When sitting at rest, a typical "g-meter" shows a value of one, however the meter is not accelerating. In orbit, it shows zero, but is under constant acceleration. The actual force of gravity in both situations is almost identical, so what exactly is being measured?

  • 1
    $\begingroup$ Suppose an astronaut on board the International Space Station (ISS) sticks a "g-meter" to the wall with sticky tape. I think that astronaut would disagree with you about whether or not the g-meter was "at rest." At rest with respect to what? The Earth? The ISS? The Sun? The center of the Milky Way galaxy? Some other point in the Universe? FYI: Any coordinate frame that is attached to the Earth is an accelerated coordinate frame. A coordinate frame attached to the ISS is much closer to being an inertial frame. $\endgroup$ Jun 5, 2022 at 20:13
  • $\begingroup$ Could you re-phrase that Question to justify the claims that at rest, a "g-meter" shows a value of one? In orbit, it shows zero, but is under constant acceleration. The actual force of gravity in both situations is almost identical, so what exactly is being measured? $\endgroup$ Jun 5, 2022 at 22:35
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – called2voyage
    Jun 8, 2022 at 16:06
  • 1
    $\begingroup$ The really simple answer is that it measures weight. $\endgroup$
    – JimmyJames
    Jun 8, 2022 at 18:00

7 Answers 7


Technically, an accelerometer does not measure acceleration, it measures the difference in acceleration between the "sensing body" inside the sensor and the sensor's exterior casing. That means that if you have an accelerometer sitting still on a desk, it will measure somewhere around $9.81 \frac{\text{m}}{\text{s}^2}$ or 1G pointing downwards because while gravity is pulling on the "sensing body", the desk below is pushing upwards and preventing the casing from moving (see Newton's laws).

If you then flip the sensor on it's side, it will measure the same magnitude of acceleration, but pointing sideways instead, and someone who is just watching the numerical output change on the computer wouldn't be able to tell if the sensor had just been rotated on its side or if the direction of gravity had changed.

This also means that if the casing of the sensor is experiencing the same acceleration as the actual measuring tool inside, the sensor will read "zero gravity" because the difference is zero. You can test this by downloading an app that logs acceleration data on a smartphone and then (gently and carefully!) throwing it into the air. Afterwards you will see that the moment the phone leaves your hands, the acceleration jumps to zero along all three directions (XYZ typically) because the phone, which is connected to the sensor's casing, is itself accelerating downwards at 1G and thus the difference becomes zero despite the fact that seen from an external reference frame, it is accelerating downwards.

This is why your g-sensor will read zero and you will feel an apparent "zero-g" in orbit, because you are effectively in a free-fall all the time. You are constantly accelerating but because the internals of the sensor are too, the difference reading is zero.

  • 7
    $\begingroup$ 'it measures the difference in acceleration between the "sensing body" inside the sensor and the sensor's exterior casing.' – Huh? That doesn't seem to make any sense. If I have an accelerometer sitting on my desk, then the acceleration of the sensing body is 0, and the acceleration of the sensor's exterior casing is also 0, so the difference between the two is 0. What am I misunderstanding? $\endgroup$ Jun 5, 2022 at 19:17
  • 1
    $\begingroup$ @TannerSwett, You are right about one thing: The "sensing body" of an accelerometer does not accelerate with respect to the "case" when the acceleration is constant. But, when I set my phone on the desk, the accelerometer doesn't say zero. It says that my phone is accelerating upward at 1G. That's as it should be because any coordinate frame that's rigidly attached to the Earth is an accelerated coordinate frame. My phone's accelerometer only says zero when it is stationary within an inertial coordinate frame (i.e., when the phone is in free-fall.) $\endgroup$ Jun 5, 2022 at 19:44
  • 22
    $\begingroup$ @Dragongeek, An accelerometer does not measure "difference in acceleration..." It measures the force between the "sensing body" (a.k.a., "proof mass") and its supporting structure. $\endgroup$ Jun 5, 2022 at 19:52
  • 1
    $\begingroup$ An ideal accelerometer in general relativity measures acceleration with respect to a co-located, co-moving inertial frame. (Inertial frames in general relativity are not the same as inertial frames in Newtonian mechanics.) On the surface of the Earth, that's a stream of free-falling apples. In space, a non-thrusting vehicle that is not subject to drag, radiation pressure, etc. is a local general relativistic inertial frame. Real accelerometers are fairly close to ideal accelerometers in general relativity. $\endgroup$ Jun 6, 2022 at 2:15
  • 1
    $\begingroup$ @Ben: You can't measure hypotheticals. Acceleration is well-defined, it's the derivative of velocity. The velocity of the accelerometer might not be zero, depending on your choice of reference - see the example of an accelerometer in orbit, considered from an observer on the ground. Taking the derivative drops the reference term. In any case, the proof mass stays within the accelerometer, so its velocity must be on average identical, and its acceleration too. $\endgroup$
    – MSalters
    Jun 7, 2022 at 11:31

