What is the condition of true weightlessness?

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True weightlessness occurs only when an object is not subjected to any gravitational force, which theoretically means being far from all celestial bodies where gravitational acceleration (g) equals zero. In contrast, apparent weightlessness, as experienced by astronauts in the ISS, results from being in free fall, where gravitational forces still act but are countered by the acceleration of the spacecraft. The discussion highlights confusion between these concepts, with some participants questioning the existence of true weightlessness and the relevance of terms like "apparent" versus "true." The distinction between weightlessness and free fall is emphasized, noting that weight is a contact force influenced by gravity. Ultimately, the conversation reflects ongoing debates about the definitions and conditions surrounding weightlessness in physics.
  • #31
hmmm27 said:
Was the term "true" relevant contextually to the question ?
yes . because the part before this part is asking this question:
explain why an astronaut in a spacecraft orbiting the Earth appears to be weightless
 
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  • #32
shk said:
yes . because the part before this part is asking this question:
explain why an astronaut in a spacecraft orbiting the Earth appears to be weightless

"appears to be weightless" from whose point of view ? The astronaut isn't going to feel the negligible tidal forces acting on their body. Or, do you mean "Hey look, why do I not seem to be falling from this great height ?" which one would hope an astronaut would not need to ask.

EDIT: It just seems that while, whether you're going to hit the planet or not is very very important, and navigation through gravity wells is not a straightforward thing, "real" and "apparent" weightlessness don't seem (to me) to be a metric that isn't greatly overshadowed by others.

A couple of fringe cases :

Using tidal forces to move around in space

Flying fighters in space battles
 
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  • #33
shk said:
so how would you differ true weightlessness and apparent weightlessness?

I reckon that there cannot be any formal definition of "apparent" in physics. Something either has a property or not. "Apparent" suggests an unspecified situation where for an unspecified reason it may not be clear whether something has the property or not.

For example, if I define property X to be "is wearing shoes". Then it's clear what property X is. But, what would be the definition of "apparently" wearing shoes? As opposed to "really" wearing shoes. Does it mean you can see what appear to be shoes? Or, wearing shoes on other than your feet?

It's a waste of time worrying about what "apparently" wearing shoes might mean. And, it's a waste of time wondering what "apparently" weightless might mean. You're either weightless, according to the definition, or you're not.
 
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  • #34
shk said:
explain why an astronaut in a spacecraft orbiting the Earth appears to be weightless
I guess this means to ask why he floats around relative to the space craft, and doesn't fall to the floor. The answer is that he and the spacecraft are both in free fall, so there is no relative acceleration between them.

I guess "apparent weightlessness" indicates here, that there is a substantial Newtonian gravitational force (also called "weight") acting on him. So he not floating because he escaped the Earth's gravitational pull.
 
  • #35
As I would define the terms:

In the Newtonian model, gravity is a force and there is a notion of coordinate acceleration that is invariant across all inertial frames. A condition of "true weightlessness" would indicate a place where the coordinate acceleration of a freely falling object is zero with respect to an inertial frame.

By contrast, a condition of "apparent weightlessness" would indicate the the use of an accelerated reference frame in which a freely falling object remains at rest.

In the model of general relativity, gravity is not a force and the notion of globally inertial reference frames is discarded. There is no longer any grounds for a distinction between "true" and "apparent" weightlessness. Instead, there are notions of coordinate acceleration and proper acceleration.

Back in the Newtonian model... For any finite and static collection of gravitating masses, it seems clear that there must be at least one point where the acceleration of gravity is zero and "true weightlessness" would apply. For example, in a hypothetical Earth-moon system with Earth and moon somehow fixed in place there would be three such points. One inside the Earth, one inside the moon and a third somewhere in between.

Edit: to make this last a theorem I think we'd need to add a caveat for point masses where instead of a point where gravity is zero, you'd otherwise get a point where gravity is undefined. I think a rule that if there are any point masses, that there must be at least two would handle that corner case.
 
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  • #36
Dale said:
Then please provide a scientific reference that explains this concept of “apparent weightlessness”. It is a term I have not seen.
Could you clarify please if there are any similar terms that are commonly used. I think this or similar questions have come up before, and I've been surprised by this apparent confusion. Are not "true weight" and "apparent weight" common and well defined terms in physics?
Drakkith said:
I wouldn't use either term. I would just use 'weightlessness', and I would define it as the absence of a normal force derived from gravitation.
I guess my concern would be that if "true weight" and "apparent weight" are commonly used terms, then "weightless" must be a form of one of those two and should have a modifier...unless because "true weight" is never zero, only "apparent weightless[ness]" is needed.
 
  • #37
PeroK said:
I reckon that there cannot be any formal definition of "apparent" in physics. Something either has a property or not. "Apparent" suggests an unspecified situation where for an unspecified reason it may not be clear whether something has the property or not.

For example, if I define property X to be "is wearing shoes". Then it's clear what property X is. But, what would be the definition of "apparently" wearing shoes? As opposed to "really" wearing shoes. Does it mean you can see what appear to be shoes? Or, wearing shoes on other than your feet?
Whether you are wearing shoes or not is not frame dependent. Your weight is.
 
  • #38
russ_watters said:
Could you clarify please if there are any similar terms that are commonly used. I think this or similar questions have come up before, and I've been surprised by this apparent confusion. Are not "true weight" and "apparent weight" common and well defined terms in physics?

I guess my concern would be that if "true weight" and "apparent weight" are commonly used terms, then "weightless" must be a form of one of those two and should have a modifier...unless because "true weight" is never zero, only "apparent weightless[ness]" is needed.
I have seen "weightless" as what he was calling "apparent weight" in my old physics textbook. I cannot remember a "true weightless" term, but perhaps it was called "zero g" or something.

