# Does weightlessness make no difference to free fall?

I looked up a lot of articles and questions about free fall motion and weightlessness. And I know there are a lot of similar questions.

In conclusion, I understand that there is no way to distinguish between free fall motion and zero gravity. (Those two are said to be perfectly identical.)

But the following points make me confused. To feel exactly the same is simply an observer's feeling, and I think it's more to do with psychology than physics (there's a strict definition of physics, but it's probably something I don't know).

From the standpoint of a free-falling observer, no force is felt on the observer. This would be the same for observers in zero gravity.

Let's assume an external observer fixed to the ground. Although the free-falling observer and the weightless observer feel the exact same thing, the two are different movements for the ground observer.

Calculated by an observer fixed to the ground, the net forces acting on these two are different.

In other words, an observer who is in free fall or weightlessness cannot distinguish between these two situations, but an observer who is stationary on the external ground can distinguish between these two situations.

Does it depend on which system the observer observes, whether free fall and zero gravity can be distinguished or not? (Does net force have nothing to do with distinguishing the motion of an object?)

I know it's very arrogant to try to understand these situations without a lot of physical knowledge. But on this great site, I'm asking if someone can help me.

P.S.: Now let's assume that the spacecraft is accelerating the observer inside the spacecraft to g, the same direction and magnitude as the acceleration of gravity. The observer at rest on the ground will calculate the net force acting on the observer equally in mg, both free fall motion and this motion. In addition, both observers inside the spacecraft and those who are free to fall will calculate the net force acting on them as zero. However, the same net power does not mean that these two situations are the same. Obviously, the person in the spaceship and the person in the free fall feel a completely different thing. This is said to be the difference between the characteristics of gravity, which is a non-contact force, and the forces that are not. But it's not easy to understand something by itself. How do I understand this?

An observer who is stationary on the external ground can distinguish between these two situations.

Only if they assume that they are stationary. But they don't know that. The person on the ground knows two things:

1. Relative to them, the person in free-fall is accelerating at $$9.8$$ m/s/s
2. The ground is exerting a normal force on them of $$9.8$$ newtons per kilogramme

Both of these facts have two alternative explanations:

1. They are stationary and there is a gravitational force of $$1$$ g acting both on them and on the person in free-fall.
2. There is no gravitational force, they are being accelerated upwards at an acceleration of $$9.8$$ m/s/s, and the person in free-fall is stationary.

(In fact, there are other explanations which combine gravity of different strengths with different rates of acceleration, but these two are the simplest explanations).

There is no local measurement that the person on the ground can make to distinguish between these two explanations. So their situation is just as ambiguous as for the person in free-fall.

• thx! I think I understand. Even observers who assume that they are stationary outside cannot tell whether they are free fall or stationary. So, you mean you can't tell if an observer is free-falling toward a stationary object or if the object is free-falling toward a stationary observer. Any observer in any system can state that it is free-falling or stationary, depending on the point of view. Feb 20 at 10:43