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?

  • $\begingroup$ This is discussed in some detail in the what-if forum as well. The responses are basically in line with Kyle's answer. $\endgroup$ Jan 16, 2015 at 21:42

1 Answer 1


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.

The other possibility is that the volume of the bowling ball changes (shrinks), causing it to displace less water and thus decreasing the buoyant force, so it would sink faster. I'm not sure about the compressibility of bowling balls, but unless there are air pockets in the resin (most balls are resin), it's unlikely that the balls are much more compressible than water.

So the rate of descent will be nearly constant over the entire descent; I'd be more worried about any motion of the water in the vertical direction than changes in buoyancy. I see no problem with Mr. Munroe's calculations.

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.

  • $\begingroup$ Thanks for the calculations. It does makes sense that the difference would be small. But what would the magnitude of the difference be in terms of travel time? How much longer/shorter would the descent be between "assuming the buoyancy is constant" and "factoring in the slight changes in buoyancy"? $\endgroup$ Jan 16, 2015 at 18:11
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    $\begingroup$ @DavidCoffman Well, the answer is still "quite small". Getting an estimate would involve a simple integral, but I'd confidently guess less than a minute (of the ~2.5 hour descent mentioned in the article), possibly only a few seconds. $\endgroup$
    – Kyle Oman
    Jan 16, 2015 at 18:20
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    $\begingroup$ There is something called the bulk modulus which describes the relative change in volume under hydrostatic pressure. It is $$K = \frac{E}{3 \left(1-2 \nu \right)}$$ where $E$ is the youngs modulis and $\nu$ is the Poisson's ratio. $\endgroup$ Jan 16, 2015 at 19:28

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