zero) net force, and perpendicular to the plane, because the net force in that direction is zero (the forces in that direction are in balance). However, in the present situation, the acceleration is not parallel to the plane. The car is not sliding down (or up) the plane but is turning in a horizontal circle, which means there is a centripetal force - the net force - acting horizontally toward the center of the circle. It turns out it is much better in this case to resolve the forces in the horizontal and vertical directions, because the net force is acting horizontally and forces are balanced in the vertical direction. The coordinate system used is, after all, your choice. The analysis could be done resloving the forces along any pair of perpendicular axes, but the most convenient pair is horizontal and vertical.

The diagram on the right of Figure 6.9 shows the forces resolved into their horizontal (x) and vertical (y) vector components. Only the weight, mg, is acting along an axis, directed down the negative y axis. The normal force is resolved into a component, N sinθ, in the negative x direction and a component, N cosθ, in the positive y direction. Finally, the force of static friction, Fs, divides into two components, Fscosθ in the negative x direction and Fssinθ in the negative y direction. (Recall that frictional forces always act parallel to the surfaces involved, so in this case the friction is acting down the plane.) The car will be in equilibrium in the vertical (y) direction so that

(6.14) N cosθ - mg - Fssinθ = 0.

There is net force in the negative x direction, however, the centripetal force which causes the centripetal acceleration and thus the turning. It's F = ma equation is

(6.15) N sinθ + Fscosθ = mv2 / r,

where the formula for centripetal acceleration is used.

One final point about Eq. (6.15) before we apply it in the next section. Figure 6.9 is a snapshot of the car and the forces acting on it at some instant in time. Since the car is turning, the forces shown will have changed direction when the next snapshot is taken. However, as long as the car maintains the same speed and turns at the same rate, the banking angle remains the same, and the properties of the road surface don't change (a wet spot, for example), the magnitudes of the forces will remain the same and they will always be directed toward the center of the same circle. Hence, Eq. (6.15) will hold for every position of the car along the curve it is taking.

Highway Robbery: Designing a Properly Banked Curve

A couple of years ago, I posed a problem to my freshman engineering students that had to do with deciding if a designed banking angle was suitable for a certain highway turn. The students were charged with analyzing a statistical study to see if the safety of the proposed curve was acceptable. The projected traffic on the curve was something like 100 cars per minute, and the probability that a car entering the curve would have a fatal accident, given the traffic speed and curve design parameters, was something like 0.1%. More than a few students decided that this was quite acceptable, not stopping to think that, at 100 cars per minute, there would be a fatal accident every 10 minutes on the curve.

I was quite impressed with this and recommended them for jobs with the state highway department, where cool heads and deliberate inaction are highly sought after. Highway department people are not the type to panic and rush to redesign a stretch of highway just because a few hundred people have lost their lives on it. (In my resident town, Texarkana, this would be the U. S. 71-I 30 interchange.)

"Officials say I-20 safe despite rash of fatal accidents." - Associated Press, Pecos, Texas, July 28, 2002. (disclaimer)

Imagine that you are a highway department engineer assigned to investigate this curve that has a fatal accident every 10 minutes. This is not an easy task. In addition to the average of one fatal accident every ten minutes, there are usually several non-fatal accidents and fender-benders every minute. Trying to survey and take data amidst piles of wreckage, blaring sirens, an armada of tow trucks, and occasional corpses overlooked by the ambulances, is a stressful experience. Nevertheless, you pause from time to time to ponder the marvel of engineering that has made all this possible. It seems a shame, but the curve must be redesigned because of tremendous public criticism.

Your survey shows that, in lulls between horrendous carnage, the average car negotiates the curve at 70 mph. The aim is to make the banking angle, θ, just the right number of degrees such that a car traveling 70 mph experiences no frictional force on its tires. When this is the case, only the horizontal component of the normal force will act as the centripetal force. The surface of the road could be black ice and still you would turn safely. (Or not, since not everything is being taken into account in this analysis.) Since Eqs. (6.14) and (6.15) describe, mathematically, the action of the forces, you set the frictional force to zero (Fs = 0) and solve for the angle that will satisfy this condition. Equations

(6.14) N cosθ - mg - Fssinθ = 0,

and

(6.15) N sinθ + Fscosθ = mv2 / r,

become

(6.16) N cosθ = mg,

and

(6.17) N sinθ = m v2 / r.

You divide Eq. (6.17) by (6.16) to eliminate N and get

(6.18) tanθ = v2 / gr    (= ac / g).

Note that the mass cancels out as well as the normal force, N. The radius of the curve is 500 m and v = 31.3 m/s (70 mph). Therefore, the banking angle must be θ = tan-1{31.32/[(9.8)(500)]} = 11.3°. Unfortunately, there is insufficient funds to reconstruct the curve to this banking angle, because the Federal government has given those firms constructing the highways tax breaks, which has reduced Federal income, leading to cuts in highway construction money. The only obvious solution is to put out a whole bunch of orange barrels and reduce the speed limit to 25 mph.

Just as you are about to deploy the orange barrels, word comes that the funding has been restored. Luckily for you, you work for the state highway department of West Virginia, and Senator Robert Byrd, the undisputed king of pork-barrel spending (at least before the "bridge to nowhere"), has used his influence to not only rebuild the curve, but also to construct ten interstate highways across West Virginia and build the world's first inland naval base. "The sea-based attack on the U.S.S. Cole has proven that we must protect our fleet by stationing it safely inland," said Senator Byrd at a national press conference.

Sadly, the funds were not forthcoming to dig a canal through Virginia that would have been necessary to restation the fleet. It seems as if the entire population of Virginia, upon hearing of the plan, threatened to defect to al-Qaeda, thus assuring defeat of the proposal. Nevertheless, millions of dollars to study the feasibility of the canal were appropriated, establishing a "Canal Research Institute" in Wheeling, while construction of the base itself was initiated in hopes that the canal would eventually be realized. To hedge his bets, Senator Byrd has also pushed through funding for a gigantic cargo plane designed to airlift ships. "If we can put a man on the Moon, surely we can airlift a battleship to West Virginia," declared the Senator.

With money in hand, you donate the orange barrels to Houston, which can always find places to put them, and set about your next task, finding what maximum speed the curve will allow before a car loses control due to skidding. The applicable equations are, once again, (6.14) and (6.15), which desribe the forces acting on a car rounding the curve. If the car is about to skid, the frictional force is given, according to the friction model elaborated in Chapter 3, by

(3.11) Fs,max = μs N,

where the maximum possible force attainable by static friction is proportional to the normal force, with the constant of proportionality given by the coefficient of static friction, μs, a number that must be found by experiment.

Equations

(6.14) N cosθ - mg - Fssinθ = 0,

and

(6.15) N sinθ + Fscosθ = mv2 / r,

become

(6.19) N cosθ - mg - μs N sinθ = 0,

and

(6.20) N sinθ + μs N cosθ = mv2 / r.

You do a study and find that, for most vehicles driving in dry weather, the coefficient of static friction, μs ≅ 0.7. You are interested in what the speed of the vehicle must be to satisfy the "about to skid" equations, (6.19) and (6.20). This requires you eliminate the unwanted unknown, N. First, from Eq. (6.19), you see that

(6.21) N = mg / (cosθ - μs sinθ)

You now substitute the expression for N in Eq. (6.21) into (6.20) to eliminate N. This gives you

(6.22) N(sinθ + μs cosθ) = m v2 / r  →  [mg / (cosθ - μs sinθ)](sinθ + μs cosθ) = m v2 / r.

The mass of the car, m, apparently cancels out, allowing you to solve for v as

(6.23) v = [rg(sinθ + μs cosθ) / (cosθ - μs sinθ)]1/2.

Your numbers for this equation are θ = 11.3°, μs = 0.7, and r = 500 m. These numbers give a speed of 71.6 m/s, or about 160 mph. So, for dry conditions, there is no danger of skidding, even at quite excessive speeds.

What about when the coefficient of static friction is reduced to, say, 0.2 due to a very slick roadway? Running the numbers again gives v = 45.2 m/s = 101 mph. You are confident that you have designed a curve that is very safe, even under adverse road conditions. You get the Senator Byrd Award for meritorious service to West Virginia and are appointed chief engineer in charge of dry dock facilities at the new naval base. As dry dock chief for an inland naval base, you can spend your time from now to retirement playing video games on your office computer.

(6.4) Cosmic Gyrations: A Primer in Orbital Mechanics

Let's return to your bad experience of achieving orbit by launching yourself from Newton Tower only to find that, after one orbit, you are apparently about to slam into the opposite side of the skyscraper. No you aren't! You forgot that the Earth is turning on its axis, so when you complete the first orbit, the building is no longer there! What a relief! By gazing at the stars as you complete orbit after orbit, you become aware that, as far as the distant stars are concerned, your orbit is not changing. You see the same stars in the same positions every orbit you make. It is the Earth below that is turning with respect to the stars as it rotates about its axis. This is an instance of the first law of motion, in that there is no force to change your orbit, so you maintain it. However, you are not going in a straight line, even though, being in circular orbit, your speed is constant. Why is this?

Given what you have learned earlier in this chapter and recalling Eq. (3.5) of Chapter Three,

(3.5) F = G (m1m2)/r2,

which gives the gravitational force of attraction between two masses, m1 and m2, separated by distance, r, you now have the means to find out.

First, you recall that an object (in this case, you) moving in a circle at a constant speed must have a centripetal force acting on it. Contrary to what many have surmised, seeing astronauts "floating" in their spaceships seemingly unaffected by any force (only because they are seen in a freely falling reference frame), this centripetal force is gravity. Gravity is not zero above the Earth. Instead, for an object in a circular orbit, it is playing the role of the centripetal force and therefore must equal your mass times the centripetal acceleration, that is, mv2/r, where v is your speed and r is the radius of your orbit.

Strictly speaking, r is the distance between your center of gravity and the common center of gravity between you and the Earth. Practically, it is the distance between you and the Earth's center of gravity, since the Earth is much larger than you are (unless you are Edward Kennedy). The center of gravity of the Earth, as you might expect, is pretty much its center.
Symbolizing the Earth's mass by M and your mass by m, the centripetal force equation is

(6.24) GMm / r2 = mv2 / r.

Note that the satellite mass (yours) appears on both sides of the equation and therefore divides out. Solving for the speed of the satellite, v, you get

(6.25) v = [GM / r]1/2.

