I am going to explain what I think I know and how that leads up to my question - please correct any false claims I make or conclusions I draw.

I initially thought that $G$s were defined purely by acceleration. This confused me though, because when a body is at the surface of the Earth, even without change in velocity, it is said to be experiencing 1g. Furthermore, when in free fall, a body is said to experience 0g.

One answer I got on Quora said that $G$s are actually defined by the resistance to acceleration. This seems to make sense, when in freefall I am not resisting any acceleration, and when I am standing on earth I am resisting 1g.

However, this started to confuse me when I began to add other forces. For example a body in a complete vacuum is acted on by a force. The only resistance to this force is the body's inertia, which resists acceleration. That makes sense to me. However, what happens when I add resistive forces? Do you only take the resultant force? That seems an attractive option, because the resultant force is the only part of the force that induces acceleration. But, if I go back to earth, when I stand on its surface, I have gravity pulling me down, and I have the force of the earth's surface (normal) resisting my motion. If I am not accelerating, that means there is no resultant force, and yet I still experience 1g.


Note: G and g are different. g and g are also different.

How G and g are different

G is the (universal) gravitational constant. g, however, is the gravitational acceleration.

G is a constant. It's used in the formula for the Law of Universal Gravitation (look it up if you need), which gives the gravitational force between any two objects, even you and a star thousands of light-years away.

g is the gravitational acceleration and is different depending on which planet you are on. On the surface of Earth, it is around 9.8 m/s². The definition actually is purely acceleration.

Now for why "g" is different from "g"

Normally, in physics, g is the acceleration you would have if you're free-falling in a vacuum.

But when you say "experience one g", the "g" is not the same "g". Here it actually means G-force, which basically means the force of gravity. So if you experience 2g it means you are experiencing a force twice as strong as the gravity of the Earth.

The reason you would experience more than one G-force is because of acceleration, for instance, if you're in a rocket that is propelling itself and speeding up at a fast rate. Because F = ma (Newton's Second Law), having a larger a makes the force on you larger.

To recap, gravitational acceleration, g, is defined just by the acceleration an object would have in free-fall, which is about 9.8 m/s² on the surface of Earth. When you experience 1g it means G-force instead of the gravitational acceleration. You can experience more than 1g from gravity because of acceleration.


When forces act on a body, there are three main effects:

  1. the body may accelerate as a whole
  2. the body may be squeezed or stretched or otherwise distorted as a whole
  3. the body may be caused to rotate
  4. internal parts of the body may be forced to rearrange themselves

When we want to measure the level of discomfort experienced by a pilot in an aircraft, or similar situations, it is item 4 here that is the significant one.

Let's apply this to the cases you mentioned.

First: someone standing still on planet Earth. They have $mg$ gravity force pulling them down, and a force of the same size pushing upwards on their feet. Therefore they are being squeezed. But this squeezing is quite gentle compared to the squeezing effect of atmospheric pressure, so it has no physiological effects. The main effect of gravity is on the motion of fluids in our bodies, especially blood flow and also eyeballs, brain and ears. Gravity is pushing our blood towards our feet. The veins and the heart need to provide a compensating pressure, and this is the scenario known as "1g".

Now consider an aircraft pulling out of a steep dive. In this situation the force on the pilot's body is $mg$ from gravity (where $m$ is the mass of the pilot) and also a larger force $F$ from the seat. The acceleration of the pilot is therefore $F - mg$. There are two effects on the physiology of the pilot. First the two forces $F$ and $mg$ acting in different directions provide the same squeezing effect as would be obtained from a pair of forces of size $(F+mg)/2$ acting from above and below for a pilot at rest. But a body can resist this squeezing quite easily. We are already being squeezed by atmospheric pressure much more than that.

What matters to the experience of the pilot is the other effect: the blood being pushing towards his feet, and things like that. To calculate that, first note what motion of the pilot would result in no such force at all. It is the freefall motion (when he is falling down, acceleration downwards at $1g$). His actual motion is acceleration upwards with acceleration $(F/m - g)$. So his acceleration relative to the one which would give no blood pressure effects is $$ ((F/m) - g) + g = F/m $$ So the "number of G's" is $F/(mg)$.

The same result can also be calculated another way.

Consider an artery or vein somewhere in the body. The artery wall is accelerating upwards at acceleration $a = ((F/m) - g)$. A blob of blood of mass $m_b$ located inside the artery experiences a force $m_b g$ from gravity, and also a pressure force from the surrounding fluid, so that its acceleration matches that of the rest of the pilot's body. So the pressure force $F_p$ must be given by $$ F_p - m_b g = m_b a $$ therefore $$ F_p = m_B (a + g) = m_b (F/m) $$ This means that the blood has to behave as it would if there were no gravity and the acceleration was $(F/m)$. So again the number of G's is $F/(mg)$.

The overall conclusion is that the force which counts here is the non-gravitational force which is causing the motion of the body to be different to the one it would have in free-fall. (And by the way, by that insight you also gain the beginnings of an intuition about the way gravity is understood in general relativity).


How do we define $G$s?

The misnamed g-force, or "$G$s", on an object is the component of acceleration that result from every external force except gravity, scaled by 9.80665 m/s^2. I wrote "misnamed" because $G$s have units of acceleration, not force. I wrote "component" because just as one can break down the net force acting on an object into individual components, the net acceleration on an object can be broken down into individual components by dividing the individual force components by mass.

The reason for excluding acceleration due to gravity is that nothing can sense that acceleration. The gravitational acceleration on astronauts and cosmonauts in the International Space Station is about 89% of the value on the Earth's surface. Those astronauts and cosmonauts are said to be in a zero-g environment because they cannot sense that gravitational acceleration. No local experiment can sense that acceleration. This is a direct consequence of the equivalence principle.

On the other hand, an object at rest on the surface of the Earth is pushed upwards at 1 g by the normal force from the Earth's surface. While the object cannot sense acceleration due to gravity, it can sense that upward acceleration due to the normal force. An accelerometer at rest on a table will register an acceleration of 1 g upward. An accelerometer at rest in the International Space Station will register close to nothing.

The concept of "$G$s" is important because it represents the acceleration that humans can feel.


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