The Physics of Superheroes: Spectacular Second Edition (15 page)

A moment ago I referred to the reason why surface tension arising from even a thin layer of water in the lungs does not kill us as “soap.” While not technically correct, in that pulmonary surfactants are not soaps, the converse is true, in that soaps are surfactants, with water-attracting and water-repelling chemical groups at either end of long skinny, chain-like molecules. Soap helps one clean up by reducing the surface tension of water, so that it can make direct contact with the dirt. That is, surfactants make water wetter, and help us breathe easy as well.

In his first adventure in 1941’s
More Fun Comics # 73
, Aquaman is shown in the first oversized panel (which is commonly referred to in comic books, appropriately enough, in this case, as the “splash panel”) swimming in the ocean, rescuing a woman from the lifeboat in his right arm while he deflects a shell fired from a German U-boat with his left. We can calculate how tough Aquaman’s skin must be by considering the force necessary to change the momentum of this missile (as discussed in Chapter 3). A one-pound shell traveling at 500 feet per second would require Aquaman to generate a force of 30,000 pounds to deflect it in a time of one millisecond (one thousandth of a second). This force is applied over a very small area, let’s say three square inches. The pressure (force per area) on Aquaman’s forearm is consequently 10,000 pounds per square inch, which is more than 660 times atmospheric pressure and comparable to the strength of cast iron. A year later, in
More Fun Comics # 85
, at the North Pole the Aquatic Ace tosses a full-grown polar bear at a group of poachers from more than 100 yards away as if the ursine projectile were a football (a seemingly odd thing to do with one of nature’s creatures when battling poachers, but the polar bear was unharmed in the incident). It is surprising that the King of the Seven Seas relies on fish to save him from tight spots or to help capture malefactors as often as he does, for he clearly has the stuff that’s tough enough!

The standard explanation for Aquaman’s superstrength is that it is a natural consequence of his being able to withstand the pressures at the ocean’s floor. As in the case of breathing underwater, Aquaman’s great strength in being able to support the weight of an ocean is merely different in degree, rather than in kind, from your own great strength, Fearless Reader. After all, every time you stand up from a chair, you are lifting up an ocean as well—only in your case it is an ocean of air. But if Aquaman has to contend with a half mile of water pressing on his shoulders, we must deal with approximately fifteen miles of air above us. The source of this pressure, for both Aquaman and us, is basically Newton’s laws of motion.

Air molecules move at the speed of sound, roughly 700 miles per hour. As we discussed earlier, any object in motion, whether a Mack truck or an oxygen molecule, will remain in motion until acted upon by an outside force. That outside force can come into play when the oxygen molecule collides with the surface of your shoulders; the force the shoulders exert on the molecule to deflect its trajectory is balanced by an equal and opposite force (Newton’s third law) that the molecule exerts on the shoulder. The greater the number of molecules that strike your shoulder per second (that is, the denser the air), the larger the net force that these collisions exert on your shoulders. Every single square inch is subjected to a force of nearly fifteen pounds from these atmospheric impacts. We never really notice this pressure, for basically two reasons. The first is that we have lived our entire lives under the weight of our atmosphere, and have evolved a skeleton structure, muscles, and skin capable of withstanding this pressure. The second is that the force exerted by collisions of air molecules is fairly uniform in all directions. Air is pretty dilute, with the spacing between molecules roughly ten times the diameter of any given molecule. Thus, while it is technically true that the entire weight of miles of air presses down upon us, the number of collisions coming up from below is essentially identical to the number pressing down from above, and we never experience an unbalanced force from atmospheric pressure under normal circumstances.

While a pressure of fifteen pounds per square inch seems pretty high, it pales compared with the pressures experienced by Aquaman beneath the ocean’s surface. Oxygen molecules consist of two oxygen atoms and are heavier than water molecules (which are comprised of only one oxygen atom and two hydrogen atoms). The momentum of a typical air molecule is 60 percent greater than for a water molecule in liquid (the water is moving roughly as fast, on average, as the air molecules, but has a smaller mass). Nevertheless, the pressure exerted by collisions from liquid water is much greater than from the atmosphere, as the liquid state is over eight hundred times denser than the atmosphere. The water molecules are essentially in direct physical contact thanks to the weakly attractive van der Waals force (which is also responsible for surface tension) described above. Unlike gases, liquids are essentially incompressible, and the more fluid above you, the greater the weight of water pressing down on you.

The force exerted when you are underwater is larger than that on dry land, and the further below the water’s surface you are, the heavier the water’s weight presses down on you. In fact, the pressure you experience is directly proportional to how deep under the surface you are:

The depth is measured from the water’s surface (which is why, if the depth is zero, the pressure you experience is just that of the atmosphere). The density of water (one gram per cubic centimeter) multiplied by the acceleration due to gravity is just another way of stating the weight of the water per volume. When multiplied by the depth, the units of the second term on the right-hand side of the equation above represent weight per area, which is the same as the pressure.

One eighth of a mile beneath the surface of the ocean, the pressure is equivalent to twenty atmospheres, and across Aquaman’s broad shoulders, with an area of approximately ten square inches, lies a weight of three thousand pounds. To stand under this crushing pressure requires an equivalent force supplied by the Aquatic Ace’s legs—just as Superman’s muscles and skeletal structure had to be strong enough to withstand the higher gravity of Krypton. We proposed in Chapter 1 that Superman would weigh roughly three thousand pounds on the planet of his birth. Aquaman’s legs have to be at least this strong, as he can swim much deeper in the ocean. The King of the Seven Seas must be roughly at least as strong as the Golden Age Man of Steel!