Let's assume Newton Tower tops out at 200 km. With a mean Earth radius of 6370 km, that puts your r at 6570 km. The mass of the Earth is about 5.98 × 1024 kg. Substituting these numbers into Eq. (6.25), (remembering to convert km to m and that G = 6.673 × 10-11 N·m2/kg2) you are able to conjure up your speed, a breezy 7.79 km/s.

There are quite a few footnotes to this speed, however. This speed assumes a perfect spherically-symmetric Earth. In reality, the Earth is slightly pear-shaped, presumably because it's at middle age. Also, this ignores the effects of varying mass distribution, especially near the Earth's surface, where the proximity to your orbit gives the variations in mass more effect. (Recall that the influence of gravity falls off as one over the distance squared, meaning, e.g., a mass twice as close has an effect four times as large.)

On the real Earth there are mountain ranges and ocean trenches, the former often regions of excess mass on the surface and the latter regions of mass deficiency. Your speed and altitude will change by minute amounts as you cross these features. You will speed up and decrease altitude over mass excesses and slow down and increase altitude over mass deficits. This effect allows the mass distribution in the Earth to be mapped by tracking satellite motion and is even used to map mass distributions on other solar system objects. This is how the dense concentrations of mass ("mascons") were found beneath the lunar surface.

Even as I write these words a pair of satellites, the Gravity Recovery and Climate Experiment (GRACE), circle the Earth in polar orbits, one following the other at a separation of several hundred kilometers. The speeding up and slowing down mentioned just above is monitored by instruments that use microwaves to measure distance changes between the satellites (and serve a second purpose by providing passing astronauts a means to heat their burritos). When the lead satellite encounters a mass excess, it speeds up and the distance between the satellites increases. The opposite happens when the lead satellite encounters a mass deficit. Monitoring changes in the mass distribution on the Earth may help track such things as changes in water distribution on and near the Earth's surface.

Safely in orbit and now understanding the cosmic force that keeps you there, you can hardly take it all in. The beauty and excitement of orbiting the Earth, the thrill of learning about the mechanics of orbiting satellites, and the sheer exhiliration of soaring over the Earth at such a fantastic speed was well worth all your effort, wasn't it? (Should I tell you that as the Earth inevitably turns on its axis, your building will eventually reappear in your path?)

Reach for Your Highest Potential! (Even Though It's Zero.): Generalizing Gravitational Potential Energy

In the last chapter, you learned that, since gravity is a conservative force, depending only on position, you could associate a potential energy with it. This potential energy must be defined with respect to an arbitrary zero level. Establishing a vertical axis, y, and letting y = 0 be the zero level, the potential energy for gravity near the surface of the Earth, symbolized by U, is given by Eq. (5.22),

(5.22) U = mgy.

The key phrase here is "near the surface of the Earth". What about gravitational potential energy in "the final frontier"? Well, what you have to do here is to back up a few steps and recall how potential energy was defined in the first place. There are two equivalent definitions:

(1) The potential energy between two objects subject to a conservative force equals the work done against the conservative force in moving them from the arbitrary zero level to their current positions. (You actually only have to move one of the two objects depending on how you define the zero level. For example, you can define the zero level between the Earth and a book as where the book lies on top of a table. Then the work you do in lifting the book up against its weight equals the potential energy between the book and the Earth.)
(2) The potential energy between two objects equals the negative of the work done by the conservative force when they are moved from the arbitrary zero level to their current positions. (In the above example, gravity does negative work when you lift the book - the book's weight is directed downward while the motion is upward - and the negative of this negative work equals the positive work done by your lifting and therefore equals the increase in the potential energy.)
The latter definition is generally more useful.

All right! All you have to do to find the gravitational potential energy of a satellite, for example, is (1) choose a zero point and (2) calculate the work gravity does on the satellite when moving it from that point to wherever you want to place it. Taking the negative of that work gives you the gravitational potential energy. Slight problem. The force of gravity is given by Eq. (3.5),

(3.5) F = G (m1m2)/r2,

which is not constant due to its dependence on r.

Recall that, when finding the gravitational potential energy near the surface of the Earth, you just set r = the radius of the Earth in the Eq. (3.5) and treat the force as constant, ignoring minor changes due to ups and downs near the Earth's surface. When you have a constant force doing work, you can use the "force parallel to the motion times distance" equation, Eq. (5.2) where the component of the force parallel to the motion is given by Eq. (5.3).
The calculation of the work done by gravity in the case where the force is not constant due to the separation r varying significantly is a job for Techs-Mechs (see below). The result, however, is quite simple, given by

(6.26) U = -G m1m2 / r + U,

where U is the gravitational potential energy at infinite separation of m1 and m2. Obviously it is easiest to choose U = 0 at r = ∞, because then U = 0. This is the almost universal choice for the zero level, making the gravitational potential energy equal to

(6.27) U = -G m1m2 / r.

There is a price to pay, unfortunately, by taking the zero point at infinity and thereby getting rid of the constant in Eq. (6.26): The potential energy becomes negative everywhere. It's largest value, zero, occurs when the two masses are separated by an infinite distance. (Practically, speaking, this is just a very large distance.) As the masses are placed closer and closer to each other, the potential energy grows more negative. This often causes students problems. To say the potential energy is zero means, to many of them, there is no potential energy. However, that's like saying if the temperature is zero celsius, there is no temperature. In fact, zero potential energy is the largest amount you can have with the usual choice of zero level. All other potential energies are negative and therefore smaller.

Techs-Mechs
Here is how you arrive at Eq. (6.27). First, retrive Eq. (5.40) from the last chapter.

(5.40) W = dW = F•ds.

This equation indicates the process necessary to calculate the work done along a particular path by a variable force. Your program is to integrate this equation, starting when the objects are separated by an infinite distance to some final separation r. You start at infinity, because that is your chosen zero level, the point at which you take the gravitational potential energy to be zero. Note that "at infinity" means all directions. That is, if m1 is here, m2 can be at an infinite distance in any direction. So, you set m1 at the center of your coordinate system, at r = 0, and put m2 infinitely far away. Now you move m2 in toward m1 until they are at a separation, r, calculating the work done by gravity while this is done (Figure 6.10).

The following integration, Eq. (6.28), describes this calculation. Note that the generic infinitesimal path vector ds has been replaced by dr, since r is the variable for this calculation, and that the limits of integration have been emplaced. For now, I am pretending that we haven't yet chosen infinity as the zero level, so I am setting W as the lower limit, assuming we start out calculating the work done from infinity to r with some work, W, already being done to bring m2 from the zero point to infinity.

(6.28)  W W dW =  r F•dr.

The next item on the agenda is to change F•dr into a form amenable to integration. Recalling the definition of the dot product, F•dr = Fcosθdr, where θ is the angle between F and dr. Since the vector r points away from m1 and F points toward m1, θ is 180° and cosθ = -1. Therefore,

(6.29) F•dr = -Fdr = -G(m1m2)/r2dr,

where the expression from Eq. (3.5) has been substituted in for F. The integral (6.28) becomes

(6.30)  W W dW = -G m1m2  r r-2dr.

This looks like something most any techie should be able to integrate, recalling the polynomial integration formula

xndx = xn+1 / n+1.

The result is

(6.31) W - W = -G m1m2  [r-1 /(-1)]r  = G m1m2 / r.

Eq. (6.31) shows the work of gravity is positive, which it must be, since m2 is moving in the same direction as the force. By definition, the potential energy is the negative of this work. Therefore, setting U = -W and U = -W, you get

(6.32) U - U = -Gm1m2 / r.

This equation is the same as (6.26). Eq. (6.27) follows by setting the zero level at infinity; that is, setting U = 0.

Man, Hohmann, Hohmann!: Energy and Orbits

In this subsection you are going to learn (please do! I need the personal validation!) about the different types of possible orbits. Energy-wise, there are three types. In what follows I narrow the focus to the case where one body (the small body) is very much less massive than the other (the large body).

• Bound Orbits: A satellite in a bound orbit does not have enough energy to escape the object it is orbiting. It moves around the large body in an oval-shaped curve called an ellipse. (A circle is a special form of an ellipse.) If you take the gravitational potential energy between the satellite and the large body to be zero at infinity, the energy associated with a bound satellite is less than zero. (It has positive kinetic energy, but the potential energy is negative with a larger magnitude.)

• Unbound Orbits: An object in an unbound orbit follows a path called a hyperbola, which is an open, rather than closed, curve. Unless something happens to rob the object of some energy, it will move away from the large body and will gradually slow down to something close to a constant speed. (Its speed would actually continue to decrease but at a slower and slower rate. If it were possible for the body to move to infinity, it would have a constant speed.) At infinity, the potential energy is zero, but the non-zero speed means the object has kinetic energy. Hence, its energy is positive, and this is the requirment for a body to be unbound: its total energy must be greater than zero.

• A Barely Unbound Orbit: This is more of mathematical than practical interest. If the total energy associated with the smaller body is zero, then, mathematically, it would leave the larger body along an open path called a parabola, slowing down gradually toward zero speed (but never getting there, because it will never reach infinity). At the zero level of infinite separation with zero speed, the total energy associated with this orbit is zero.

Time to get serious about orbital dynamics. You, a leading astronaut and space hero, have been launched into low Earth orbit as part of a desparate rescue effort. A female astronaut piloting an orbital shuttle, has accidentally run into a homeless alien. The creature is now stuck on her viewing port. When she tries to pilot her craft, all she can see are these suckers that hold the creature's appendages to the window, as if it were some sort of bizarre Garfield auto decoration. "Sheesh", she thinks to herself, "back home in Texas all I had to worry about were bugs on the windshield." (Please note that this is pure fantasy. For one thing, suckers don't work in space due to the lack of atmospheric pressure. They work as post docs instead.)

Your problem is to reach the higher orbit of the female astronaut. Your orbit is only 500 km above the Earth's surface, while hers is at a 1000 km altitude. Both orbits are circular. How can you change your orbit to match hers? The technique favored by Hollywood space movies, that of zooming over as if you were piloting an F-16, doesn't work in orbit, as you would quickly find out if you tried. You would have to have a rocket thrust significantly greater than the weight of your spacecraft to even attempt this. Fortunately, you are well-trained for this mission and realize that the best thing to do is to execute a maneuver called the "Hohmann transfer".

The Hohmann transfer will get you to the higher circular orbit by two steps.

