Re: How much does buoyancy in air affect the weight of a person?

We can do a few quick back-of-the-envelop calculations to get a rough idea of these effects. To find the buoyancy effect, you need to determine the densities of the fluid (air) and the object in the fluid (human body). The density of air varies with altitude, pressure, temperature, and humidity, but dry air at sea level and 20 degrees C is about 1.2 kg/cubic meter (see NASA calculator reference). The density of human bodies varies depending on percentages of fat, muscle and bone, but published values of about 1,010 kg/ cubic meter seem reasonable when you consider the human body generally is close to floating in water, and so has a similar density (water at body temperature, aprox 37 degrees C, has a density of about 993 kg/cubic meter).

So, air has only about 0.12% the density of the human body (divide the density of air by the density of human body = 1.2 / 1010 x 100% = 0.12%), which means dry air at sea level will create about 0.2 lbs of buoyancy for a 170 lb person (0.12% x 170 lbs = 0.2 lbs).

The density of air on Mount Everest can be roughly approximated using the �equation of state� or ideal gas laws, where the density is equal to the pressure divided by the temperature and gas constant. We can find the pressure from various sources, but NASA has a nice calculator for this (see references below), which gives a pressure of roughly 31.3 K-Pa. If you use the equation previously noted above (or the NASA link) you can calculate the density to be roughly 0.48 kg/cubic meter.

So on Everest (29,002 feet or 8840 meters) air has only about 0.048% the density of the human body (divide the density of air at 29,002 ft altitude by the density of human body), which means air on Everest will create roughly 0.08 lbs of buoyancy for a 170 lb person (0.048% x 170 lbs = 0.08 lbs).

The gravitation effect can be approximated by understanding the gravity decreases at a rate of 1 divided by the distance squared (from the object creating the gravity). So, If we take the approximate distance to the center of the earth at sea level and square that, then divide it by the approximate distance to the center of the earth from the top of Mt Everest squared, that will give us a % difference which we can multiply times your weight to get a rough idea of the reduced gravity effect on Everest compared to sea level.

This calculation shows that the gravity on Everest is roughly 99.72% of that at the equator, so it has been reduced by roughly 0.28%. Multiply by 170 lb human body and you get 0.47 lbs.

Summary:
Buoyancy effect of air at sea level on a 170 lb human is roughly 0.2 lbs (0.12%).
Buoyancy effect of air on Mt Everest on a 170 lb human is roughly 0.08 lbs (0.048%)
Reduced gravity effect on Mt Everest on a 170 lb human is roughly 0.47 lbs (0.28%)

These are rough approximations, but should give an idea of the relative effects for the different conditions. See links below for additional information and if you want help doing the calculations yourself.

NASA - gas laws; http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html
NASA - pressure calculator (input altitude); http://www.grc.nasa.gov/WWW/K-12/airplane/atmosi.html
NASA - air properties; http://www.grc.nasa.gov/WWW/K-12/airplane/airprop.html
NASA - earth properties; http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
Density of water at various temperatures; http://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html