this post was submitted on 26 May 2026
140 points (99.3% liked)

xkcd

16507 readers
538 users here now

A community for a webcomic of romance, sarcasm, math, and language.

founded 3 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
[–] homes@piefed.world 28 points 1 day ago* (last edited 1 day ago) (1 children)
[–] spittingimage@lemmy.world 9 points 1 day ago (2 children)

There was no depth where it floated? Interesting.

[–] akwd169@sh.itjust.works 4 points 22 hours 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

[–] tal@lemmy.today 23 points 1 day 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.

[–] addie@feddit.uk 16 points 1 day 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

[–] Duallight@lemmy.today 8 points 1 day 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.