Why is microgravity experienced on the ISS despite the presence of gravity?

[fog37]

Hello Forum,

It is said that astronauts on the International Space Station (ISS) experience microgravity and not complete zero gravity. The acceleration of gravity at the ISS altitude is still pretty significant (~8.7 m/s^2), far from zero.

However the free fall motion of the ISS and its passengers produces the sensation of zero weight (weightlessness), i.e. the apparent weight becomes zero but the actual weight is not zero. The sensation of our weight disappears.

Why is it called microgravity? Where does the tiny gravity effect come from? Shouldn't the free fall motion provide that complete zero gravity sensation and feeling?

thanks
fog37

 

[fresh_42]

I assume that ≈350 km height isn't enough to be in a 'good' vacuum. ISS loses height over time, it has to be lifted from time to time.
Therefore it's not a perfect free fall. (Maybe also the moon's gravity comes into play.)

 

[fog37]

I just read that because the ISS is big, as big as a football field, there is a variation in gravitational acceleration much as 14 micro-g between the bottom and the top of the ISS, i.e. there is a gravity gradient.

I am not clear why this gravity gradient would cause microgravity. Aren't both the top and the bottom of the ISS free falling? at the same time? What does this micro gradient cause on an object that is inside the ISS?

Also, I would say that the microgravity could be caused by the mutual attraction between the various objects inside the space stations (astronaut to astronaut, space station to astronaut, etc.)

thanks

 

[fog37]

Thanks fresh_42. I guess you mean that because it slows down it start getting out of orbit (constant distance from the Earth surface) so it needs to be lifted back at the right altitude and speed and that involves forces which break that zero gravity/free fall balance...

 

[russ_watters]

Aren't both the top and the bottom of the ISS free falling? at the same time?

But not at the same rate nor at the same gravitational acceleration. The top is further from Earth than the bottom, so it is being accelerated less by gravity. And it is moving faster, so it is being accelerated more by centripetal acceleration.

Just a little.

What does this micro gradient cause on an object that is inside the ISS?

I'll give you a hint: it's in the title of the thread and it also contains the word "micro"...

 

[fog37]

thanks russ_watters.

The bottom part of the ISS feels a downward acceleration of gravity g_bottom and the top part an acceleration g_top.

g_bottom > g_top

The ISS moves downward as a rigid body. The role of the centripetal force is played by the force of gravity mg itself. The bottom part seems to require a larger centripetal force than the top part of the ISS because it need to follow a more curved path than the top part. However, the top part travels at a faster tangential speed so it would seem to require a larger centripetal force...I am a little confused on which part (top or bottom) experiences the largest centripetal force...

And how would this create a microgravity environment on the objects inside the ISS?

 

[fog37]

In the meantime, I found this

http://www.spaceflight.esa.int/impress/text/education/Microgravity/Question_Microgravity_008.html

I will study it and report back

 

[russ_watters]

And how would this create a microgravity environment on the objects inside the ISS?

The ISS is rigid, so it is under tension due to these forces. Two people floating on opposite sides of the station are not rigidly connected: so they drift apart.

 

[fog37]

Thanks. I didn't know that (i.e. two people could drift apart).

So if an experiment is carried out inside the ISS, it should be carried as close as possible to the center of mass of the ISS where the microgravity is the least. Otherwise, the components of the experiment will feel these gravitational type of forces.

The point about experiments in microgravity is to eliminate all possible forces acting on the experiment (force of gravity included, as much as possible)

 

[ZapperZ]

Thanks. I didn't know that (i.e. two people could drift apart).

So if an experiment is carried out inside the ISS, it should be carried as close as possible to the center of mass of the ISS where the microgravity is the least. Otherwise, the components of the experiment will feel these gravitational type of forces.

The point about experiments in microgravity is to eliminate all possible forces acting on the experiment (force of gravity included, as much as possible)

But if everything is "falling" at the same acceleration, there is no difference between that, and being in zero g.

