# Tag Info

## New answers tagged free-fall

0

It is because general theory of relativity. As it says, one person is in a lift suddenly someone cut the cable, then he will experience no gravity as both the lift and person are falling at same rate.

1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.

1

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2}$ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2}$ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...

1

There is another aspect somehow overlooked by the other answers. Consider a pile of iron filings accelerated towards a magnet. If you were to arrange so that they all have the same magnetic force per unit mass they would appear to experience no force relative to each other while being accelerated towards the magnet, and if you had weak bonds holding them ...

1

We don't need to appeal to relativity to explain why you don't feel any force in free fall. Plain old Newtonian mechanics predicts that too. What you actually feel when you feel a force being applied to you is that the external force applies only to a small part of your body (the soles of your feet if you're standing up and feel the normal force from the ...

0

Why does a free-falling body experience no force despite accelerating? Because there is no force acting upon it. If you look at some pictures of the principle of equivalence, you will find that they typically depict a guy in a rocket accelerating through space. There's a force on his feet, he can feel it. They also depict a guy standing on the surface ...

39

Before telling you why an observer in free fall does not feel any force acting on him, there are a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like the position, velocity, and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the ...

5

Well, everything depends on what you mean by "to experience a force". I suspect that you are thinking of some psycho-physical idea. Indeed both floating in space and freely falling we perceive similar sensations. The reason is simply due to the fact that, in both situations, all particles of our body moves with the same speed (due to a spatially uniform ...

1

You need a coordinate system to decide a body’s position, velocity, acceleration, momentum or force on it. Assume the body is in free fall near the Earth. 1) First consider a coordinate frame (3 perpendicular rods and a clock) with its origin in free fall near the free falling body. By the equivalence principle we know the rods are falling in unison with ...

23

It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...

3

falling in a gravitational field is physically indistinguishable from floating in interstellar space Yes. Indeed, this is one of the founding principles of general relativity and is (one of the forms of) the equivalence principle. Your argument is that we can feel acceleration, and gravity makes you accelerate, so shouldn't you feel acceleration while ...

1

I fail to understand the question. What is the idea behind the question ? The distance the rock travels has to be h, in reality. But how does taking photos change the average distance ? This seems like an exercise in mathematical methods of the probability theory, there is not really any physical reality behind it. What Griffiths is probably referring ...

0

The object's center of mass will fall straight downward, accelerated at g. All objects accelerate at the same rate in a vacuum, regardless of their weight, so neither end will accelerate faster than the other. That said, with air resistance, both ends of the object will feel the same drag force, so the end of the object with less mass will feel a greater ...

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It looks like this picture on the right from Misner-Thorne-Wheeler: It's almost a straight line. But can I add this: I don't like Kruskal-Szekeres coordinates at all. Take a look at the picture on the left. That shows the object's path on the Schwarzschild diagram. Note how it's truncated vertically? The vertical axis is the time axis. In order to ...

0

The sign is simply a convention: a convention is being used which identifies motion away from the large, gravitational body (against gravity) as having a positive sign. Thus motion in the direction of gravity is negative. This convention is useful because when we perform work on an object by raising it against against gravity, we increase its gravitational ...

0

There is a subtle difference between displacement and position. Position is a vector measured relative to a defined origin and is often designated by the vector symbol $\vec{r}$. In a Cartesian coordinate system, $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},$$ where $x$, $y$, and $z$ are the individual coordinate values, positive or negative, relative to the ...

0

Suppose I throw a ball from my chest-height straight up into the air: it travels upwards for a time $\tau$ before it reaches its maximum, then it falls down for a time $\tau$ too. We know that it achieves the height $h = \frac 1 2 ~g~\tau^2$ above my chest-height in this time where $g \approx \text{9.81 m/s}^2$, so if we take my chest-height as the value ...

1

Displacement is the vector version of distance, so it has a magnitude (the distance) and a direction. If the motion is only in one dimension, as in free fall, then the direction manifests only as positive and negative, or up and down but you are free to define whether up is positive or negative (and similarly for down) as long as you are consistent within ...

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