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Wizards, Aliens, and Starships: Physics and Math in Fantasy and Science Fiction

Page 10

by Adler, Charles L.


  Free fall is a subtle concept. There are two ideas we need to discuss here:

  • The idea that a falling person would not feel his or her own weight, and

  • Why an object that is falling need not hit the ground.

  7.3.1 Why a Falling Person Doesn’t Feel His or Her Own Weight:

  I … who have made the test, assure you that a cannonball that weighs one hundred pounds, or two hundred, or even more, does not anticipate by even one span the arrival on the ground … [of one] weighing half as much when dropped from a height of 200 braccia.

  —GALILEO GALILEI, DIALOGUES CONCERNING TWO NEW SCIENCES

  The above quotation concerns a realization by Galileo that has become fundamental in mechanics: objects acting under the force of gravity have an acceleration independent of their mass. If we take two cannonballs and drop them from a high tower, they will hit the ground at almost exactly the same time, even though their weights may differ by an enormous amount. (This is ignoring any effects that air resistance may have on them.)

  Here’s a thought experiment: in many amusement parks there are “drop towers”, rides in which the people are lifted to some large height above the ground and then dropped toward the ground, only slowed to a stop scarily close to the ground. Now, imagine going on one of these rides with an apple in your pocket. As you are lifted up, remove the apple and then, once the fall begins, release it from your hand.

  When you let it go, it falls toward the ground, of course. However, you are also falling to the ground with exactly the same acceleration as the apple; from your point of view, the apple simply hangs in the air. If you push on the apple, it moves away from you on what seems to be a straight-line path. This is from your point of view; to someone standing on the ground, it appears to move in a parabolic arc. You doesn’t feel the force of gravity acting on you because all objects around you are falling with exactly the same acceleration. What we think of as the force of gravity isn’t really the force of gravity: the heaviness we feel standing up is simply the force of the floor pushing up on us, preventing us from falling through the floor. A man falling off a roof doesn’t feel his own weight, nor do astronauts in a space station that is continually falling under the influence of gravity.

  7.3.2 Why a Falling Object Doesn’t Have to Hit the Ground

  Let’s say we take a cannonball, drop it from a high tower, and measure how far it falls during the time it drops. We’ll find that it falls a distance y in time t, given by the formula

  where g = 9.8 m/s2 is the acceleration of gravity near the surface of the Earth. In 1 second the cannonball will fall a distance . We then try throwing the cannonball as we drop it so that our throw is perfectly horizontal—that is, the cannonball travels parallel to the surface of the Earth. Surprisingly, the distance it falls is exactly the same no matter how fast we throw it, as long as the throw is perfectly horizontal. This is because there is nothing pushing or pulling it parallel to the surface of the Earth; the only pull is toward the center of the planet. Most people have an intuitive idea that the faster the horizontal motion is, the less far it will fall in a given time, like Wiley Coyote running off a cliff in the Road Runner cartoons. Like most cartoon physics, this isn’t the way things work.

  If the Earth were flat, as it appears to be over short distances, then the cannonball would simply hit the ground after a while. But because the Earth is a sphere, the ground is dropping away a bit as the cannonball flies out. Move horizontally far enough and you’ll find yourself above the surface of the Earth. How far does the cannonball have to move horizontally before it is 4.9 m above the surface? We can find this through a little geometry: let x be the distance the cannonball moves horizontally and y be the distance from the surface of the Earth, while R is the radius of the Earth (6,400 km, or 6.4 × 106 m). From the Pythagorean theorem,

  Eliminating R2 on both sides of the equation, we are left with

  using the fact that y is much smaller than R. From this,

  Therefore, if the cannonball travels a horizontal distance of 7,920 m in 1 second, it will drop a distance equal to the distance that the horizon “dips.” Another way of saying this is that if any projectile near the surface of the Earth is given a horizontal velocity of 7,920 m/s, it will never hit the Earth. In 1 second it will fall exactly the same distance that the Earth’s surface falls away as it moves horizontally. The motion of the projectile will be circular, with a radius the same as that of the Earth.

  Note that this tells us exactly the same information that equation (6.3) did; in fact, it tells us a little bit less than that equation did, because we have specified what the acceleration of gravity is. However, with a little work, the reasoning given above can be used to derive the circular velocity for a satellite at any distance from the planet.

  7.4 ARTIFICIAL “GRAVITY” ON A SPACE STATION

  A Space Shuttle astronaut has told me that the first few days in space feel like falling on an endlessly long roller coaster. He referred to this as the “inertia of the viscera,” which results from the small displacement of one’s internal organs owing to the absence of apparent weight. This is one of the medical effects of weightlessness, and astronauts who spend long periods of time in space tend to have medical problems associated with weightlessness, as the human body evolved in a world with a relatively high gravitational field.