Or perhaps even stronger. In
Justice League of America # 200
, Aquaman must battle an alien creature that has assumed an artificial body composed of pure glass. Now, glass is actually pretty tough, and in its pure form is stronger than many metals. However, scratches and cracks, inevitable when fighting the Justice League, will severely weaken glass’s ability to withstand compression. Even so, it would take a pressure of more than a ton per square inch to shatter such glass. Using the previous equation, we conclude that roughly one mile beneath the ocean’s surface, the pressure is indeed 2,600 pounds per square inch. Aquaman is able to withstand such forces—the alien invader, not so much.

At the water’s surface, there is obviously no weight of water overhead, and the pressure is just that of the atmosphere. Since the ocean of air is the same over the surface of the planet, any body of water will have the same height, regardless of shape or size. In fact, two bodies of water, if they are connected by an underground channel, will by necessity have the same height, which we refer to as “sea level.” What if there was no atmosphere—what would be sea level—the height of the ocean’s surface—then? This is not such an academic question, and in fact we make practical use of this equation, which describes the height of a column of water in the absence of, or at least reduction in, atmospheric pressure, whenever we use a drinking straw. We will now show that physics places severe limits on the long-distance thirst-quenching ability of Superman!

Fig. 14.
The Aquatic Ace dispatches an alien creature that has assumed a body composed of glass by dragging him to the ocean’s depths in
Justice League of America # 200.

The name of the inventor of the drinking straw is lost to the mists of time, but there is evidence that its use dates back to ancient Mesopotamia, the region that is now part of modern-day Iraq. The reeds that grow on the shores of the Tigris and Euphrates rivers (an area known as the Fertile Crescent, one of the founding sites of agriculture) were useful not just for sipping a tasty beverage, but when pressed edge first onto wet clay (also found in abundance at these rivers’ shores) could be used to form wedge-shaped characters that left permanent markings when the clay had dried and hardened. The Sumerians of about 4000 B.C.E. are credited with being the first to use writing to record and transmit information, and their distinctive “letters” are referred to as “cuneiform,” which is Latin for “wedge-shaped.” Drinking straws have therefore been around nearly as long as writing, though the twisty “crazy straw” had to await innovations in injection plastic forming.

When a straw is inserted into a glass of water, the two “bodies of water”—that is, the water inside the straw and outside the straw—are connected by the fluid in the rest of the glass. We’ll assume that the straw is not firmly pressed against the bottom of the glass, and the water can move between these two regions freely. The atmospheric pressure on the water outside the straw is the same as that inside the straw, when first placed in the glass. The force on the liquid inside the straw is thus the same as for the liquid outside the straw, and the liquid inside and outside the straw will be at the same level. When we place our lips around the end of the straw and suck, we form an airtight seal that enables us to reduce the pressure inside the straw. The atmospheric pressure outside the straw is unchanged, and there is consequently a pressure difference between the “two bodies of water.” This pressure difference results in an unbalanced force on the incompressible fluid, the result being that the normal air pressure on the drink outside of the straw pushes the water inside the straw to a greater height.

What is the limit on the height that the water can rise up such a straw? The lower the pressure inside the straw, the larger the pressure difference between the water outside the straw and inside the straw. Suppose Superman (in particular, the Silver Age Man of Tomorrow who can fly and has super-breath) had an extremely large straw that he dipped into a large body of water. Is there a length of straw that would be too long for even Superman to be able to sip from? Yes! The greater the pressure difference between the outside and the inside of the straw, the larger the unbalanced force that pushes the water up the straw. But the pressure on the outside of the straw cannot be larger than the atmospheric pressure of fifteen pounds per square inch. When we suck on a soda straw, we can possibly decrease the interior pressure to ten or five pounds per square inch. Superman can obviously do much better, but even he can’t improve on a perfect vacuum. The lowest possible pressure inside the straw is zero pounds per square inch, achieved when there is not even one single air molecule remaining inside the interior of the straw. To find out how high the water could rise under this extreme suction, we use the equation relating pressure and depth on page 84, only now the pressure on the left-hand side of the equation is zero, and we’ll calculate a negative depth, which is equivalent to a positive height above the water’s surface. The pressure difference, between fifteen pounds per square inch and zero, can push the water inside the straw to a height of thirty-four feet, regardless of the width of the straw. Which is quite high, but also quite finite. If Superman tries to drink from a straw a “mere” thirty-five feet long, he will come up—literally—h igh and dry!

Of course, in the spirit of the somewhat goofy Silver Age adventures of Superman, it’s always possible to concoct a mechanism by which the Man of Steel can drink from any length straw he needs. As stipulated, he cannot decrease the pressure inside the straw to below zero pounds per square inch. But perhaps he can increase the pressure on the water outside the straw to a value greater than normal atmospheric pressure. If he were to give a quick puff of super-breath directed down toward the body of water he was attempting to drink from immediately before suctioning on the straw, he could induce a much larger pressure difference, and consequently bring the water up the straw to any height that the story might require.

A large pressure imbalance can lead to more dramatic effects than simply enabling one to sip a cool drink. In the above discussion, we have assumed that the straw is constructed from some sort of high-strength metal that will not collapse due to this pressure differential. A lower pressure inside the straw means that there are fewer molecules striking the inner wall of the straw per second than there are outside the straw. That is, there is an unbalanced force on the walls of the straw, just as the unbalanced force on the liquid led to the water rising up the straw to our lips. If this force is large enough, it can cause the straw to collapse along its length—as anyone who has drawn too forcefully on a paper straw can confirm.