And that’s it—all of Newton’s laws of motion can be summarized in two simple ideas: that any change in motion can only result from an external force (F = ma), and that forces always come in pairs. This will be all we need to describe all motions, from the simple to the complex, from a tossed ball to the orbits of the planets. In fact, we already have enough physics in hand to figure out the initial velocity Superman needs to leap a tall building.
IN A SINGLE BOUND
Superman starts off with some large initial velocity (fig. 4). At the top of his leap, a height h = 660 feet above the ground, his final velocity must be zero, or else this wouldn’t be the highest point of his jump, and he would in fact keep rising. The reason Superman slows down is that an external force, namely gravity, acts on him. This force acts downward, toward the surface of the Earth, and opposes his rise. Hence, the acceleration is actually a deceleration, slowing him down, until at 660 feet, he comes to rest.
Imagine ice-skating into a strong, stiff wind. Initially you push off from the ice and start moving quickly into the wind. But the wind provides a steady force opposing your motion. If you do not push off again, then this steady wind slows you down until you are no longer moving and you come to rest. But the wind is still pushing you, so you still have an acceleration and now start sliding backward the way you came, with the wind. By the time you reach your initial starting position, you are moving as fast as when you began, only now in the opposite direction. This constant wind in the horizontal direction affects an ice-skater the same way gravity acts on Superman as he jumps. The force of gravity is the same at the start, middle, and highest points of his leap. Since F = ma, his acceleration is the same at all times as well. In order to determine what starting speed Superman needs to jump 660 feet, we have to figure out how his velocity changes in the presence of a uniform, constant acceleration g in the downward direction.
Fig. 4. Panel from Superman # 1 (June 1939) showing Superman in the process of leaping a…well, you know.
As common sense would indicate, the higher one wishes to leap, the faster the liftoff velocity must be. How, exactly, are the starting speed and final height connected? Well, when you take a trip, the distance you travel is just the product of your average speed and the length of time of the trip. After driving for an hour at an average speed of 60 mph, you are 60 miles from your starting point. Because we don’t know how long Superman’s leap lasts, but only his final height of h = 660 feet, we perform some algebraic manipulation of the definition of acceleration as the change in speed over time and that velocity is the change in distance over time. When the dust settles, we find that the relationship between Superman’s initial velocity v and the final height h of his leap is v × v = v2 = 2gh. That is, the height Superman is able to jump depends on the square of his liftoff velocity, so if his starting speed is doubled, he rises a distance four times higher.
Why does the height that Superman can leap depend on the square of his starting speed? Because the height of his jump is given by his speed multiplied by his time rising in the air, and the time he spends rising also depends on his initial velocity. When you slam on your auto’s brakes, the faster you were driving, the longer it takes to come to a full stop. Similarly, the faster Superman is going at the beginning of his jump, the longer it takes gravity to slow him down to a speed of zero (which corresponds to the top of his jump). Using the fact that the (experimentally measured) acceleration due to gravity g is 32 feet per second per second (that is, an object dropped with zero initial velocity has a speed of 32 feet/sec after the first second, 64 feet/sec after the next second, and so on) the expression v2 = 2gh tells us that Superman’s initial velocity must be 205 feet/sec in order to leap a height of 660 feet. That’s equivalent to 140 miles per hour! Right away, we can see why we puny Earthlings are unable to jump over skyscrap ers, and why I’m lucky to be able to leap a trash can in a single bound.
In the above argument, we have used Superman’s average speed, which is simply the sum of his starting speed (v) and his final speed (zero) divided by two. In this case his average speed is v/2, which is where the factor of two in front of the gh in v2 = 2gh came from. In reality, both Superman’s velocity and position are constantly decreasing and increasing, respectively, as he rises. To deal with continuously changing quantities, one should employ calculus (don’t worry, we won’t), whereas so far we have only made use of algebra. In order to apply the laws of motion that he described, Isaac Newton had to first invent calculus before he could carry out his calculations, which certainly puts our difficulties with mathematics into some perspective. Fortunately for us, in this situation, the rigorous, formally correct expression found using calculus turns out to be exactly the same as the one obtained using relatively simpler arguments, that is, v2 = 2gh.
