Centripetal force, weight, stress and the Earth's equatorial bulge.

by Donald E. Simanek

Disclaimer: This document, like many others on this website, attempts to get across a complex bit of physics in language as simple, conceptual and non-mathematical as possible. This is, of course, the hard way to do it. If we were to use full-blown vector calculus the arguments could be a lot more compact, and far more precise. The reader is strongly urged to consult such treatments in textbooks.

Centripetal Force.

The acceleration of a body is the change in its velocity divided by the time duraton of that change, a = D V/D t. Since velocity is a vector, any change in its size or direction, or both, requires acceleration. A body moving in a curved path is accelerating, even if its speed is constant in size. Therefore, from Newton's law, we know that the net force on it is non-zero.

The diagram shows a body moving with constant speed, V on a curved path of radius R. Two positions are shown, during which the body has moved a distance S along an arc. At the beginning of the time interval the body's velocity is V1. At the end of the interval it is V2. We give them distinguishing subscripts because they have different directions, even though they have the same size. During that time the body moves through angle a. At the right we show a vector diagram of the relation between these velocities and their vector difference, D V.

Now consider the limiting case as the time interval gets very small, approaching zero. The angle approaches zero also. The diagrams have two similar, very skinny triangles. We can write:

D S / R = D V / V

So: V D S = R D V

and: V (D S / D t ) = R (D V / D t ).

But V = D S / D t, and a = D V / D t, so we can write:

V2 = R a, which becomes a = V2/R .

This is the size of the centripetal acceleration vector. The direction of the centripetal acceleration vector is inward toward the center of the arc at any instant, and is therefore perpendicular to the velocity vector at that instant, which is always tangent to the curve.

We can associate this acceleration with the inward (radial) component of whatever net force happens to be acting on the body (of mass m). We call that component the centripetal force, with size Fcentripetal = m a = mV2/R. It, too, is a vector directed radially inward.

Centripetal force is not some new kind of force, but just a convenient name for the radial component of the sum of all of the real forces acting on the body.

A fuller treatment would show that this definition of centripetal force is useful for any kind of motion along a curve of any sort. It is not restricted to circles. Nor is it restricted to constant speed along the path. This works because any physical path is such that a small enough portion of it approximates a circlular arc very well. In calculus, we take the limit as the arc becomes zero length, and speak of the relation between the instantaneous radius of the arc at that point on the curve, the instantaneous velocity at that point and the instantaneous acceleration there. The relation still turns out to be a = V2/R. And if the net force has a component tangent to the arc, that causes an increase or decrease in the body's speed along that arc.

For a calculus treatment of this and many other physics topics, consult Jess H. Brewer's excellent physics tutorial, The Skeptic's Guide to Physics. See the celestical mechanics chapters.

Stress.

Elastic bodies can deform under applied forces. The bonding forces which hold molecules in an equilibrium position in solid and liquid bodies act something like springs. If a spring is stretched by pulling molecules farther apart, these bonding forces increase in size. They also increase in size when the molecules are pushed closer together. We can model this behavior as shown in the figure, with a spring representing such forces acting between chunks of matter. For our limited purposes, we may assume that the spring is approximately Hookian in behavior, that is, it obeys Hooke's law. The law says that the spring tension T changes by amount D T = - KD L where k is the spring's elastic constant and D L is the change in its length, measured from its initial position. If the tension increases, the spring's length decreases. If the tension decreases, the spring's length increases.

We will assume the mass of the spring itself is negligible compared to the mass of the two blocks of matter. Therefore the spring tension exerts forces of equal size on each of them, as shown.

F2 and F1 represent the gravitational force on upper and lower blocks. In equilibrium the lower block must experience an upward force of F1 + F2 (from whatever supports it from below). If these were equal in size, then for equilibrium, the tension T would also be that same size.

Now suppose that the upper block were made smaller in mass. The tension in the spring must get proportionally smaller to achieve static equilibrium. Therefore the spring must increase in length, separating the two blocks a bit.

The same thing would happen if some additional non-uniform gravitational field were applied to the system, such that it exerted greater downward force on the lower block than it did on the upper one. The tension on the spring would decrease and its length would increase. But what if such an external field caused a stronger upward force on the upper block than on the lower one? Again, the tension on the spring would decrease and its length would increase. This becomes important when one discusses Earth tides due to non-uniform gravitational fields from the Moon.