Zz.

 

[fog37]

Thanks ZapperZ.

I wasn't clear. I guess I would say we can simulate the absence of the force of gravity by using free fall, i.e. making everything fall with the same acceleration (~8.7 m/s^2)

What interested me in this topic was the fact that it is not a perfectly zero gravity environment and wanted to know the causes of that...

 

[ZapperZ]

Thanks ZapperZ.

I wasn't clear. I guess I would say we can simulate the absence of the force of gravity by using free fall, i.e. making everything fall with the same acceleration (~8.7 m/s^2)

What interested me in this topic was the fact that it is not a perfectly zero gravity environment and wanted to know the causes of that...

But I still don't understand your problem.

You DO know that in a uniform circular motion, the object making that circular motion is in a constant "free fall" towards the center, don't you?

If ISS is in "zero g", then it will NOT move in a circular orbit! The fact that it is and not flying off to some random direction means that it is still tethered to the Earth's gravitational field. Yet, I claim that this is the same as being "weightless", and that this weightlessness is no different than being in "zero g".

Zz.

 

[fog37]

I agree with what you are saying:

weightlessness is a little of a misnomer since the force of gravity is there providing the centripetal force for the ISS to move into its orbit. What there is absence of are the effect of gravity: regardless of the presence of gravity we are able to simulate an environment where the effects of gravity are not present (only in small part)

 

[ZapperZ]

I agree with what you are saying:

weightlessness is a little of a misnomer since the force of gravity is there providing the centripetal force for the ISS to move into its orbit. What there is absence of are the effect of gravity: regardless of the presence of gravity we are able to simulate an environment where the effects of gravity are not present (only in small part)

Well, I disagree with what you're saying. "Weightlessness" is the more accurate term than "zero g", because g isn't zero in this case. Weightlessness refers to the fact that there is no "normal reaction force" that we teach students in intro physics when they have to draw a free-body diagram. So the object "sense no weight", and thus, weightlessness.

But if you do a proper treatment of this right out of intro physics, "zero g" environment and "weightlessness" is no different, the same way you can't tell if you're moving with constant velocity or stationary. So that is why I do not understand the problem here.

Zz.

 

[fog37]

Ok, let me try to be more clear:

1) The ISS experience a force of gravity downward equal to (M_iss)*(8.7 m/s). This force is far from being zero.
2) The force (M_iss)*(8.7 m/s) plays the role of the centripetal force
3) Weightlessness is the lack of the perception of weight, i.e. the lack of the sensation of this force (M_iss)*(8.7 m/s)
4) We produce weigthlessness, i.e. the sensation of weight and its effects too, by free falling around our planet
5) the point of making experiments on the iss is to produce conditions un which the objet being investigated does not feel the effects of gravity

 

[ZapperZ]

5) the point of making experiments on the iss is to produce conditions un which the objet being investigated does not feel the effects of gravity

But you haven't shown the difference between "zero g" and "weightlessness", and HOW, in terms of mechanics, that those two would be any different! If you are inside a closed box, can you construct an experiment to distinguish between the two, i.e. can you determine if you are really in "zero g" or just "free falling"?

Zz.

 

[A.T.]

3) Weightlessness is the lack of the perception of weight, i.e. the lack of the sensation of this force (M_iss)*(8.7 m/s)

Weightlessness is the lack of contact forces that support you against gravity. Gravity itself is acting approx. uniformly on your body, so it doesn't cause any "sensation".

 

[fog37]

Ok,

I guess I should say that the feeling of zero gravity is due to the absence of a contact support force (normal force) on our body. The contact may be there but not the force itself...

 

[ZapperZ]

Ok,

I guess I should say that the feeling of zero gravity is due to the absence of a contact support force (normal force) on our body. The contact may be there but not the force itself...

You haven't answered my question. Can you device an experiment to distinguish the two?

Zz.