How does Superman achieve this initial velocity of more than 200 feet/sec? As illustrated in fig. 5, he does it through a process that physicists term “jumping.” Superman crouches down and applies a large force to the ground, and the ground pushes back (since forces come in pairs, according to Newton’s third law). As one would expect, it takes a large force in order to jump up with a starting speed of 140 mph. To find exactly how large a force is needed, we make use of Newton’s second law of motion, F = ma—that is, Force is equal to mass multiplied by acceleration. If Superman weighs 220 pounds on Earth, he would have a mass of 100 kilograms. So to find the force, we have to figure out his acceleration when he goes from standing still to jumping with a speed of 140 mph. Recall that acceleration describes the change in velocity divided by the time during which the speed changes. If the time Superman spends pushing on the ground using his leg muscles is ¼ second,12 then his acceleration will be the change in speed of 200 feet/sec divided by the time of ¼ second, or 800 feet/sec2 (approximately 250 meters/sec2 in the metric system, because a meter is roughly 39 inches). This acceleration would correspond to an automobile going from 0 to 60 mph in a tenth of a second. Superman’s acceleration results from the force applied by his leg muscles to get him airborne. The point of F = ma is that for any change in motion, there must be an applied force and the bigger the change, the bigger the force. If Superman has a mass of 100 kilograms, then the force needed to enable him to vertically leap 660 feet is F = ma = (100 kilograms) × (250 meters/sec2) = 25,000 kilograms meters/sec2, or about 5,600 pounds.
Fig. 5. Panels from Action Comics # 23, describing in some detail the process by which Superman is able to achieve the high initial velocities necessary for his mighty leaps.
Is it reasonable that Superman’s leg muscles could provide a force of 5,600 pounds? Why not, if Krypton’s gravity is stronger than Earth’s, and his leg muscles are able to support his weight on Krypton? Suppose that this force of 5,600 pounds is 70 percent larger than the force his legs supply while simply standing still, supporting his weight on Krypton. In this case, Superman on his home planet would weigh 3,300 pounds. His weight on Krypton is determined by his mass and the acceleration due to gravity on Krypton. We assumed that Superman’s mass is 100 kilograms, and this is his mass regardless of which planet he happens to stand on. If Superman weighs 220 pounds on Earth and nearly 3,300 pounds on Krypton, then the acceleration due to gravity on Krypton must have been fifteen times larger than that on Earth.
So, just by knowing that F = ma, making use of the definitions “distance = speed × time” and “acceleration is the change in speed over time,” and the experimental observation that Superman can “leap a tall building in a single bound,” we have figured out that the gravity on Krypton must have been fifteen times greater than on Earth.
Congratulations. You’ve just done a physics calculation!
2
DECONSTRUCTING KRYPTON—NEWTON’S LAW OF GRAVITY
NOW THAT WE HAVE DETERMINED THAT in order for Superman to leap a tall building, he must have come from a planet with a gravitational attraction fifteen times that of Earth, we next ask: How would we go about building such a planet? To answer this, we must un
derstand the nature of a planet’s gravitational pull, and here again we rely on Newton’s genius. What follows involves more math, but bear with me for a moment. There’s a beautiful payoff that explains the connection between Newton’s apple and gravity.
As if describing the laws of motion previously discussed and inventing calculus weren’t enough, Isaac Newton also elucidated the nature of the force that two objects exert on each other owing to their gravitational attraction. In order to account for the orbits of the planets, Newton concluded that the force due to gravity between two masses (let’s call them Mass 1 and Mass 2) separated by a distance d is given by:
where G is the universal gravitational constant. This expression describes the gravitational attraction between any two masses, whether between the Earth and the sun, the earth and the moon or between the Earth and Superman. If one mass is the Earth and the other mass is Superman, then the distance between them is the radius of the Earth (the distance from the center of the Earth to the surface, upon which the Man of Steel is standing). For a spherically symmetric distribution of mass, such as a planet, the attractive force behaves as if all of the planet’s mass is concentrated at a single point at the planet’s core. This is why we can use the radius of the Earth as the distance in Newton’s equation separating the two masses (Earth and Superman). The force is just the gravitational pull that Superman (as well as every other person) feels. Using the mass of Superman (100 kilograms), the mass of the Earth, and the distance between Superman and the center of the Earth (the radius of the Earth), along with the measured value of the gravitational constant in the previous equation, gives the force F between Superman and the Earth to be F = 220 pounds.