There's at least one other answer here that is correct in all its details, but it gives a lot of details. Here's the simple version:

An accelerometer measures the force that's needed to keep a "test mass" stationary with respect to its supporting structure, and then it divides that number by the nominal weight of the test mass to give a reading in "g" units.

As @Barmar said in a comment on another answer, the mechanism of an accelerometer is the same as the mechanism of a bathroom scale. The only difference is in how the measurement is interpreted. With the scale, the mass is unknown, and the acceleration is assumed to be 1G ($9.8\frac{m}{s^2}$ at sea level on Earth). With the accelerometer, the mass is the known quantity, and the acceleration is the unknown.

  • 1
    $\begingroup$ A lot of answers are missing the point and you caught it, so I beg you to emphasize it more: "accelerometer" is a misnomer because it does not measure acceleration, it measures force. It shows 1g when stationary on a desk because there's (1 times the standard Earth force of gravity) pulling/pushing the test mass and the wall of the superstructure closer together (and an equivalent force keeping them apart; springs, the normal force, etc.) Toss it in the air and catch it, and it will reflect the lack of internal forces, followed by considerable forces required to decelerate. $\endgroup$ Jun 6, 2022 at 1:21
  • 2
    $\begingroup$ @SirTechSpec In my opinion, both approaches have merit and are somewhat correct. Accelerometer measures force, but that force is equal to $F = ma$ and since the mass $m$ of the accelerometer's damped load component is constant, the only measurable parameter that is variable is $a$. Therefore, in a way the accelerometer measures both $a$ and $F$, they are differing only by the scaling factor. $\endgroup$
    – user47149
    Jun 6, 2022 at 1:41

What is measured with g-forces?

In Newtonian mechanics, it's acceleration (scaled by $9.80665\,\text{m/s}^2$) due to real forces, except for gravitation. Accelerometers do not measure accelerations due to fictitious forces such as the fictitious centrifugal and Coriolis effects as those are not real forces. In general relativity, it's acceleration due to real forces (also scaled by $9.80665\,\text{m/s}^2$) as gravitation is a fictitious force rather than a real force in general relativity.

The general relativity explanation is perhaps easier, at least conceptually. (The mathematics of general relativity is a rather difficult.) That gravitation is a fictitious force in general relativity is a consequence of the equivalence principle. No local experiment (an accelerometer is close to being a local experiment) can detect acceleration due to gravitation.

From a Newtonian mechanics perspective, accelerometers do not detect acceleration due to gravitation because every accelerometer has a test mass inside it. That test mass is somewhat free to move with respect to the accelerometer case. The accelerometer uses some device such as a spring to keep the test mass inside the accelerometer case. In the case of a spring, the accelerometer reading is the extension or compression of the spring.

Consider an accelerometer orbiting the Earth. The gravitational acceleration of the case is the same as the gravitational acceleration of the test mass. The accelerometer reads nothing.

Next consider an accelerometer at rest on the top of a table which in turn is at rest with respect to the Earth's surface. The accelerometer is subject to gravitation plus the normal force exerted by the tabletop. The test mass is subject to gravitation plus the spring force (or whatever other mechanism is used to keep the test mass inside the case) -- but not the normal force. The acceleration of the test mass due to the spring force (spring force divided of the test mass's mass) has to be equal to the acceleration of the case due to the normal force. The accelerometer registers 1 g up.

