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I find the term "microgravity" to be misleading, how was it coined?

NASA provide this definition:

Microgravity is the condition in which people or objects appear to be weightless. The effects of microgravity can be seen when astronauts and objects float in space.

Presumably the word "micro" is not being used in its mathematical sense, and is being used to express something that is small. However, the above article goes on to state that:

The International Space Station orbits Earth at an altitude between 200 and 250 miles. At that altitude, Earth's gravity is about 90 percent of what it is on the planet's surface.

I would not describe 90% as small.

It seems a poorly constructed term.

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    $\begingroup$ Well, the gravity that you appear to have is pretty close to zero, which I'd call micro. Free-fall isn't about the actual gravity, but the one the body appears to have in relation to surrounding bodies. $\endgroup$
    – Polygnome
    Commented Oct 6, 2020 at 7:00
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    $\begingroup$ @MarkMorganLloyd: The use of "micro-" to mean "very small" long predates its use as an SI prefix for "one millionth". The word itself comes from Ancient Greek μικρός (mikrós) = "small". $\endgroup$ Commented Oct 6, 2020 at 12:51
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    $\begingroup$ No room in your physics laboratory for microscopes, microphones, and microcomputers, then? (-: $\endgroup$
    – JdeBP
    Commented Oct 6, 2020 at 13:28
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    $\begingroup$ @MarkMorganLloyd From now on, I'm referring to magnifying glasses as "milliscopes" $\endgroup$ Commented Oct 6, 2020 at 23:27
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    $\begingroup$ @JdeBP Now, should we rename "microprocessors" to "nanoprocessors" for SI-accuracy? Nobody uses 1-micron manufacturing process for processors anymore... $\endgroup$ Commented Oct 7, 2020 at 1:14

8 Answers 8

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Microgravity is used because zero gravity is inaccurate. The ISS, at 400 km, experiences an average atmospheric density of 4 nanograms per cubic meter. It's frontal area varies from 700-2300 square meters. At 1000 m$^2$, the drag force is $ \frac 1 4$N. With a mass around 250,000 kg, that's $10^{-6}$ m/s$^2$, or 0.1 $\mu$g.

Hence: microgravity, literally.

If you leave an object at the back of the space station, it will fall forward, falling 100 m (the length scale of ISS) in 4 hours, with an impact speed of $\sqrt 2$ cm/s.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Oct 7, 2020 at 17:43
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    $\begingroup$ Why are there so many upvotes for an answer that is so overly simplistic that it misses the mark? Drag is a rather small contributor of the ISS microgravity environment. Vibrations from rotating machinery, crew bumping into walls and riding exercise bicycles, gravity gradient, and centrifugal effects are contributors that can be multiple orders of magnitude larger than drag. $\endgroup$ Commented Oct 8, 2020 at 10:09
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    $\begingroup$ @DavidHammen: Most of the things you mention average out to 0 over time due to conservation of momentum. They can cause vibrations, but they won't cause a long-term drift of objects in a specified direction. Gravity gradient and centrifugal force also don't have a specified direction (they push away from the center), and according to my calculations, are maybe just 1 order of magnitude larger than drag. $\endgroup$ Commented Oct 8, 2020 at 10:34
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    $\begingroup$ @MeniRosenfeld - That they average out is irrelevant to some very delicate experiments performed on the ISS. What is relevant is the peak acceleration, which at least on early configurations of the ISS exceeded 10000 micro gs -- outside the microgravity range of a hundredth to a millionth of a g. $\endgroup$ Commented Oct 8, 2020 at 11:47
  • $\begingroup$ @DavidHammen So I worked on a remote sensing instrument that is mounted to the ISS, and I asked the experts at NASA: what's the drag force on the ISS (because the data was sensitive to altitude)? I just repeated the overly simplistic mark missing answer that they gave me. $\endgroup$
    – JEB
    Commented Dec 16, 2022 at 19:37
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Microgravity doesn't mean that the gravitational field is negligible, just that the system in question does not feel its effects due to being in free-fall. The interior of a freely-falling elevator is a microgravity environment - at least until it hits the ground.

