Astronomical Games: July 2002

Of Matter's Weight and Fall

The key to weightlessness is falling as freely as possible

By his gates of breath / There lies a downy feather which stirs not: / Did he suspire, that light and weightless down / Perforce must move.

King Henry IV Part 2, Act IV, scene v

PEOPLE GENERALLY attach more significance to a saying when someone well-known or important says them. That's human nature, of course, but all the same it's a little dangerous. There's always the possibility that the sayings don't really convey what the quoted people meant very well, either because they misspoke or because the sayings are too easily misunderstood.

For example, in an interview in the Saturday Evening Post, nearly three-quarters of a century ago, Albert Einstein was quoted as saying, "Imagination is more important than knowledge." What he apparently meant was that anyone could accumulate currently known facts about physics, or biology, or whatever, but the scientists who advanced the field the most were those who were able to combine that knowledge with creative speculation, and that obviously requires an active imagination.

Inevitably, however, Einstein's quote has been taken out of context, so that various folks who evidently cannot conceive of more than two levels of importance (all or none) take it to mean that imagination is all you need, and to heck with knowledge. I wouldn't be surprised if it's been used to excuse poor academic performance. After all, Einstein said you didn't really need to know anything, all you needed was your imagination.

Well, he didn't say it, but that's what many people think he said, and so this idea that knowledge isn't important spreads quickly via bumper sticker and marketing pamphlet.

That's one example, here's another. In A Brief History of Space and Time, the British physicist Stephen Hawking (b. 1942) defines weight as "the force exerted on a body by a gravitational field." He goes on to say that it is proportional to but not equivalent to its mass.

A few people have written to me, claiming that Hawking's definition shows that there is no such thing, really, as being weightless. Why not? Because there is no place in the universe where one is truly free of all gravitational fields, so one's weight must always be some positive (though possibly vanishingly small) amount. I wouldn't disagree with Hawking when it comes to gravity, would I?

Well, far be it for me to dispute Hawking on a matter of theoretical physics, but the plain fact is that weight is not a theoretical concept, and I have as much right as the next person to challenge a definition I think isn't completely thought through.

And, of course, a right isn't worth much without exercising it, so let's get started.

Of the two concepts, weight and mass, weight must be, by far, the earlier of the two to be thought up. Mass was only developed as a separate concept by the English physicist Isaac Newton (1642–1727) in his masterpiece, the Principia Mathematica, whereas even pre-humans must have been aware, at least vaguely, that various objects were harder to pick up than others, not because they were slippery, or too large to handle easily, but because they seemed to "want" to fall down very much, and this "desire" to fall down is one of the first things that we called "weight."

Moreover, weight is a persistent property of an object. Large rocks are heavy. Left to themselves, they remain heavy. It's never the case that you come back later, and what used to be a heavy rock has suddenly become light. It is a property of the object, in other words, not a happenstance of time and space. We might say that it is an invariant of the object. Prehistoric man's definition of weight, then, might have gone something like this:

Definition 1. An object's weight is how hard you have to pull/push/lift in order to pick it up.

When I write "you," I mean anybody. The fact that a strong man can lift a small tree over his head that a small child can't budge, however, does not mean that the tree is somehow lighter for the man than it is for the child. It means that the man can pull harder on the tree than the child can. To put it in Newtonian terms, it means that the man can exert more force than the child. To be a bit more precise, then, we should rephrase Definition 1 as follows:

Definition 2. An object's weight is the minimum amount of force required to lift it up.

Why minimum? Because if you can lift the object up with a certain amount of force, then you can certainly lift it with twice the force; the only difference is that you'll lift the object up faster in the latter case. We can reason out this minimum force in the following way. When you try to lift an object, you are exerting a force on it, which is in direct conflict with the force of gravity. The net upward force on it, therefore, is therefore the force with which you pull up on the object, minus the force with which gravity pulls down:

Fnet = FpullFgrav

Since Newton's second law of motion tells us that F = ma (see "The Grand Illusion"), and the acceleration due to gravity is denoted g, Fgrav = mg, so

Fnet = Fpullmg

In order to lift the object, the upward force on it must be positive, so the minimum force you need to apply must be just enough to counteract the pull of gravity. In that case, we have

Fpull = mg

and since Definition 2 equates that minimum pull with weight, we can write more quantitatively:

Definition 3. An object's weight is equal to its mass (that is, the amount of matter it contains) times the acceleration due to gravity.

