Re: How much weight change is there from sea level to the top of Mt. Everest

It is true that the acceleration of gravity is less at the top of Mt. Everest. This is primarily due to the fact that on its summit one is farther from the center of the earth than one is at sea level, by about 8850 meters. (height of Everest) There is a really good explanation at Cornell that goes into the details of why this is so. According to this site a pendulum clock at the top of Mt. Everest would be in error by about 2 minutes per day due to the difference in gravitational acceleration.

The acceleration of gravity on Mt. Everest is computable using the usual equation for gravitational acceleration such as given in the reference above (at Cornell). There are some very very small anomalies due to density differences in the Earth's crust, but they are, as I said, very very small. You can see some of the measurements that have been made of these anomalies here. We will not make any corrections for the anomalies because they are almost insignificant.

The acceleration of gravity, a_g, is
a_g = GM_E / R_E²
where G is the famous gravitational constant, M_E is the mass of the Earth, and R_E is the (mean) radius of the Earth. The values are
G = 6.67259E-11 (SI units [meters, kilograms, seconds])
M_E = 5.975E24 kg
R_E = 6.3712E6 m
If you make the calculation, a_g = 9.822 m/s², which is good enough for the accepted value for a_g.

If one now changes the distance from the Earth's center to include the height of Mt. Everest, R = 6.38005E6 m, and a_g = 9.795 m/s².

For interest, let's estimate a_g at the top of Mt. Everest based on the fact that a pendulum clock would be about 2 minutes off during a day. The period of a pendulum depends only on the length of the pendulum and the local acceleration of gravity, the equation being
P = sqrt(l/a_g)
where P is the period, l is the length, and "sqrt" is the squareroot. Doing some algebra we can easily find that
a_g = l/P²
So, if the period is a ratio r longer than the period at sea level, the gravitational acceleration will be 1/r² smaller than the standard 9.822 m/s². A day is 1440 minutes, so we will take, at the top of Mt. Everest, r = 1442/1440 = 1.00139, and r² = 1.00278, so that on Mt. Everest a_g = 0.997228 times the "standard" value, so that a_g = 9.795 m/s² at the top of Mt. Everest using the information that a clock is off by 2 minutes during a day, in agreement with the calculations made above based on the height.

Your question concerns how much weight change there would be. If your weight is W Newtons (or, if you are in the United States and use the awful units of pounds, you can do so!), your weight on the top of Mt. Everest will be 0.9972 W. Not much difference! (0.9972 = 9.795/9.822)

John Link, MadSci Physicist