Re: Altitude effects on body weight

Hi Bill,

Before we attack your question per se, let's define weight. Weight is really the force of gravitational attraction between two objects and is given by Mr. Newton's formula for Gravitation: F=GmM/r² (I guess if you really feel like torturing yourself, you could try deriving the appropriate formula in Einstein's theory of general relativity :). Anyway, the radius of separation here would be the distance from the center of the Earth to whereever you are. So let's try this first for Cincinnati and then for Puebla.

Some Preliminary Data:

r is the radius of the Earth: 6378137 m

G is the gravitational constant: 6.672 59 × 10-11 m^3 kg^-1 s^-2

M is the mass of the Earth: 5.9736x10^24 kg

m is your mass (let's assume 50kg)

Cincinnati: 870' above sea level (approx. 265.176 meters). Therefore, for Cincinatti, r is (radius of Earth + 265m = 6378402) which gives a gravitational force of approximately:

F = GmM/r² = 489.87N = 110.22lbs

Puebla: 7,500' above sea level (approx. 2286.00 meters). Hence, for Puebla, r is (radius of Earth + 2286m = 6380423) and the gravitational force would be approximately:

F = GmM/r² = 489.56N = 110.15lbs

This is a percent decrease of (110.22 - 110.15)/110.22 * 100 = .06%. That's quite a small change in weight but if you were keeping careful track, or weighed quite a bit, then I guess it would be noticeable. There's a way to make this formula a little more tractable. You might have noticed that the GmM doesn't change in this formula (in fact, the g in W=mg is just GM/r²); so you could take that out as a constant and just compute the % change in 1/r² and that would also give the percent change in weight. Let's test this out:

For r₁=(r + 265m) and r₂=(r + 2286m), r₁²=40684012073604.00 and r₂²=40709797658929.00; therefore, (1/r₁² - 1/r₂²))/(1/r₁²) * 100 = (2.457968005197809e-14 - 2.45641112829424e-14)/2.457968005197809e-14 * 100 = 0.0633399987419193 which is approximately .06%, as predicted.

So, putting all this together, if you need a formula to see how much your weight changes when you go from one locale to another, all you need to do is:

1. Get your current altitude, add that to the radius of the Earth, and call it r₁
2. Get your new altitude, add that to the radius of the Earth, and call it r₂
3. Finally, plug your values into the following formula: [(1/r₁² - 1/r₂²)/(1/r₁²)] * 100

Btw, in preparing this answer, I used a HTML online scientific calculator, this unit conversion calculator, and got the GeoPhysical data and physical constants here and also here . This information was very easy to find using the appropriate keywords at my favorite search engine, Yahoo. Well, I hope that helped answer your question... it was certainly very observant of you to notice such a slight change!

Regards,

Rick.

Jason Goodman comments:

While the difference in altitude makes some slight difference to your weight in Puebla vs. Cincinnati, the *latitude* of Puebla is even more important. The Earth is an ellipsoid, not a sphere: it bulges out at the equator like a beach-ball when someone sits on it. The distance from the center to the equator is 6372 km, the distance from the center to the north or south pole is 6350 km. To first order, there's a 22-km-high "mountain" at the equator. This makes the gravity at the equator less than at the pole, via the process Ricky described. Puebla's a lot closer to the equator than Cincinnati.

Using the calculations at this site and plugging in 39 degrees for Cincinnati's latitude and about 10 degrees (a guess -- I don't know the exact figure) for Puebla, you get 9.8128 m/s^2 for Cincinnati and 9.7893 for Puebla. Your mass remains the same, so since weight = mass * gravity, your weight will differ by .24% -- an effect four times greater than the altitude effect Ricky described.

Adding both the altitude effect and the latitude effect, we see that your weight should be .3% less in Puebla -- a difference of 1/3 of a pound for every 100 pounds you weigh. Even this much larger effect would be very difficult to observe -- I'm impressed that you noticed.

An interesting sidenote: if you had used a doctor's scale -- the tall upright ones with the weights which slide along rods -- the scale would measure the same in both places. See if you can figure out why.

And Ricky Sethi replies:

Since the derived formula is still valid, all that's needed are revised radii. However, the 22km difference that was suggested is invalid as that is the difference between the polar radius and equatorial radius (a separation of 90 degrees), whereas Cincinnati and Puebla suffer a separation of only (approximately) 30 degrees. Thus, we need another formula to get the correct radii. One way is to use the regular formula for an oblate spheroid, which, according to my edition of Math Methods for Physicists, is as follows:

r(l) = s{1 + e^2/(1-e^2)[sin^2(l)]}^-1/2

where l is the latitude of the locale and e is the square root of [(a^2-c^2)/a^2] and a is the equatorial radius and c is the polar radius. Plugging all this in gives:

1/r1^2: 2.5991048167074418e-14 1/R2^2: 2.5332996604926712e-14 [(1/r1^2 - 1/r2^2)/(1/r1^2)]*100 = 2.597606482999843%

This gives a final %difference of approximately 2.6%, which is significantly greater and certainly much easier to notice.