11

How to Get Anywhere in About Forty-Two Minutes

Some rainy afternoon, when you have little to do, you might want to spend some time looking up the following rather interesting information. If you dug a hole straight downward from where you live, through the center of the earth, where would you come out?

If you live quite close to the intersection of Montana, Alberta, and Saskatchewan you are quite fortunate. You will come out on dry land, even though it is bleak and dreary Kerguelen Island in the south Indian Ocean.

If you live anywhere else in the continental United States, Alaska, or Canada, you'd better be ready to build a dike at the other end to keep the Indian Ocean out.

A dike is also needed if you live in Great Britain. The exit of your tunnel will be in the Pacific Ocean, south of New Zealand.

However, you are indeed lucky if you live in Honolulu. As you punch out the hole at the other end you will find yourself in west central Botswana in southern Africa.

It is good to know these things. They can be very useful for (a) initiating conversations, (b) changing the subject, and (c) eliminating awkward periods of silence at dinner parties, or even job interviews. They might also prove to be beneficial if you find yourself on Jeopardy! sometime.

FIG. 11.1

Definition sketch for the motion of a vehicle in a shaft through the earth.

Well, after a few discussions and some serious thought, you and I decide there may be real commercial possibilities here. All we need is a well-constructed shaft of length equal to the world's diameter and a vehicle to carry passengers wishing to go from Hawaii to Botswana and vice versa. So we form a company, raise some cash, and hire some engineers and scientists to devise the necessary technology.

The first thing we need to do is carry out the basic analysis of the problem of our vehicle moving through the shaft. We decide to neglect air resistance, wall friction, centrifugal force due to the earth's rotation, and things like that. It is also assumed that the earth is a homogenous mass and that all thermal problems can be ignored. Actual conditions and circumstances involve details the engineers can work out.

A definition sketch of our problem is shown in figure 11.1. We start with the case in which the shaft passes through the center of the earth (i.e., α = 0°). Recall that Newton's second law of motion is

in which is the summation of all the forces acting on the vehicle, m is its mass, and a its acceleration. Neglecting air resistance, wall friction, and centrifugal force, the only force involved is the weight of the vehicle, that is,

As we saw in chapter 10, the gravitational force g inside the earth is given by the expression where is the gravitational force at the earth's surface, r is the distance from the center of the earth, and R = 6.37 × 10⁶ m is the radius of the earth. In addition, we have the relationships

where t is time, U is the velocity of the vehicle, and a is its acceleration. The weight of the vehicle, acting in the negative r direction, is

Utilizing these relationships, we obtain the expression

This is a simple example of a so-called differential equation. We seek its solution, r = f(t). An excellent reference for problems of this kind is the book by Boyce and DiPrima (1996). They provide the following solution to this equation:

where c₁ and c₂ are constants and is the so-called frequency of oscillation. The values of c₁ and c₂ are determined from the “initial conditions”: (a) r = R when t = 0 and (b) U = 0 when t = 0. It is easily established that c₁ = R and c₂ = 0. Accordingly, the answer is

in which is the period of oscillation. This equation describes simple harmonic motion.

Substituting the numerical values of R and g₀ into this expression, it is determined that the period of oscillation is T = 5,060 seconds = 84 minutes 20 seconds. This is the time needed for a complete oscillation, that is, the round trip of our vehicle. The duration of the one-way journey is T/2 = 42 minutes 10 seconds.

The following expressions for the velocity U and acceleration a are obtained by using equations (11.2) and (11.5):

It is observed that the maximum acceleration (or deceleration) is g₀ (or –g₀) and occurs at t = 0 (or T/2). The maximum velocity is experienced when the vehicle passes through r = 0 at t = T/4; its value is The average velocity during the journey is

This is really fast! In fact, assuming that the speed of sound in air is C = 340 m/s, we will be moving at about Mach 23 as we pass through the center of the earth. But do not forget that there is no air in the shaft.

In any event, if a tour group from Botswana wants to spend a surfboarding weekend in Hawaii, they can get there in about 42 minutes. Likewise, if people want to escape the high cost of living in Honolulu, they can be in Botswana after a 42-minute trip—not even time for a nice nap.

Startling Piece of Information 1

It turns out that the period of oscillation, is the answer to our problem whether or not the shaft passes through the center of the earth. With reference to figure 11.1, suppose that the shaft is inclined at some angle α connecting any two points on the earth's surface. The period is still the same and, regardless of the angle α, the one-way journey is always 42 minutes 10 seconds.

On this point we shall not go through the details of the analysis; however, you may want to do the computations. Suffice it to say that the length of the shaft is L = 2R cos α and so the journey is shorter. However, as we could show, the average velocity, is less because the average value of the gravitational force is less. Accordingly, with the cosine terms cancel and the period is the same as before.

