The Answer Is Forty-Two — Many Mechanics Problems, Only One Answer

Contents

• BODY OF ARTICLE
• Grazing Satellites and Earth Tunnels
• A Very Long Pendulum
• Summary
• Acknowledgments
• REFERENCES
• About the Author
• FIGURES

The answer is 42, actually 42 minutes, but what was the question? There are, in fact, a number of physics questions, not at first glance closely related to one another, to which 42 minutes (or perhaps 84 minutes) is the answer. This paper was prompted by one such question, a Figuring Physics item in a recent issue of this journal,¹ which called for a description of the motion of a block released on a flat frictionless plate tangent to the Earth's surface. The correct answer given in Figuring Physics was that "the block will oscillate to and fro,"² but I was disappointed to see that the time of oscillation was not called for. So let us refine the question: "For small-amplitude oscillations, how long does it take for the block to travel from one extreme to the other, i.e., what is the half-period of the motion?" This is the question (actually just one of the questions) to which the answer is 42 minutes, as we can readily calculate. Those who are familiar with The Hitchhiker's Guide to the Galaxy [Douglas Adams (Harmony Books, New York, 1979)] will understand why I chose to ask initially for the half period (42 min) rather than asking the more obvious question, "What is the period?"³

Throughout this paper I will assume a complete absence of friction, air resistance, and other nuisances from the real world that tend to spoil physics homework problems. I will also take the Earth to be stationary, nonrotating, and perfectly spherical, i.e., without hills or other imperfections. (Any other assumption, e.g., reality, would spoil the fun.) The mass distribution of the Earth will also be taken to be spherically symmetric. Thus, the gravitational force on any object at or above the surface is equal to that exerted by a point object of mass M located at the center of the Earth, with M the total mass of the Earth. For the most part, it will not be necessary to assume a homogeneous Earth. When we come to consider the motion of objects in tunnels inside the Earth, then at that point we will take the Earth's mass distribution to be not only spherically symmetric but also homogeneous — contrary to fact. Further, in any such "tunnel problem," the tunnel itself will be assumed to be of negligible diameter; removal of dirt to dig the tunnel will have a negligible effect on the gravitational force on an object in the tunnel. (Just as the gravitational force on an object on the surface is primarily due to the distant parts of the Earth — the amount of distant stuff is so large that it overwhelms the nearby stuff in spite of the inverse square law — likewise in a tunnel, the nearby material is of no importance relative to the distant parts of the Earth. The only exception occurs when the motion is restricted to the immediate neighborhood of the exact center of the Earth.)

Any one-dimensional oscillatory motion (well, almost any) is simple harmonic for small amplitudes, i.e., the physics is characterized by a restoring force proportional to the displacement from the equilibrium position. The canonical example is the motion of a mass on a spring, F = –kx, with a resulting period:

To calculate the period, all we need is the coefficient of x in the force law, together with the mass. It may be simpler in some cases to find the period by using the equivalent language of energy. For small displacements, the potential energy will be proportional to x², and if the energy is written in the form

then the period is again found from the ratio of the coefficients in the kinetic and potential energy terms.

Now look at Fig. 1. With M and R denoting the mass and radius of the Earth and m the mass of the block, the gravitational force is (very nearly) GmM/R², and to find its component toward x = 0, we need only multiply by sin x/R. Thus:

and so the analog of the spring constant, k, is (GmM/R³), and the period of oscillation is

a half period of 42 minutes.

Figure 1.

Since g = GM/R², one can also write = 2, the period of a hypothetical simple pendulum in a uniform gravitational field of strength g, whose string is of length R. (See section "A Very Long Pendulum" for a treatment of a "real" pendulum whose string is of length R and whose bob grazes the Earth's surface.)

Here is another question, familiar and much easier. What is the orbital period of a "grazing satellite," i.e., one whose orbital radius is equal to R, the radius of the Earth? The centripetal force is just the gravitational force, F = GmM/R², and equating this to the mass times the centripetal acceleration gives:

and thus = 2/ = 2 = 84 min. Recall that the period of the first Sputnik, a low-altitude satellite, was only slightly greater, about 90 minutes.

