▲ 138 ▼ What if you dropped a bowling ball in the Mariana Trench? (www.youtube.com) submitted 5 months ago by schnurrito@discuss.tchncs.de to c/xkcd@lemmy.world 13 comments fedilink hide all child comments
[–] spittingimage@lemmy.world 8 points 5 months ago (4 children) There was no depth where it floated? Interesting. permalink fedilink source parent hideshow 4 child comments replies: [–] tal@lemmy.today 23 points 5 months ago (2 children) 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. permalink fedilink source parent hideshow 2 child comments replies: [–] addie@feddit.uk 14 points 5 months ago 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 permalink fedilink source parent [–] Duallight@lemmy.today 9 points 5 months ago 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. permalink fedilink source parent [–] akwd169@sh.itjust.works 4 points 5 months ago 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 permalink fedilink source parent
[–] tal@lemmy.today 23 points 5 months ago (2 children) 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. permalink fedilink source parent hideshow 2 child comments replies: [–] addie@feddit.uk 14 points 5 months ago 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 permalink fedilink source parent [–] Duallight@lemmy.today 9 points 5 months ago 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. permalink fedilink source parent
[–] addie@feddit.uk 14 points 5 months ago 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 permalink fedilink source parent
[–] Duallight@lemmy.today 9 points 5 months ago 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. permalink fedilink source parent
[–] akwd169@sh.itjust.works 4 points 5 months ago 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 permalink fedilink source parent