 

[fog37]

Well, the typical experiment is an elevator in free fall. We are inside the elevator with a scale under our feet. The scale reads zero. The"apparent weight" is zero even if our actual weight is still mg...

But that apparent weight being zero has real effects: our body parts don't feel the same type of compression it would if the elevator was not in free fall.

any better?

 

[ZapperZ]

Well, the typical experiment is an elevator in free fall. We are inside the elevator with a scale under our feet. The scale reads zero. The"apparent weight" is zero even if our actual weight is still mg...

But that apparent weight being zero has real effects: our body parts don't feel the same type of compression it would if the elevator was not in free fall.

any better?

No, because if I put that elevator in space with zero g, you will NOT be able to detect any difference!

Zz.

 

[fog37]

Well, first of all we need to assume we are inside a gravitational field that exerts a downward attractive force.

As you mentioned, if everything is "falling" at the same acceleration, there is no difference between that, and being in zero g.

I am genuinely not seeing your question...

 

[ZapperZ]

Well, first of all we need to assume we are inside a gravitational field that exerts a downward attractive force.

As you mentioned, if everything is "falling" at the same acceleration, there is no difference between that, and being in zero g.

I am genuinely not seeing your question...

And I'm genuinely not see why you don't understand what I asked.

This is the foundation of Relativity, and General Relativity, that you can't tell the difference between "accelerating upwards" and being in a gravitational field with the same g, that there is no experiment that you can construct to distinguish between the two! So if you can't tell if you are accelerating, or in a g-field, how can you tell if you are "free falling" versus "in zero g"? No experiment that you can do will get you to distinguish that!

So therefore, why would any experiment that you do on ISS matter that you are not in "zero g", as long as everything is free falling?

Zz.

 

[russ_watters]

ZZ, my read of the OP's question is that he was confused about the difference between "zero" and "micro": that he didn't realize that the prefix "micro" is used because the acceleration of an object in the space station with respect to the space station actually isn't necessarily quite zero.

I think your post #11 misses that point and implies a disagreement/error where none exists: he was correct that the acceleration of an object with respect to the space station is lowest at the center of mass.

 

[fog37]

"...So therefore, why would any experiment that you do on ISS matter that you are not in "zero g", as long as everything is free falling?..."

Well, let me think

--Everything inside the ISS (ISS included) is falling downward
-- There is no possible experiment that can tell us if we are in free fall or in a zero g environment
-- Isn't an experiment conducted on ISS in "microgravity" purposely conducted in that setting so the effect of that gravity would cause are not included?
-- From my reading, tidal forces, small attractive forces between other objects having mass inside the ISS, and other effects cause the environment to not be completely gravity free.

Did you have a chance to click on that link I found last night?

http://www.spaceflight.esa.int/impress/text/education/Microgravity/Question_Microgravity_008.html

that is what sparked my very initial question...

 

[ZapperZ]

ZZ, my read of the OP's question is that he was confused about the difference between "zero" and "micro": that he didn't realize that the prefix "micro" is used because the acceleration of an object in the space station with respect to the space station actually isn't necessarily quite zero.

I think your post #11 misses that point and implies a disagreement/error where none exists: he was correct that the acceleration of an object with respect to the space station is lowest at the center of mass.

Sorry, but I disagree. I think you overestimated what the OP understood. Based on Post #21, it appears that the OP does not fully understand the simplest idea to start with. And that was what I was trying to established FIRST.

Zz.