1. Change to an elliptical orbit where your perigee (point of your orbit nearest to the Earth) is the same as the altitude of your present circular orbit and your apogee (point farthest from Earth) is the same as the altitude of female astronaut's orbit.

2. Change this elliptical orbit into a circular orbit with an altitude equal to the apogee of the elliptical orbit. This orbit will be the same as that of the astronette in distress.

The Hohmann transfer is based on two conservation principles, one old and one new:
• Conservation of energy.

The energy of an object of mass m in orbit about a much larger object of mass M is given by the kinetic energy of the small object plus the gravitational potential energy of the interaction between the two objects. If the large object is sufficiently huge, you can take it to be stationary with zero kinetic energy (ignoring the common speed the objects share, such as the Earth-Moon system orbiting the Sun). Actually, it will move a little due to the pull of the smaller object during its trip around the larger one, but this is so minute you can ignore it. At a given time the satellite has a speed v and is a distance r from the center of the larger object (actually from the common center of gravity, which is essentially the same thing for very small objects orbiting large ones). Thus the energy of the system of the two bodies is very close to, using Eq. (6.27),

(6.33) E = K + U = (1/2)mv2 - G Mm / r,

where M is the mass of the larger object (the Earth, say), m is the mass of the smaller object (e.g. a spacecraft in orbit), r is their separation, and v is the speed of the smaller object. Since gravity is a conservative force, the quantity E must remain the same for the entire orbit, and from orbit to orbit. This is true whether the orbit is circular or not. (Gravity's conservative nature does not depend on the orbit's shape.) If r changes, thus changing U, v must change to adjust K so that K + U remains constant.

• Conservation of angular momentum.

Here is a new concept for you - a concept to be developed more fully in a future chapter. At this point you only need to concern yourself with the angular momentum of a point particle, which is defined thusly (Figure 6.11).
Consider a point mass, m, traveling at a velocity, v. It therefore has a linear momentum of p = mv. The amount of angular momentum an object has depends on the position from which the object is viewed. If the object is not coming right at you or moving directly away, you will see its angular position change. The angular momentum of this mass, therefore, must be taken with respect to some point, for example, point O in the figure. Choose a different point, and the angular momentum of m will be, in general, different. Extend the direction of the velocity of m as shown in the figure, then draw a perpendicular, r, from point O to the line defined by the velocity. The magnitude of the angular momentum about O, L O, is given by the magnitude of the linear momentum times r, that is,

(6.34) L O = rmv.

Like linear momentum, angular momentum is a vector. It's direction is not given by that of v, however, but by a so-called "right-hand rule". Point the forefinger of your right hand in the direction of r (away from O) and your middle finger in the direction of v. Your thumb now points in the direction of the angular momentum (out of the page in the case of the figure).

It turns out that angular momentum of a particle is conserved about a point (such as O) so long as the forces acting on the particle are directed either at that point (attractive) or away from that point (repulsive). Since the gravitational force acting on a satellite is attractive toward the center of the larger body, the angular momentum of the satellite about that point is conserved (remains the same from one moment to the next).

Your original orbit is in a circle of ro = 500 km + 6370 km = 6870 km. The female astronaut's orbit is at r = 1000 km + 6370 km = 7370 km. First, you need to put your craft in an elliptical orbit with a perigee of 6870 km and an apogee of 7370 km (Figure 6.12). You can do this by firing your rockets at that point in your orbit you want to be the perigee of the elliptical "transfer orbit". By how much must you increase your speed?

I know you will be thrilled to do the Hohmann transfer analysis using algebra. (Algebra - from the Arabic "al-jabr", literally "death to the infidels" - was invented, of course, by an Arab terrorist organization for the sole purpose of driving Western students to despair.) If the radius of your original orbit is ro and your speed vo, then your energy is

(6.35) E = (1/2)mvo2 - GMm / ro,

where M is the Earth's mass. Using Eq. (6.25) for your speed, Eq. (6.35) becomes

(6.36) E = (1/2)m[GM/ro] - GMm / ro = -(1/2)GMm / ro.

Note that your total energy is negative, and, interestingly enough, equal to half your potential energy. (This is because your kinetic energy is one-half the absolute value of your negative potential energy.) Your total energy is negative, because your negative potential energy is greater in magnitude than your positive kinetic energy.

You are thus in a bound orbit, that is, you can't escape the Earth unless you pick up more energy. To see this note that the largest bound orbit you can have is when the radius of your orbit gets very large (often referred to in textbooks as "approaches infinity" - actually a meaningless phrase; which I confess I sometimes use myself - better expressed as "grows larger indefinitely"). When ro approaches an extremely large value, your potential energy approaches zero because you are "approaching" the zero level. Your kinetic energy approaches zero because, by Eq. (6.25), your speed approaches zero. Hence your total energy approaches zero. You therefore must have at least zero energy to get totally away from the Earth.

Orbiting objects with negative energy are bound to the Earth. Objects with positive energy can move away from the Earth indefinitely and never slow to zero speed. The knife edge is zero energy, where the object, if moving away from the Earth, will slow down toward a speed of zero as it gets farther and farther away.

A total energy of zero is more of theoretical than practical interest. If an object has enough energy to get far away from the Earth, for example, it will come under the influence of the Sun's gravity (assuming it misses the Earth's Moon). It may have escaped the Earth only to assume a solar orbit. To escape the solar system, the object would have to have zero energy taking into account the potential energy interaction with the Sun. Even then, it would not be beyond the combined gravitational influence of all the matter that constitutes the Milky Way galaxy. To escape the galaxy, it would have to have at least zero energy taking into account all the galactic matter. In all likelihood, an object will not have exactly zero energy with respect to another body, but either more than enough or too little to escape entirely.
Objects with non-negative energy are said to be unbound. Of course, as explained above, this description ignores the possibility that the orbiting objects might encounter a body other than the Earth on their journey - most notably, the Sun, whose gravity will dominate when the object moves far enough away from the Earth.

More sophisticated mathematical analysis (we're talking differential equations here - mathematics well beyond what is used in introductory physics courses) shows that

• bound (negative energy) orbits are ellipses,
• orbits with positive energy (unbound) are hyperbolas,
• and zero energy orbits are parabolas.
Introductory physics primarily deals with circular orbits. A circle is just a special form of ellipse, as already noted.

This division of orbits according to energy sheds some more light on the subject of comet collection by the planet Jupiter, discussed in Chapter Four. A comet plunging toward the Sun from a great distance in the outer solar system will have almost zero energy and be a barely bound object. Its orbit will be such an elongated ellipse it will appear to be a parabola when among the planets. If the comet encounters a large planet, such as Jupiter, it can either gain energy from or lose energy to the planet. In the former case it may well gain enough energy to leave the solar system forever, since its energy is almost zero in the first place. If it loses orbital energy, its energy becomes more negative. The more negative the energy, the more tightly bound the comet. Hence it may be captured into the inner solar system, losing matter on each close encounter with the Sun. Having lasted for over 4 billion years, it will now dissipate into a long train of rubble, mostly dust, after several dozens to hundreds of orbits. If its orbit crosses that of the Earth, it will be reborn as a yearly meteor shower.

Getting back to executing the Hohmann transfer, when you give your spacecraft a boost in your direction of motion, you increase your speed to, say, v'o. If you do this very quickly, your craft will be at the same altitude as before but traveling faster. Your energy will now be

(6.37) E' = (1/2)mv'o2 - G Mm / ro,

that is, you have the same potential energy as before but an increased kinetic energy. Your increased energy is the result of work the boosters have done on the spacecraft. You want to reach the point r = 7370 km, the apogee of your new orbit. Since the energy E' must remain constant after the boost due to the fact that only conservative gravity is now acting on your craft, a larger r means increased potential energy, so your kinetic energy, and hence your speed, must decrease as you approach apogee (Figure 6.12).

You have two unknowns, the boost speed v'o and the speed when you reach apogee, v'. Unfortunately, energy conservation is but one equation. However, there is another quantity besides energy that remains constant between perigee and apogee, your angular momentum. This gives you two equations. Conservation of energy requires the energy at perigee equal the energy at apogee, or,

(6.38) (1/2)mv'o2 - GMm / ro = (1/2)mv'2 - GMm / r.

Conservation of angular momentum requires the angular momentum at perigee equal the angular momentum at apogee, or,

(6.39) mv'oro = mv'r.

Note that at perigee and apogee the velocity of the satellite is perpendicular to r, so that r of Eq. (6.34) is equal to the perigee and apogee, respectively, at these points on the orbit. The speed at apogee, v', is seen, from Eq. (6.39), to depend on your boost speed as

(6.40) v' = v'o(ro / r)

Substitute this expression for v' back into Eq. (6.38) then do a little college algebra manipulation. (Can you manage it?) The result is

(6.41) vo' = {2GMr / [ro(r + ro)]}1/2

Let's see what your numbers yield. From Eq. (6.25),

(6.25) v = [GM / r]1/2,

your speed in your original orbit was [(6.673 × 10-11 N·m2/kg2) (5.98 × 1024 kg) / (6,870,000 m)]1/2 = 7620 m/s = 7.62 km/s. Then, Eq. (6.41) requires your boost speed to be [2(6.673 × 10-11) (5.98 × 1024)(7370000) / (6870000) (7370000 + 6870000)]1/2 m/s = 7750 m/s. Therefore, to get into the elliptical orbit you desire, you must increase your speed by 130 m/s.

You might legitimately wonder why this burst of speed doesn't change your perigee. Why would you return to the same lower altitude you had before you fired thrusters? The answer is that, for your energy and angular momentum to remain the same after the short thruster firing, you must go through the point in your orbit where you fired the thrusters. This is the only point on your original orbit consistent with both your new energy and new angular momentum.

A more mathematical way of looking at this involves solving the so-called differential equation of motion of a satellite. This differential equation is conceptually nothing more than your old nemesis F = ma, where the acceleration is expressed as the time rate of change of the time rate of change of the position, making it what techies call a "second derivative of position with respect to time". The solution shows that an orbit is determined by the initial altitude and velocity of the satellite, and the satellite must return to that initial position and velocity. Techies may recall that you have to integrate a second derivative twice to undo the double rate of change, and each time you pick up a "constant of integration". In the case of the equation of motion for the satellite, one of these is the initial position, and the other is the initial velocity.)

