The Physics of Superheroes: Spectacular Second Edition (7 page)

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 = v
²
= 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 v
²
= 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 v
²
= 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, v
²
= 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/sec
²
(approximately 250 meters/sec
²
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/sec
²
) = 25,000 kilograms meters/sec
²
, 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 understand 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)/d
²
. 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/sec
²
= 32 feet/sec
²
.

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/sec
²
—about one sixth as large as on Earth.