And that's the definition that Hawking gives in his book. Weight is no longer an invariant property of an object, but depends on two other things that might or might not vary from time to time and place to place. In everyday experience, of course, neither mass nor gravitational acceleration changes enough to notice. An object's mass can only be changed by changing the object itself, either by adding mass to it or taking it away, and in both cases it can be argued that it's no longer the same object.

What about gravity? Before Newton discovered the law of universal gravitation, no phenomenon existed that suggested that the acceleration due to gravity (g = 9.8 meters per second squared) was anything less than a constant—at least, not in those places where it applied at all. Once you factored out other effects such as air resistance, objects fell just as fast at the top of the highest mountain as they did at the bottom of the deepest valley. To be sure, objects such as the Moon, and the Sun, and the other planets and the stars—these all moved as though they were not subject to gravity, but then, that was expected. Gravity was understood to be an earthly phenomenon, not a celestial one.

But then came Newton, and his discovery that the gravitational force between two point objects depended only on the masses of the two objects and the distance separating them. Of course, there's no such thing in real life as a point object, but Newton took care of that, too, by inventing the calculus and showing that his law could still be used for objects like planets and stars provided that they satisfied some simple conditions, like radial symmetry. (A planet may be composed of different substances at different depths as you dig down from the surface to the center, but it doesn't matter very much where you start digging.) Let's start with Newton's formula:

Fgrav = Gm1m2/d2

In earthly scenarios, we're talking about the gravitational force between the Earth and some other, much smaller object (such as an anvil, or a basketball). Since the Earth is so much larger, let's denote its mass by M, and that of the other object by m. In that case, the above formula can be rewritten as

Fgrav = GMm/d2

If the Earth's gravity is the only force acting on an object, we can obtain an expression for g by using Newton's second law of motion:

mg = Fgrav = GMm/d2

and dividing by m on both sides gives us

g = GM/d2

where G is the universal constant of gravitation, M is, again, the mass of the Earth, and d is the distance between the center of the Earth and the object. The universal constant of gravitation is, as far as we know, constant, and the Earth's mass is very nearly constant (in relative terms), but the distance separating an object from the center of the Earth isn't quite constant. If only we could raise ourselves high enough, we should be able to easily detect a change in the acceleration due to gravity.

However, the Earth is very large in human terms, so that its radius, which is the distance between us and its center, is just about 6,375 km (about 3,960 miles). We would have to go up about 64 km just to effect a 1 percent change in the distance between us and the Earth's center, and a 2 percent change in the acceleration due to gravity. No wonder that g was considered constant—for all ordinary purposes up through the 20th century, it was.

But then humans landed on the Moon, with its dramatically smaller size and mass. How does that change things? The mass of the Moon is only 1/81 that of the Earth, while its radius is about 3/11 that of the Earth. Equation 2 tells us that the Moon's surface gravity, then, is only (1/81)/(3/11)2, or about 1/6, times that of the Earth. If you weighed 150 pounds on the Earth, you would weigh only 25 pounds on the Moon.

This illustrates dramatically that an object's weight (by Definition 3) depends not only on the mass—the amount of "stuff" the object contains, but also on how hard gravity pulls down on the object. On the Earth, we are used to a certain correspondence between mass and weight. If an object weighs 10 pounds, it takes 10 pounds of force to hold it up against the force of gravity. It also takes 10 pounds of force to accelerate it by a certain amount (9.8 meters per second squared) in a sideways motion, even though this motion isn't affected by gravity. We are so used to this correspondence that we don't even think about it.

If we ever establish permanent bases on the Moon, however, it is something that we must train ourselves to think about. The fact that an object that weighs 150 pounds on the Earth only weighs 25 pounds on the Moon does not mean that the mass of the object has been reduced by a factor of six. You could carry a 150 pound-mass bag of rocks in one hand on the Moon, since it weighs only 25 pounds, but if one strikes you in the knee at even 5 m/s, it will be as though a 150 pound bag hit you, since the bag's momentum is dependent in part on its mass, not its weight, and you will be severely injured.

Of the eight major planets besides the Earth, only four—Mercury, Venus, Mars, and Pluto—have any solid surface to speak of, but let's pretend that each of the four gas giants—Jupiter, Saturn, Uranus, and Neptune—have solid surfaces just under the top of their cloud layers, How much would a 150 pound person weigh on each of the planets?