Finally, you are ready to fire thrusters at apogee to complete the Hohmann transfer. Once again, this point on your new elliptical orbit will be the only one consistent with your energy and angular momentum after firing thrusters the second time. You want your new speed to be that of a circular orbit at 7370 km. Eq (6.40) determines your speed at apogee in the transfer orbit. It is (7750 m/s)(6870 km) / (7370 km) = 7220 m/s. From Eq. (6.25) the speed you need for the circular orbit will be
[(6.673 × 10-11 N·m2/kg2) (5.98 × 1024 kg) / (7,370,000 m)]1/2 = 7360 m/s. So, to get into the circular orbit you have to increase your speed by 140 m/s.

You might wonder, "Why not increase the speed by 270 m/s at the first thruster firing and be done with it?" A moment's reflection may indicate why this won't work. You will still have to go through the point where you fired your thrusters, whether increasing your speed by 130 m/s or 270 m/s. If you increase your speed by 270 m/s with a single firing, you will have the same perigee that you had when you fired for an increase of 130 m/s. This means you have an elliptical orbit with the same perigee as before, but an apogee that takes you well beyond the altitude of the female astronaut. You have expended more energy than the Hohmann transfer but are still not in the correct orbit. (Recall that kinetic energy increases as the square of the speed.)

The Hohmann transfer illustrates some of the basics of orbital mechanics. For a small body orbiting a much larger one, you may ignore the kinetic energy of the much larger object by assuming it to be at rest. Therefore the total energy of the system is given by Eq. (6.33). This energy is constant due to the conservative nature of gravity (unless the body is a spacecraft firing booster rockets and ignoring the minute drag of the tenuous high atmosphere if you are in low-Earth orbit). Also, since the smaller body is pretty much orbiting the center of the larger one, the orbital angular momentum of the system is just that of the smaller body about the center of the larger one. This angular momentum is also constant (in the absence of booster firings) due to the central location of the source of the gravitational force (as proved by Sir Isaac using calculus to sum the gravity of all the atoms of the larger body). Circular orbits of greater radius have greater energy, according to Eq. (6.36). However, this energy is due to the increase in potential energy, since the kinetic energy has decreased due to the smaller speed (Eq. (6.25)). Orbital mechanics of a small object orbiting a larger one is useful, not only for man-made satellites and spacecraft, but also for comets and asteroids orbiting the Sun, and even for planets, since even Jupiter is pretty puny when put up against the mass of the Sun.

What about bodies of comparable size in mutual orbit, such as Pluto and its Moon Charon, binary star systems, or even the Earth and its relatively large moon? For these systems you must count the kinetic energy of the larger object in addition to that of the smaller one. Adding their mutual gravitational potential energy completes the energy picture. Also, each body is orbiting a common center of gravity, which would be halfway between the two objects were they the same mass. So, you must take into account the angular momentum of each object around this common center of gravity. Nevertheless, the total energy remains constant as well as the sum of the angular momenta of the two bodies, which gives astronomers and astrophysicists somewhat to work with.

As always, there are problems with the beautiful energy and momentum conservation described above. Gravity may on average act from a center of gravity, but in reality every atom attracts every other atom. This gives rise to tidal forces which, in the case of the Earth and Moon system, cause these bodies to flex. Since neither body is perfectly elastic, some of the orbital energy is lost in the internal friction due to the distortion of the two bodies, being transformed ultimately into heat.

In the short run this energy loss is no big deal. In the long run (billions and billions of years), our system is losing orbital energy. Were this loss to continue uninterrupted by more profound cosmic events (such as the death of the Sun), the Earth and Moon would end up keeping the same face toward each other. One side of the Earth would never see the Moon and the other would always see it. In this configuration internal friction dissipation ("tidal dissipation") would be at a minimum, and one day on the Earth would be the length of a lunar month. This is already the case for Pluto and Charon.

"Bye, Bye, Miss American Pie!": How to Escape the Earth

Well, as luck would have it, the female astronaut happens to be the same one who dumped you during training. Upon learning this, you decide to abort the rescue and just fly off into outer space, never to return. How do you fire thrusters to achieve this? You recall that to escape the Earth, your energy must at least be zero (taking potential energy to be zero at infinity). Eq. (6.35) gives your current energy as (1/2)mvo2 - GMm / ro, which is a negative number due to the fact that the potential energy has a greater negative value than the value of the positive kinetic energy. To get to zero energy, you must fire thrusters to increase you kinetic energy until your total energy becomes at least zero. The speed that gets you to zero total energy is called the "escape speed". You can find your escape speed by setting the energy in Eq. (6.35) to zero. Then vo becomes your escape speed, vesc. Solving for vesc gives you

(6.42) vesc = [2GM / ro]1/2.

With this speed you will sail off into the blackness of space. If the Earth were the only body in the solar system, you would gradually slow down, toward a limit of zero speed at an "infinite" distance from Earth. However, there is this small detail of the Sun, which contains almost all the mass in the solar system. Hence what you would actually do is launch yourself into a solar orbit. And since you launched from the position of the Earth, your orbit would inevitably cross the Earth's path again (although the Earth most likely would not be at that location when you got back).

Eventually, of course, as you made orbit after orbit, you would happen to cross the Earth's path when the Earth was there. The female astronaut would then rescue you and introduce you to her husband, Sucky, the homeless alien who has found love at last. Too late, you realize that you had to apply Eq. (6.42) to the Sun once you had moved sufficiently away from the Earth's influence if you wanted to escape the solar system, making M the Sun's mass and ro your distance from the Sun. Escaping the Earth does not mean you escape the Sun as well.

What if you had overcome your childish impulse to launch yourself morosely into the depths of unrequited despair? You would find yourself, after your two burns, in the same orbit as the distressed female astronaut. Let's say you are several minutes behind her. How would you catch up? You might be tempted to fire forward thrusters to increase your speed. However, orbital mechanics shows that this will actually boost you into a higher, slightly elliptical orbit of greater energy (due to the positive work done by the thrusters), whose average speed is less than that of your previous orbit. Refer to Eq. (6.25), which shows that the greater the orbital radius, the smaller the speed. With a smaller speed and a larger orbit to traverse, you would find yourself falling behind the female astronaut if you fired forward thrusters.

What if you, contrary to all common sense, fired reverse thrusters? This would put you into a lower, less energetic, orbit with a faster speed! (Once again see Eq. (6.25).) You would catch up to the female astronaut, coming up on her from a lower altitude. Once you caught up you would fire forward thrusters to increase your altitude and slow your speed, bringing you into docking position. Without this training, but rather flying by the seat of your pants, you might never get into a position to complete the rescue.

(6.5) "Let Me Take You Down, 'Cause We're Going to [Gravitational] Fields": Gravity Field Theory

Newton's theory of gravity, embodied in Eq. (3.5),

(3.5) F = G (m1m2)/r2,

works great for almost every situation involving gravitational interactions you can imagine. It guides our spacecraft unerringly and safely to their destinations, so long as you don't get feet and meters confused:

### MARS CLIMATE ORBITER FAILURE BOARD RELEASES REPORT, NUMEROUS NASA ACTIONS UNDERWAY IN RESPONSE

Wide-ranging managerial and technical actions are underway at NASA's Jet Propulsion Laboratory, Pasadena, CA, in response to the loss of the Mars Climate Orbiter and the initial findings of the mission failure investigation board, whose first report was released today.

"The 'root cause' of the loss of the spacecraft was the failed translation of English units into metric units in a segment of ground-based, navigation-related mission software, as NASA has previously announced," said Arthur Stephenson, chairman of the Mars Climate Orbiter Mission Failure Investigation Board. "The failure review board has identified other significant factors that allowed this error to be born, and then let it linger and propagate to the point where it resulted in a major error in our understanding of the spacecraft's path as it approached Mars.

- NASA news release (emphasis mine)

Ahem... Newtonian theory explains the orbits of the planets (well, except Mercury). It explains why the Earth has tides, why apples fall to Earth, and why Gypsy Rose Lee could say she has everything she used to have, just a few inches lower. However, there are a few areas where Newton's theory falls short (pun not entirely unintended). It doesn't satisfactorily explain "gravitational lensing", that is, the bending of light around massive objects. Nor does it describe the gravitational redshift (clocks running slower when subject to stronger gravity). It is not a good theory to comprehend the phenomena of black holes, and, as mentioned above, it fails to predict the fact that the elliptical orbit of Mercury gradually changes orientation (precesses) about the Sun.

"Action at a Distance" Now Only Refers to On-Line Sex

With all that said (and more yet to say), Newton's theory suffers from a great philosophical problem. How can a mass like the Sun, 93 some-odd million miles away from us, reach out over all this space and influence the Earth? I mean, it's not like there's this cable connecting the two bodies. No wires, no strings, no smoke, no mirrors - even David Copperfield should be jealous. Newton himself recognized this as a problem and "solved" it by the concept of "action at a distance", which was no real solution at all. The phrase "action at a distance" basically means, "Hell if I know." Newton was well aware of the shortcomings of the "action at a distance" proposal, but he couldn't think of anything else. The Sun reaches out and pulls on the Earth because... well, because it just does, that's why! It took the insight of Michael Faraday to envision that space can have properties, one of which is to support a "field".

Just what is a field? A property that has a value at all points in a region of space is said to constitute a field in that space. Figure 6.13 illustrates several different fields.

The figure is a weather map from mid-February "pirated" from Aviation Digital Data Service. Now, this is a real weather data map, not the glitzy, gee-whiz weather maps you are used to seeing on the TV weather shows. Actually, it is a type of weather map that shows the weather at all reporting stations, the so-called METARS (meteorological reports from airports). Each circle represents a weather-reporting station. For example, TXK is the Texarkana airport, TYR is Pounds Field in Tyler, AUS is Austin, etc.

Note the numbers in red to the upper left of each circle. These are current surface temperatures. Every point on Earth has a surface temperature, and the ones on this map are merely representative of the surface temperatures of the region. Therefore, these temperatures represent a field: the field of surface temperatures. With weather balloon soundings, the temperature field could be extended upward as high as the balloon rises, into the lower stratosphere. A temperature field is an example of a scalar field, since temperature is a scalar. The number in green to the lower left of each circle is a dewpoint temperature. Dewpoint temperature is another scalar field. The final scalar field depicted is the surface pressure (adjusted to sea level): the gray numbers to the upper right.