Here is a somewhat more challenging problem. What is the period of oscillation of an object dropped into a tunnel that passes through the center of the Earth (the proverbial "Hole to China"⁴)? The equilibrium point is at the exact center of the Earth. Small-amplitude oscillations will again be simple harmonic, but if we now assume that the Earth, in addition to being spherically symmetric throughout, is also homogeneous, the motion turns out to be simple harmonic even for an amplitude as large as R, with period (once again, of course) equal to 84 minutes. We invoke the usual arguments that depend on spherical symmetry as well as on the basic inverse square nature of the gravitational force to show that at any point in the tunnel, a distance x from the center, the total gravitational force on the object is as if all the mass closer to the center than x were replaced by an equal point mass at x = 0, with all the mass farther than x from the center contributing nothing. If we now assume a homogeneous Earth, the amount of mass in (0, x) is just M(x/R)³, and this mass acts as if it were a distance x away. Thus,

giving simple harmonic motion with = 84 min. (As with any simple harmonic motion, the period of an object oscillating in this tunnel is amplitude-independent, for all amplitudes up to 4000 miles.)

But this tunnel result is more general. Consider any straight tunnel with its two ends anywhere on the Earth's surface, say New York and Paris or perhaps Hopkinton, Mass., and Boston. (More about the latter tunnel in a moment.) Once again (see Fig. 2) we let x denote the distance from the equilibrium position, now at the center of the tunnel, let P denote the position of the object at any instant, and s the distance from P to the center of the Earth. The amount of mass closer to the center of the Earth than P is M(s/R)³. Thus, the gravitational force on our object is

and to get the component along the tunnel, we must multiply by sin = x/s. Now a miracle occurs. The various factors of s cancel, and

Once again the period of oscillation (no small-amplitude approximation needed) is 84 minutes, independent of where and how far apart the two ends of the tunnel are! If the tunnel is a long one (e.g., along a diameter), the initial acceleration is large ( 9.8 m/s²). If the tunnel is shorter, the initial acceleration is smaller, but just enough smaller so that the time to reach the other end is the same, 42 minutes.

Figure 2.

Thus we have a wonderfully impartial mail delivery system. A letter dropped into the New York end of either the New York-Paris or the New York-Boston tunnel (or even the New York-Hoboken tunnel) will take exactly 42 minutes to reach its destination. Even better, this result leads to my sure-fire strategy for winning the Boston Marathon next year. I will dig a straight tunnel from Hopkinton to Boston and jump into it at noon as the starting gun goes off, serenely arriving sweat-free in Copley Square 42 minutes later, while the supposedly elite runners are barely entering Natick at the eight-mile mark. (I will pass them a bit after the one-mile mark.) Though my initial (and maximum) acceleration will be only about 0.03 m/s², my average pace for the whole trip will be 1:36 per mile (three times faster than marathon winners, and extremely easy on the knees to boot), and at the midpoint of the course I will be "running" one-minute miles (about 26 m/s), at which time I will be about 35 m below the pavement in Wellesley. One might worry that I will be disqualified for taking a shortcut. I do not think that will be a problem, however; I leave it as an exercise for the reader to estimate how much shorter my race course will be than the official distance of 26.2 miles. (The answer is not 42, by the way, at least not in any set of units, SI or otherwise, with which I am familiar.) Only the most hardhearted official could disqualify me for such a trifling amount of cheating.

Here, saved for last, is the most difficult question in this collection: "What is the small-amplitude period of a simple pendulum of length l?" Nothing difficult about that — everyone knows that the period is 2. Well, not so fast. There is an assumption that is always made in doing that calculation, an extraordinarily well hidden (and little known) assumption, that the length of the string is much less than the radius of the Earth. But what if we relax that assumption? (Don't ask who or what is holding the string; remember, this is physics, not engineering.)

Consider Fig. 3 and use as the coordinate to describe the motion. We consider only small displacements from equilibrium, and thus both and are small. In addition, the magnitude of the gravitational force on the pendulum bob is very nearly GmM/R², no matter what the relative sizes of l and R may be. The tricky part is finding the component of F along the arc of the circular path, toward = 0. That component of F is given by –F cos , and a bit of work with the relevant triangle shows that = (/2) – – . We can also see from Fig. 3 that (again, small angles are assumed) l R. Thus,

and so,

Equating this to ml immediately gives the small-amplitude period as

For l << R, = 2 as usual, but as l is increased, gradually approaches an l-independent limit:

Notice, by the way, that when l= R, = (84 min)/ = 59 min.

Figure 3.

It is easy to see why the usual small-angle treatment of a simple pendulum breaks down as l becomes very large. In the conventional approach, we take the Earth's gravitational field to be everywhere in the same direction, parallel to a line from the support point to the center of the Earth. But as l , the bob's trajectory becomes a horizontal straight line (just like the flat plate in Figuring Physics), and thus the restoring force, the component of the gravitational force along the trajectory, would go to zero and the period would become infinitely large. In both these problems, the crucial consideration that provides a small restoring force is the fact that the gravitational force on the object has a different direction when the object is not at the equilibrium point.

