Micro-Gravity Challenge

I'd think she'd spend a long, long time getting the balls to stop moving within the jar... and just when she did, she'd sneeze, and have to start over. But as long as the jar is at the ISS center of mass, then we could ignore the ISS's gravity. The Earth's gravity is balanced by the ISS orbit speed and arc.

So you'd expect the balls to behave like everything in the ISS - just drifting around. However, the Moon's gravity should cause a tide effect, so that over time, the balls would organize slightly according to the moon's relative position. I'd ignore the influence of the other planets, because calculating their overall effect would give me headaches, and because their effect would not be visible, or measurable, by anything I have sitting around.

In addition to the forces of the Moon, since the orbit of the ISS decays because of atmospheric friction, the jar would be unable to remain in any ISS center of mass as the friction of the atmosphere would have no effect on the jar and little balls so they would cntinue as they had been left, but the ISS would drop down towards the gravitational center of the Earth, but then to be returned to it's original point of equilibrium at some later time by the firing of 1 or more thrusters. As the ISS moved towards Earth, the gravity of the ISS for a short period of period of time would be greater on the heavier of the jar or the heavier balls pulling them away from the lighter balls, but then pulling them back together as the ISS is pushed back into orbit. I believe that the jar and heavier balls may approach their original positons in line but they may never totaly return there as the angle of the pull of the ISS would have to be from above the original orbit in order to return the balls and jar to exactly the same relative position as at the start of the experiment. There may be other forces at work as well. This is not intended to be the end all statement as I have to give somebody else a chance and go to bed myself.

Not zero gravity. This is due to the small drag exerted on the ISS by the atmosphere up there. It is being decellerated. Eg. the lead balls will migrate forward toward the direction of travel, and the floaters will go aft. Simple!

Since, in the absence of any corrective boost, the ISS is falling toward the earth, and given that the balls are free to move in the liquid, the distribution of the balls would mimic the behavior of the distribution of the oceans on Earth. That is, the balls, irrespective of mass (lead vs. plastic) that are to the Earth side of the center line of the jar would migrate toward the Earth since they will fall ever so slightly faster than the others due to their nearer proximity (greater gravitational effect) and the balls on the away side of the center-line will fall more slowly (due to a slightly lower gravitational effect) and consequently migrate to the away side of the jar.

I believe I overlooked the fact that the water, being more dense than the plastic balls will manifest a buoyant effect that will result in the plastic balls being forced to the center of the jar while the lead balls will end up being equally distributed at either end of the jar.

Being no external gravity (canceled out) the balls will attract each other forming a bigger ball some where in the jar. The inner will be formed of lead balls and the outer should be plastic balls. That's all.

I would agree with this. However, the system must be in an ideal state where there is no gravity or forces from any outside source that is significant to disturb the system.

What I am not clear about is what micro effects would be introduced by orbiting a planet (i.e., Earth) and that orbit would need to be a true circle and not an ellipse about the center of gravity of the planet. It seems that this is a more complex environment than the ideal state mentioned above.

My guess is that the water at the center of the jar is in free fall and weightless; but the water lower down is orbiting too slow to be weightless and so its density increases towards the Earth. On the other hand the water above the center is orbiting too fast and so centrifugal force causes an increasing density outward from the center. The result is that the plastic balls collect at the center of the jar. The lead balls collect at the two extrimities.

So far, less than half the replies got anywhere near the correct answer! Not to 'spoil the fun' for latecomers, lets just say that the 'weightlessness' in orbits do not mean zero gravity - like has been said: the operative word is 'micro-gravity'!

Okay, my understanding would be that the micro-gravity that the system experiences would cause the three items in the jar to stratify with the heaviest (lead) "falling" towards the center of the Earth, followed by the water, then the plastic balls. Of course the water would also encapsulate the lead balls, since the lead does not pack perfectly together without space between them. The surface tension of the water would probably add a dimension to the experiment. I would not expect the water to simply fill the jar as if it was sitting on a picnic table in park!

Wait, I think the stacking order will be reversed with the more massive lead gravitating away from Earth.

My thinking is that the system will under a centrifuge effect as it orbits.

Close, but not quite! One can ignore the surface tension of the water if the jar was filled and sealed as specified. What cannot be ignored is pressure differentials inside the jar...

Also see my reply to Blink on: "The answer is simple, or not…"

The lead balls will form a spherical nucleus in the center, with the plastic balls floating along the walls of the jar in single layer, as far from the nucleus as the geometry will allow.

Here's why:

The point of the "perfect conditions", it seems, is to achieve true "weightless" conditions. Given that assumption, with only three materials that can move or have an effect on each other, the only factors are gravitational effects of the materials relative to each other and relative buyancy caused by varying densities of the materials. Two of the materials are solid, which cannot change density, and one is liquid, the water.

The water, being liquid can have different densities, in a gradient increasing towards the center of mass. However, in that small quantity, we can assume a relatively homogenous density throughout, relative to the larger and smaller densities of the other two materials.

Normally, gravitational forces between such small quantities of matter would be very slight. However, given the absence of any other gravitational forces, and this is key, the "over time" in which the small forces can act, the tendency would be for the more massive (more density for the same volume) lead balls to attract each other and form a dense spherical nucleus. Similarly, the water and the lighter plastic balls would be attracted to the lead nucleus. However, the bouyancy of the plastic balls in water would make them "float" along the walls of the jar, since they cannot "sink" in the water towards the lead nucleus.

Q.E.D.

I like this STL's explanation, although it makes me think we would have to do a little arithmetic, which would be contrary to the instructions in the question.