Now, what are those funny looking little lines extending from each station with the tails that sort of look like the back end of an arrow? These show the direction from which the wind is blowing. The "ticks" on these lines indicate the wind speed. A short tick is 5 knots (5 nautical miles per hour, a nautical mile being 1852 m, as things stand now), a long tick is 10 knots, a short plus a long is 15 knots and so forth. So, at the time of this map, the wind in Texarkana was about 10 knots out of the NW. The wind at Gage, OK, (GAG) was out of the N, at 15, gusting to 24 knots. Finally, the wind at Tucumcari, New Mexico, (TCC) was NE at 5 knots. The wind consists of both a speed and a direction, so the wind constitutes a vector field. This is the kind of field Faraday envisioned for electricity, which can also be applied to gravity.

Since the force of gravity has both direction and magnitude, you might conclude, correctly, that gravity must form a vector field. However, the force of gravity itself can't be the field, because objects with two different masses will experience two different gravitational forces at the same point in space. On the other hand, the force of gravity must somehow depict the gravitational field. The way out of this difficulty is to define the gravitational field as the force of gravity per unit mass. Conceptually, this involves placing a very teeny "test mass" at a point in space where you want to measure the gravitational field. (The mass needs to be very small, that is, infinitesimal, so that you don't disturb the distribution of mass causing the gravitational field.) You measure the force acting on the test mass and divide that by the test mass. The force of gravity on an object divided by the mass of the object is none other than the acceleration of the object due to gravity. Since this acceleration is independent of the amount of mass of the object, we have just the thing to represent the gravitational field: The gravitational field at a point is the acceleration due to gravity at that point. Figure 6.14 illustrates this concept. The magnitude of the gravitational field at a distance r from an object of mass m is given by

(6.43) g = G m/r2.

The figure only depicts the acceleration of gravity due to the Earth. In reality every body with mass, including you, creates a gravitational field around it. Here on the surface of the Earth, the gravitational field is about 9.8 m/s2 directed toward the Earth's center. However, both the Sun and the Moon create gravitational fields around them, and these slightly modify the Earth's field by adding to it vectorially. The word "slightly" is due to the fact that, although the Sun is much more massive than the Earth, it is very distant. Eq. (6.43) shows that the field drops off as 1/r2, so that the field of the Sun at our location is much less than 9.8 m/s2. The Moon is much closer than the Sun but, compared to the Earth, has a very small mass. Nevertheless, the gravitational fields of the Sun and Moon are responsible for our tides.

Get the Ocean in Motion: Why Are There Tides?

Out of idle curiosity, calculate the magnitudes of the gravitational acceleration of the Moon at the Earth's surface. It will not be exactly the same on the near and far sides of the Earth, so you'll have to compute it for both. It is this difference that is responsible (sort of) for the tides. Using the semimajor axis of revolution for the Moon's orbit (which is sort of an average distance) in these calculations and putting all units in SI, you get

gMoon,near = G MMoon / (r - RE)2 = (6.673 × 10-11)(7.354 × 1022) / (3.84403 × 108 - 6.379 × 106)2 = 3.434 × 10-5 m/s2.

Note that you have to subtract the radius of the Earth, RE from the Earth-Moon distance, r, to get the distance to the near side of the Earth. The far-side acceleration of gravity is

gMoon,far = G MMoon / (r + RE)2 = (6.673 × 10-11)(7.354 × 1022) / (3.84403 × 108 + 6.379 × 106)2 = 3.213 × 10-5 m/s2.

The acceleration of gravity due to the Moon is obviously much smaller than that due to the Earth. No surprise here. However, on the side near the Moon, this quantity opposes the Earth's acceleration and makes the ocean weigh slightly less there than it does in surrounding regions. [Recall weight = mg, so that weight = m(gEarth - gMoon) on the side of the Earth toward the Moon.] The result, according to the usual explanation found in introductory textbooks, is that the ocean rises on the side of the Earth near the Moon, which is at "high tide". That is, the Moon pulls the ocean toward itself slightly, so high tide is fairly easy to understand. The water that piles up on the side of the Earth toward the Moon has to come from somewhere, so low tide in the region of the oceans surrounding high tide is not hard to understand either.

Well....except what I have just described, although the explanation you get in most elementary textbooks, is not exactly the whole story. In fact, it is not even really true. And it doesn't explain at all the fact you get a high tide on the side of the Earth opposite the Moon. The typical explanation there is that the Moon pulls harder on the solid Earth than it does on the ocean on the far side, because the solid Earth, on average, is closer. This is also an incomplete explanation.

To understand what is really going on, imagine that you have a whole mess of very unusual particles. These particles do not have gravity of their own, but they do respond to the gravity of other bodies. Take a bunch of these particles and form them into a sphere. Now find a large celestial body and launch this sphere around it at some distance r so that the center of the sphere has the correct speed [Eq. (6.25)] such that it will move about the celestial body in a circular orbit.

All of the particles initially have the same speed. However, Eq. (6.25),

(6.25) v = [GM / r]1/2,

which gives the speed necessary for a circular orbit of radius r, shows that a particle closer to the celestial body than the center of the sphere needs a faster speed to execute a circular orbit, whereas a particle farther from the body needs a slower speed. [Note the speed necessary changes as (1 / r)1/2 so that it is faster if the radius, r, of the orbit is smaller and is slower if the radius is larger.]

So what happens to these particles? They can't hold together because there is no mutual gravity (and no epoxy or cement). The ones nearest the celestial body, not having enough speed to maintain a circular orbit, take up an elliptical orbit, moving closer to the celestial body and speed up. The ones farthest from the celestial body, having too much speed, also take up an elliptical orbit, but move away from the celestial body and lose speed. The other particles that don't happen to be the same distance from the celestial body as the center of your sphere begin to make similar moves. Your original spherical distribution of particles begins to elongate toward and away from the celestial body. Since the particles aren't held together by any force, your sphere gradually comes apart.

The argument for a solid object (but deformable - all real objects are deformable) is pretty much the same as that above. The particles (atoms) that make up the body are held together by gravity and cohesive forces (that are fundamentally electrical in nature), but they would still like to take the paths taken by the imaginary particles described above if they could. So a body tends to elongate toward and away from the object it is orbiting. (The same is more or less true of the larger body, depending on the size differential.) Gravity is indeed involved in this, but without orbital motion, there would be no tides as we know them. If the Earth and Moon were stuck on pins like bugs in a collection, the Moon would not produce a high tide on the opposite side of the Earth.

The Moon, however, is not alone in producing ocean tides. The Sun, though much farther away, pulls on the Earth with a force about 177 times that of the Moon due to its great mass. Performing a calculation to find the gravitational field at the Earth due to the Sun, similar to the one done above for the Moon, gives 0.00590 m/s2, compared to about 0.0000332 m/s2 for the Moon. You might think the tides produced by the Sun would be so much greater that those produced by the Moon that the lunar tides should hardly be detectable. What's going on?

What's going on is that tides depend on the difference in the gravitational field across the body, because this has to do with the difference in the orbits particles on either side of the body would take if they were free of the forces holding them to the body. The Moon's field decreases (look at the calculation several paragraphs above) by about 2.2 × 10-6 m/s2 across the Earth's diameter, whereas the Sun's field decreases by about 1.0 × 10-6 m/s2. This comparison helps to explain why the tides produced by the Moon are about three times as large as those produced by the Sun.

However, there is another aspect to the production of tides. The Earth and Moon are actually in a mutual orbit about each other, orbiting a point between them, their common center of gravity. Since the center of gravity is essentially the same as the center of mass unless the situation is exceptional, you can use Eq. (4.11),

(4.11) xcm = (m1x1 + m2x2 + m3x3 + ... + mNxN) / (m1 + m2 + m3 + ... + mN),

to find the center of gravity of the Earth-Moon system. There are only two masses involved, so Eq. (4.11) becomes, using rcm for xcm, which is more consistent with the usual notation,

(6.44) rcm = (MEarth(0) + MMoon rMoon) / (MEarth + MMoon),

taking the center of the Earth to be the origin. This evaluates as

rcm = (7.354 × 1022 kg)(3.844 × 108 m) / (5.979 × 1024 kg + 7.354 × 1022 kg) = 4.67 × 106 m.

This is rcm / (Earth radius) = 4.67 × 106 m / 6.37 × 106 m = 0.73 (or 73%) of the radius of the Earth. Hence the Earth and Moon orbit a common point 27% of the way to the Earth's center. During one revolution of the Moon, the Earth swings around this point like an off-center load of laundry. The oceans on the far side of the Earth are also swinging about this point. By the first law of motion, they would like to keep going in a straight line were it not for the force of gravity. Being farther from this common center of revolution than any other part of the Earth, the tendency to "pull away", like a ball on a tether, is likewise stronger than for any other part of the Earth. (Of course, its the first law of motion that makes it seem that the oceans are trying to pull away; they are really just trying to go straight.)

This explanation for the high tides on the side of the Earth away from the Moon is equivalent to the one discussed above. (See Figure 6.15.) The rotation of the Earth about its center rotates the oceans into two high tides and two low tides every day. (A little slower than this, actually, since the Moon is orbiting in the same direction as the Earth's rotation.) As has been pointed out by others, the tides don't really "come in" or "go out". The Earth rotates into the tidal bulges. I might also point out that the solid Earth itself flexes in response to the same tidal forces that affect the ocean, but the distortion of the solid Earth is much less than that of the ocean.

The Sun, as well as the Moon, affects the tides, as already noted, causing high tides to be extra high and low tides extra low when on the same or opposite side of the Earth as the Moon, when the Moon is in the new or full phase, and causing the tides to be muted in amplitude when at right angles to the Moon, when the Moon is in first or third quarter. The enhanced tides are called spring tides (no relation to the season) and the reduced tides are called neap tides. There are two spring tides and two neap tides every lunar month.

(6.6) I Don' Need No Steenkeeng Gravitational Field!: Introduction to the Modern View of Gravity

All parts of the Earth (except its rotational axis) experience centripetal acceleration due to the Earth's roation. An object falling at a latitude away from the poles falls at a slower rate than at the poles. This is because part of the force of gravity is producing the centripetal acceleration, an effect that is maximum at the equator. This acceleration is present for the object even if it isn't falling. That leaves "less acceleration" to be manifested in the fall of the object, to put it in rather pedestrian terms. This causes problems in defining the gravitational field, a problem eventually addressed by Albert Einstein.

Let's say you are on the equator, all excited about measuring the gravitational field. You have been training several years for this very moment, running 5 miles a day, spending hours at the gym, studying karate and oriental philosphy. (Only later did you learn none of this was necessary.) But you are in great shape and are psyched up for the most profound experience of your life: dropping an object and measuring its rate of fall. Yes, like skydivers and snow boarders, you've always been that type of person who lives on the edge, and the utlimate thrill is at hand.