I leave the following as an exercise for the reader. Drill an 8000-mile hole through the center of the Earth. Suspend a pendulum bob from a point at the surface, this time on a rigid massless rod of length l (0 < l 2R). Show that the pendulum's small-amplitude period is = 2, for all values of l up to 8000 miles, where g is the gravitational acceleration at the surface of the Earth, 9.8 m/s². (A string would be acceptable for values of l up to 4000 miles; beyond that, the rod is needed.) For this one, assume again that the Earth is homogeneous, and — if the bob is at or very close to the center of Earth (l R) — assume that the cavity you scoop out to let the pendulum swing is filled with a totally fictitious material of earth density that offers no resistance to the swinging pendulum. Notice, by the way, that when l R, the period is once again equal to 84 minutes; explain physically why, if l R, this small-amplitude pendulum has the same period as the "Hole to China Problem."

In retrospect, it is not surprising that the flat plate (the original Figuring Physics question), the pendulum with l >> R, and a very short tunnel (just below the surface) all give the same result. In all these problems, the object in question only samples the gravitational field at or very close to the Earth's surface. But the objects in longer tunnels sample much more of the Earth's field, and I still find it astonishing that all tunnels through the Earth give the same result for the period (subject, remember, to the simplifying assumption of a homogeneous Earth) and that this period is the same as those of the grazing satellite, the infinitely long pendulum, and the flat plate. Were it not for the fact that the basic force law is inverse square, none of those tunnel results would be as they are; indeed, for any other force law, even the simplest tunnel problem seems to become extraordinarily difficult and probably not soluble in terms of well-known functions. As the Ukrainian philosopher Gregory Skovoroda wrote in the 18th century, "We must be grateful to God that He created the world in such a way that everything simple is true and everything complicated is untrue."

I should add that I have provided no references other than The Hitchhiker's Guide to the Galaxy, because most of the problems discussed in this paper are in the repertoire of most physics teachers. The two problems discussed in the penultimate section are exceptions, though they may well have appeared in 19th-century textbooks and perhaps on the Cambridge Tripos examinations.

I thank Dudley Herbert Towne, David Sumner Hall, Larry Russell Hunter, and Jagu for helpful comments and discussions.

References

1. Paul Hewitt, Phys. Teach. 40 (8), 456 & 501 (Nov. 2002). first citation in article
   
2. In the original Figuring Physics question, it was stated that the block might be released with some initial speed (but less than the escape velocity). If the initial velocity were not precisely toward or away from the point of tangency, then the block would go into orbit on the plate (for small displacements an ellipse with center at the point of tangency) rather than "oscillating to and fro" in a degenerate ellipse. Throughout this paper, I will assume that the object is simply released at rest, thus giving one-dimensional ("to and fro") oscillatory motion. first citation in article
   
3. For those who find that choice obscure, I quote brief excerpts from The Hitchhiker's Guide (pp. 178–180): "Seventy-five thousand generations ago, our ancestors set this program in motion, and in all that time we will be the first to hear the computer speak." "An awesome prospect, Phouchg," agreed the first man . "We are the ones who will hear," said Phouchg, "the answer to the great question of Life !" "The Universe !" said Lonquawl. "And Everything!" "Tell us!" "All right," said Deep Thought. "The Answer to the Great Question " "Yes!" "Of Life, the Universe and Everything" said Deep Thought. "Yes!" "Is" said Deep Thought, and paused. "Yes!" "Is" "Yes!!!?" "Forty-two," said Deep Thought, with infinite majesty and calm. first citation in article
   
4. This may be referred to in Argentina or Chile as the "Hole to China Problem," but in the United States it should properly be called the "Hole to the Indian Ocean Problem." In China, it is presumably referred to as the "Hole to Argentina Problem." first citation in article
   

1. Kepler's Third Law Without a Calculator
   Michael J. Ruiz, Phys. Teach. 42, 530 (2004)

Robert H. Romer is now an "emeritus" (i.e, unpaid) professor at Amherst College. From 1988–2001, he was editor of the American Journal of Physics.Physics Department, Amherst College, Amherst, MA 01002; rhromer@amherst.edu

Full figure

Fig. 1. The block on a flat plate. Our theory is restricted to small-amplitude oscillations ( << 1). First citation in article

Full figure

Fig. 2. An off-center tunnel through the Earth. The motion is simple harmonic with / 2 = 42 min for all amplitudes, and the tunnel need not pass through the center of the Earth. First citation in article

Full figure

Fig. 3. The long-string pendulum. Our theory is restricted to small-amplitude oscillations (, << 1). First citation in article

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