For example, Mercury has a radius of about 0.383 Earth radii (Re), and and a mass of about 0.0553 Earth masses (Me). Its surface gravity, by Equation 2, is (0.0553)/(0.383)2, or 0.377 times that of the Earth. A 150-pound person (such as myself), therefore, would weigh 150 pounds, times 0.377, or 56.5 pounds, on Mercury.

Without leading you through the straightforward but tedious computations, here are the corresponding results for all of the major planets in the solar system.

Planet Radius Mass My Weight
Mercury 0.383 Re 0.0553 Me 56.5 lb
Venus 0.949 Re 0.815 Me 136 lb
Earth 1.000 Re 1.000 Me 150 lb
Mars 0.533 Re 0.107 Me 56.8 lb
Jupiter 11.2 Re 318 Me 380 lb
Saturn 9.45 Re 95.2 Me 156 lb
Uranus 4.01 Re 14.5 Me 136 lb
Neptune 3.88 Re 17.1 Me 171 lb
Pluto 0.178 Re 0.00213 Me 10.0 lb

Note that surface gravity, and hence weight (according to Definition 3) is not a simple function of mass, nor of size, nor even of density (as it turns out). The Earth is the densest of the major planets in the solar system, but my weight on Earth is merely fourth greatest. However, low density and small size certainly imply a low surface gravity—the gravity on the surface of Pluto is only about 40 percent of what it is on even the Moon.

Now let's take a different tack that will, for a while, seem unrelated.

The atmospheric pressure at the surface of the Earth—about 14.7 pounds per square inch—is due to the weight of the atmosphere. That is, if you were to take all the atmosphere in a square-inch column all the way from the ground to the top of the atmosphere (which isn't well-defined, but the atmosphere above about a couple of hundred kilometers is negligible), it would weigh about 14.7 pounds.

Naturally, since the amount of air above you diminishes as you rise in altitude, so does the atmospheric pressure, so that every 5.6 km or so, near the surface of the Earth, the pressure drops by half. In other words, at an altitude of 5.6 km, the atmospheric pressure is only 7.35 pounds per square inch.

The Earth's atmosphere is sufficiently dense, especially at the surface, that the difference in pressure over small changes in altitude can be detected, in the form of buoyancy. Imagine a 1 foot cubical helium balloon (enveloped in some hypothetical weightless and inelastic material, so that the elasticity of the balloon doesn't pressurize it). At standard temperature and pressure, this balloon weighs about 5 grams, or about 0.011 pounds.

Now, if the bottom face of the balloon is on the ground, the top face is at an altitude of 1 foot. The atmospheric pressure at an altitude of 1 foot is less than it is at the surface by the weight of 12 cubic inches of air per square inch, or 0.00056 pounds per square inch. Since the top and bottom faces of the cubical balloon have a surface area of 122 or 144 square inches, the atmosphere pushes up on the top of the balloon by 0.00056 times 144 or 0.080 pounds more than the top pushes down. There is also air striking the balloon from the sides, but since the air pressure at any given altitude is the same, the effects of the air striking the sides balances out to zero. Since the helium in the balloon only weighs 0.011 pounds, there is a net lift on the balloon of 0.069 pounds. This is the reason that helium balloons rise. (Of course, in practice, even with inelastic mylar balloons, the mylar itself has some mass, and the net lift is less than it would be otherwise.) This is the buoyancy force, and it's equal to the weight of the displaced air. The only thing working against buoyancy is the weight of the displacing helium, and since helium is lighter than air, buoyancy lifts the balloon against the force of gravity.

Something very similar happens with floating objects. For example, a given piece of iron might weigh 5 pounds. We can measure this by hanging the iron on a spring. The weight of the iron deforms the string, stretching it out. The amount of stretching depends only on the weight of the iron and the so-called spring constant of the spring, not on any other details about its shape or construction. This property of springs was first worked out by the English physicist Robert Hooke (1635–1703).

However, if we submerge both the iron and a spring scale, and attach the iron to the spring underwater, we find that the spring doesn't stretch as far as it did above the water, so that it might appear that the iron weighs somewhat less—perhaps about 4.3 pounds. This is because the water directly underneath the iron pushes up on it more than the water above it pushes down. Furthermore, the difference in the applied forces is equal to the weight of a volume of water equal to that of the iron.

For this reason, many people count buoyancy against weight. Again, if you take a piece of wood that weighs 2 pounds above water, and try to weigh it underwater, you'll find that it no longer weighs 2 pounds. In fact, it doesn't stretch the spring at all; it compresses it, as the buoyancy force is enough to make the wood begin to rise, a rising that is stopped only by the compressing spring. This is because the buoyancy force—equal to the weight of the displaced water—is actually greater than the force of gravity on the wood. For a given piece of wood, the buoyancy force might be 3 pounds, creating a 1 pound upward force on the wood.

