Of course they do. That’s how we produce microgravity on Earth. In fact objects in orbit are constantly falling. But how about a standing person who passes out or otherwise just tips over?. The feet are solidly at 1 G while the head is in (more or less) free fall. Think of the consequences!
Of course people do not weigh less when they are falling. Weight is the result of the effect of gravity on a mass. That is acting on the mass whether they are falling or not. I am not sure what you are calling microgravity on Earth, but free-fall is not microgravity.
Disregarding factors such as air resistance, you weigh nothing at all when you are falling. If you had a scale with you and could somehow step onto it and weigh yourself in mid-descent, the display on the scale would read zero.
But your mass always stays the same, at least under everyday Newtonian conditions.
Weight /= mass.
It most certainly is. We have two microgravity facilities at NASA Glenn Research Center. The 2 second Drop Tower and the 5 second Zero Gravity Facility (ZGF) where I was Facility Manager. At the ZGF we dropped research vehicles in a vacuum with experiments in a sealed chamber on board. The split instant we released the vehicles there are falling at 1/100,000 G. That’s lower then most low Earth orbit conditions. I believe it was Einstein who proved that in a closed “room” there is no way to distinguish free fall from zero G.
For a more perplexing thought consider how we produced 10 seconds of microgravity. The experiments were installed in a sphere about 4 feet in diameter. It was placed atop a hydraulic piston at the bottom of the facility, 500 feet below ground. The piston rapidly accelerated the package upwards and let it loose. It shot to the top of the vacuum chamber 500 feet up, where it reversed direction and fell back down. It is microgravity for the whole ride with NO sensation of “going over the top”.
While it is true that a scale that is in freefall (which is equivalent to what people refer to as “microgravity”) will not register any force, and thus no weight in that reference frame, there is a virtual force (per the Lagrange–d’Alembert principle, often just referred to as the “d’Alembert force”) that is applied within the reference frame of the fixed ground that can be observed as causing the body to accelerate. In essence, this force is doing the virtual work that adds kinetic energy to the body (again, in the Earth reference frame). If you could instantaneously freeze the acceleration of the scale in mid-drop (while retaining its acquired velocity and momentum), it would register the weight of the body just as if it were standing on the surface.
Of course, when the body hits the surface, the impulse that would be registered by a scale will be much larger (transitorily) than just the weight of the body because it has acquired a large amount of kinetic energy relative to the ground frame. The characteristic of that impulse depends on how elastic and ‘squishy’ (deformable) the body is; after all of that energy is expended (through damping, deformation, and other energy-absorbing processes), the force measured by a scale will return to the standard weight in that frame, assuming you can collect all of the body parts and fluids together into a bucket and put them back on the scale.
This statement is ambiguous. Anything in the Earth gravitational field near the surface is going to be accelerating downward at ~1 g = 9.81 meters/sec2 (not “G”, which is the gravitational constant, even though many people misuse it in this fashion) as observed by someone in the ground frame. However, as long as it is in a chamber with essentially no air resistance, and there is nothing else interfering with its motion, an accelerometer the vehicle will register no acceleration, and at the apex, there will be virtually no velocity (other than the components due to the rotation of the Earth…if you could suspend the vehicle for long enough these components would become apparent but are basically inconsequential for most terrestrial purposes).
Stranger
If we assume a person standing on a scale, and model the person as a rigid rod, and then tip the rod over, then while the rod is tipping over, the scale will read some nonzero amount, but less than the full weight of the person, and this reading will change as the angle of the rod changes.
“The split instant we released the vehicles there are falling at 1/100,000 G”
I don’t think it is ambiguous at all. Our release mechanism bends a brittle rod so that it snaps without imparting any vibration to the vehicle. Careful measurements of acceleration show that the drop vehicles abruptly transition from 1 G to zero G when the mechanism is no longer touching the vehicle. There is no transition “slope” from 1 G to zero G. . So it has technically not yet moved but is at zero G. Likewise when the 10 second vehicle reaches the top of its travel and descends it does not have any gravity even though the velocity is 0 for an instant.
I like to think of weight as a measure of non-free fall motion. When standing up on a scale, you weigh your mass (M). When in true free fall (in a vacuum), you weigh zero. Falling through air or water, you weigh something between zero and M - until you reach terminal velocity, at which point you weigh M exactly. As @Chronos suggests, by this definition when you are collapsing to the ground, you weigh less than M during the fall, until you end up resting entirely on the floor - which is what a scale would tell you. An accelerometer (which measures deviation from free fall) tells you the ratio between weight to mass whether you are in free fall, falling through the atmosphere or through the ocean, and only falls to give a reasonable answer for the “collapsing” case, because parts of you weigh the usual amount and parts less.
If we define “weight” as “the force exerted on a mass by gravity” then your weight at a given point is the same whether you are standing on a surface fixed in relation to Earth or whether you are falling. I think that this restates is layman’s terms what @Stranger_On_A_Train described.
