# [Physics] Bowling Ball Sinking Rate – Variable or Constant

buoyancynewtonian-mechanicspressure

In today's (Jan. 16, 2014) xkcd What If?, Randall Munroe discusses how fast a bowling ball will sink to the bottom of the Mariana Trench. He calculates the drag on the ball, and then plugging in the weight of the ball, comes up with the time it will take for the ball to sink to the bottom. However, he never explicitly states that he takes into account the changing pressure of the water as the ball sinks. It seems to me as though the increasing pressure of the water would change the rate of descent. Is this the case? Are Mr. Munroe's calculations correct?

Buoyancy is dependent on the density difference between the object and the surrounding fluid. Though the pressure goes up in the ocean with increasing depth, the density remains more or less constant because water is almost incompressible. The pressure in water increases by about $100\,{\rm kPa}$ for every $10\,{\rm m}$ of depth, and the Marianas trench is about $10\,{\rm km}$ deep, so the pressure change from surface to bottom is about $10^8\,{\rm Pa}$. The compressibility of water is a few times $10^{-10}\,{\rm Pa}^{-1}$, so the change in density from top to bottom is a little less than one part in a hundred. The buoyancy force is directly proportional to the density, so the change in buoyancy is also a few parts in a hundred.
For reference, the buoyant force is: $$B = \rho V g$$ Where $\rho$ is the density of the fluid, $V$ is the volume of the falling object and $g$ is the local gravitational acceleration.
Incidentally, the gravitational acceleration also changes with depth, but over a drop of $10\,{\rm km}$ from the surface, the change is only a few parts in a thousand, so the effect is secondary to the change in density of the water.