But this is just Superman’s weight on Earth, which is measured when he steps on a bathroom scale on Earth. The cool thing is that these two expressions for the gravitational force on Superman are the same thing! Comparing the two expressions for Superman’s weight = (Mass 1) × g and the force of gravity = (Mass 1) × [(G × Mass 2) /(distance)2], since the forces are the same and Superman’s Mass 1 = 100kg is the same, then the quantities multiplying Mass 1 must be the same; that is, the acceleration due to gravity g is equal to (G × Mass 2)/d2. Substituting the mass of the Earth for Mass 2 and the radius of the Earth for d in this expression gives us g = 10 meter/sec2 = 32 feet/sec2.
The beauty of Newton’s formula for gravity is that it tells us why the acceleration due to gravity has the value it does. For the same object on the surface of the moon, which has both a smaller mass and radius, the acceleration due to gravity is calculated to be only 5.3 feet/sec2—about one sixth as large as on Earth.
This is the true meaning of the story of Isaac Newton and the apple. It certainly wasn’t the case that in 1665 Newton saw an apple fall from a tree and suddenly realized that gravity existed, nor did he see an apple fall and immediately write down F = G (m1 × m2)/ (d)2. Rather, Newton’s brilliant insight in the seventeenth century was that the exact same force that pulled the apple toward the Earth pulled the moon toward the Earth, thereby connecting the terrestrial with the celestial. In order for the moon to stay in a circular orbit around the Earth, a force has to pull on it in order to constantly change its direction, keeping it in a closed orbit.
Remember Newton’s second law of F = ma: If there’s no force, there’s no change in the motion. When you tie a string to a bucket and swing it in a horizontal circle, you must continually pull on the string. If the tension in the string doesn’t change, then the bucket stays in uniform circular motion. The tension in the string is not acting in the direction that the bucket is moving; consequently, it can only change its direction but not its speed. The moment you let go of the string, the bucket will fly away from you.
Back to the case of the moon. If there were no gravity, no force acting on it, then the moon would travel in a straight line right past the Earth. If there were gravity but the moon were stationary, then it would be pulled down and crash into our planet. The moon’s distance from the Earth and its speed are such that they exactly balance the gravitational pull, so that it remains in a stable circular orbit. The moon does not fly away from us, because it is pulled by the Earth’s gravity, causing it to “fall” toward the Earth, while its speed is great enough to keep the moon from being pulled any closer to us. The same force that causes the moon to “fall” in a circular orbit around the Earth, and causes the Earth to “fall” in an elliptical orbit around the sun, causes the apple to fall toward the Earth from the tree. And that same gravitational force causes Superman to slow in his ascent once he leaps, until he reaches the top of a tall skyscraper. Once we know that in order to make such a powerful leap, his body had to be adapted to an environment where the acceleration due to gravity is fifteen times greater than on Earth, that same gravitational force informs us about Krypton’s geology.