If we detach the wood from the spring, we find that the wood accelerates upward to the surface. But what happens when the wood reaches the surface? It certainly doesn't keep accelerating into the air and out into space. Rather, it stops rising, because if it goes into the air, the buoyancy force is no longer the weight of the water displaced by the wood, but the weight of the air displaced by the wood. Since air is about 750 times lighter than water, the buoyancy force on the wood is correspondingly less in air. If we were to weigh the wood in a perfect vacuum, we would find that it to weigh more than it does under normal circumstances, but the effect would be so small as to be unnoticeable without precision instruments.

For that matter, the wood doesn't rise entirely out of the water, either. It only rises until the weight of the displaced water equals the weight of the wood on dry land. If the wood is two-thirds submerged and one-third above water, we can easily calculate that the displaced water would weigh two-thirds of 3 pounds, or 2 pounds. Since that perfectly counteracts the dry weight of the wood, the wood is now in equilibrium and floats, more or less placidly, on the surface of the water.

We can turn it around: We can say that if an object sinks, it must be completely submerged, and therefore displaces its entire volume in water. On the other hand, if it floats, it only displaces as much water as is needed to counteract the force of gravity on the wood, and therefore a floating object displaces its weight in water. This is the central principle of buoyancy, first glimpsed by the Greek scientist, Archimedes (287–212 B.C.).

However, I have always found this counting of buoyancy against weight to be a little artificial, and I'll explain why. If we do count buoyancy against weight, then a floating piece of wood is weightless, since it is in equilibrium and the net force on it is zero. We can check that by attaching the wood to an unstretched spring, and noting that the spring remains unstretched.

But what happens if we lift the spring scale and the wood up from the water? The spring surely doesn't continue unstretched. No, the moment the wood is lifted from the water, buoyancy effectively is removed from the equation, and the wood is observed to weigh its full amount of 2 pounds. (We'll assume that it hasn't gotten waterlogged.) In fact, if you lift the whole contraption slowly enough, you'll see the weight slowly increase from 0 pounds to 2 pounds as the wood displaces less and less water, until at last it is fully out of the water. In other words, this kind of weight isn't a persistent property of the wood.

In still other words, the wood isn't weightless unless it doesn't move at all, which means that counting it as weightless is only good when one is doing statics—that is, the physics of motionless objects under force. And if one is doing statics, one can simply separate the forces into two parts: the weight of the wood, and the force of the buoyancy. There's no need to combine them and call the wood weightless.

What's more, one doesn't feel weightless when floating in water, say, unless one remains perfectly still. The moment you try to get up out of the water, you are no longer in equilibrium. You can't effortlessly slide out of a swimming pool onto the deck because the water pressure underneath you is no longer sufficient to counteract the force of gravity. You have to drag yourself by climbing a ladder, or pushing down on the edge of the pool, or something.

Then, too, buoyancy is something that can only be provided by a fluid—either a gas or a liquid. A solid cannot provide buoyancy. But then, consider the following scenario. A piece of wood is floating on a body of water. This wood is counted as weightless. Then the water slowly cools and freezes. First, a thin skin of ice forms on the surface of the water and the underside of the wood; eventually, all the water is frozen into ice. But ice is an ordinary solid; it cannot provide any buoyancy. At what point does one say the wood is no longer weightless? When the skin of ice first forms? When it first resists the movement of the wood? When the entire body of water is frozen solid? Even when all the water becomes ice, the water molecules are not stationary; they are still vibrating around. They simply vibrate less, and most of them are locked into a crystalline lattice. And when the water thaws out, at what point does the wood become weightless again?

There are only four types of forces in nature. Two are nuclear forces and can be safely ignored so long as we are talking about large objects like pieces of wood or bodies of water. That leaves gravity and the electromagnetic force. Gravity alone can only pull the wood down; it can't pull it up. That means that if the wood isn't moving, regardless of whether it's lying on ice or in water, the force that's keeping it in equilibrium must be electromagnetic in nature. Well, if the force that the H2O exerts on the wood is electromagnetic and equal in magnitude and direction no matter if it's liquid or frozen, why should the wood be counted as weightless in one case but not in the other?