Weight.

Weight may be defined as the force required to keep an object at rest relative to its surroundings. This definition is consistent with most colloqual interpretations of the word (surprise!).

The figure shows a person of mass m standing on a bathroom scale on the surface of the Earth. The scale exerts a force of size W upward on his feet. We call this scale reading the "weight" of the man. If the Earth were stationary, the man would be in equilibrium, and we would have W = mg, where mg is the gravitational force on the man.

Think of the scale as like the spring between masses. It responds to the stress between feet and floor.

Effect of Earth rotation on weight.

But if the Earth is rotating, the man, scale and ground under it are moving with speed V in a circular path of radius R, where R is the radius of the Earth. These objects are no longer in static equilibrium. They all have an acceleration a = v2/R toward the center of the Earth. This centripetal acceleration changes the direction of velocity, but doesn't change its size, because the acceleration vector is always directed toward the center of the circle, and is therefore perpendicular to the body's velocity at any time.

The size of the man relative to the Earth's radius is greatly exaggerated in this diagram. The relative sizes of the forces are also.

Newton's law tells us that when a body of mass m has an acceleration a, the net force on the body must be F = ma. So the size of the net force on this man must be F = ma = mv2/R.

The figure shows the force vectors W and mg, which are the only forces acting on the man. The vector F is their sum. W is directed along the radius of the Earth. Being the radial component of the net force (it is the net force in this case), its size is a = v2/R (the centripetal force). Now compare these two cases. On the non-rotating Earth the man's weight was of size mg. Remember, the weight of an object is the force required to support it, i.e., the force exerted upward by the weighing scale. With the Earth rotating, that force is smaller than before. The contact force between the man's feet and the scale is reduced. But all other such stress forces are reduced as well, within the man, within the scale's springs, within the body of the Earth itself. This causes a slight decompression of these materials, a relaxation of the spring in the scales. In fact, the entire body of the earth expands slightly and the man and scale move outward from the axis of rotation slightly, until forces come into balance with the requirements of rotational stability at the new radius. This is the reason for the equatorial bulge of the Earth due to its own axial rotation.

As we said, the diagrams are exaggerated. Simple calculation shows that the centripetal acceleration is only 0.3% of g. So the net force is only 0.3% of W. The man's weight (registered on the scale) is 0.3% smaller than it would be at one of the Earth's poles. The resulting relaxation of stress in materials is the reason for the equatorial bulge of the Earth, making the equatorial radius 43 kilometer greater than the polar radius.

The "cause and effect" dichotomy.

Here we can get into a tricky dilemma of interpretation. Is the weight reduction at the equator simply due to the smaller gravitational force at the greater radius? Or is it simply due to the reduction in stress between his feet and the scale? This is another unfortunate consequence of using "cause and effect" language in something that is too complex for such a simplistic description. It is also a trap for teachers who would pose multiple choice questions on exams.

Suppose someone says that the reduction in weight is solely due to the increased radius of the equator and the slightly smaller gravitational field at the larger radius. Then if the earth were prefectly rigid and there were no equatorial bulge, you would logically conclude from this that since the radius doesn't change, then the weight (registed on the scale) doesn't change. But this happens to be false. The weight would decrease, as it must, from the fact of the radial acceleration due to the circular motion and Newton's law F = ma.

If someone says that the reduction in weight is due to the decreased stress caused by the stretching of the materials of the Earth, scale and man, we must agree, for that's what the scale mechanism measures, and that's our definition of weight. But this stress reduction is also the reason for the stretch of the earth, and for the increased radius at the equator. And this in turn does decrease the gravitational force due to the Earth at that larger distance. The bottom line is that the gravitational force decreases and the stress decreases, but not in proportion. To say just one of these things is the "cause" of any one of the others is just too simplistic to be useful. And to try to "get by" explaining the equatorial bulge without mentioning stress reduction in materials is somewhat of a cheat.

Is "centrifugal force" necessary here?

Notice that in this discussion we never had to introduce the term "centrifugal force". That term is only useful when one chooses to analyze this problem in a rotating non-inertial coordinate system.

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