One consequence of Newton’s law of gravitation—which states that as the distance between two objects increases, the gravitational pull between them becomes weaker by the square of their separation—is that all planets are round. A sphere has a volume that grows with the cube of the radius of the orb, while its surface area increases with the square of the radius. This combination of the square of the radius for the surface area with the inverse square of the gravitational force leads to a sphere being the only stable form that a large gravitational mass can maintain. In fact, to address the astrophysical question of what distinguishes a very large asteroid from a very small planet, one answer is its shape. A small rock that you hold in your hand can have an irregular shape, as its self-gravitational pull is not large enough to deform it into a sphere. However, if the rock were the size of Pluto, then gravity would indeed dominate, and it would be impossible to structure the planetoid so that it had anything other than a spherical profile. Consequently, cubical planets such as the home world of Bizarro must be very small. In fact, the average distance from the center of the Bizarro planet to one of its faces can be no longer than 300 miles, if it is to avoid deforming into a sphere. However, such a small cubical planet would not have sufficient gravity to hold an atmosphere on its surface, and it would be an airless rock. Since we have frequently seen that the sky on the Bizarro world is blue like our own (and shouldn’t it be some other color if it is to hold true to the Bizarro concept?), this would imply that there is indeed air on this cubical planet. We must therefore conclude that a Bizarro planet is not physically possible, no matter how many times we may feel in the course of a day that we have been somehow instantly transported to such a world.
Back to normal spherical planets like Krypton. If the acceleration due to gravity on Krypton gK is fifteen times larger than the acceleration due to gravity on Earth gE, then the ratio of these accelerations is gK/gE = 15. We have just shown that the acceleration due to gravity of a planet is g = Gm/d2. The distance d that we’ll use is the Radius R of the planet. The mass of a planet (or of anything for that matter) can be written as the product of its density (the Greek letter ρ is traditionally used to represent density) and its volume, which in this case is the volume of a sphere (since planets are round). Since the gravitational constant G must be the same on Krypton as on Earth, the ratio gK/gE is given by the following simple expression:
where ρK and RK represent the density and radius of Krypton and ρE and RE stand for the Earth’s density and radius, respectively. When comparing the acceleration due to gravity on Krypton to that on Earth, all we need to know is the product of the density and radius of each planet. If Krypton is the same size as Earth, then it must be fifteen times denser, or if it has the same density, then it will be fifteen times larger.
Now if, as we have argued at the start of this book, the essence of physics is asking the right questions, then it is as true in physics as it is in life that every answer one obtains leads to more questions. We have determined that in order to account for Superman’s ability to leap 660 feet (the height of a tall building) in a single bound on Earth, the product of the density and radius of his home world o
f Krypton must have been fifteen times greater than that of Earth. We next ask whether it is possible that the size of Krypton is equal to that of Earth (RK = RE) so that all of the excess gravity of Krypton can be attributed to its being fifteen times denser than Earth. It turns out that if we assume that the laws of physics are the same on Krypton as on Earth (and if we give up on that, then the game is over before we begin and we may as well quit now!), then it is extremely unlikely that Krypton is fifteen times denser than Earth.
We have just made use of the fact that mass is the density multiplied by volume, which is just another way of saying that density is the mass per unit volume of an object. Now, to understand what limits this density, and why we can’t easily make the density of Krypton fifteen times greater than Earth’s, we have to take a quick trip down to the atomic level. Both the total mass of an object and how much volume it takes up are governed by its atoms. The mass of an object is a function of how many atoms it contains. Atoms are composed of protons and neutrons inside a small nucleus, surrounded by lighter electrons. The number of positively charged protons in an atom is balanced by an equal number of negatively charged electrons. Electrons are very light compared to protons or neutrons, which are electrically uncharged particles that weigh slightly more than protons and reside in a nucleus. (We’ll discuss what the neutrons are doing in the nucleus in Chapter 16.) Nearly all the mass of an atom is determined by the protons and neutrons in its nucleus, because electrons are nearly two thousand times lighter than protons.
The size of an atom, on the other hand, is determined by the electrons or, more specifically, their quantum mechanical orbits. The diameter of a nucleus is about one trillionth of a centimeter, while the radius of an atom is calculated by how far from the nucleus one is likely to find an electron, and is about ten thousand times bigger than the nucleus. If the nucleus of an atom were the size of a child’s marble (a diameter of 1cm) and placed in the end zone of a football field, the radius of the electron’s orbit would extend to the opposite end zone, 100 yards away. The spacing between atoms in a solid is governed essentially by the size of the atoms themselves (you can’t normally pack them any closer than their size).
The Physics of Superheroes: Spectacular Second Edition Page 5