I do admit that it has been more than a couple of decades since I did introductory physics so it may have been defined in my textbook and then just mentally discarded due to the uselessness of the "true weightless" definition.
 
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  • #39
Dale said:
I have seen "weightless" as what he was calling "apparent weight" in my old physics textbook. I cannot remember a "true weightless" term, but perhaps it was called "zero g" or something.

I do admit that it has been more than a couple of decades since I did introductory physics so it may have been defined in my textbook and then just mentally discarded due to the uselessness of the "true weightless" definition.
I think my question was too long, so you kinda skipped over it. What I really would like to know is:

Are "true weight" and "apparent weight" common and well defined terms in physics?

[negative wording reversed]
 
  • #40
shk said:
so you're saying that when g is zero weight is zero. correct ?
Also in the case of the baseball, the person throwing it will need to supply basically the same force to it in order to throw it at a speed of 60 m.p.h. whether he is on Earth or in an orbiting space ship. The force needed is such that it accelerates the ball from being motionless to a speed of 60 m.p.h. before it leaves the thrower's hand. The ball still has mass. ## \\ ## Unless it is explained properly, someone not familiar with the idea of mass may think that weightless implies the baseball is light as a feather when it comes to putting it in motion. The ball is weightless, but its mass is still present.
 
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  • #41
russ_watters said:
I think my question was too long, so you kinda skipped over it. What I really would like to know is:

Are "true weight" and "apparent weight" common and well defined terms in physics?

[negative wording reversed]
I wouldn't say they were common or well defined terms used by Physicists. I would say that they can probably be found in not very well founded descriptions of the effects in free fall and in orbit but not in credible sources. After all, Weight is what is sensed or measured when an object is is contact with another object. There is not much confusion on Earth but even though the forces in space travel are very small they are still Weight.
 
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  • #42
A bit of googling about and "true weight" concerns only mass and subjective (real)gravity ; "apparent weight" includes centrifugal force, buoyancy, acceleration other-than-gravity, etc.

But, I still fail to see the utility in differentiation, outside of simply noting that it can be differentiated.
 
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  • #43
Okay, I thought of a situation where "real" and "apparent" weight could be useful.

So, you're on Mars, where you've set up a lawnchair in a centrifuge, which spins fast enough to mimic Earth gravity. The lawnchair - designed for ground level on Earth - says "Load limit 200 lbs". You have no clue what your current mass is but it could swing either way. So, you get out a spring scale to find your "real" (local) weight, then multiply that by 3 (or whatever) to get "apparent" weight in the centrifuge.
 
  • #44
Dale said:
The condition of weightlessness is usually that a device which measures weight (a scale) read 0.

russ_watters said:
I think this or similar questions have come up before, and I've been surprised by this apparent confusion. Are not "true weight" and "apparent weight" common and well defined terms in physics?

There is not consistency in the definition of "weight" in, e.g., first-years text. See my post

https://www.physicsforums.com/threa...rk-in-general-relativity.262703/#post-1912465

and the subsequent post by @jtbell .
 
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  • #45
George Jones said:
There is not consistency in the definition of "weight" in, e.g., first-years text. See my post

https://www.physicsforums.com/threa...rk-in-general-relativity.262703/#post-1912465

and the subsequent post by @jtbell .
Thanks!

One problem with Google is that anything you google you will find, but that doesn't tell you how prevalent the thing you are searching for is in reality or even if Google is creating reality as you search for it. :wideeyed:
 
  • #46
russ_watters said:
Are "true weight" and "apparent weight" common and well defined terms in physics?

Common among those introductory physics textbook authors who use them. :smile:

I'd say, off hand, somewhere around half.

In this vocabulary true weight is gravitational force and apparent weight is what you get when you weigh an object. I would say that they are well defined. The former, obviously, is well defined in Newtonian physics. The latter can be well defined as the force needed to make the object accelerate at a rate equal to the local free fall acceleration.

I prefer a different vocabulary. I call the former the gravitational force and the latter the weight force.

Here's an example. Suppose you have an object and you measure its mass to be 100.00 kg. On Earth's equator, where the free fall acceleration is 9.78 m/s2, the weight force is 978 N, and indeed it takes a force of 978 N to support that object, neglecting buoyancy. But the gravitational force is 981 N. The difference being due to Earth's spin.

Using that other vocabulary they'd say the true weight is 981 N and the apparent weight is 978 N.
 
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  • #47
George Jones said:
There is not consistency in the definition of "weight" in, e.g., first-years text. See my post

https://www.physicsforums.com/threa...rk-in-general-relativity.262703/#post-1912465

and the subsequent post by @jtbell .

Well, in those posts you're using definitions you find in introductory physics textbooks. Definitions adopted by official organizations may or may not lead you to different conclusions. One thing I've found is that when it comes to this topic there are a rich variety of conclusions, all reached by trained physicists, from the same set of premises.

By the way, if you use ##mg## as the magnitude of the gravitational force, and you use the local free fall acceleration magnitude for ##g##, that constitutes an inconsistency. A very common one.
 
  • #48
Weightlessness can be determined by placing a drop of electrically neutral liquid in the center of a transparent sphere. If it remains in place, everything stationary relative to the drop is weightless. If the drop appears distended, there is a tidal influence present, indicating that there is a significantly steep gravitational gradient straining the molecular bonds of the drop; but if the center of the drop's mass remains in place, the drop is nonetheless weightless.* Note that this condition can be satisfied if the sphere is falling toward a massive body, in an orbit, or floating amid the gravitational influences of a number of planets, stars, and galaxies. It applies to any condition of weightlessness -- there is no "apparent" weightlessness.

*Note: If there is a severe tidal gradient, otherwise weightless bodies in a rigid container can become weighted, pressing against the opposite walls of the container along the axis of gravitation.
 
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