For instance, although the center of the jar is at the center of mass of the ISS, every other point in the jar is not. The center of mass of the ISS is an imaginary point, where there may be very little actual mass. Thus, you'd expect the balls to be attracted to the various significant masses in the station, so that the distribution of the heavier balls might, given time, reflect the mass distribution of the ISS (which would change with people movement, etc.)

For example, suppose two balls are one gram each and 5cm apart. 100 cm away is a person of 90,000 grams. Both balls would be pulled to the person with a force much greater than their mutual attraction. (g*m1*m2)/d^2 Similar calculations could be made for other mass concentrations in the station.

So, to know where the balls would go, we would need to know the mass distribution of all the items in the station, and ask everyone to sit still. But before we were done with the calculating, wouldn't someone have to get up to grab a snack?

And then there are the inertial concerns mentioned by others... one orbit adjustment... footsteps vibrating the jar... all that would throw off our calculations. The effect of some would be orders of magnitude greater than microgravitational effects. Perhaps not so simple, after all.

Blink wrote ".... Perhaps not so simple, after all."

Yep, there are some uncertainties. One thing that is certain though (ok, I admit I used my calculator for this, but since you threw in some values...) is that the gravitational pull of a 90 kg person at 50cm range is about one order of magnitude less than the micro-gravity caused by Earth at 10cm below or above the ISS centre of mass (COM). (It's roughly 25 vs. 2.5 nano-g ).

As far as the distribution of mass inside the ISS goes, it largely cancels out around the COM. Granted, it's not a spherical symmetrical mass distribution, but as can be learned from the 90kg person at 50cm, the effects will be very small.

So what else stands in the way of the 'feeble' micro-gravity doing its business?

A good article that summarizes some of the perturbations in microgravity that interfere with experiments/demos like the one we have been discussing is here:

www.problems-to-tasks.com/people/stefan/publicatio ns/orbiting/hhamacher-rjilg-herichter-sdrees-19980 8.pdf

It's in plain english and pretty pragmatic. The various vibrations you can imagine (thrusters, people moving around, water dumps, machines starting up) are generally of far more influence than Earth-caused microgravity gradient effects.

Given that the ISS interior is apparently kept at a comfortable temperature, I'd want to assess the effect of Brownian movement too.

So how can one measure the height of a building, using a barometer. (Obviously, drop it off the top, and time its fall.)

OK, maybe it's not so simple... but it's a great question, because it got us thinking.

I could not get the address you gave to open, but I have some comments anyway.

I do think things like thrusters, water dumps, etc, i.e. one-time acceleration effects, can have a detrimental effect on experiments like the one under discussion. Vibrations will probably cancel out in this case.

One of my relativity friends pointed out that another problem in my described scenario might be surface effects between the balls and between them and the glass. They might just stick and the acceleration might be too small to move them. But, then, with only 40 or so small balls and idealized to be evenly distributed over the length of a jar, may this effect be negligible?

It looks like the ideas now dried up. There were actually two spot-on replies, #7142 and #7146. Here is the answer as sound boarded against a real relativist:

Ignoring surface effects (or coat all the balls in some hydrophilic substance), the lead balls end up at the top and bottom of the jar, with the plastic balls near the center. The reason is that only the center of mass (COM) of the ISS is on a perfectly natural elliptical orbit. Balls starting the experiment below the COM are traveling a bit too slow and try to go into a lower orbit. Balls starting the experiment above the COM are traveling a bit too fast and try to go into a higher orbit. The water molecules suffer the same fate, so the center of the jar ends up with a little less pressure than the top and bottom, hence buoyancy moves the plastic balls to the center.

BTW, the balls can theoretically migrate to the extremities within 20 minutes, ignoring ones that may get stuck due to surface effects. So if the crew does not use thrusters or dumps water in this period, all should be fine for the 'experiment'. Further, we can quite safely ignore mutual gravity between the balls and between balls and other objects - see post #7203 above.

Okay, here is my question. I have no idea how long a laboratory jar is so I can't confirm your solution. However, I would think that you have some issues to overcome. The first one concerns gravity of the lead balls themselves. Early experiments showed two lead balls attracting each other suspended on arms that swing freely. So, there must be some gravitational attraction between the balls that would be significant.

Second, I would expect that the radius of the orbit is pretty large and since no dimension of the length of the jar is given (only the material inside is homogonously dispersed), it is difficult to determine if the distance between the lead balls is wide enough to prevent significant interaction and clumping of the lead balls. Second, the distance over which the balls are spread along the axis pointing toward the center of the Earth's mass is long enough to allow segregation above and below the center of mass of the space station (AKA point of orbit).

In other words, I can see a scenario where the balls are close enough to each other that they conjugate together and simply fall as a single mass to one end of the jar or another. It would seem to me that there is a critical length to the jar where one scenario would happen and not the other due to interaction between the balls.

What do you think?

Recall my reply above (to Ken): ".... the gravitational pull of a 90 kg person at 50cm range is about one order of magnitude less than the micro-gravity caused by Earth at 10cm below or above the ISS centre of mass (COM). (It's roughly 25 vs. 2.5 nano-g )."

I think the mutual attractive acceleration of small lead balls at a few mm range must be orders of magnitude less than 25 nano-g, but I must confess I did not check that numerically.

The apogee of the ISS is roughly 400km altitude and I worked on a jar of 20cm length. As my relativist friend said, the main problem may be surface effects, if the balls come into contact with each other or the glass walls. If one idealistically start the balls well separated from each other and the walls, the outcome should be as posted above

Thanks! It was a fascinating learning lesson!