Spyke

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13 replies

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lemmy.today

Water's not compressible, so the density doesn't change with depth. Either the bowling ball is denser than water or less dense than water.

23

Water is compressible; it has a bulk modulus of about 2.2 GPa. So at the 1086 bar at the bottom of the Mariana trench (~109 MPa), it'll have compressed about (109 / 2200) ~= 5%. Materials with a different bulk modulus to water may start to float at sufficiently high depths.

https://en.wikipedia.org/wiki/Bulk_modulus#Selected_values

18

Water does change density with temperature, so it is denser the deeper you go. I doubt there's a normal bowling ball weight that would have the right density for it to float at some random depth though.

9

because water is fairly incompressible, the density of sea water doesn't change much as you go down

Therefore no there isn't a depth where the weight of the seawater above compresses the water to a density equal to that of the bowling ball

4
lemmy.world

It would probably end up right next to the tape measure I've been looking for since Sunday.

21

I was more interested if it would compress than where they floated

3

I think these calculations could be wrong, considering that the ball would either absorb water and decrease in density or implode / break

1

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What if you dropped a bowling ball in the Mariana Trench? | Spyke