 

[fog37]

I would like to review, for my own sake, some basic ideas:

The ISS is about 100 meter wide and it is at an altitude of about 400 meter. The ISS is in continuous free fall towards Earth but manages to remain at a fixed distance away from the Earth surface. This is possible because the ISS has a specific speed (4.76 miles/s). If the ISS was at a higher altitude the required speed would be lower.
The force that allows the ISS to remain in its orbit is the attractive gravitational force F=G (M_earth)* (M_ISS)/(R_earth+400)^2. This force is the centripetal force (M_ISS)*v^2/(R_earth+400) where v turns out to be the 4.76 miles/s

this discussion assumes that the attractive gravitational force F=G (M_earth)* (M_ISS)/(R_earth+400)^2 is applied to the center of gravity of the ISS. Those parts of the ISS that are either higher or lower than the center of gravity of the ISS experience a different attractive gravitational force because g is slightly different at those locations. The ISS is a rigid body so everything moves together. So the parts that are lower and higher than the center of mass feel some nonzero contact force.
For instance, the higher part requires a centripetal force that is larger than what the single gravitational force mg can offer. So a contact force contributes to provide the needed centripetal force.

Astronauts or other objects that are not linked to the ISS will possibly drift around inside the ISS due to the fact that their individual gravitational force may not match the required centripetal force. Everything and everybody is still in free fall toward planet earth. But an astronaut above the center of mass of the ISS will drift toward the top part of the ISS while an astronaut below the ISS center of mass will drift toward the lower part of the ISS...

right or wrong? :)

 

[jbriggs444]

I read #21 as indicating that one can distinguish between an elevator going down the shaft in free fall and an elevator stopped on the second floor at Macy's. (quickly stand on a scale and look at the reading). It's not wrong. It's just not responsive to a question about how to distinguish between "weightless" and "zero g".

 

[russ_watters]

Sorry, but I disagree. I think you overestimated what the OP understood. Based on Post #21, it appears that the OP does not fully understand the simplest idea to start with. And that was what I was trying to established FIRST.

I think you misread post 11 - read something that isn't there - and have continued your error with post #21.

Your response to post 11 implies you think post 11 is saying that a person in the ISS may accelerate at 9 m/s^2 with respect to the ISS. That isn't what he's saying/what the discussion is about.

 

[russ_watters]

I read #21 as indicating that one can distinguish between an elevator going down the shaft in free fall and an elevator stopped on the second floor at Macy's. (quickly stand on a scale and look at the reading). It's not wrong. It's just not responsive to a question about how to distinguish between "weightless" and "zero g".

Agreed. So my point was that for the OP's question, there is no need to make that distinction, nor is there an incorrect understanding implied. I'm quite certain he knows that the elevator and person are both accelerating toward the Earth at g.
Edit: He even provided the value for g at the ISS in the OP.

 

[fog37]

One more question about weightlessness:

The acceleration of gravity on the Sun is 273.7 meters/sec^2 (huge compared to 9.8 m/sec^2). As we mentioned, while we are in free fall toward the Earth we don't perceive our weight (even if the weight force is surely nonzero). All the internal organs, bones, etc. of our body falls down at the same acceleration of 9.8 m/sec^2 and cause no pressure on each other.

Forget for a moment that the Sun is a ball of fire, what would happen if we free fell under the sun's gravitational pull 273.7 meters/sec^2? I think we would still free weightless since the same argument would apply: all the internal organs, bones, etc. of our body falls down at the same acceleration of 9.8 m/sec^2 and cause no pressure on each other.

However, on earth, when we experience huge accelerations, i.e. large g-forces, we can pass out. Those forces produce acceleration that are surely much smaller than 273.7 meters/sec^2 which is about 28 times larger than g. People start passing out at 4 or 5 g.

Why would't we pass our while we free fall toward the sun with that huge acceleration? Or would we?

thanks

 

[fog37]

I think g forces larger than one become an issue only when our human body is in contact with a support.

In free fall, no matter how large the acceleration is, we will feel the same as weightlessness...

 

[russ_watters]

All the internal organs, bones, etc. of our body falls down at the same acceleration of 9.8 m/sec^2 and cause no pressure on each other.

Forget for a moment that the Sun is a ball of fire, what would happen if we free fell under the sun's gravitational pull 273.7 meters/sec^2? I think we would still free weightless since the same argument would apply: all the internal organs, bones, etc. of our body falls down at the same acceleration of 9.8 m/sec^2 and cause no pressure on each other.

