I was studying about $g$-forces. It is basically non-gravitational accelerations imparted to a body by forces acting on it, other than gravity. I came to know that standing still on Earth's surface means we are experiencing $1g$. When we are in free fall, we are experiencing $zero$ $g$ because there is no non-gravitational acceleration, gravity is the only force that is accelerating us. Similarly, if we are accelerating upward in an elevator at $2$ $\frac{m}{s^2}$, we have a net non-gravitational acceleration of $12$ $\frac{m}{s^2}$, we are experiencing $1.2g$. These are a few examples.

So from what I see, $g$-force tells us about our non-gravitational acceleration, and it has nothing to do with force, it is obvious. Coming back to the elevator accelerating upward at $2$ $\frac{m}{s^2}$, let's say there are two people inside the elevator, and their mass is $50$ $kg$ and $60$ $kg$, respectively. Since both are accelerating at the same rate, i.e $2$ $\frac{m}{s^2}$, it clearly means both are experiencing Normal reaction of different magnitudes, $600N$ and $720N$ respectively. But both are experiencing a $g$-force of $1.2g$.

So doesn't $g$-force tell us about acceleration? What does the word $force$ in $g$-force mean?


2 Answers 2


When you are in free fall you are subjected to the force of gravity. But you do not "feel" or "sense" the force. In other words you are not subject to contact forces. In practice g forces are surface-contact forces between objects expressed in multiples or fractions of the force of gravity on your mass. When you stand on the ground you experience a contact force (on your feet) of 1 g (a force of mg divided by your mass m). Essentially, the g force is the force per unit mass on your body (F/M, or Mg/M = g).

In free fall you experience no contact force, so you are experiencing zero g.

Hope this helps.

  • $\begingroup$ It did help, thanks a lot. The problem is, when I googled what g-force was, different websites explained it in different ways. Some say that g-forces can be thought of as non-gravitational accelerations imparted by non-gravitational forces (hence my post). While others say that g-forces are non-gravitational forces due to object's acceleration caused by circular or linear accelerated motion...etc. Though they are right in their understanding of g-forces, it would confuse those who don't know what g-forces are, and are trying to understand, which is what happened with me. $\endgroup$
    – 4d_
    Nov 24, 2019 at 16:58
  • $\begingroup$ The way you explained them in terms of surface-contact forces between objects clears it up for me. Can I say that the Earth does not experience any g-force when going around the Sun because of Sun's gravity? Likewise, moon does not experience any g-force due to Earth's gravity? Because they are not in contact? $\endgroup$
    – 4d_
    Nov 24, 2019 at 17:01
  • $\begingroup$ @πtimese Re earth sun moon analogy, hadn't thought about it. But I suppose if the earth and moon could "feel" things then I guess you are correct. $\endgroup$
    – Bob D
    Nov 24, 2019 at 17:13
  • 1
    $\begingroup$ This discussion of "weightlessness" centers around what a person "feels" or senses. I just meant the earth and moon don't feel things. But orbits are essentially free falls about the center of mass of the orbiting objects. Thankfully, they do not involve "contact" forces between the sun and earth and moon and earth. Hope that clarifies $\endgroup$
    – Bob D
    Nov 24, 2019 at 17:29
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    $\begingroup$ @πtimese Yes you can. I never heard of zero-g planes until you mentioned it, but I looked it up in Wikipedia. It says while following the parabolic path the aircraft and its payload are in free fall at certain points of the path and that it is used in this way to demonstrate to astronauts what it it is like to orbit the earth. When a satellite orbits the earth it is essentially in continual free fall about the earth. The same applies to the moon orbiting the earth and the earth orbiting the sun. So, yes, the earth and moon experience 0 g just like the zero g plane and astronauts. $\endgroup$
    – Bob D
    Nov 25, 2019 at 3:03

It's definitely measuring force as opposed to acceleration.

For example, in freefall, we say there are $0$ $g$'s acting on the body; but it's accelerating.

Conversely, a person standing on Earth's surface has no vertical acceleration, only a vertical force equal to $1g$; yet we say it experiences $1 g$ in the vertical direction.

G-force is a measure of force per unit mass acting on the object. This happens to have the same units as acceleration, but that does not make $g$-force an acceleration on it's own. It still measures the force; just normalized for the mass experiencing the force.

It's very useful. For example, on Earth we know all masses should experience 1 $g$ of weight (force) when supported by Earth's surface; it's a constant relationship between mass and force. When traveling in elevators, or maneuvering in planes (a couple examples), the movement will cause different forces to act; but these forces may change by a predictable amount per unit mass - this is where the concept of $g$-force is useful.

  • $\begingroup$ This is a useful distinction, but still a little confusing. After all, an object in free fall is accelerating AND experiencing a force. $\endgroup$ Nov 24, 2019 at 15:07
  • $\begingroup$ @JahanClaes I was trying to think of a way to work that in without making it more confusing; but I figured starting with the obvious examples of acceleration with $0 g$'s and no acceleration with $1 g$ might clear up OP's confusion. If they do bring up that there are forces in freefall, I could definitely elaborate; but I felt it may do more harm than good if I add too much information that they didn't directly ask about. $\endgroup$
    – JMac
    Nov 24, 2019 at 15:13
  • $\begingroup$ @JMac "G-force is a measure of force per unit mass acting on the object, there's why happens to have the same units as acceleration". Thanks, I was confused why it had the same units as acceleration, it cleared my doubt. If I go back to the example in my original post, the guy whose mass is 50 kg experiences a normal reaction of 600 Newtons, that's 12 N per kg, which is 1.2g or 12m/s². Similarly the 60 kg guy is experiencing 720N/60kg = 1.2g, or 12 m/s². $\endgroup$
    – 4d_
    Nov 24, 2019 at 17:12

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