142
What if you dropped a bowling ball in the Mariana Trench?
(www.youtube.com)
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 26 May 2026
142 points (99.3% liked)
xkcd
16514 readers
215 users here now
A community for a webcomic of romance, sarcasm, math, and language.
founded 3 years ago
MODERATORS
There was no depth where it floated? Interesting.
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
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.
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
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.