6

Is it possible to counter-act g-force for a jet-pilot, by him putting on a scuba-diving suit and filling the cockpit with water? On earth we are constantly pulled down, or accelerated with one g. In this situation, if we put the jet-pilot in a pool, he would neither sink nor float.

If we could increase Earth's gravity to say 9 g's, the pilot in the pool would float even more.

Is this correct? Thanks...

Malabarba
  • 5,061
freeside
  • 543

4 Answers4

4

If you fill the cockpit with water, the pilot will feel a buoyant force. Humans have about the same density as water, so ignoring the scuba suit, the pilot will feel a buoyant force about equal to his own weight.

The plane's maneuvers don't change this result much. By the equivalence principle, when the plane accelerates, the water in the cockpit and the pilot both get heavier. The buoyancy changes in the same way the pilot's weight does. If he feels weightless before takeoff, he'll continue feeling weightless during maneuvers. If he's supported by buoyancy before takeoff, he'll continue to be supported by buoyancy during maneuvers. (This assumes that the jerk, or time derivative of acceleration, is low enough that the water can remain in hydrostatic equilibrium during the maneuver.)

Although basic buoyancy works the same, there will be a pressure change during accelerations. If the plane is accelerating up at 9g, he'll feel pressure as if he's under water ten times as deep as he really is dense as the water really is. As Bruce pointed out, this is a big pressure gradient, and the pilot will still realize that the force beneath him pushing up is stronger than the force above pushing down. He will essentially feel heavy.

Ignoring compression, changing Earth's gravity won't change whether something sinks or floats. Something more dense than water sinks. Something less dense floats. If we multiply gravity by some constant, an object immersed in water has the net force on it also multiplied by that constant. So if you have something that floats and has a net force on it of $1N$ upward under normal gravity, if you double the gravity, it will experience a net upward force of $2N$ and have double the acceleration (ignoring drag).

Community
  • 1
Mark Eichenlaub
  • 52,955
  • 15
  • 140
  • 238
  • 1
    In addition to the buoyant effect, you'd also reduce the effects of acceleration on the pilot by adding a whole bunch of extra mass to the plane (all the water to fill the cockpit), which would cut into its acceleration... – Chad Orzel Jan 03 '11 at 13:25
  • I would clarify (as a final point) that the method won't work as a way to counteract the acceleration force, and the pilot will feel g-forces just as if there was no water. – Malabarba Jan 03 '11 at 13:29
  • 1
    @Bruce I would disagree that he "feels g-forces just as if there was no water." There is a net force on him just as if there were no water, but what that net force feels like is different than if there were no water. – Mark Eichenlaub Jan 03 '11 at 13:57
  • @Mark Sure, I just didn't elaborate in the comment. – Malabarba Jan 03 '11 at 14:03
  • 1
    I suppose the jerk --and also a bit the impracticality of flying in a diver suit in a waterfilled cockpit-- was what necked the engineer who ever proposed it. – Raskolnikov Jan 03 '11 at 14:54
  • I'd like to emphasize, he will not feel weightless. He will feel the water accelerating him up, and pretty strongly. – Malabarba Jan 03 '11 at 15:05
  • The reason we normally feel weightless underwater is because the buoyancy force is distributed through a large surface (your body) instead of just your feet. This gives the impression that we are weightless underwater, because the pressure gradient underwater is so low that our body doesn't detect it, but that's biological, not physical. If the pressure gradient were stronger (which is what happens when you increase $g$) our body would be able to feel the water holding us up, much like can we feel the ground holding us up. – Malabarba Jan 03 '11 at 15:19
  • @Bruce I'm not sure we have any disagreement. He would feel whatever he would feel if he went swimming on a planet with high g, agreed? – Mark Eichenlaub Jan 03 '11 at 16:48
  • @Bruce After thinking about it a little more carefully, I realized that what you said is totally right. Thanks for straightening me out on this. I think I was thrown off by thinking of swimming on Earth as feeling weightless, but with pressure. That's not right, I now realize. I edited my answer. – Mark Eichenlaub Jan 03 '11 at 17:16
  • @Mark He would feel whatever he would feel if he went swimming on a planet with high g, agreed? Agreed, the problem is what happens on a planet with high g. On a planet with high $g$, the human body would float just the same, but the pilot would feel the force lifting him (as opposed to what happens on earth, where you don't quite feel the force because your body can't detect such a small pressure gradient). My point is that the pilot would not feel weightless. He would feel a very strong force keeping him afloat. – Malabarba Jan 03 '11 at 17:23
  • @Mark: Ok, you fixed it while I was responding. You still left a "weightless" in the third paragraph, don't know if it was intentional or not. I'm removing my down vote anyway :) – Malabarba Jan 03 '11 at 17:27
2

Your logic is mostly precise, but it won't counteract g-forces.