Given the mass of the Earth and your distance from its center, you compute that the acceleration of gravity at your location should be, to three significant figures, 9.80 m/s2. Now it is time to verify this calculation. You confidently measure the acceleration of a falling object and, to your utter horror, find that the object fell at 9.77 m/s2! What could have gone wrong? You measure the acceleration again and again, but get the same result, to three significant figures, each time. Your great moment has turned to tragedy! Is there experimental error? Alien influence? Has Kansas passed a law redefining gravity? No! It's none of these! You have been scammed by the rotation of the Earth! This rotation mascarades as a gravitational field acting oppositely to the "real" gravitational field of the Earth. The free-body diagram of Figure 6.16 illustrates the Newtonian view of this situation.

This is a snapshot of you just as you release the ball whose acceleration you are measuring. As the figure shows, the only force acting on the ball is its weight, mg. However, because the ball is traveling in a circle (you, too) along with the Earth's rotation, there is a centripetal acceleration in addition to the acceleration arising from the rate of change of the speed of the falling ball. Recall that centripetal acceleration is due to the change in the direction of motion of the ball.

The two types of acceleration are depicted by the double pinkish arrow in the figure. What you measure is the rate of change of the speed of the ball in the radial direction, ar. Because your centripetal acceleration is the same as that of the ball, you don't measure this when you take your data. You are in a rotating reference frame and have a (centripetal) acceleration and therefore are not in an inertial reference frame. You cannot measure an object's true acceleration if you are in an accelerating frame of reference. The value of the centripetal acceleration at your location is given by Eq. (6.4), where the speed is that of a point along the Earth's equator, that is, the distance around the equator divided by the time of one day, about 465 m/s. Eq. (6.4) gives

ac = (465 m/s)2 / (6.4 × 106 m) = 0.03 m/s2.

Well, you think to yourself, this is just the amount by which my measurement was off. The total acceleration is indeed 9.80 m/s2. 9.77 m/s2 of it is the rate of change of the speed of the ball toward the center of the Earth, which is what I measured. The other 0.03 m/s2 is the centripetal acceleration, which I was not able to measure. The force of gravity, mg, is responsible for the sum of the two accelerations. (They sum together like scalars since they are both directed toward the center of the Earth.) This makes gravity appear to be weaker at my location, as if there were a negative gravitational field acting upward in opposition to that of the Earth. So, there is no problem after all, you conclude. The "real" gravitational field is 9.80 m/s2. What I measured was the apparent gravitational field. You are satisfied and relieved.

Well, you are satisfied at least for a while, until you learn that the Earth is also revolving about the Sun. This is another circular motion giving rise to another (much smaller) centripetal acceleration directed toward the Sun. But this starts you thinking: I am just assuming that, if the Earth were to become perfectly stationary, the gravitational field I measure (ignoring that of the Moon, Sun, etc.) would be the actual field of the Earth. How do I know the Earth is not still turning? The distant stars appear to be stationary, so the Earth doesn't seem to be turning with respect to them, but how do I know they aren't turning? Then I would be back to the same problem of trying to figure out what is gravity and what is acceleration due to some other motion. It was thoughts like these that troubled Albert Einstein. For one thing, the acceleration due to gravity and that due to "motion" were philosophically difficult to separate one from the other. For another thing, Newton's theory of gravity did not jibe with Albert's special theory of relativity.

The Decline and Fall of Newtonian Gravity

To illustrate the difficulty of separating the acceleration of gravity from other acceleration, imagine that you are a contestant on a game show, a cross between "reality TV" and a popular quiz program, called "Who Wants to Risk Death to be a Millionaire". In the show airing a week before yours, the contestant was a woman, terrified of heights, who had to do a parachute jump to prove herself worthy of a fortune. In a humorous twist, her husband was employed to thwart her. If he succeeded, he won the million instead.

She jumps out of an airplane accompanied by a cameraman and a "buddy" sky diver provided by the show. She is scared to death, and her terror is captured on video for all the viewing audience to enjoy. But the best is yet to come. She pulls her cord and out comes, not a parachute, but her husband's dirty laundry! Ha, ha, what a riot! She is falling, terrified out of her mind, and streaming behind her is a comet's tail of boxer shorts that had been stuffed into her pack by her husband!

Thankfully, she has a backup chute. Her companion glides over to her and points it out. As she pulls the cord to this chute, her companion opens the visor of his helmet. It is her husband! The backup chute is full of plastic Wal-Mart bags that fly off into the wind, and her husband cackles! Is this great footage or what! With a mock goodbye kiss blown toward her, he pulls his cord as the ground approaches. But, get this, out comes his wife's dirty laundry! It's hilarious! He falls tumbling and screaming with bras and panties flying out of his pack in all directions! (They were put there by the show's producers when they realized there was no prize money available for that episode.)

That episode was so successful, the show easily earned enough money to award a million dollars on your episode. Your particular phobia is being alone, so a sound-proof container is prepared for you without your knowledge. As you visit with the show's host, a drug is slipped into your water. When you awake, you find yourself in this container, shaped like a cylinder. There are what appear to be a few portholes that are all shut tight. Other than that there is nothing but a muffled roar and the remote video camera that follows your every move.

The container is vibrating, which, with the roar, gives you the sensation of motion. It feels like you are accelerating at about the same rate as gravity, since you seem to weigh about the same as before. But aren't you just on the surface of the Earth? What is going on? Just then the portholes begin to open, and what you see fills you with utter horror! You see stars, and, in the distance, the planet Earth! You have been shot off into space, thousands of miles from Earth, accelerating away from the planet at 9.8 m/s2! This is your worst nightmare come true! You are completely, utterly alone with no hope of ever returning to family and friends. What a delicious sight for the viewing audience!

Wait just a second, you think to yourself; it costs millions of dollars to launch a rocket into space, and this cheap-dip show is not going to shell out that kind of money. You suspect you are not really in space accelerating away from the Earth at all, but are in some sort of simulator. You try to think of a way to prove that you are in the gravitational field of the Earth, not picking up speed in outer space. However, as much as you try, and even though you made a B in freshman physics, you can't think of any test that would distinguish between the two situations.

Suddenly, you recall the laser pen in your pocket. If I shine a laser beam across the room, you muse, it should continue in a straight line if I'm at rest on the Earth. However, if I am actually accelerating and picking up speed, it should appear bent. That is, if the width of the cylindrical room is L, it will take a time equal to L / c for the light to reach the other side, where c is the speed of light. In the same time, you will have traveled a distance that can be found using Eq. (2.16):

(2.16) x = xo + vot + (1/2)at2.

To apply this equation you identify "a" as your accelertion (9.8 m/s2), the arbitrary speed at t = 0 as vo = 0, t = L / c, and x - xo = Δx = the distance you travel while the light travels across the room. (Since both the spacecraft and the laser beam are traveling with speed vo at t = 0, you can take this speed to be zero.) Figure 6.17 illustrates your theory.

In this figure the light is emitted from the right side of the container toward the left. Because the accelerating spaceship has moved a distance Δx with respect an object traveling at constant speed (the light beam) in the time it takes the light to traverse the distance L, the light strikes the wall a distance Δx below where it started, as seen by you traveling with the container.

You set up the experiment and count on your calibrated eyeballs to detect the minute distance the beam has been deflected. (In reality, it would be impossible to detect this deflection, even with sophisticated equipment. If your "spaceship" were 3 m across it would take 3 m / (3 × 108 m/s) = 10 ns (nanoseconds) for light to make it to the opposite side. This would amount to a deflection of (1/2)(9.8 m/s2)(10 × 10-9s)2 = 4.9 × 10-16 m, which is about the size of an atomic nucleus! For this experiment, however, we'll let you be superman.)

Oh no! You do detect a slight deflection of the light! So it's true! You're heading off into the depths of space all alone, never to be heard from again! After the audience has fed on your despair until they are beginning to get bored (plus, it's time for commercials), a hatch is opened and, behold! You were indeed in a simulator on the Earth's surface the whole time! You are now able to make good use of your million in winnings by seeing a battery of expensive psychiatrists. And, you burn your physics textbook with its chapter on Newtonian gravity.

"Gettin' Down" with Albert

OK, so why did the light appear to bend even though you were "stationary" on the Earth? This is answered by Einstein's Equivalence Principle, which states that all experiments performed in a gravitational field will give the same results when repeated in a spacecraft in deep space accelerating at the same rate as the gravitational field. That is, light will bend in a gravitational field just like it does in an accelerating spacecraft. This was verified in 1919 when an expedition to observe a total solar eclipse recorded a change in the apparent position of stars whose light passed near the Sun to what they were when the light paths were far from the Sun. The Sun had bent the paths of the light traveling past it by an amount that was in accord with the Equivalence Principle as elaborated in Einstein's General Relativity theory. The Equivalence Principle states that a gravitational field is an acceleration and an acceleration is a gravitational field. (However, see this caveat.)

This is where the equivalence between inertial and gravitational mass comes into the picture in Einstein's formulation of gravity. You can't tell whether you are standing on the Earth or accelerating at 9.8 m/s2 in a region far from a source of gravity. If you can't tell the difference, perhaps there is really no difference at all. Gravitational mass and inertial mass are the same. However, if you take this view to its ultimate conclusion, it would seem to imply that the surface of the Earth must be accelerating away from the center of the Earth at 9.8 m/s2. We don't see any such acceleration, so does that mean the Equivalence Principle is incorrect?

As a student in high school you learned all manner of theorems, lemmas, postulates, and axioms of what is called Euclidean geometry. Now, the name "Euclid" may have never crossed the lips of your geometry teacher. If your teacher was like mine, these lemmas (not to be confused with a South American pack animal) were absolute. The circumference of a circle was πr2. The hypoteneuse of a triangle was the square root of the sum of the squares of the two sides. The angles of a triangle added to 180°. And that was all there was to it. I'll bet it never occurred to you to actually measure the circumference of a circle to see if it really did equal π times the diameter. In high school geometry exists on a flat sheet of paper.

Now, what if high school geometry were taught on the surface of a sphere? Draw a circle on the sphere and measure its radius. Then measure its circumference. The circumference does not equal 2πr! In fact, it is less than 2πr. Now draw a triangle on the sphere. Measure its angles. They do not add to 180°; they add to less than 180°. If two-dimensional space can have different geometric properties, like the flat sheet of paper on the one hand and the spherical surface on the other, what about three-dimensional space?

Consider the following scenario. You live in a three-dimensional space. You are the only being in this space. In fact, you are the only object in this space with the exception of a piece of bubble gum. You shrug your shoulders at this situation and put the gum in your mouth. After getting the gum to the right consistency, you begin to blow a bubble. This is really great bubble-blowing gum, so you can blow a bubble which continues to grow and grow, and its surface becomes less and less curved. At some point you notice that the surface of the bubble no longer appears round at all. It appears as a flat plane which seemingly extends forever. Again you shrug and continue blowing. Now you notice that the surface of the bubble appears to curve inward, as if you were on the inside of the bubble, not the outside. You shrug and continue blowing. You become aware that you really are inside the bubble, because you can see it surrounding you. Were you to continue blowing, the bubble would shrink until it coated your body! What the heck is this all about?

You are not in Euclidean space.You are in a curved space, in particular, in a spherically curved space. In two dimensions this would be the space of the surface of a sphere, and you would be a spot on the sphere. The surface of the sphere is all you are aware of; you are not aware of the radial dimension that extends both above you, away from the center of the sphere, and below you, towards its center. When you blow your bubble, it is a circle on the sphere. Refer to Figure 6.18.

At the time corresponding to circle 1, everything looks pretty much as you would expect. You see the curved "surface" of the bubble (actually a curved line in two-dimensional space). You keep blowing and the bubble gets bigger. By the time circle 4 exists, you see a straight line, which is actually a "great circle" on the sphere. (The equator, for example, is a great circle on the surface of the Earth.) As the blowing continues, the bubble goes beyond the great circle and becomes a smaller and smaller circle, which eventually surrounds the blower (circle 7). The surface of a sphere is the two-dimensional analog of three-dimensional spherical space. If you lived in a spherical space, triangles would have vertices summing to close to 180° if they were small enough, but less than that if they were large. Large circles would have a circumference less than 2πr. Note that the surface of a sphere has no edge. You can travel all over the sphere and never find the end of your two-dimensional space. Nevertheless, it has a finite area; it is not infinite. Similarly, a spherical three-dimensional universe would be finite but "unbounded" (no edge).

If you set off in what you thought was a straight line, you would eventually return to where you started, just as a world traveler, following a great circle on the Earth, would come back to where he began his trip. If you could see far enough, you could view yourself in the distance, because the light rays reflected by you would follow the curved space and eventually return to you. Actually, you would view yourself in all directions, since each light ray that reflected off you would travel in a great circular path (as viewed from a "flat" 4-dimensional space, just as the surface of a sphere is obviously curved when viewed from three dimensions) and eventually return to its point of origin. You would see a giant image of yourself, only it would be your backside! (All this assumes the space is small or you are a very long-lived being.)

Spherically curved space is only one possible universe. Other possibilities could be infinite, such as the "saddle" universe. In this universe large circles would have circumferences greater than 2πr, and large triangle would have angles that summed to greater than 180°. Such a universe would be unbounded and infinite. The universe might even be a "flat" space. That is Euclidean geometry holds everywhere. This space would also be infinite and unbounded. (The latest cosmological data indicate the universe is, in fact, very flat.)

What does this have to do with gravity? According to Einstein's view, gravity is not a field superimposed on a space that is already there. Gravity is part and parcel with space and time. Mass warps space and time (spacetime) in its vicinity. Release tiny "test masses", and they will follow "direct paths" (termed geodesics) in the warped spacetime. To a three-dimensional creature, the paths will looked curved, because we do not perceive the dimension of time in the same way we perceive the dimensions of space. The masses will appear to us to be accelerating, as viewed from our three-dimensional perspective, when they are not actually accelerating in spacetime. A body in spacetime that is allowed to fall freely is said to be following its "worldline", which is its most "natural" motion (natural in the same sense motion in a straight line at a constant speed is the most "natural" motion in Galileo's universe).

Contrast Newton's view of gravity (with the addition of the "field" idea) versus Einstein's for your orbit after launching yourself from Newton Tower. In Newton's universe the Earth sets up a gravitational field around it. Just what this field is is not clear, but it has a magnitude and direction at every point in space, which is the acceleration of gravity. The gravitational field produces the gravitational force you experience. This force acts as the centripetal force which holds you in your orbit. In the universe of Einstein, you follow a "geodesic" in spacetime, conveyed on your way by spacetime's curvature.

Now you can't really plot motion in full spacetime, due to the inability to depict four dimensions, but you can restrict the spatial dimensions to two (say, "x" and "y") and plot a perspective diagram with time being the third dimension - except your time axis isn't just time. To have spatial units you have to mark off the time axis in units of ct, where c is the speed of light and t the time. Your orbits in spacetime might look like Figure 6.19.

This picture is in a sense highly misleading. With an orbital radius of 6570 km, the orbit has a circumference of 2πr = 41,280 km. Traveling at 7.79 km/s, you complete one orbit in 88.3 minutes. However, in that same 88.3 minutes light has traveled (3 × 105 km/s)(5299 s) = 1.59 × 109 km. Therefore, the ratio of the length of the helical figure to its diameter is (1.59 × 109 km) / [(2)(6570 km)] = 1.21 × 105:1. One turn of the helix would be 121,000 times as long as it is wide. This is not very curved. The plot of the helix would look like a straight line. For real curvature you have to be near a really massive object, such as a neutron star or black hole.

(6.7) The Dark Side Of Newton: Gravity and "Chaos"

The idea of a planet such as Jupiter majestically circling the Sun in an elliptical orbit was one of the most gracious scientific gifts that Newton could have bequeathed on his and future generations. However, Newton knew that the view of planets orbiting endlessly in stable orbits - a solar system of timeless order and regularity - was not part of his gravitational theory. The problem is not that of the gravitational attraction between a given planet, Mars say, and the Sun. Taking this attraction alone, Mars will maintain its current orbit and rotational orientation indefinitely. The problem is that the Sun is not the only body that Mars has to deal with. Mighty Jupiter also pulls on Mars, and the pull is especially strong when Mars is closest to Jupiter in its orbit. In fact, all the planets pull on one another.

In some cases the interaction is negligible, as is the case for Mercury and the outer planets. Mercury is so close to the Sun and so far from the outer planets (Jupiter, Saturn, Uranus, and Neptune) that there is essentially no effect of these planets on Mercury's orbit. The same is not true in other cases. The farther out in the solar system you go, the farther the planets are from each other, but also the farther they are from the Sun. Therefore the interaction between planets grows relatively greater in importance. Each planet perturbs the orbit of its neighbor just a little each time they are relatively near one another. Newton knew that this situation was inimicable to a well-ordered solar system. Only God, he thought, could keep the solar system stabilized by occasionally intervening to make the orbits right again.

Newton had come to the precipice of the physics of chaos and turned back. Given the state of scientific knowledge of the time, that is really all he could have done. At least he appreciated the perils of his ideas. This was, alas, not always true of his disciples. The brilliant French mathematician (no, that is NOT a double oxymoron) Laplace was concerned about this question. He performed difficult calculations that, he thought, showed the solar system was stable after all. Unfortunately, he had to make approximations (no supercomputers then) that led him to the wrong conclusion. Isaac probably turned in his grave as the seeming timeless order of the solar system was ascribed to his laws. Newton had known better.

After Newton an era of "Newtonian certainty" came to hold sway in physics. A large number of physicists thought that the universe was solved. If, at a given time, you could know the position and velocity of every particle in the universe, you could know what would happen indefinitely into the future. "All you have to do" is to take these positions and velocities, compute the forces acting between the particles, calculate the particles' accelerations, and from these come up with new positions and velocities an instant later. Repeat this process over and over, predicting by computation the future positions and velocities indefinitely into the future. Or, calculate where the particles were an instant earlier, then an instant before that, and continue this until you have reconstructed the past as far back as you wish to go.

Now, physicists realized this was, practically, an impossible task. Nevertheless, with a deterministic view of the laws of Newton, it was possible to imagine doing this. The only being capable of actually carrying out this agenda was God. So God knows (or at least can know) where every particle in the universe will be at any given time in the future. (This "God" could be no more than the ultimate supercomputer! Not a pleasant thought.) The future, in this view, depends only on where the particles are now and their velocities and nothing else. The future is completely determined, a lead-pipe cinch. There is thus no free will, no chance to alter the future - to make it better or worse. "What will be will be." Predestination is thus not a religious doctrine, but a fact of science. It has already been determined whether your new diet will work, whether you will get that raise, whether you will pull at least a C in physics, when you will bite the big one, etc., etc. The dye is cast. The fix is in. The fat lady has sung.

This "scientific predestination" was pretty much what a lot of physicists believed until the advent of quantum theory. A fellow named Werner Heisenberg, one of the founders of quantum theory, discovered theoretically that it is not possible to know both the position and velocity of a particle simultaneously. This, if true, would eliminate the knowledge necessary to assign positions and velocities to all the particles in the universe at the start of a computation designed to predict what will happen in the future.

At first, many thought this was merely a matter of the imprecision of measurement. You have to poke a particle (for example with photons of light to illuminate it) in the act of measurement, and that poke will change either the particle's position or velocity, or both. If you had a better measuring device that wouldn't poke so hard, you could refine the measurement. Create better and better measuring devices, and you can eventually narrow down both the particle's position and velocity with arbitrarily good precision. However, further development of quantum theory has shown that the imprecision discovered by Heisenberg is not a matter of technology but a fundamental law of nature. A loose analogy is the speed of light. Relativity theory shows that exceeding the speed of light (at least "locally") is not possible because of the fundamental properties of space and time, not just because we don't have the technology to do it yet. Heisenberg's Uncertainty Principle trounced scientific predestination at the atomic level.

Well, OK for the atomic level, you say, but we live on the level of "macroscopic objects" - objects made up of Avogadro's number of molecules plus. Can it be that events are pretty much pre-determined at our level? For example, can't we just predict the weather indefinitely into the future, given kick-butt supercomputers, highly refined computer models of the atmosphere, and sufficient atmospheric data?

I'm glad you asked that question, because the answer to the possibility of predicting the future at our level was actually answered by someone trying to predict the weather. Edward Lorentz in the early 1960s, using a computer that today would lose a computational contest with Mr. Coffee, found that he could actually come up with reasonable weather forecasts for a day or so in the future. However, he also found if he let the computer run out to a forecast of a few days, then changed the input data ever so slightly and reran the program, the predicted weather would be dramatically different. The weather that was predicted was incredibly sensitive to small changes in the input data.

