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As you probably know, the two next Olympic games will take place in London and Rio. Randall, author of xkcd, did an interesting observation (that I had never thought of) in http://xkcd.com/852/ , arguing that the difference in latitude may influence comparisons between different world records in pole vaulting. As I started considering other results that might be affected by a small change in the local g, I could not find an interesting model for swimming. Does anyone know a model to determinate if the swimming results would be better or worst in a gravity-increased scenario?

i.e., would it be easier to swim on Jupiter or on the moon?

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You can estimate that someone swimming in water has a Reynolds number of about $10^6 - 10^7$; what counts is that this number is $\gg 1.$ In that case, you're dealing with the drag equation: http://en.wikipedia.org/wiki/Drag_(physics). If we assume that our swimmer has the same power $P$ on the moon and on Jupiter, his velocity $v$ scales as $$v \propto \left(\frac{P}{\rho A}\right)^{1/3}$$ (modulo a dimensionless constant), where $\rho$ is the density of water and $A$ the size of Michael Phelps' torso. In particular, this doesn't depend on gravity, so it should be the same everywhere.*

*This is somewhat of a simplification of the very complicated physics behind the problem, but I hope that it's easy to understand. [edit: fixed the HTML, removed the low-Reynolds thing.]

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Which doesn't mean that gravity has absolutely no effect on swimming, but since the variation in g is less than $10^{-2}$ anyway, the variation in finishing times is only going to be $10^{-2}$ if there is a direct, first-order effect. Here there might still be some indirect effect from gravity, but the resulting variation in finishing times has got to be $10^{-5}$ or less, which won't really affect any records. –  Keenan Pepper May 19 '11 at 0:28
    
Far more important has got to be the oxygen partial pressure of the air, which depends on altitude. –  Keenan Pepper May 19 '11 at 0:29
    
An example of such an effect that I can think of: water has a compressibility of ~ $10^{-10}$ /Pa. Since the characteristic length scale of pressure change in the atmosphere is ~ 10 km, a height difference of 100m induces a pressure difference of 1% ~ 1 kPa, which corresponds to a density change of 1 part in $10^{-7};$ for every 50 meters you swim, this effect alone causes you to lose ~ 10 microseconds in London! –  Gerben May 19 '11 at 9:38
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