(I'll assume the jet flyes straight up, for simplicity, but it doesn't make any difference.) Like you said, if the pilot was originally in equilibrium inside the water and we accelerate the jet at 9g, the buoyancy would get 10 times as strong (9+1). However, the pilot will still feel a force accelerating him. The only difference is that, instead of being pushed by the jet seat, he will be pushed up by the water itself.

So he will still feel the g-force, only a little differently. Instead of the force being concentrated on the area of contact between him and the seat, it will be distributed through the area of contact between him and the water (approximately half of it, to be more precise). But the net force will be the same.

The final effect will be that he'll float inside the water. The water is accelerating up with the jet, and he'll accelerate up as well (because of all the extra buoyancy).
If he were submersed in a heavier fluid to start with, the effect would be the same, except he would float partially above water.
If he were submersed in a lighter fluid to start with, then he would be able to touch his jet seat, and the upward acceleration would come partially from the seat and partially from the liquid.

Malabarba
  • 5,061
  • @Bruce I agree with your physics, but saying "it won't work" is equivalent to telling someone who wants to go swimming to take the load off that that won't work, either, since the net force on them is the same. – Mark Eichenlaub Jan 03 '11 at 13:55
  • @Mark I'll be more specific – Malabarba Jan 03 '11 at 14:01
  • @Bruce Cool. I think I agree with what you've said. +1 – Mark Eichenlaub Jan 03 '11 at 14:04
  • @Bruce : you say that the force will be " distributed through the area of contact between him and the water". so the jet-pilot / diver will not get pushed into his underwater seat, but rather just feel the water pushing more on him, the sensation being the same as diving deeper? i mean the pilots surface is a closed one, so if the water would push vertical on any given point of the surface /skin the sum is zero? assuming the pilot is a round ball, the pressure felt at each point is a vertical to the tangent at that point. – freeside Jan 03 '11 at 14:14
  • @freeside: not quite. If you push a balloon underwater, its surface is closed, pressure is distributed through its entire surface, but the net force is not zero. Pressure below the balloon is stronger than the pressure above it, that's what causes buoyancy. In the case of the jet pilot, it's the same thing. Pressure is applied to his entire surface, but the pressure above him is smaller than the pressure below him, so he feels a net force upward. The only difference is that the jet pilot is being pushed up by a force 10 times as strong (which is why he accelerates). – Malabarba Jan 03 '11 at 14:24
  • @freeside: and yes, the jet-pilot will not get pushed into his underwater seat, but rather just feel the water pushing more on him. But he would have to dive to a depth were water was 10 times as dense to experience that effect simply by "diving deeper". – Malabarba Jan 03 '11 at 14:28
  • @bruce assuming he is only 10 cm underwater and at equilibrium, not sinking and not rising, doing 10g, would give him the feeling of being 1 meter underwater? – freeside Jan 03 '11 at 14:30
  • 1
    @freeside no. Diving depth odes not affect buoyancy the way you're thinking. Visit http://en.wikipedia.org/wiki/Buoyancy#Forces_and_equilibrium and see that $B=\rho V g$. Note that water depth does not enter this equation. The effect that the pilot feels is the same as multiplying $g$ in that equation by 10. Simply diving deeper would not reproduce that effect. – Malabarba Jan 03 '11 at 14:37
  • @freeside: buoyancy force does not increase with depth, unless you consider the variation of water density with depth (which is very slow). – Malabarba Jan 03 '11 at 14:38
  • 1
    If the pilot were uniformly the same density as the water he would feel no effect (except changes in the external pressure as the product of his depth times the effective force of gravity changes). Of course real humans have density variation, especially air filled lungs. If we filled the lungs with a liquid of near water density we could greatly reduce this. We do have a fluid that was developed for artificial blood, and a human immersed in it with lungs full of it won't drown. But, I think it is heavier than water. – Omega Centauri Jan 03 '11 at 15:37
  • 1
    @Omega: No, he would feel the effect. Just look at it from a reference frame in the ground. The pilot is accelerating upwards with the jet, thus he is suffering an (pretty strong) upwards force. If he is suffering that very real force (it's not a gravitational force), then he is feeling it. Since he is only in contact with the water, then this force must have been exerted by the water. – Malabarba Jan 03 '11 at 17:14
  • @Omega: All of the above is assuming the pilot has approximately the same density as water. The increase in overall water pressure also happens, but what really matters is the increase in the pressure gradient (that's what causes the increased buoyancy). – Malabarba Jan 03 '11 at 17:17
  • i'm just considering what happens to a jelly fish that has a nice form in water, but collapses completely if you remove it from water. going the other way, human beings are pretty solid outside of water and would get even more "solid" if submerged in water, and hence if inside a water bubble could survive many many more G-forces then outside of the water bubble ... – freeside Jan 03 '11 at 17:40
  • @freeside: A jellyfish is not more solid underwater, he is just as fragile as he is outside the water. The reasons he keeps a nice form is because he is floating (the water sustains him with an evenly distributed force), not because he is stronger. – Malabarba Jan 03 '11 at 18:20
  • Bruce, the pressure gradient is cancelled out by the hydrostatic pressure gradient. Or maybe I should say the pressure gradient is the hydrostatic pressure gradient, and is cancelled out by the force of gravity (gravity in the general sense incluing aceleration). So the individual molecules of our ficticious pilot (or Jellyfish) are always in an equlibrium position. Internal stresses do not build up. A cloud experiences as large range of pressures top to bottom, but is not ripped apart by them. – Omega Centauri Jan 03 '11 at 19:53
  • @Omega: Yes, the pressure gradient I was talking about is the hydrostatic pressure gradient. This gradient results in an upward force of 10mg (where m is the pilot's mass). Unlike gravity, this force is not applied to every molecule, only to the ones on the surface. Consequently, it forces the molecules out of their equilibrium position, and its effects are measureable from within the cabin. – Malabarba Jan 03 '11 at 20:01
  • @Bruce maybe a different approach. consider a planet with 100 (hundred) g on the surface. if we could teleport there we would be instantly crushed. but let's assume we had a bucket that could withstand 100 g, then fill the bucket with water and a jellyfish. we then teleport that bucket of water / jelly fish to the 100 g planet. what happens to the jellyfish? – freeside Jan 03 '11 at 20:26
  • @freeside It keeps floating inside the water, but gets crushed as well. It gets crushed by the overall hydrostatic pressure $\rho g h$. Even you don't consider the overall pressure (lets say the jellyfish is really close to the surface (and there's very little atmosphere)) than it would be crushed by the buoyancy force (pressure gradient) pushing it up. – Malabarba Jan 03 '11 at 21:30
  • @bruce one things for sure then. if i had an alien friend on the 100g planet and he would want to know how a jellyfish looks like, i'd better send it to him in a bucket of water. without it my alien friend would think jelly fishes are pizzas ... – freeside Jan 03 '11 at 22:22
0