At first he thought this was due to the imprecision of the computer. Further investigation showed that it was due to the physics of the atmosphere instead. This meant that predicting the weather beyond a couple of days was difficult to impossible. If you don't believe me, check out the seven-day forecasts (but not when the weather pattern is stagnant). And, while you're at it, take those seven-day precipitation probabilities to Las Vegas and see it you can break the bank - or break even for that matter. (By contrast, here in Texas, we have an accurate 90-day forecast. It is issued in June and extends through August. Spring is a different story. The forecasts in spring in Texas are often not even good for the next five minutes.)

The source of Lorentz' woe was the "nonlinear" equations by which atmospheric physics works. Nonlinear equations mean that what you are solving for appears in the equation with some power other than one. Some nonlinear equations, for example the familiar quadratic equation, are easy to solve. The equations of the atmosphere, however, involve not just nonlinear quantities, but also rates of change of those quantities in both spatial directions and over time. (For example, the temperature varies from one location to another and changes as the day goes on.) These equations are a bear to solve (takes supercomputers if you want timely results) and tend to wander off into la-la land if you let the computers run long enough.

The National Centers for Environmental Prediction run several computer models of the atmosphere to try to predict the weather. The models will never agree with each other exactly. Sometimes the agreement is fairly good, and the models are all good-natured and chummy. Sometimes the models square off in the alley and try to beat the living pressure anomalies out of each other. However, given enough run time, even the same model will predict totally different weather if given slightly different initial atmospheric data. Lorentz dubbed this sensitivity, "The butterfly effect."

The name "butterfly effect" comes from Lorentz' contention that a butterfly in Brazil, flapping its wings, could cause a killer tornado in Texas a few days later. It is appalling that nothing has been done about this! I say we go down to Brazil, find that butterfly, and stop it before it kills again! But then I read where there are lots and lots of butterflies in Brazil. Also an awful lot of coffee. But back to the butterlies, how do we find the one responsible for our tornadoes? And does it work for a terrorist organization? Clearly, John Ashcroft needs to drop his pursuit of New Orleans' whores and get on this immediately. (Update: Unfortunately, John resigned before he could implement the "War on Butterflies". And to the date of this writing neither the Brazilian butterfly nor Osama Bin Laden have been captured.)

You may be very skeptical that some butterfly, flapping away in the rain forest from flower to tropical flower, could possibly cause the F5 that destroyed parts of Jarrell, Texas. And you would be (mostly) right. One of the characteristics of a chaotic system is that its future development can be very sensitive the exact properties it initially has.

Say, as an omnipotent being, you have the power to determine the positions and velocities of all the particles of such a system (the Earth's atmosphere, say) at a certain time. You also have the power to rewind time, try new positions and/or velocities, and see how the system evolves as a result of these changes. First you choose positions and velocities and run the system for a certain amount of time. Then you rewind, change the position of only one particle just a tiny little bit, and press Start. At first it looks like everything is going to develop exactly as before. You, of course, expect some changes to eventually set in, but not big changes. As time unfolds you are astounded that the system becomes more and more unlike the behavior of the system when the position was not slightly altered. Eventually, you have behavior that is totally unexpected, given your knowledge of the behavior of the system without the small change you made in the position of one particle. See Figure 6.20 where small differences in the Brazilian butterfly's movements lead to widely divergent outcomes.

It's tough to make predictions, especially about the future. - Yogi Berra

What you have just observed is a chaotic system that is on the razor's edge between different possible futures. The "butterfly effect" does not mean that there is a strict cause-and-effect relationship between the Brazilian butterfly's flight and the tornado in Texas. What it is saying is that the atmosphere in this instance is on a razor edge between different futures, some of these involving a tornado developing in Texas. That is, a tornado at a certain location in Texas is a possible future event in the atmosphere if a small amount of air in Brazil moves one way and not another. Along comes the butterfly, moving the air in a way that will lead to the evolution of atmospheric conditions that produce the Texas tornado. Had the butterfly chosen to flit to a different flower, the "initial conditions" would have changed and the tornado would not have occurred.

In a chaotic system, small perturbations can lead to entirely different futures. No matter how precisely you define the initial conditions of such a system, the inevitable imprecision, however small, will prevent you from predicting how the system will behave indefinitely far into the future. Add to this the quantum uncertainty embodied in Heisenberg's principle, and you might as well pack away the tea leaves and supercomputers, hit the beach, and enjoy your freedom from inevitability.

The future ain't what it used to be. - Yogi Berra

It's Good to be Mooned, But Bad for Uranus

Somewhere above I mentioned the precarious position of Mars, which has to deal with the perturbative pulls of both Jupiter and the Earth (and the other planets to a much less extent). The Earth and Jupiter are the two planets that come closest to Mars (Jupiter in the next orbit beyond Mars and the Earth in the orbit just inside that of Mars). The main player is Jupiter, whose pull on Mars when they are at closest approach is about 60 times that of the Earth. Mars has been calculated to have undergone chaotic motion in the past, both in its orbit and in the tilt of its rotational axis in relationship to the plane of its orbit.

The important perturbations that move Mars to erratic behavior have to do with "resonances". Basically this is when the period of one repeating event is related mathematically to that of one or more other repeating events. As a simple example of a "two-body" resonance, the Moon always displays the same side to Earth because it rotates about its axis with a period equal to that of its revolution about the Earth. In other words, the Moon rotates just fast enough to keep the same side facing us, a motion similar to that of a parent circling the remains of his child's latest mishap. This is called a one-to-one (1:1) resonance and, in the case of the Earth and Moon is caused by tidal action.

Resonances can enhance stability and nonchaotic behavior in a system or can be the source of chaotic behavior. It has been calculated that a resonance between the precession of Mars about its axis (like the precession of a spinning top) and the change of its orbital plane with respect to the other planets (Jupiter in particular) has caused the inclination of Mars' rotational axis to move all over the place. Today, Mars' axis is inclined at an angle of 25.2° from being perpendicular to its orbital plane, an angle which gives Mars seasons much like those on Earth. For part of the year the north pole of Mars is tilted toward the Sun and the south pole away, and you have summer in the northern Martian latitudes and winter in the southern Martian latitudes. Half a Martian year later, the south pole is tilted toward the Sun and you have southern hemisphere summer and northern hemisphere winter.

This very Earth-like behavior may not be entirely typical of the Martian past, however. Calculations based on the period of the precession of Mars and the periodic motions of the planets indicate that the angle of inclination of Mars to its orbit has drastically changed over time. A choatic behavior is indicated in that the tilt angle oscillates slowly back and forth about a certain mean for a while, then goes through a major change to another mean angle. You might get the same effect if you twirled a top on a "shake table" (a table designed to simulate the motion of the ground in an earthquake) and set the table to oscillate up and down at a frequency to which the top was sensitive (due to one of those "resonances" previously mentioned). Because the intensity of the seasons (or even whether or not they exist) depends on this angle, the climate of Mars must have undergone dramatic changes which have undoubtedly affected the Martian landscape we view today.

The Earth spins on its axis and is subject to all the types of pulls Mars is, so, the question arises, why hasn't the Earth's axis migrated all over the place in the past? Had it done so, the history of life on Earth would have been profoundly different - that is, if there even was a history. Had the Earth's rotational axis ever been nearly parallel to its orbital plane, as that of the planet Uranus is today, you, as an organism at that time, would have experienced long periods of darkness and long periods in light, the actual amounts depending on your latitude.

If you were a green plant, tough patootie. No photosynthesis for months on end. Instead, you would suffer months of dark winter. On the other hand, for much of the year the Sun would never set and no relief from the heat could be found. In effect, all points on the Earth would have a seasonal variation similar to that experienced currently by the regions above the arctic and antarctic circles. Except that the temperature extremes would be greater, especially near the poles where the Sun would be nearly overhead for months in the summer and on the opposite side of the Earth for months in the winter. Were life to adjust to months of dark, freezing winter alternating with months of searing summer, it would be very different from what we see now.

Then, just as life settles down to this regime, the orientation of the Earth's spin axis changes again and life has to get used to, say, very little seasonal change due to the axis being nearly perpendicular to the orbital plane. It is not clear life could have survived these drastic swings. How has the Earth escaped this dire fate?

Hats off to the Moon. Because of our large moon, which Mars lacks, we have a stabilizing influence. Our friend in the sky exerts a torque on the Earth's bulge, which makes for a relatively fast precession (26 000 years as contrasted with around 170 000 years for Mars) and prevents a resonance between that period and period of the wobble of the Earth's orbit, keeping the Earth in line. The Earth cannot stray from its rotational orientation, because the Moon's influence keeps it to the straight and narrow. Sort of like being married. No more raucous nights out with the boys. No, that sort of behavior could have you spinning in the wrong direction. The Earth-Moon-Sun three-body system is a success story of resonances that, coupled with the absence of any influence by Donald Rumsfeld, hold the system together and prevent destructive excursions into the evils of chaos.

Even more dramatic than Mars is the plight of poor Uranus. Not only does it have a name that is the butt of countless puns, but it is being bullied by the likes of Jupiter and Saturn. These planets have a three-way mutual resonance that, over long periods of time, induces Uranus to change its orbit. Saturn causes the orbit of Jupiter to change its eccentricity (degree of "ovalness") over time through a 2:5 resonance. (Jupiter makes about five orbits for every two made by Saturn.) Jupiter, ticked off, affects the orbit of Uranus through a 1:7 resonance. (Jupiter makes about seven trips around the Sun for each one of Uranus.) Eventually (not in our lifetime), Uranus may get into an orbit that causes it to stray too close to Saturn. When this happens, Saturn will exclaim, "Get Uranus outta here!" and poor ol' Uranus will be ejected from the solar system by gravitational interaction with Saturn.

Disclaimer: Apparently, I-20 is no worse than any other interstate in Texas, but this quote was too good to pass up.

Caveat: As I understand it, the Equivalence Principle is only strictly true in a uniform gravitational field. Einstein's first attempt to calculate the bending of light by the Sun apparently took the warping of time by matter into account but not space. Hence his original calculation predicted only one-half the bending of light that was actually measured in the solar eclipse experiment. Luckily for him, he found the error and corrected it before the experiment took place.