People react oddly. After I stood up, I completely ignored my badly sprained (and possibly broken, though it later turned out not to be) right wrist imd lifted my untouched left wrist to my ear.
What worried me was whether my wristwatch were still running.
In fact, I have been told that infants have an instinctive fear of falling, and that this arose out of the survival value of having such an instinctive fear during the tree-living aeons of our simian ancestry.
We can say, then, that gravitational force is the first force with which each individual human being comes in contact. Nor can we ever manage to forget its existenc6, since it must be battled at every step, breath, and heart beat. Never for one moment must we cease exerting a counterforce.
It is also comforting that this mighty and overwhelming force protects us at all times. It holds us to our planet and doesn't allow us to shoot off into space. It holds our air and water to the planet too, for our perpetual use. And it holds the Earth itself in its orbit about the Sun, so that we always get the light and warmth we need.
What with all this, it generally comes as a rather sur prising shock to many people to learn that gravitation is not the strongest force in the universe. Suppose, for in stance, we compare it with the electromagnetic force that allows a magnet to attract iron or a proton to attract an electron. (The electromagnetic force also exhibits repul sion, which gravitational force does not, but that is a detail that need not distress us at this moment.)
How can we go about comparing the relative strengths of the electromagnetic force and the gravitational force? . Let's begin by considering two objects alone in the uni verse. The gravitational force between them, as was dis covered by Newton, can be expressed by the following equation (see also Chapter 7):
Gmr amp;
Fg = (Equation 1) d2 where F, is the gravitational force between the objects; m is the mass of one object; the mass of the other; d the distance between them; and G a universal "gravitational constant."
We must be careful about our units of measurement. If we measure mass in grams, distance in centimeters, and G in somewhat more complicated units, we will end up by determining the gravitational force in something called "dynes." (Before I'm through this chapter, the dynes will cancel out, so we need not, for present purposes, consider the dyne anything more than a one-syllable noise. It will be explained, however, in Chapter 13.)
Now let's get to work. The value of G is fixed (as far ,as we know) everywhere in the universe. Its value in the units I am using is 6.67 x 10-8. If you prefer long zero riddled decimals to exponential figures, you can express
G as 0.0000000667.
Let's suppose, next, that we are considering two objects of identical mass. This means that m = m'. so that mm' becomes mm, or M2. Furthermore, let's suppose the parti cles to be exactly I centimeter apart, center to center. In that case d = 1, and d2 = 1 also. Therefore, Equation 1 simplifies to the following:
F, = 0.0000000667 m2 (Equation 2)
We can now proceed to the electromagnetic force, which we can symbolize as F,.
Exactly one hundred years after Newton worked out the equation for gravitational forces, the French physicist Charles Augustin de Coulomb (1736-1806) was able'to show that a very similar equation could be used to deter mine the electromagnetic force between two electrically charged objects.
— Let us suppose, then, that the two objects for which we have been trying to calculate gravitational forces also carry electric charges, so that they also experience an electro magnetic force. In order to make sure that the electromag netic force is an attracting one and is therefore directly comparable to the gravitational force, let us suppose that one object carries a positive electric charge and the other a negative one. (The principle would remain even if we used like electric charges and measured the force of clec tromagnetic repulsion, but why introduce distractions?)
According to Coulomb, the electromagnetic force be 102 tween the two objects would be expressed by the foflo ' wmg equation:
F. (Equation 3) d2 where q is the charge on one object, q' on the other, and d is the distance between them.
If we let distance be measured in centimeters and elec tric charge in units called "electrostatic units" (usually abbreviated "esu7'), it is not necessary to insert a term analogous to the gravitational constant, provided the ob jects are separated by a vacuum. And, of course, since I started by assuming the objects were alone in the universe, there is necessarily a vacuunf between them.
Furthermore, if we use the units just mentioned, the value of the electromagnetic force will come out in dynes.
But lefs simplify matters by supposing that the positive electric charge on one object is exactly equal to the nega tive electric charge on the other, so that q = q,* which means that the objects . qq = qq = q2. Again, we can allow to be separated by just one centimeter, center to center, so that d2 = 1. Consequently, Equation 3 becomes:
Fe = q2 (Equation 4)
Let's summarize. We have two objects separated by one centimeter, center to center, each object possessing identi cal charge (positive in one case and negative in the other) and identical mass (no qualifications). There is both a gravitational and an electromagnetic attraction, between them.
The next problem is to determine how much stronger the electromagnetic force is than the gravitational force (or how much weaker, if that is how it turns out). To do * We could make one of them negative to allow for the fact that one object carries a negative electric charge. Then we could say that a negative value for- the electromagnetic force implies an attraction and a positive value a repulsion. However, for our pur poses, none of this folderol is needed. Since electromagnetic at traction and repulsion are but opposite manifestations of the same phenomenon, we shall ignore signs.
this we must determine the ratio of the forces by dividing
(let us say) Equation 4 by Equation 2. The result is:
F, q2 (Equation 5)
F, 0.0000000667 M2
A decimal is an inconvenient thing to have in a denomi nator, but we can move it up into the numerator by taking its reciprocal (that is, by dividing it into 1). Since 1 di vided by 0.0000000667 is equal to 1.5 x 101, or 15,000, 000, we can rewrite Equation 5 as:
F, _ 15,000,000 q2 (Equation 6)
Fg m2 or, still more simply, as:
F,. = 15,000,000 (VM)2 (Equat;on 7)
Since both F, and F, are measured in dynes, then in taking the ratio we find we are dividing dynes by dynes.
The units, therefore, cancel out, and we are left with a "pure number." We are going to find, in other words, that one force is stronger than the other by a fixed amount; an amount that will be the same whatever units we use or whatever units an intelligent entity on the fifth planet of the star Fomalhaut wants to use. We will have, therefore, a universal constant.
In order to determine the ratio 0 the two es, we see from Equation 7 that we must first determine the value of qlm; that is, the charge of an object divided by its mass. Let's consider charge first.
AJI objects are made up of subatomic particles of a number of varieties. These particles fall into exactly three classes, however, with respect to electric charge:
1) Class A are those particles which, like the neutron and the neutrino, have no charge at all. Their charge is 0.
2) Class B are those particles which, like the proton and the positron, carry a positive electric charge. But all particles which carry a positive electric charge invariably carry the same quantity of positive electric charge what ever their differences in other respects (at least as far as we know). Their charge can therefore be specified as +I.
3) Class C are those particles which, like the electron and the anti-proton, carry a negative electric charge.