If you suspended a jellyfish in a bucket of water and accelerated the bucket upwards at say 100g, the jellyfish will remain stationary relative to the water. IE the jellyfish will not experience any acceleration, it is weightless as it normally is. All of the additional forces due to acceleration are imparted on the buckets bottom and sides.

So by this rational the human body should be able to withstand very high g forces when submerged in water, as long as all air cavities were filled with fluid of equal density.

keith
  • 1
-1

I came upon this discussion while musing about the effects of spinoculation on cells in culture (e.g., see http://www.ncbi.nlm.nih.gov/pubmed/21795326). Forgive me, I'm a virologist and not a physicist, though I did teach 8th grade physical sciences for a while and was often challenged to go far beyond the text and understand things more deeply than the most curious and intelligent student in the class. It seems with respect to the pilot that both Connor and Eichenlaub may be correct in their own way, depending upon what one means by "feeling." We tend to "feel" acceleration via our inner ear which, being encased in a bony homeostatic case, is not subjected to the equilibrating pressures of the surrounding water. On the other hand, the pilot's posterior would feel relatively fine, and he would presumably not suffer blackout from acceleration because (as with a pilot's pressure suit), blood would not be depleted from his brain. So, assuming the pilot was breathing (necessarily pressurized) air during a long acceleration, would he (she) get the bends when acceleration ceased? Sounds like an "ask NASA" question. It all might be relevant to spinoculation, in which cells do survive but change.

Alan
  • 1