It sounds like OP and some others are defining “weight” as “the force exerted by a body of mass against a surface that is fixed in relation to Earth.” Weight is not the same as mass, but it is the force of gravity acting on a body, not the force exerted by that body on an arbitrary surface as a result of gravity.
This is begging for an agreement on what the OP meant by “weight.”
Relativity screwed up our understanding of weight so the physics community decided the standard definition of weight would be the gravitational acceleration acting on a mass, or something like that. The wording varies depending on where you look. The equation is: w=mg, weight equals mass times gravitational acceleration.
To make w less while falling, it would require m or g to go down, but they don’t so weight doesn’t go down either.
To be 100% accurate, g actually goes up as a person falls toward earth because gravity is stronger the closer you are to the center of earth. So a person would actually weigh more when falling.
I’m kind of confused about the topic. The OP, who manages a NASA facility that studies something close to this very question, answered it in the first two words and I don’t presume they’re going to budge from their own answer. What enlightenment do you seek from random internet people?
No. You still weigh your mass times the strength of gravity acting upon you. Otherwise you wouldn’t accelerate. And feeling the subjective feeling of weightlessness is not microgravity or relevant regarding your actual weight. Microgravity is a very low, relative to the surface of Earth, gravitational field.
By this definition, spacecraft in orbit around the Earth, such as the ISS, are not in microgravity either. They’re still so close to the centre of the Earth that Earth’s gravity is not much less than on the surface (still almost 90% of surface gravity). Instead, the weightlessness on board such spacecraft comes from the same effect as in the OP, namely, that they’re in continuous free fall.
It’s surely possible to define “microgravity” in such a way, but that’s not the commonly used meaning of the word.
[quote=“CookingWithGas, post:9, topic:1005069”]
Weight is not the same as mass, but it is the force of gravity acting on a body…[/quote]
This is correct, at least for scientists and engineers. But it really does depend on who’s talking.
So for the purposes of this thread, the OP has to define what they mean by weight.
Yeah, the word “microgravity” is inherently confusing, because its actual meaning isn’t anywhere remotely close to what one would logically think it means. When people say “microgravity”, what they mean is “zero gravity”. I think that what happened was that someone, somewhere, said that there was some sense in which it wasn’t quite correct to refer to objects in orbit as being in “zero g”, and then someone else misinterpreted that to mean that gravity in orbit was really, really small. It’s not really small: It’s either zero, or nearly full, depending on how you define it.
I won’t belabor the point further, other than to say that you need to be subject to gravity to even be in orbit.
and to get fully pedantic (physics really really sucks), g is also affected by the gravitational pull of other objects. So, if you’re falling off of the Empire State Building at night during a full moon, the moon is actually exerting a very small gravitational pull in the opposite direction. As you fall, that force lessens, and the gravitational pull of the earth increases. Your weight when you jump is less than your weight when you land. Not by much; probably not different enough to be measured (if there was a way to actually accurately measure it in mid-air vs. at splat time) because the height of the Empire State Building is inconsequential when compared to the gravitational pull of the Earth.
If you’re in the reference frame implied by that statement, then yes, that’s the case where gravity is nearly full. It can also be sensible to work in an inertial reference frame, in which case you’re not “subject to gravity” because there isn’t actually any such force, and you’re just following a spacetime geodesic. In that reference frame, gravity is zero. And you could, if you really insisted, come up with some reference frame where the gravitational force exists but is small, but that would be a silly reference frame, and there would be no reason whatsoever to consider it. There’s no sensible reference frame where the astronauts in the ISS are in “very low gravity”.
Note: By the definition I’m using weight is what’s measured by a spring scale - on the ground, it reports the full effect of gravity so you weigh 220 pounds while you mass 100 kg, in a free falling elevator or at the ISS, you weigh 0 pounds, but still mass 100 kg. While in a falling elevator with some friction slowing your fall (but still accelerating downward), you weigh somewhere between 0 and 220 pounds (mass unchanged), and while collapsing to the floor you weigh somewhere between 0 and 220 pounds until you are lying on the floor (and at the top of a ladder that stands as tall as the ISS is high, you weigh a little less than 220 pounds).
The reading on a scale isn’t the definition of weight. By the normally accepted definition of weight your weight is constant as you fall or not. Where does the force to accelerate you come from? Your weight. That doesn’t magically appear and disappear due to the presence of a scale.
I would disagree with this. People usually say that people onboard the ISS experience microgravity. And the ISS does experience residual acceleration from tidal, centrifugal, and aerodynamic forces. As it happens, these forces are on the order of a millionth of a gee. Since all acceleration is equivalent, calling this microgravity is perfectly reasonable.
There are some satellite experiments where a test mass in orbit is shielded from the outside and the body of the craft uses thrusters to maintain its position around the mass. That is true zero-gee, or at least as close as we can reasonably get.