  • 6
    $\begingroup$ This sound similar to what a bathroom scale measures -- the compression of a spring. But rather than calculating the mass of the object based on the known gravitational acceleration, it calculates the acceleration based on a known mass. $\endgroup$
    – Barmar
    Jun 5, 2022 at 19:13

It is the amount of acceleration relative to the conditions of free-fall -- that is what an accelerometer measures. Acceleration defined as relative to the conditions of free-fall is referred to as "proper acceleration". Here is how the relevant Wikipedia article introduces us to this concept:

[...] proper acceleration is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is thus acceleration relative to a free-fall, or inertial, observer who is momentarily at rest relative to the object being measured. Gravitation therefore does not cause proper acceleration, since gravity acts upon the inertial observer that any proper acceleration must depart from. A corollary is that all inertial observers always have a proper acceleration of zero.

In your question, you are correct to notice that an accelerometer sitting at rest on the surface of the Earth shows $1\ g$ despite not accelerating. However, let's consider a different frame of reference, where we agree to set the accelerometer's scale in a way that $0\ g$ means "the conditions the accelerometer would be experiencing during free-fall". Then, we arbitrarily assign "zero acceleration" to that state -- in other words, we define an "accelerational standstill" to be the state of any object experiencing free-fall. From the viewpoint of an object being in the state of free-fall, an accelerometer sitting at rest on the Earth's surface does indeed accelerate; in specific, it experiences $1\ g$.

Meanwhile, being in an orbit IS being in freefall. The experiences are equivalent and an accelerometer reads $0\ g$ in those cases.

An analogous example would be to consider a hypothetical scenario where we agreed that, in a certain context, we will be using a specifically modified measuring tape. The modification is such that the tape does not start at the point where it reads $0\ cm$; instead, it extends another $10\ cm$ in the opposite direction, and that "duplicated" $10\ cm$ mark is where it starts. Going from the start, the numbers go down until they reach zero, then start increasing again as they would do in a normal tape:

enter image description here

Going along with that, one could ask a perfectly valid hypothetical question similar to yours: if I am measuring something of zero length, why does the tape read "$10\ cm$" instead? If I am measuring the tree log illustrated above, which clearly has non-zero length, why does the measurement read "$0\ cm$"? Answering those, hopefully trivial, questions and applying analogous thought process to your initial question could help the concept of accelerometer's spirit truly click in for you.

† g-forces could be really small, but are never exactly zero; the small-but-not-zero values could differ a little. For all intents and purposes, however, the experiences are equivalent, and the value of zero could be used as if it was exact.

  • $\begingroup$ Re, "being in orbit...never exactly zero," In a non-rotating coordinate system whose origin is in orbit around the Earth, the proper acceleration exactly at the origin is zero, but it's non-zero everywhere else. Its magnitude increases with distance from the origin. We call the force needed to hold a particle at a fixed location within that coordinate frame, "tidal force." Within a spacecraft the size of the International Space Station, tidal force is small enough that astronauts aren't continually aware of it, but it's large enough to be measured. $\endgroup$ Jun 7, 2022 at 14:16

An accelerometer may be built from a small mass held by a spring. The force on the mass is measured by the deformation of the spring.

According to Einstein's very famous thought experiment with the elevator you can't decide with measurements made only inside the closed elevator box what is the reason of the measured acceleration. It may be a gravitational field only or an accelerated movement or a combination of both.

Zero gravity may be measured when the elevator is very far away from any star or planet. But a free falling elevator may cause zero gravity, too. You can't measure with the accelerometer if it is a circular or elliptical orbit around Earth or a linear, parabolic, or hyperbolic trajectory.

A Lagrange point is another possibility for zero gravity.