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  • $\begingroup$ The ISS interior does experience gravitational effects from the Moon, Jupiter and (to a much lesser extent) every other particle of matter in the solar system and beyond. These gravitational effects are not negated by freefall, but they also aren't very large in the first place, so we mostly ignore them and they mostly don't make a difference. $\endgroup$
    – Kevin
    Commented Oct 6, 2020 at 1:26
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    $\begingroup$ @Kevin Sure, there are tidal effects. $\endgroup$
    – J. Murray
    Commented Oct 6, 2020 at 1:40
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    $\begingroup$ The interior of a freely-falling elevator may be microgravity for a second or three but then it certainly will be macrogravity over a few microseconds. $\endgroup$
    – Stian
    Commented Oct 6, 2020 at 8:23
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    $\begingroup$ One can also experience microgravity for longer periods in parabolic flights inside the atmosphere. $\endgroup$
    – Ruslan
    Commented Oct 6, 2020 at 10:24
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    $\begingroup$ @Kevin There are tidal effects, but not straight gravitational effects. "Being in orbit" or "being in freefall" means you move along with every gravitational pull that is acting on you. The ISS in its entirety is pulled on by the Moon, the astronauts inside are moving along with the same speed as the outer walls. So they don't feel any gravity from the Moon when they look around. Only if the ISS were to follow some circular path around the Earth, like on a giant set of rails, they'd feel gravity from the Moon (and Jupiter, etc.) $\endgroup$
    – Emil Bode
    Commented Oct 8, 2020 at 8:52
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You are not the only person who has problems with the term microgravity. I doubt there is any record of how or why the term was coined.

My guess is that NASA introduced the term microgravity because it sounds more scientific that "zero gravity" (which is just plain wrong) but less scary than "free fall", which is accurate but might suggest to the man in the street the ISS was about to fall out of the sky.

This article suggests that due to active stabilisation the environment on the ISS is actually closer to "nanogravity" than microgravity.

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    $\begingroup$ cosmicopia.gsfc.nasa.gov/qa_gp_gr.html : But oxygen and the rest of the atmosphere (mostly nitrogen) just gradually fade out and extend hundreds of miles over the surface of the Earth. Even where the shuttle and space station are (greater than 400 miles up), there is enough air resistance that there is apparent acceleration of about a millionth of that on the Earth's surface. This is why experiments there are called "microgravity" experiments, not zero G experiments. Dr. Eric Christian $\endgroup$
    – eps
    Commented Oct 6, 2020 at 2:30
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    $\begingroup$ @eps units are messed up there. ISS is at altitude of about 400 km, not miles. It's about 250 mi, certainly not "greater than 400". Still, indeed, air resistance isn't negligible there, although it's way above Kármán line. $\endgroup$
    – Ruslan
    Commented Oct 6, 2020 at 10:31
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The ISS is subject to vibration from many sources.

  • astronauts moving around, exercising etc.
  • mechanical equipment (motors running etc)
  • thermal effects (as the station moves in and out of Earth's shadow(
  • visiting spacecraft docking, undocking.
  • reboosting
  • aerodynamic drag

The gravity environment on the ISS is monitored by the Principal Investigator Microgravity Services.

The requirements for the gravity environment on the ISS are:

for 0.01 ≤ f ≤ 0.1Hz: a ≤ 1.6 µg
for 0.1< f ≤ 100 Hz: a ≤ f x 16 µg
for 100< f ≤ 300 Hz: a ≤ 1600 µg

That document also has measurement results. A few highlights:

  • the gravity environment is not the same across the entire station. Experiment cabinets can be isolated from the station, leading to lower vibration in the cabinet
  • Events like reboosts, dockings cause lots of vibration, mostly at times predictable enough that they can be planned around.

Regular reports show the environment in practice:

enter image description here

The quasi-steady state regime is the lowest frequencies (≤ 0.01 Hz).

These low-frequency accelerations are associated with phenomena related to the orbital rate, primarily aerodynamic drag. However, gravity gradient and rotational effects may dominate in this regime, depending on various conditions and an experiment’s location relative to the vehicle's center of mass (CM).

You can see the term 'microgravity' is appropriate. Most of the accelerations in these reports are in the 1-1000 micro-g range.

There have been plans for unmanned space stations with a better quality vibration environment (MTFF), but these were not proceeded with.

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  • $\begingroup$ The linked presentation states that the vibratory environment on the ISS is on the order of tens to thousands of micro gs rather than one 𝜇g. $\endgroup$ Commented Oct 8, 2020 at 14:25
  • $\begingroup$ How is drag “gravity”? Drag is a force like gravity but is not gravity. $\endgroup$
    – xaxxon
    Commented Sep 4, 2023 at 2:15
  • $\begingroup$ Drag means the station is not completely in freefall, as JEB's answer shows. $\endgroup$
    – Hobbes
    Commented Sep 6, 2023 at 12:17
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The Space Station is in orbit, i.e., in free fall. In free fall, gravity is not felt. If you are unfortunately in a falling elevator and you let go of an object it will remain where you left it and not fall to the floor. In the elevator frame of reference there is no force of gravity because it is in free fall. Of course, to an observer in the building there is gravity.

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    $\begingroup$ For me, this explanation motivates the term "zero gravity", but not "micro gravity". $\endgroup$
    – M. Winter
    Commented Oct 6, 2020 at 8:49
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    $\begingroup$ @M.Winter ISS is still subject to atmospheric drag, however small it is (compare ISS's orbit of mere 400 km with Kármán line of 100 km). So microgravity is a better term. $\endgroup$
    – Ruslan
    Commented Oct 6, 2020 at 10:27
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    $\begingroup$ @Ruslan Absolutely, and the answer of JEB convinced me of this. But drag is not part of this answer, which, I would argue, does not answer the question of OP. $\endgroup$
    – M. Winter
    Commented Oct 6, 2020 at 10:34
  • $\begingroup$ Re In free fall, gravity is not felt. Gravity is not felt anywhere. (Okay, almost anywhere. Very large objects such as the ISS can feel differential stresses and strains due to the gradient in gravitational acceleration, and people who get too close to a neutron star). $\endgroup$ Commented Oct 8, 2020 at 14:15
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There is gravity, there is just no Normal force that would exist if you were on a surface on Earth. You're constantly falling so there little upward resistance; you experience weightlessness because of this.

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The greater the distance from Earth's surface, the less the gravitational force. As the space station is in orbit, not stationary, the 90% g is the centripetal force on it. This is equaled by the centrifugal force of its rotation around the Earth. At the stations center of gravity these forces are equal so there is no net gravity from Earth. Above or below (relative to Earth) the center of gravity there are net microgravity effects, which could be termed tidal effects. For the small distances involved these are usually negligible. Other causes of microgravity effects on the station could be from the small amount of air resistance, and small net gravitational pulls from the Sun, Moon, and other objects.

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    $\begingroup$ It is irrelevant if the so-called centrifugal force equals the force of gravity or not. If the station had been in a highly elliptic orbit, it would still have microgravity, but the altitude would be successively increasing (from perigee to apogee) and decreasing. In the extreme case where a space station was free-falling in a straight line towards the center of the Earth, at altitude 400 km there would still be microgravity, so the centrifugal argument is wrong. (Such a station would later hit the Earth and be destroyed by great forces, but that is another story.) $\endgroup$ Commented Oct 7, 2020 at 11:32
  • $\begingroup$ @JeppeStigNielsen But the ISS is not highly elliptical and I was explaining how the 90% g was offset $\endgroup$ Commented Oct 7, 2020 at 12:22
  • $\begingroup$ I thought centripetal force and centrifugal force were two ways of looking at the same force, but you say deviations from the centre of mass produce tidal effects. What is the formula for this, please? $\endgroup$
    – spraff
    Commented Dec 11, 2020 at 9:06
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Any point-body in free fall in a vacuum would experience zero gravity. The ISS differs from this in all defining aspects:

  • It is not in permanent free fall due to corrective manoeuvres every now and then
  • It is not in vacuum because there is still enough atmosphere to cause a tiny amount of drag
  • It is more than 100m larger than a point so that the inhomogeneity of the gravitational field comes into play.
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    $\begingroup$ I think the asker understands that. Let us assume no maneuvers were ever made, assume there was a perfect vacuum at the altitude, and assume the station was a point-body. Then we could use the term "zero gravity". But the asker would ask the same question: Why do they call it zero gravity when the (centripetal) acceleration due to gravity is 90% of 1 G? The question is not so much about whether the free fall is perfect, or only almost free. It is about whether the terms "zero gravity" and "microgravity" are poor terms for describing (almost) free falls in nonzero gravity fields. $\endgroup$ Commented Oct 7, 2020 at 11:41
  • $\begingroup$ Can't a point be affected (accelerated) by the inhomogeneity of the gravitational field? $\endgroup$ Commented Oct 7, 2020 at 16:35
  • $\begingroup$ @PeterMortensen "Equivalence Principle says inertial and gravitational masses are indistinguishable, ideally there's no way to tell whether a force comes from gravity or acceleration." "But there is, a gravitational field's strength varies at different points due to inverse square law, but acceleration does not. One can use a gravity gradiometer to check that." "cough cough EP is purely theoretical, so we say it's only valid at a point (dx, dy, dz)." "But a single point's motion is still affected by a gravitational gradient." "it's (dx, dy, dz, dt) then." $\endgroup$ Commented Jun 3 at 12:16

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