  You will find the idea in old science fiction novels, such as Robert Heinlein’s The Moon Is a Harsh Mistress, that low gravity or weightlessness will confer longer life on humans because the heart doesn’t need to pump as hard to move blood around the body [119]. This doesn’t seem to be true; prolonged life in free fall leads to loss of appetite, lengthening of the body by a few inches, immune system problems, muscle and bone atrophy, and increased flatulence. None of the effects is serious in the short term, but we don’t know what living in free fall for many years at a stretch will do. It’s better if we can provide some sort of artificial gravity for the inhabitants of the space station.

  The easiest way to do this is to spin the space station. The TV show Babylon 5 was set on board a rotating space station five miles long in the shape of a cylinder. It rotated on its long axis to provide the sensation of gravity, as do many other space stations in science fiction. Among these are the already mentioned Venus Equilateral, the space station in Robert Heinlein’s juvenile novel Space Cadet, and many, many others. Let’s say that the space station is a long, hollow cylinder spinning around its (long) axis. If the cylinder radius is R and the rotational speed of the cylinder is v, the acceleration of an object moving around in the circle is

  The acceleration is centripetal, that is, directed toward the center (see appendix 1). The acceleration must be caused by a force; for a person of mass M, the force is equal to Mv2/R and is supplied by the push of the station’s hull against her feet.

  Another way to look at this is to view this force from the point of view of someone inside the station, tied the rotating station. From her point of view, there is a “force” pushing her to the outside of the station: the force produces an acceleration outward from the center of the station whose magnitude is

  She will attribute this to a (fictitious) “centrifugal” force pushing her outward against the hull. Therefore, a person in a rotating space station will seem to feel her normal weight if we rotate the station fast enough. If the station has a radius R = 1 km, we need a rotational speed

  to have an “acceleration of gravity” on the station the same as at the surface of the Earth. Using these numbers, the time for one complete rotation of the station is just over one minute. However, all is not quite the same as on Earth.

  7.4.1 Thrown for a Loop

  There are effects resulting from the rotation of the station that are interesting and produce different results from what one would see on Earth. The rotation gives rise to two “pseudo-forces”: in addition to the centrifugal force pushing against the hull, there is also a Coriolis force acting on moving objects. The Coriolis for
ce also exists on Earth but is relatively weak compared to the force of gravity; its major action is on very large dynamical systems such as hurricanes, which it causes to spin one way north of the equator, and in the opposite direction south of it.1 On a space station, however, the Coriolis effect is very strong because of the relatively small size of the station and the relatively large spin rate.

  Imagine standing inside the station holding an apple in your hand and letting it go. From your point of view (rotating along with the station), the apple falls to the “ground” right next to your feet because of this “centrifugal” force pushing outward. However, to someone outside the station looking in, the situation is very different. Let’s take two people, Susan and Mike. The names of our two observers are taken from the 1990s TV show Babylon 5, set in the eponymous O’Neill colony; Mike Garibaldi was the head of security for the station and Susan Ivanova was the second in command. To date, it remains one of the few TV shows that have tried to get the space science reasonably accurate, apart from having sound in space. Let’s say that Susan is in a space suit floating in space outside the station, looking in. Mike is inside holding an apple; he is about to open his hand and let the apple “fall.” What he sees the apple doing is very different from what she sees the apple doing. This is because if the station is large enough, Mike will not notice the rotation.

  Mike and the apple are rotating along with the rest of the station. When Mike lets go of the apple, however, it doesn’t rotate; instead, because of its inertia, it moves in a straight-line path tangent to the circle it was just moving on with the speed it had just before being released. This is the key point in going from one point of view to the other: the reason why Mike sees the apple fall is that his rotation along with the rest of the station means that the apple is moving. When it is free from his hand, Newton’s first law takes over and it moves in a straight-line path. It hits the hull near the point where Susan’s feet are rotated to as it falls. To Susan, it moves in a straight-line path at constant speed; to Mike, in the rotating spaceship, it appears to drop from his hand and accelerate to the hull. The apparent acceleration is due to the fact that he isn’t moving along a straight line.

  But because the apple is in Mike’s hand when it is released, it is slightly closer to the axis of the space station than his feet are. It travels on a smaller circle around the center of the station than his feet do, but it travels that smaller circle in the same amount of time. Because it moves less far in the same time, it is moving with a slightly smaller speed than his feet are. If he drops it from a height of 2 m, it is traveling 0.2% more slowly than his feet. This means that it won’t hit the station hull exactly where his feet are but slightly in the direction opposite the direction in which the station is spinning.

  This is the Coriolis effect. (I caution that we are discussing only one aspect of the Coriolis effect here. For a complete description, see any advanced mechanics textbook, such as Classical Mechanics, by Herbert Goldstein [99, pp. 177–183].) If the station is rotating in a clockwise sense, Mike seems to see a force deflecting the falling apple counter clockwise as it falls. For objects that are either moving quickly inside the space station or projected up near the central axis, the trajectories are just plain odd. If you throw a ball inside the station, it seems to act under the influence of an external force. This is illusion! Once it leaves your hand, no forces act on it. From the point of view of someone outside the station looking in, it moves in a straight line. The appearance of force comes only from the rotated motion of someone inside the station. Now, even though the spin is supposed to simulate gravity, there are big differences between the trajectories of particles in the rotating reference frame and in a real gravitational field.

