Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Weight: Mountain vs Valley
Name:  Alyson R.
Status: other
Age: 20s
Location: N/A
Country: N/A
Date: 2000-2001

Pertaining to gravity: If I were standing in a valley would I weigh the same as if I were standing on a mountain? Assume that the earth is perfectly round and not rotating and same relative location on the earth of mountain and valley.

What is the formula that is the basis for the answer? Is it F=GMm/r^2 where 'r' is greater when standing on the mountain and therefore gravity less? But what about in the valley - if the mountains were large enough and, say, on all sides of you, wouldn't the mass of the mountains have a net gravitational effect of pulling you up thereby somewhat negating the pull of the rest of the earth towards its center?

Can we answer the simple mountain vs valley situation first? Then address the example of the mountains surrounding the valley.

I, and a small group of my friends, thank you in advance.

Alyson -

Your formula is correct as is your logic. The r is measured from the center of mass of the objects. Yes, you would weigh more in the valley than on the mountain... in theory. But there are many confounding factors. As you mentioned, a mountain next to you would exert a force on you - likely sideways... but so would the tree next to you and the roof over your head and the person standing next to you.

The key to conceptualizing this is that of magnitude. The mass of the earth below you is much greater than any of the other objects and will therefore have an overwhelming effect.

Don't forget that the sun and moon are both pulling continuously in what ever direction they happen to be. Again, magnitude is important. Your distance to the earth's center is much less than to either of these celestial objects.

One more thought about magnitude... the percent difference in r for a valley and a mountain top is small.

r(valley) = 4000 miles
r(mountain) = 4001 miles

percent difference = 1/4000 or about .025%

The difference in your weight (disregarding the confounding factors - trees, moon, etc.) would be about .004 ounces per pound.

By the way, if you want to make this even more difficult to consider... remember that the atmosphere has a buoyant effect on your body. This effect is related to the density of the air which decreases with altitude.

Interesting to ponder, intriguing to calculate, but the variation in weight is likely of little significance in real life.

Larry Krengel

Yes, any object with mass attracts any other object with mass by gravitation. If you stand on a tall mountain, yes, the downward pull of gravity will be greater than if you were in a valley. This outweighs (pun intended) the effect of increasing the distance from the center of the earth, which as you noted would reduce the gravitational force.

However, this effect is very small. The earth is very big; by comparison, mountains are very small. That is not to say that the effect can't be measured, however. There are now portable gravity detectors sensitive enough to detect magma moving under volcanoes.

Your question about the effect of nearby mountains working against the gravity of the rest of the earth if you stand in a deep valley is interesting. The effect of course is small. However, if you could drill a shaft to the center of the earth, the downward pull of gravity would decrease down to the center. It would not be a perfect linear decrease, though, because the core of the earth is denser than the minerals of the crust or mantle.

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois

You would weigh less on the mountain than in the valley, and you have the right formula. Also, in the valley the mountains would exert a net upward force on you, further decreasing your weight. The differences are very very small, of course.

Tim Mooney

Ques. 1: Perfectly Round, Point, Inelastic Earth Case:

In this approximation, the force of gravity depends strictly on 'r' and 'm'. It doesn't matter whether 'r' changes because you are on a mountain or in a balloon, F obeys an inverse square law. Even in this case the 'r' has to be measured from your center of gravity, because strictly speaking the force of gravity is less at your head than at your feet. Practically, of course, this only matters if you are approaching a black hole.

Ques. 2: Corrections to the Force of Gravity:

Your second question opens up several factors that alter the force of gravity from an exact inverse square law. Some of these, but not all, are:

A. Latitude Correction: The earth is not a perfect sphere, it is slightly oblate (wider at the equator than at the poles). This correction assumes uniform terrain, i.e. no mountains or valleys, and tide free oceans. Then: Fa = Fe[ 1 + c1*sin(a) + c2*(sin(2a))^2] where Fa = force of gravity at latitude "a" and at sea level, Fe = force of gravity at the equator = 978.0404 gals, and c1 = +0.0052954, and c2 = -0.00000585.

B. Atmosphere Correction: There is a small correction due to the the amount of air above/below the data plane that is a result of the differing amounts of air between the observation and the center of the earth.

C. Bouguer Correction: This is a correction for the attraction due to differing, but uniform, amount of material between the measuring station and the reference point. In a simplified picture, this is a correction due to lateral variations in the density of the earth.

D. Tidal Corrections: There are both ocean and land tides due to the gravitational attraction of the moon and the sun (~1/2 the effect of the moon). These can be substantial -- of the order of 0.2 gals depending upon the location of the measuring device with respect to the moon and sun.

Modern instruments for measuring the force of gravity are VERY sensitive, so the earth-model gets much more complicated because these and other small effects become apparent. Even these aren't all the corrections. There are other small deviations for which there is no clear explanation.

Vince Calder

On a Mountain: You have the mass of the earth plus the mountain under you, so you would weigh more than if you were on the surface.

In a Valley: Imagine that the earth was an orange, and the valley you are in is peel-deep. The gravity you will feel is that of the earth without its peel (i.e. less than if you were on the surface). This is because at that depth, gravity from all points of the peel would cancel each other out (remember that the peel right above you is closer, but the peel on the other side of the earth has more mass). This is as if the peel were not there.

The valley-in-the-mountains scenario is the same as in the valley case, but with an additional upward pull from the mountain.

-Wil Lam

Right. You have the right formula and are applying it correctly.

For the case of the valley surrounded by mountains -- let us consider the extreme case in which the valley is really just a very deep hole. In fact, let us go down into the hole with a supply of food, water, and air and have a friend fill the hole in on top of us, promising to dig us out after an appropriate period. Clearly, the mass of dirt above us has a gravitational effect pulling us up while the rest of the earth (below us) is pulling us down. The net result is that we are somewhat lighter than if we were standing on the surface of the earth (but at the same distance from the earths center of mass).

Now in the very extreme case we can consider that the hole is deep enough to put us at the center of the earth. Now the masses "above" & "below" us are the same -- in fact the distribution of mass all around us is essentially the same. This leads to the sensation of weightlessness. In effect, you are in freefall with the earth, in orbit around the sun.

Greg Bradburn


You would weigh a little less at the top of a tall mountain than at the bottom of a valley. At the top of Mount Everest, you would weigh about a half of a percent less than in a valley next to the mountain. This is in fact due to the relation F=GMm/r^2. The value of r is the distance from the center of the Earth, M is the mass of the Earth, and m is your mass.

As for the mountains pulling, they would no have a noticeable effect. Compared to the mass of the entire Earth, a mountain would have practically no pull. Most of the tiny pull it does have would be to the side, in no way countering the downward pull of the Earth.

Dr. Ken Mellendorf
Illinois Central College

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory