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Name: Nancy
Status: educator
Grade: 6-8
Location: MN
Country: N/A
Date: 1/17/2005

We have been discussing the difference between mass and weight. We know your weight on the moon would be 1/6 of that on Earth. We've also talked how weight would change between the lowest valley and the highest mountain. Could you give an example of about how much your weight would change in these locations?

That is a very interesting and not at all trivial question. To get an idea, I did some rough calculations on my model of Mount Everest which I took to be a perfect cone 5 miles high and 10 miles in diameter at the base with a density of 3000 kg/m^3 (3 times that of water). A 200 lb person standing on top of this mountain would find his weight increased by about 1 lb due to the attraction of the mass in the mountain. On the other hand, he would find his weight decreased by about 0.5 lb (0.25%) due to his being 0.13% further from the center of the earth (8 km out of 6400 km). Most people think the mass of the mountain is negligible compared to the mass of the earth and so will have no effect. However, they neglect the fact that the mountain is MUCH closer (several kilometers rather than 6400 km) and squaring that ratio makes the mountain have an appreciable effect.

The truth of the matter, however, is that it could go either way. In an earth in perfect equilibrium, all points on the surface of the earth would feel the same gravity. Places where the gravity is larger will be pulled down and vice versa. For example Sweden is still rising after having been pushed down by the miles of ice sitting on Sweden during the last ice age. It will continue rising (other things being equal) until equilibrium is reached again.

Where tectonic plates have pushed up mountains, gravity will be larger due to the mass of the mountains pushed up. On the other hand, negative gravity anomolies (reduced gravity) can be found where a tectonic plate is being pushed under another.

Best, Dick Plano...

Mount Everest is about 30,000 feet up, about 5 miles. There are no valleys that go very low; they are only a few hundred feet below sea level. Call it 0. But if you went down to the deepest part of the ocean in a very thick-walled submarine, that would be 30,000 feet down. So the largest difference you can get is 60,000 feet, or 12 miles. The radius of the earth is about 4000 miles, so the ratio of distances is (4000+5)/(4000-5) = 1.0025, or 0.25% difference in distance. The inverse-square law for gravity may (or may not) apply, so the largest attainable weight difference would be 0.5%. Valley to mountain, as you asked, would be 0.25% of weight.

Jim Swenson

Your question is more complicated than it might appear. Gravity, as best is known, is always attractive. Every particle in the Universe attracting every other particle in the Universe. Newton's law: F = G * M * m / R^2 where 'F' is the attractive force, 'M' and 'm' are the masses of two objects and 'R' is the distance separating the two objects. The constant of proportionality 'G' = 6.67259x10^24 N*m^2/Kg^2. The AVERAGE radius of the earth is ~ 6.38 x 10^6 meters and its AVERAGE mass is such that: F = g * m where 'g' is the collection of AVERAGE constants G*Mearth/(Rearth)^2. has the familiar value 9.8 m/s^2. Among many web sites you can find the math worked out at

Now this is the value used in introductory physics to work problems. In actual fact the force due to gravity varies with separation and the density of the matter between the two bodies. Geologists use sensitive instruments that measure the change in the force of attraction at various places on the earth. For the purpose of your calculation, you only have to consider the 1/R^2 dependence to calculate the change in weight between a valley and a mountain, but "the real" calculation would have to take into account the fact that as you climb the mountain you are putting more "mass" between you and the center of the earth, so things are more complicated.

Vince Calder

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