PCHRISTMAN is correct. There is weight in free fall. There's just no way to measure it.

http://www.astro.uiuc.edu/classes/ar...iscussion.html

 

I assume you are talking about this passage:

Q:

On question 31 of hw 3, the question was whether or not the weight of an astronaut in freefall would be zero and the answer is that it is NOT zero although the scale will read zero. Why would ones weight not be zero during free fall?
A:

This question is getting at what free fall and weight really mean. As defined in this course, "weight" always means "the force of gravity on an object." In other words, weight is a force. So by this definition, one is only "weightless" when the force due to gravity is zero. But you know that gravity reaches out into space (the rule is that the force is F=GMm/R2, which means that F is nonzero for any R smaller than infinity!). Indeed, if astronauts were really weightless (= no gravity acting), then nothing would keep them in orbit, and they would fly away in a straight line. However, the certainly do have weight, and remain in orbit. Similarly, when you jump into the air, you are not weightless, though you are not on the ground--indeed, your weight is what guarantees you will return to the ground.

But why does the scale read zero if the astronaut is not weightless? Because both the astronaut and the scale are in free fall. This means they both have the same acceleration, moving in lock step. This means that the astronaut is not accelerating towards the scale, and so exerts to force on the scale, and thus it reads zero. So it can be true that there is still a force acting (gravity) but in this case, a scale reads zero (which is why people sometimes speak of astronauts as "weightless"--though they aren't the way the term is defined in this course).

Again, this situation is like jumping in the air. Before you jump, you can feel pressure on the bottom of your feet, which is due to them transmitting your weight to the ground (and the ground's pushing back on you). But while you are in the air, you don't feel pressure on the bottom of your feet, and if a scale were slid under your feet, it would read zero--until you hit the ground!

I realize that the distinction here is subtle, but that's why this is a bonus question!


Here's the thing: Replace the astronaut with the bucket of water, and the scale with the cork. They both accelerate at the same rate. So, the cork would stay in the same place in the bucket.

 

All right, now this is getting interesting…

The link is correct but misleading and not relevant to the cork problem.

Lets digress for a minute on weightlessness. We have all seen the nice shows NASA puts on with astronauts floating cheerfully in mid air and lounging comfortably against the walls and/or ceiling with total disregard for the fixed up and down the rest of us are stuck with. Does everyone agree these astronauts are weightless?

This begs the question of “how or why are they weightless?”

Gravity decreases as an inverse square of the distance from a mass so even in an orbit 100 miles up the astronauts are still subject to 0.99 g’s. So, how do these brave astronauts float around in the air?

The forces are balanced. Orbital velocity and gravity create a centripetal force of 9.8m/s^2 canceling each other. The astronaut still has mass but no weight because the forces are equal.

The same holds true of the bucket of water and the cork except in this case it is simple linear acceleration rather than centripetal canceling the gravity. Acceleration at 9.8m/s^2 straight at the ground and the forces are neutralized.

 

If absolute pressure gauges were place in the mythical bucket such that one was just below the water surface, and one was just above the bottom of the bucket, the gauge at the bottom would indicate a higher pressure before the drop. Would the gauges both zero out during the fall, indicating that no pressure existed within the water? Or, if they are equally affected by gravity during the fall, will they each maintain their original readings? Or will they go to neutral corners and wait for the bell?

 

Great question…. OK, off the top of my head -- water will float unsupported in the air as a sphere (ends up round presumably due to surface tension of the water) in a weightless environment so my answer would be that both gages would read the ambient air pressure.

 

Lets use the same analogy for another example. You are in orbit on the shuttle with a large sphere of water floating in the air in front of you. If you push a cork into the center of the sphere does the cork come popping back out and if so in what direction will it pop?

 

If ex-astronaut Bubba were sitting on a bathroom scale that indicated he weighed 200 pounds, and then some magic contraption released him and the scale, at the same time, to fall gleefully through space, would the scale change its reading to zero (until landing)? If bubba and the scale were released at the same time, but not in contact with one another, the scale would read zero. If Bubba then swam through space and sat on the scale, it would continue to read zero. If he held it between his big meaty hands and squeezed it, the reading would change to whatever he could muster (say, 62.4 lb). In each and every case, the force of gravity would be acting, in the same direction, on all the parts and people involved, as well as on any onlookers who wandered by.

 

If you loosely chain two superduty's together (facing the same direction), same motor, drivetrain, exact same weight, same everything, and you hit the gas in BOTH at the same time, does the chain tighten, loosen, or stay the same?

If you said stay the same, you're right. And for the same reason, that cork would stay where it was in the bucket at the moment it was dropped.

 

The linear acceleration you mention is due to the force of gravity, which means that gravity is canceling itself out. Nope.

 

Yes, it would read zero while in contact
Yes, it would read zero out of contact
Yes, it would read zero if "BUBBA" sat on it
Yes, it would read register whatever grip force Bubba could muster
Yes, the force would be with him

Oh, and he would still have the same mass and still be weightless....


Poor old Bubba weighs 200 pounds on the Earth, 33 pounds on the moon, a few ounces on an asteroid and nothing in freefall.

 

Ok, so Bubba is sitting on a scale that says he weighs 200 lb, when suddenly he and the scale are released into mid-air at the same time. If the scale then goes to zero, why? Where does the force come from that causes it to move from 200 to 0?

 

Poor old Bubba is taking a pounding, but the fact remains that the scale will not show Bubba’s weight while he is in free fall. In freefall Bubba is “weightless”.

How do you explain the astronauts floating in the much lighter and less dense air if measurable weight (as shown in your example) is present?

 

Oh I forgot to answer your question. The combined forces compressing the scale springs or loadcells are only present when the scale is moving less than 9.8m/s^2. When compressed against the floor by the 9.8m/s^2 gravity acceleration you see Bubba 200 pound weight. If it moved upward at 9.8m/s^2 rate the scale would read 400 pounds, and if it moves downward at a given rate it would read less.

 

Physics - just as fun as ever. I remember getting a 35 out of 100 on my final exam in Physics 211 in college - and getting a B+!

 

Actually it is fun, it’s just sometimes difficult to get a grip on it and understand why things you think are going to happen don’t.