Let’s first state our assumptions:
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Very thin membrane, the balloon skin cannot be compressed more than it already is
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Zero trapped air
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Isothermal conditions, water temperature at the bottom of the ocean is the same as the top, that way we don’t need to deal with thermal expansion
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Balloon is a perfect sphere
Calculations:
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Pressure on the balloon is densitygravitydepth ~= 1000 x 10 x 100 = 1 Mpa or 10 atmospheres
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The bulk modulus of water is 2.2 Gpa so (change in volume)/(initial volume) = (change in pressure)/(bulk modulus) = 0.0046% decrease in volume
0.0046% decrease in volume is like a 0.015% decrease in radius
So yes the balloon does get a little bit smaller
is there a /m/theydidthemath ?
/m/theydidthemonstermath
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Liquid Water does not change it’s density much with pressure, but yes.
Water doesn’t compress (for all intents and purposes), so no. I.E. the water pressure of the ocean would indeed push inward against the balloon much more strongly, but the water inside the balloon would push back by the same amount.
Nothing special would happen if you pop the balloon, the rubber would just rapidly contract, basically the same as it would above the surface.
Water is incompressible.
Not it’s not, common myth/misunderstanding.
From Wikipedia (couldn’t find a way to link directly to that part of the page):
https://en.wikipedia.org/wiki/Properties_of_water
Go to the “Physical properties of water” and scroll down for the part in the image.
Ok, so for all intents and purposes, water is incompressible.
So what’s the threshold to say when something that compresses is compressible?
Everything is compressible with enough force. I would say that threshold is when that compressibility is relevant to the discussion. So in this case, a water ballon under water. There is no depth in the ocean where the pressure will be high enough to make a notable difference.
I would imagine in many applications when dealing with water and fluid dynamics, unless measurements have to be extremely tight and calculated, water would be treated as incompressible. I’m guessing a lot of the time, the force applied to water is not high enough for it to matter.
When the compression is large enough to be relevant to whatever you’re calculating or discussing. Part of my job is literally to develop models that, for instance, predict the compressibility of water. For pretty much all practical purposes I can attest that the isothermal compressibility of water is completely negligible.
Virtually incompressable
Depends on how much air is in there with the water.
Water itself is incompressible so the balloon won’t shrink that much, but there’s some air in there so it would shrink a bit.
The balloon’s popping action comes from being filled and expanded. It’s an elastic, so if something breaks it, it snaps back.
The only way that wouldn’t happen is if there was mostly air in there, and you went to the appropriate depth where that air was compressed down to its original volume.
At 100 m the compression would be nil.
If you take it deep enough for there to be measurable compression and then pop it, nothing will happen because it’s already compressed.
The latex would still snap back to its original shape, which would at least look similar to a normal pop.