  Because the Earth rotates, there is a Coriolis force that acts on missiles. The mathematics describing it is complicated because of the shape of the Earth and the presence of a real force (gravity) acting on the missile. A rotating space station is much simpler: the geometry is simple enough that it is very clear what is happening to the ball, but the trajectories inside the rotating frame are extremely counterintuitive.

  When Michael Garibaldi throws a baseball in the garden section of Babylon 5, he seems to see two distinct forces act on it: the centrifugal force, which pushes the ball toward the hull of the station (mimicking gravity), and a Coriolis force, which doesn’t appear in a true gravitational field [51][63][212]. The Coriolis force is known to be appreciable when the spacecraft diameter is small. It acts only on a moving object; the magnitude of the force is proportional to the velocity and the direction of the force is perpendicular to it. However, most science fiction authors ignore the Coriolis effect when talking about rotating spacecraft; the assumption seems to be that in large enough spacecraft, the effect will be negligible [189, pp. 122–124]. Also, the treatment of this Coriolis effect in rotating spacecraft in science fiction and speculative science has been confined entirely to the description of objects moving parallel to the spin axis of the ship; objects thrown down the long axis of an O’Neill colony, for example, appear to be deflected antispinward from the thrower. If you throw an object perpendicular to the spin axis, the general idea seems to be that you can ignore the Coriolis effect. Not so! For stations less than a kilometer or so in diameter, the effects are large and very strange. Let’s look at what happens when someone in a rotating station decides to start throwing a ball around.

  Consider a cylindrical spacecraft with radius R, length L. The ship rotates about the axis of the cylinder at angular frequency Ω. We set up a coordinate system x, y, z with z pointing along the cylindrical axis of the ship. In this book, we will only be concerned with motion in the xy plane. Because x and y are coordinates fixed to the ship, they do not form an inertial coordinate system. We define a coordinate system at rest with respect to the fixed stars, X, Y, a frame that is coincident with x, y at time t = 0. At any time t, the transformation between the two coordinate systems is given by

  So, Mike stands inside the station at a point x = 0, y = −y0 (see fig. 7.1) and throws a ball with velocity vector (vx, vy) in the x, y frame. What is the trajectory of the ball?

  To Susan floating in the inertial frame X, Y, the trajectory is very simple. The ball travels on a straight line with velocity vector (vx − Ωy0, vy), because once it leaves Mike’s hand it is no longer under the influence of any forces. (Like all good physicists everywhere, we ignore air resistance.) To Susan, the ball travels on a straight line given by the equations

  To Mike, it’s not at all that simple.

  Let’s consider a few cases:

  1. The “circular orbit”: vx = Ωy0, vy = 0. Mike throws the ball in an antispinward direction at exactly his rotation velocity. To Susan, the ball stops moving, because it has zero velocity in the inertial frame. Mike, however, is rotated under the stopped ball: he sees the ball follow a circular orbit around the center of the station. Figure 7.1 shows the trajectory of the ball in both the inertial and rotated reference frames. The arrow indicates the direction of the throw in the non inertial coordinate system.

  Figure 7.1. The “Circular Orbit” Trajectory.

  This is not an unreasonable velocity for a thrown ball. Consider the cases of Babylon 5 and another fictional spacecraft, the Discovery, from 2001: A Space Odyssey. The diameter of the rotating section of Discovery is 11 m [58]. If Discovery is rotated to produce 1 g, the hull velocity is ∼7 m/s. In the episode “The Fall of Night,” Susan Ivanova mentions that Babylon 5’s garden section rotates at 60 mph (that is, 28 m/s), which is about right, if we remember that the space station is roughly 0.8 km in diameter and the acceleration of gravity in the garden section is 1/3 g. Even in an Island One structure as proposed by O’Neill, the hull velocity is only about 45 m/s, the speed of a fastball thrown by a major league pitcher [189].

  2. “Over the shoulder” trajectory: vx = Ωy0, vy = vx (fig. 7.2). Throwing the ball at an angle of 45 degrees in the air antispinward causes the ball to loop back over the thrower’s head and
land behind him. In the inertial frame, the ball travels along the y-axis until its trajectory intersects the hull. Because vy is relatively large, the trajectory intersects the hull before it has enough time to make one complete rotation.

  3. The “Wiley Coyote” trajectory: vx = Ωy0, vy = 2Ωy0/π (fig. 7.3). This trajectory is so named because the ball hits the thrower in the head, from the back. In this case, the y velocity is slow enough that the station makes a complete rotation before the ball intersects the hull again. One gets the feeling that golf would be a wild and rather dangerous sport on Babylon 5.

  4. The “Etch-a-Sketch” trajectory: vx = Ωy0, vy = 0.08Ωy0 (fig. 7.4). The smaller the y velocity, the larger the number of times the station rotates beneath it before it intersects the hull. Here is a very complicated trajectory whose initial conditions are very close to those for a circular orbit. Unlike the problem of two-body orbits under the influence of gravity, the circular orbit of figure 7.4 is unstable to small perturbations in the initial velocity.

 

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