# Zero Gravity or microgravity?

ZapperZ
Staff Emeritus
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.

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
Homework Helper
2019 Award
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
Mentor
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
Mentor
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.

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

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
Mentor
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.

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."

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
Homework Helper
2019 Award
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.

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
Staff Emeritus
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.

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
Mentor
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.

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
Mentor
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
Homework Helper
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
Mentor
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
Staff Emeritus
Homework Helper
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
Mentor
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 lets 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.