What does a helium balloon do on the ISS vs in open space?

[A.T.]

So, "a measurable weight" is not what is being registered by the scale.

What does "measurable weight" mean here? Do you and Dave have the same thing in mind?

 

[A.T.]

So, okay, keep the air, and let the person have a cross-sectional area normal to the direction of travel sufficient that the air imparts the same drag (upward force) on the person as the bathroom scale as they both fall through the air. What does the scale register?

I would say, once you reach terminal velocity , (no acceleration) you then weight what you weigh on the surface of the earth, just like an elevator going down.

If part of the weight is balanced by the drag on the body, the scale will not show the full weight at terminal velocity.

 

[Zeke137]

What does "measurable weight" mean here? Do you and Dave have the same thing in mind?

No, I mean that the force registered by the bathroom scale in Dave's example is simply the difference in forces exerted by the air on the person and on the bathroom scale. If the drag on the person were the same as the drag on the scale as they both fall unhindered, then no net force would be registered by the scale. As in the Einstein elevator thought experiment.

 

[Zeke137]

Try this thought experiment: increase the density of the medium, bit by bit, from the density of air to the density of ground. At what point does it go from being "merely air" to being a surface dense enough that you consider it a bona fide weight on the scale?

Both the person and the scale are descending through the air, so they are surrounded by the air. Let's make the drag on the person and the bathroom scale equal: no net force is registered by the scale. Now, let's have the person and the scale fall through molasses: assuming the drag on both are still equal in this new fluid, which surrounds them, then again no net force will be registered. Replacing the molasses with soil or rock, they'll both be surrounded by this new medium and again, no net force will be registered by the scale.

 

[zanick]

If part of the weight is balanced by the drag on the body, the scale will not show the full weight at terminal velocity.

yes , i was assuming that the person was in the wake of the scale , not experiencing any drag... so it would be similar to the elevator thought experiment. but yes, the weight of the scale itself, matched by its drag would have measurable weight, but that would be measured by the drag forces required to stop its acceleration. same with the falling person ... there would be no measurement on the scale ...

 

[A.T.]

If the drag on the person were the same as the drag on the scale as they both fall unhindered, then no net force would be registered by the scale.

This is not true, unless the scale and the person have the same mass.

 

[zanick]

I was referring to their drag equaling their respective weights.