But let me not belabor the point. I've made my case that buoyancy shouldn't count against weight and I won't argue it any further.

But if buoyancy doesn't count against weight, does anything? Or can Definition 3 stand as is?

I think there is, and Hawking of all people must know it. In "The Grand Illusion," I mentioned Einstein's equivalence principle, which states that for point particles, there is no experiment that one can run that will distinguish between a uniform gravitational field on one hand, and uniform acceleration in the absence of gravity on the other.

One can also use this principle to show that if one accelerates in the presence of a uniform gravitational field, there is no way to distinguish between that and the absence of gravity. And if there is no gravity acting on an object, then the object must be weightless.

And how must one accelerate in order to simulate the absence of a gravitational field? The answer is simple and I'm sure that Galileo (1564–1642) would have known the answer if someone had simply framed the question appropriately to him. In order to simulate the absence of a gravitational field, one must simply accelerate exactly according to the gravitational pull. That is, one need only free-fall.

Since free-falling objects of all masses fall at the same rate in the same gravitational field, a piece of wood or iron or anything at all, suspended from a spring scale when both are free-falling, will read zero weight. There is no difference between a free-falling body under the influence of a gravitational field, and a body in the absence of any gravitational field at all.

Well, that's not precisely true, in part because a real gravitational field such as the Earth's is never truly uniform and this induces tidal forces, and in part because the body itself produces its own gravitational field, and in part because there are other forces acting on the body, such as atmospheric resistance. But for all intents and purposes, an accelerating frame of reference can and should be counted against weight (or for it, if the acceleration is in the direction one is counting as up).

One doesn't encounter such scenarios for long ordinarily, of course. An object let go from a height of 1 km can only free-fall for 1 km, at which point its weight increases suddenly from zero to a very high number, much higher than its ordinary weight, since it takes less time to come to a stop than it took to fall for 1 km, so the deceleration experienced by the object is correspondingly greater.

I can hear you saying now: But why is this sudden change in weight OK, when the sudden change in weight in frozen or liquid water isn't? The difference is that in the case of the wood in water, the sudden change of weight is caused by an arbitrary stipulation that the state of matter of the substance exerting the electromagnetic forces on the wood determine whether it contributes to weight or not. The actual force doesn't change: not its amount, nor its direction.

In the case of the free-falling object, however, at one moment, there is only one force on the object: gravity. The next moment, there is another force: the electromagnetic force exerted by the ground, violently upward. There is a quantitative change in the net force applied to the object, and we shouldn't be surprised that its weight would change under those circumstances.

However, in point of fact, a free-falling object doesn't have to strike the ground. It can at least delay the impact by moving sideways a little, and in so doing, take a little longer to fall, since the surface of the Earth curves away from it. In fact, if it moves sideways fast enough (about 7 km/s near the surface of the Earth), it will continue moving sideways and never strike the Earth at all. The object is then in orbit and is still in free-fall, even though the fall never ends.

And if the object is a spacecraft, everything aboard it is in a free-falling frame of reference. Inside the spacecraft, we can say, with equal validity, that there is a force of gravity, but the frame of reference accelerates along with it; or we can say that the environment within the spacecraft lacks gravity at all. For that reason, setting aside the non-zero size of real objects, everything—the astronauts, the pencils, the experiments, etc—everything is weightless. (This is of great value in determining what the effect of zero gravity is on various biological activities, for example, even though the force of gravity is nearly as strong in low Earth orbit as it is on the surface.) This line of reasoning yields yet another definition of weight:

Definition 4. An object's weight (in a frame of reference) is equal to the total net force upon that object, minus the force on that object (in that reference frame) due to gravity.

But if we eliminate the gravitational force on an object, that leaves only the electromagnetic forces, so long as we are talking about objects far beyond the subatomic scale. That means that we can re-formulate Definition 4 in a much simpler way:

Definition 5. An object's weight is equal to the total electromagnetic force upon that object.

Definition 5 tells us that we can measure an object's weight by measuring the stretching of a spring (an electromagnetic force), the pressure on a plate (another electromagnetic force), the tug on my left arm (another electromagnetic force)—in short, any method at all, so long as one is measuring only electromagnetic forces. And that is the definition of weight that makes sense to me.

With that, I'll set aside the matter of gravity for now. (At last!) In my next essay, I'll consider the Earth's tilted axis, its non-circular orbit, and what these two things have to do with that funny figure-8 found in the South Pacific on your more exclusive globes.

Copyright (c) 2002 Brian Tung