Correct.

However, on earth, when we experience huge accelerations, i.e. large g-forces, we can pass out. Those forces produce acceleration that are surely much smaller than 273.7 meters/sec^2 which is about 28 times larger than g. People start passing out at 4 or 5 g.

Why would't we pass our while we free fall toward the sun with that huge acceleration?

No. Think about it: what, specifically is applying the force you feel when under a significant acceleration on Earth? Are you in freefall when it happens?
...
erp -- looks like you already figured it out:

I think g forces larger than one become an issue only when our human body is in contact with a support.

In free fall, no matter how large the acceleration is, we will feel the same as weightlessness...

Yep.

 

[DocZaius]

To answer the original question and support what I think you've figured out, fog37: "microgravity" is a very misleading term to use for the condition of astronauts inside the ISS. I am surprised NASA still uses it. I would recommend not using it because it can cause a great deal of confusion. I believe "free fall" is the best term to use. Although it can have some potential for confusion, too. People associate "fall" with moving straight down through the gravity gradient. But if someone asks how it can be falling if you don't hit the earth, you can tell them: "you're moving so quickly to the side that, despite falling, you keep missing the earth."

 

[fog37]

hello russ_waters,

I have a related question: I know that in free fall, inside the ISS, it is easy to move heavy object around and do somersaults, etc. Why?

I would say that the entire body is free falling, i.e. each body part is free falling at the same 8.7 m/sec^2 and they don't feel their mutual tug, pull. That said, on earth, it takes an effort to raise an arm while on the ISS is practically effortless. To be correct and precise, it still requires some force but it is much smaller than the force that should be applied on the surface of the earth...why? I guess I am trying to figure out what happens biomechanically. On the surface of the Earth the arm pull on the should. That does not happen on the ISS. But why should it be easier to lift the arm on the ISS?

Same goes for moving heavy objects around just with a finger...

 

[jbriggs444]

On Earth you need to maintain that arm at an acceleration of 9.8 meters/sec² relative to free fall. In the space station you have to maintain that arm at an acceleration of 0 meters/sec² relative to free fall.

 

[A.T.]

But why should it be easier to lift the arm on the ISS?

You need only the force for acceleration, not the one for support.

Same goes for moving heavy objects around just with a finger...

Same as above compared to carrying the object on the surface. No friction compared to pushing the object across the surface.

 

[fog37]

How does the feeling of apparent weightlessness (experienced in free fall) and the feeling of being buoyant in a fluid compare to each other? Do the two feel exactly the same?

A scale would tells us that our apparent weight is less when we are bathing in water. The buoyancy surely decreases our contact with a surface. But I wonder if that effect of reduced compression is also transmitted to the internal organs reducing their internal compression.

I know astronauts train in buoyancy environments to simulate weightlessness on earth...

In the case of skydiving, a skydiver jumps out of a plane and free falls feeling weightless for a few seconds. As the upward air resistance force develops and grows the skydiver starts feeling some sense of its weight because the air resistance acts as a supportive force.
Once air drag equalizes the skydiver weight, the skydiver should feel his weight completely as if he was belly down on a soft pillow. The wind provides the overwhelming experience.

The moral of the story is that as soon as any type of supportive force (even smaller in magnitude than our weight) develops and starts counteracting our true weight (our gravitational attraction to earth) we start gaining some sensation of our true weight because of the compressive state we start feeling.

When we are buoyant on top of the water, for example, we should still feel our weight since the buoyant force is a supportive force that matches our weight. It should be the same as when we are laying down on the floor...

Why then is being buoyant compared to weightlessness?

 

[ZapperZ]

Why then is being buoyant compared to weightlessness?

It doesn't compare!

Astronauts train in water to simulate the fact that they will be floating, and that their feet will not be anchored to the ground. It is not to simulate weightlessness, because if this is true, then training in the vomit comet would not have been necessary.