  • 3
    $\begingroup$ Lagrange points are irrelevant here, for two reasons. One is they are not points at which gravitational acceleration is zero. They are points at which the acceleration due to gravitation cancels fictitious accelerations as calculated in the synodic frame, a frame that rotates at the same rate and same axis of rotation as that of the two bodies orbiting another. Another reason they're irrelevant is because an accelerometer in orbit (with no thrust, no drag, no radiation pressure, etc.) will read zero acceleration. Accelerometers do not sense gravitation. $\endgroup$ Jun 5, 2022 at 20:13
  • $\begingroup$ @DavidHammen "Accelerometers do not sense gravitation" ? From en.wikipedia.org/wiki/Gravimetry#Gravimeter : "An instrument used to measure gravity is known as a gravimeter. For a small body, general relativity predicts gravitational effects indistinguishable from the effects of acceleration by the equivalence principle. Thus, gravimeters can be regarded as special-purpose accelerometers." If a gravimeter is a special kind of accelerometers then accelerometers do sense gravitation. The solid state accelerometer chips used in smart phones to detect the orientation do sense gravitation. $\endgroup$
    – Uwe
    Jun 5, 2022 at 21:29
  • 1
    $\begingroup$ That's wikipedia for ya. Accelerometers measure everything but gravity. A gravimeter measures the normal force. If an object is standing still on the surface of the Earth, then the upward acceleration that a gravimeter does measure is equal to the combination of the downward acceleration due to gravitation (which is a fictitious force) and the upward centrifugal acceleration (which is a fictitious force). As a side remark, geophysicists and geologists distinguish between gravitation and gravity. Gravity includes both gravitational and centrifugal acceleration. $\endgroup$ Jun 5, 2022 at 23:02
  • $\begingroup$ So the ADXL335 chip does not measure gravity? So this claim of the datasheet is wrong "The accelerometer can measure the static acceleration of gravity in tilt-sensing applications as well as dynamic acceleration resulting from motion, shock, or vibration."? See pdf.utmel.com/r/datasheets/… on page 10 $\endgroup$
    – Uwe
    Jun 5, 2022 at 23:30
  • $\begingroup$ No, it does not measure gravity. No local experiment can measure gravity. From a Newtonian Earth-centered Earth-fixed perspective, the forces on an accelerometer at rest on a tabletop are the inward acceleration due to the Earth's mass, the outward acceleration due to the fictitious centrifugal force (small, at most 0.00346 g), and the normal force. These forces cancel, and yet the accelerometer reads a value of 1 g upward. $\endgroup$ Jun 6, 2022 at 2:03

G-force is acceleration (its magnitude). Yes it is that simple! A spinning object or a non-rigid object will have different acceleration at different points, so the g-force across the object varies. Thus the crash earpieces in F1 cars.

Objects at rest on Earth's surface have zero coordinate acceleration. Meaning that they are not changing the numerical values of their coordinates (such as altitude). But they are constantly accelerating upward at $9.8 \frac{\text{m}}{\text{s}^2}$ (and towards Earth's axis at a smaller number due to Earth's rotation). We are more interested in coordinate acceleration if we want to know where we are in space so many meters subtract out the $9.8$. Since gravity varies slightly, some meters have a "sit still" calibration procedure to measure the precise gravity.

With all this acceleration, why are these objects not getting anywhere? Spacetime is curved around massive planets such that geodesics tend to converge as do "parallel" lines on the sphere. Thus you have to keep accelerating to fight this divergence and stand still.


There is no physical difference at all between being supported in a gravity field and being accelerated, or between free-falling in a gravity field and floating in (a hypothetical) gravity-free space. Spacetime, as seen by the accelerometer, is curved — or not curved — in exactly the same way, and this curvature exerts forces on masses, and these forces are measured by it.

There is a little paradox here, if you want: In a gravity field, the accelerometer actually measures the absence of acceleration as observed by a third party: After all, we are sitting still relative to our gravity well. Therefore, accelerometer is something of a misnomer because it shows acceleration where a third party observer does not see any. (I mention the "third party" because only relative to the third party an accelerometer sitting on a table in a black box is un-accelerated. The accelerometer itself, as mentioned, perceives being accelerated.)

What the accelerometer measures is the force (if any) keeping it from following a geodesic, the shortest path between two points in spacetime.

Paradoxical consequence: Since there is no gravity-free point in spacetime, all accelerometers showing zero acceleration — because they are in free-fall — are actually accelerating according to third party observers at rest relative to the source of gravity. And conversely, all accelerometers at rest relative to a near-by gravity well — like the one on your desk — actually show acceleration.

  • $\begingroup$ I deliberately avoided mentioning the word accelerometer in the question. Your bold text is helpful and concise, thanks! $\endgroup$
    – Innovine
    Jun 9, 2022 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.