Good day, student of introductory astronomy!

Since you were interested enough to inquire, I have decided to pass along the following to you. This is in regards to our discussions of the concepts of weight, "weightlessness," and gravity.

First, a little aside about the difference between weight and mass. My mass is approximately 75 kg; it is a measure of how much material in protons, neutrons and electrons composes me. My weight is a measure of the force of gravity between me and whatever planet or otherwise massive body I am standing on (usually Earth).  On the surface of the Earth, it is about 735 Newtons (or 165 pounds in the outdated English set of units).  My mass is the same no matter what I planet I am on, but my weight will differ - weight and mass are not the same thing! More on this below...

Many people have the misconception that somehow once you leave the Earth, you become weightless. But does an astronaut standing on the Moon have weight? Indeed he or she does, about 1/6th that on the surface of the Earth. This is because the acceleration due to gravity on the Moon's surface is about 1/6th that on Earth. Weight is usually defined as the local force of gravity between a mass and some other massive body: the product of an object's mass and the local acceleration due to gravity. Thus the weight of an object whose mass is m is a force (F = m x a), and the acceleration of mass m due to gravity is: GM/(R+h)2, with M being the mass of the planet, R the planet's radius, h being the height of the center of mass m above the planet's surface, and G the gravitational constant. Thus your weight is the force gravity exerted by the Earth on you: Fgrav = x GM/(R+h)2. You can diminish your weight by moving away from the center of the planet you are on (increasing h). For example, you have slightly less weight flying in a jet airliner at 40,000 feet than you do at the surface of the Earth. Or you can change your weight by going to some planet or moon whose gravitational acceleration (= GMplanet/R2planet) is different than Earth's.

But "weight" carries with it a meaning beyond just the local force of gravity between an object and its planet (or moon). As usually defined, we measure weight only in the absence of net accelerations (or at least when they are very small). That is, when you stand on the ground, the ground pushes up on you with as much force as you push down on it (due to the force of gravity between you and the Earth). Since these forces are equal and pointed in opposite directions, no net force is present, and so your center of mass does not accelerate. If, however, the force between you and the Earth (gravity) were larger than the force supporting you from below (usually intermolecular electric forces of the material you're standing upon), a net force would result and you would fall (accelerate), like a truck breaking through a rickety old bridge. When you step onto a scale, the force of gravity between you and the Earth causes you to accelerate, compressing the springs inside the scale. You continue to fall until the spring force in the scale (which increases in proportion to the amount of spring compression) exactly matches your gravitational force. Because the net force on you is then zero, your acceleration stops, the scale reading comes to equilibrium and your weight is measured.

When we say that an astronaut is "weightless," this can mean one of two things. First, that he or she is so far away from a massive object (asteroid, moon, planet, star), that the value of the force of gravity on mass m = GMm/(R+h)2 is vanishingly small (i.e., h is enormous in comparison to R). Here, the force of gravity is so small that one can say that one is "weightless." Second, that he or she is continuously "falling" around (e.g., an orbiting astronaut) or toward (e.g., a falling apple) a massive object. Now, astronauts (or moons) in orbit around a planet still exert a force of gravity. It isn't that the force of gravity is very small or zero - if it were, the moon or astronaut would not remain in orbit! It's just the fact that the astronaut (or moon) is continuously falling (accelerating) around the planet1 with sufficient angular momentum that he/she misses the planet's surface. It puzzled Newton2 why the inertial mass m in F = ma is equivalent to the gravitational mass m in Fgrav = GMm/d2.

While falling freely, a mass effectively has no weight - as discussed above, we generally measure what we call weight only in the absence of (net) accelerations of the mass in question. As a concrete example of this, imagine tying a scale onto the bottom of your feet and jumping out of the jet airliner at 40,000 feet. While standing inside the jet, you had weight, and the scale would have told you so. But during your fall, even with your feet pointed downward the scale will read zero because the springs and the scale are falling at the same rate you are, with no opposing force from below (neglecting air resistence, another force)3. This zero reading tells you that you are effectively "weightless." This result is not due to the method of measuring weight, but is the result of how we define weight. It means just a bit more than simply the force of gravity2.

thanks for your interest,

Professor Korista


1Or, without sufficient angular momentum, falling directly towards the planet. However, that results in a rather unhappy conclusion (splat!), so let's stick with falling around, i.e. orbiting, a planet.
2The arguments presented here are those that Mr. Isaac Newton would have made, inspite of the fact that he did not understand why gravitational and inertial masses are apparently equivalent. In reality, however, things are a bit more interesting than what is described above. Albert Einstein's theory of General Relativity finds that an astronaut falling freely toward or around (i.e., orbiting) a massive body and an astronaut completely isolated in space are completely equivalent  (ignoring non-uniform gravitational fields, i.e., tidal effects) - indeed, both are weightless.  Thus the effective ``force'' of gravity goes away when one surrenders completely to it! Indeed, Einstein also showed that the effects of acceleration and gravity are equivalent - an elevator accelerating upward (the direction of your head) with a constant acceleration of 9.8 meters per square second is locally equivalent to the acceleration of gravity at the surface of Earth. Or put another way, according to the theory of General Relativity gravity is not a force at all, but curvature in space-time (4-dimensional geometry) due to the presence of mass-energy.  A freely falling astronaut's (or apple's) path is the simplest one possible in curved space-time, and is also the most natural one (standing motionless at the surface of the Earth requires the electrical forces between atoms of the Earth's crust pushing you upward - the only `actual' force present!).  Einstein's equivalence principle, a primary prediction of the theory, has been tested repeatedly in experiment and so far has been shown to hold to better than 1 part in 1 trillion.  This fact provides an answer to Newton's conundrum - no forces are involved in a freely falling body (apple, Moon, astronaut, ...) of mass m, which emphasizes why it is called "free-fall".  However, the purpose of this page is to convey what Mr. Isaac Newton would say, whose predictions of nature are sufficiently accurate under most ordinary circumstances. Go here for additional discussion of this profound phenomenon.
3Meaning that the quickest way to lose weight, albeit temporarily, is to arrange to be falling. But I do not recommend this method! What you really want to do is reduce your mass.


Kirk Korista
Professor of Astronomy
Department of Physics
Western Michigan University