Based on all your questions now, this is why I disagree with Russ and why from the very beginning, I claim that you do not have an understanding of the idea of weightlessness. That should have been addressed FIRST, rather than all the extraneous issues surrounding experiments on the ISS.

Zz.

 

[fog37]

Thanks Zz.

I am open to understanding more. I still not sure which part I think I am grasping yet. Let me try to explain what weightlessness is for me:

it is the apparent feeling being without weight. This happens when we are in pure free fall because our body and its internal organs are all falling at the same rate (acceleration) and they don't "push" on each other, compress each other.

We can only experience our own weight when a supportive force enters the game. The supportive force (for example the normal force of the floor) counteracts our weight and makes us aware of it by feeling our internal organs in a mutual state of stress...

We can only become aware of our true weight (gravitational attractive force) indirectly if there is a force that counteracts it.

what do you think?

 

[mfb]

We can only become aware of our true weight (gravitational attractive force) indirectly if there is a force that counteracts it.

We can be aware of this force then, but not of gravity itself (neglecting tidal acceleration - which would bring the topic back to the microgravity).

Mainly I think there is still atmosphere and therefore friction.

That effect needs months to become notable. It is tiny compared to tidal gravity.

 

[fog37]

Zz,

I found this on Wikipedia:

"...One downside of using neutral buoyancy to simulate microgravity is the significant amount of drag presented by water.[6]Generally, drag effects are minimized by doing tasks slowly in the water. Another downside of neutral buoyancy simulation is that astronauts are not weightless within their suits, thus, precise suit sizing is critical..."

It sound that neutral buoyancy feels very similar to being weightless...What does it mean that the astronauts are not weightless within their suits?

 

[mfb]

If the suit is too large, body parts can be moved up and down in it (in their air inside them) - you still feel the weight of them.
You also don't have a realistic effect on your blood pressure, the the constant feeling of falling down is missing, and other things just because internally gravity acts on your body as normally.

 

[rcgldr]

I'm wondering about the relative magnitude of the small effect of aerodynamic drag due to the thin amount of outer atmosphere that the ISS orbits in versus the small effect of location within the ISS (closest to Earth side, farthest from Earth side).

 

[mfb]

Drag varies a lot with time. To make it worse, it does not actually reduce the speed of the station, it increases it. Starting from a perfect circular orbit, if you add air drag, the station will spiral downwards, getting faster all the time. This speed increase is due to gravity, however, so you don't feel this effect in the station.

Anyway, here is an average: Let's start with the Height as function of time. During October/November, we had a longer period without reboost, the ISS dropped from 404 to 401 km in 1.3 months, or 40 days. Let's assume the orbit is perfectly circular, the eccentricity does not change the result notably.
Specific orbital energy is GMm/(2r), the difference between the two is 0?? Thanks WolframAlpha. The difference is 13 kJ/kg. At a speed of 7.67 km/s, the ISS traveled 2.65*10¹⁰ meters. Diving specific energy loss by length gives an acceleration of 4.9*10^-7 m/s² due to aerodynamic drag.
Note: this is an average value. While in the shadow of Earth (about half of the time of the 90 minute orbit), the ISS rotates its solar panels to reduce drag. During sunlight, it rotates the solar panels to face the sun.Self-gravity is not completely negligible at those levels. 100 tons at a distance of 20 meters produce 1.5*10^-7 m/s² acceleration from the ISS mass. (Note: those values are arbitrary, I don't have a detailed mass simulation available).What about tidal gravity? To get this, we first have to check the orientation of the ISS in space: the image there is in flight direction, and the ISS rotates once per orbit so this orientation stays the same. Much larger image with a different viewing angle.

As you can see, the main structure of pressurized modules is along the orbital track of the ISS. The center of mass is roughly in this part as well, close to a US lab module. Along this path, objects will just follow each other at a constant separation. No tidal gravity.