Startling Piece of Information 2

For this we need Newton's law of gravitation, which is given by the equation

where m is the mass of our vehicle, M is the mass of the earth, r is the distance between the centers of gravity of the two masses, and newton is the gravitational constant. Now at the surface of the earth (r = R), the force exerted by gravity on the vehicle is F = mg₀. Substituting this relationship into (11.7) gives Multiplying both sides of this expression by and noting that the density of the earth, by definition, is where is the volume, we obtain the relationship,

Consequently, with we establish the identity,

or, in MKS units, where ρ is the density of the earth in kg/m³.

These two startling pieces of information tell us that the period of oscillation, T, (a) is independent of the angle of the shaft, α, and (b) depends only on the density, ρ.

Table 11.1 lists some data and computed results involving the eleven well-known bodies of our solar system. It is interesting to note that the earth has the largest density of any body in the solar system, although the densities of Mercury and Venus are only slightly less. By a substantial margin, Saturn's density is the smallest. These densities are reflected in the computed periods of oscillation, T, shown in the table. Although there may be numerous disadvantages to living on planet earth, at least we have the shortest transglobal commuting time anywhere in the solar system.

TABLE 11.1

Motion of an object in a shaft Values of various parameters for the bodies of the solar system.

Source: Audouze and Israël (1985).

Values of the maximum velocity, are also listed in table 11.1. We recall from chapter 10 that the so-called “escape velocity” is The relationship between the two velocities is interesting.

Example 1: Falling Through a Bottomless Well

This problem of a shaft through the center of the earth is discussed by Perelman (1982) in his book Physics Can Be Fun. Essentially, he describes a case in which a “bottomless well” extends from a high plateau in eastern Ecuador (elevation 2,000 m), passes through the center of the earth, and comes out at sea-level Singapore.

To elaborate on Perelman's scenario, a foreign agent jumps into the Ecuadorian end of the shaft and, about 42 minutes later, comes sailing out of the Singapore exit at a velocity of around 715 km/hr. He peaks out at a height of 2,000 m above Singapore, hastily takes video movies of the harbor defense works, and, after a fall of approximately 20 seconds, disappears into the shaft. Following another 42-minutes journey he is back in Ecuador. He climbs out of the shaft and—his mission accomplished—vanishes.

FIG. 11.2

Dimension diagram of Northeast Corridor gravity-driven maglev transportation system.

Example 2: A High-Speed Transportation System

Instead of trying to construct a shaft connecting Ecuador and Singapore or Honolulu and Botswana, we decide to start with something less complicated: a high-speed transportation system featuring a tunnel connecting Washington, D.C. and Boston.

First, what is the length of such a Northeast Corridor tunnel? With reference to figure 11.2, we can calculate the length C of the arc of a great circle from the equation

in which are the latitude and longitude of Washington and are those of Boston. The angle a is the arc angle defined in the figure. Clearly, where R is the radius of the earth. We obtained (11.10) from the section on spherical trigonometry in the comprehensive mathematics reference book by Gellert et al. (1977).

The following information concerning the coordinates of Washington and Boston is obtained from The World Almanac (1994): Washington, longitude latitude Boston, longitude latitude Substitution of these coordinates into (11.10) gives a = 5.68° and consequently C = 632 km. It is easily determined that the shaft angle α = 87.16°, and the shaft length between the two cities is L = 2R cos α = 631.23 km.

By sheer coincidence, the path of our Northeast Corridor tunnel passes directly under New York City approximately midway between Washington and Boston. The depth of the tunnel at this midway point is km.

As we pass directly under New York City, our vehicle will be moving at its maximum velocity, m/s or about 1,410 km/hr. Of course, it will not be possible to stop the vehicle at an underground station under New York. However, we could drop off some mail—assuming we have a sturdy mail catcher. The average velocity for the entire trip, The duration of the trip, of course, is minutes 10 seconds.

The journey on our high-speed vehicle will be pleasant, though perhaps a bit dull. After activating the magnetic levitation cradle and passing through the air chamber seals, our vehicle begins its trip by going down a mild slope of a/2 = 2.84°. The acceleration, is initially 0.487 m/s² (which corresponds to about 11 miles per hour in 10 seconds); this is the maximum acceleration.

Large vacuum pumps keep air out of the tunnel (to eliminate air resistance). The magnetic levitation system suspends the vehicle (to eliminate rail resistance); it is also designed to counteract Coriolis forces caused by the earth's rotation. Tunnel walls are insulated to eliminate heat and moisture intrusion.

It will be necessary to provide power for the vacuum pumps and for the magnets. For the latter, power will be minimized with the use of cryogenic superconducting materials.

However, since our maglev vehicle is “falling downhill” due to the force of gravity and then “coasting uphill,” there will be no power costs for propulsion. Further, there will be no costs associated with climate and weather and other environmental factors. Finally, since the duration of the journey is only about 42 minutes, there will be no costs involved in providing movies, headsets, or expensive meals for the passengers. Our profit should be large.