To answer the question, we certainly expect that when the distance (i.e., height) r of an object (a ball or projectile) gets larger and larger, its velocity U must approach zero. Accordingly, we set r = ∞ and U = 0 in equation (10.13). This yields the very simple expression
in which U0 is the so-called “escape velocity.” With g0 = 9.82 m/s2 and R = 6.37 × 106 m as the radius of the earth, we determine from this equation that U0 = 11,180 m/s. Thus, if an object is propelled from the earth's surface at a velocity of 11.18 km/s or more, it will escape from the earth's gravitational field.
Incidentally, the escape velocity from the moon (g0 = 1.62 m/s2 and R = 1.74 X 106 m) is U0 = 2.37 km/s. For the sun (with g0 = 274 m/s2 and R = 6.96 X 108 m), the escape velocity has the value U0 = 618 km/s.
For the region r > R (i.e., outside the earth) the gravitational force on an object is given by (10.10). By the same token, when r < R (i.e., inside the earth) the force is given by
This result is based on the assumption that the earth is homogeneous. It is well known that this is not precisely the case; however, for our analysis it is a reasonable assumption.
Analysis of Three Exciting Situations
Situation 1
In a scenario we leave to Jules Verne or other science fiction writers to explain, we are at the center of the earth (r = 0) and wish to launch our escape vehicle up an air-evacuated shaft, extending to the surface of the earth, with an initial velocity U* sufficiently large just to reach the surface (r = R). What must be the magnitude of U*?
From equation (10.15), we deduce that g = g0r/R. Substituting this into (10.11) and proceeding as before, we obtain
which gives
As required, U = 0 when r = R. Therefore, . Thus, a launch velocity U* = 7.91 km/s is needed to get our vehicle from the center of the earth to the surface of the earth. A word of caution: at the instant of arrival at the surface, make sure that the vehicle is securely attached to a landing hook; otherwise, it will fall back down the shaft.
How long will it take to go from r = 0 to r = R? From equation (10.17), with and using U = dr/dt, we have
This equation can be expressed in the form
where t0 is the transit time. Utilizing a table of integrals we determine that . With R = 6.37 X 106 m and g0 = 9.82 m/s2, the answer is t0 = 21 minutes 5 seconds.
Situation 2
Alternatively, Jules Verne plans our escape not only from the evils raging at the center of the earth but also from those in great abundance at its surface. We shall be blasted off from r = 0 to a distant galaxy, infinitely far away. What initial velocity U*0 is needed to accomplish this?
It should be clear that we can start with equation (10.17) with U*0 replacing U*. Also, we note that it is necessary to pass through the earth surface station at r = R with the escape velocity, given by (10.14), of .
With this information, it is determined that This means that with a launch velocity U*0 = 13.70 km/s, we can blast off from the center of the earth and go all the way to infinity. After the blast off, we pass through the earth surface station in 8 minutes 16 seconds, moving at a velocity of 11.18 km/s.
Situation 3
At the moment of our above-described departure from the center-of-the-earth launching station to go to infinity, a dreadful mistake is made. Instead of punching the blast-off velocity button, someone hits the button. What happens?
PROBLEM We leave the analysis of this situation to you as a homework problem. Here are the main results.
Our initial velocity, of course is 11.18 km/s.
We pass through the surface-of-the-earth station in 10 minutes 30 seconds at a velocity of 7.91 km/s. To our great dismay, this is much too slow to get us to infinity.
We reach a maximum height of r = 2R (i.e., a distance of R = 6,370 km above the surface of the earth) 34 minutes 30 seconds after passing through the earth surface station or exactly 45 minutes after blast off.
After reaching maximum height, we begin our 34 minute 30 second fall back to the earth's surface. Theoretically, we will go back down the shaft and, after another 10 minutes 30 seconds, arrive at our r = 0 launching station with a velocity of 11.18 km/s.
Fortunately, just before we reenter the shaft, a friend turns the air back on. We bail out of our capsule, parachute to safety, and then quickly subdue the evils raging on earth.
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 th
e 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.
12
How Fast Should You Run in the Rain?
There is a steady downpour of rain, you have no raincoat and no umbrella, and you must get from here to there without delay. And instinctively, you want to get the least wet as you make the excursion of specified distance.
If you walk slowly in the rain, only the top of your head and shoulders get wet; your front stays relatively dry as do yo
ur shoes and socks. However, with a slow walk you are out in the rain that much longer. If you jog at moderate speed, your front gets wet and your shoes get soggy from splashing water on the pavement; however, in this case your trip time is reduced. And if you run at full speed, the top of your head and shoulders avoid most of the rain, your front is drenched and your shoes and socks are entirely waterlogged. Even so, in this instance the time you are out in the rain is least.
So the question is asked: with what speed should you move to get least wet? This problem has been examined by Rizika (1950) and by Edwards and Hamson (1990). Both of these earlier analyses included the effect of nonvertical rainfall due to wind but disregarded pavement splashing. We shall do the reverse: neglect wind action but include splashing. This illustration demonstrates the usefulness of differential calculus in solving problems involving maxima and minima.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 9