What happens if we go up by 1 meter? Gravitational acceleration reduces, centrifugal acceleration (in the rotating frame) increases. The difference for 1 meter is 3.9*10⁶ m/s² per meter of height.

What happens if we go one meter to the side? We get a sidewards component of 1.3*10⁶ m/s² per meter sidewards.

As the station is much larger than a meter, tidal gravity wins by a good margin.

On the other hand... as you can see in the height plots, the ISS gets frequent re-boosts to keep its orbit. Those are done with accelerations of roughly 0.02 m/s². If you call them "caused by drag", then drag is responsible for the largest accelerations by far. And certainly to much more fun than tidal gravity.

 

[ogg]

Drag speeds up the ISS? Wouldn't the same logic would require Newton's apple to acquire a tangential velocity as it falls from the tree? It would be great if you (mfb) could point us to a source which derives this. IMHO, microgravity is the better term. I also think talking about "weight" is a bit pointless unless we are assuming a constant g. (or perhaps comparing g and g' between the surfaces of two planets). One old old sci-fi story had the protagonist solve the question of whether he was in orbit or weightless by placing a bunch of ball bearings onto the ceiling (or was it the walls? hmm). If you understand why that would (after perhaps days or weeks) answer the question, you understand the topic. The ONLY things which experience the exact same force (hence have no force differential) when in orbit are on the exact same circle (in the 1 dimensional meaning of the term circle). Above, or below that (assumes a spherical orbit around a spherical grav. potential) the force will be a tiny bit different, to the left or to the right, the paths (great circle) will not be parallel (they'll intersect). Also, the thing that you DON'T have in microgravity is (surface2surface) friction. Moving a large block floating on water (slowly) is nearly as easy as in space, but you still have to accelerate it (and stop it when you get to where you're going).

 

[Redbelly98]

I agree with what you are saying:

weightlessness is a little of a misnomer since the force of gravity is there providing the centripetal force for the ISS to move into its orbit. What there is absence of are the effect of gravity: regardless of the presence of gravity we are able to simulate an environment where the effects of gravity are not present (only in small part)

Well, I disagree with what you're saying. "Weightlessness" is the more accurate term than "zero g", because g isn't zero in this case. Weightlessness refers to the fact that there is no "normal reaction force" that we teach students in intro physics when they have to draw a free-body diagram. So the object "sense no weight", and thus, weightlessness.

But if you do a proper treatment of this right out of intro physics, "zero g" environment and "weightlessness" is no different, the same way you can't tell if you're moving with constant velocity or stationary. So that is why I do not understand the problem here.

Zz.

Weightlessness is the lack of contact forces that support you against gravity. Gravity itself is acting approx. uniformly on your body, so it doesn't cause any "sensation".

Some physics teachers I know refer to freefall and orbital motion as "normalforcelessness" rather than weightlessness. While they say this tongue-in-cheek, it is an instructive term to help think about what is really going on.

These same teachers don't agree on what is meant by "weight". If I recall our conversation correctly, one, who grew up in Russia, uses "weight" to mean what a scale would measure -- what others call "apparent weight". And those others use "weight" to mean "the force due to gravity". The point being, the definition of weight determines whether we can refer to objects in orbit as being weightless -- it becomes a matter of semantics.

 

[mfb]

Drag speeds up the ISS? Wouldn't the same logic would require Newton's apple to acquire a tangential velocity as it falls from the tree?

Coriolis force is doing that, but that has nothing to do with drag.

It would be great if you (mfb) could point us to a source which derives this.

Literally every textbook about orbital mechanics. Lower-energy circular orbits are faster. Drag let's the ISS lose potential energy and gain half this loss as kinetic energy.

Also, the thing that you DON'T have in microgravity is (surface2surface) friction.

You have it if you apply a normal force.
Even worse, you can have cold welding in vacuum conditions.