Read Online The Physics of Superheroes: Spectacular Second Edition by James Kakalios - Free Book Online

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 g K is fifteen times larger than the acceleration due to gravity on Earth g E , then the ratio of these accelerations is g K /g E = 15. We have just shown that the acceleration due to gravity of a planet is g = Gm/d 2 . 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 g K /g E is given by the following simple expression:
where ρ K and R K represent the density and radius of Krypton and ρ E and R E 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 of 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 (R K = R E ) 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