Weight: Mountain vs Valley
Name: Alyson R.
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.
Your formula is correct as is your logic. The r is measured from the
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
a mountain next to you would exert a force on you - likely sideways... but
the tree next to you and the roof over your head and the person standing
The key to conceptualizing this is that of magnitude. The mass of the
you is much greater than any of the other objects and will therefore have an
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 -
etc.) would be about .004 ounces per pound.
By the way, if you want to make this even more difficult to consider...
that the atmosphere has a buoyant effect on your body. This effect is
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.
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.
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.
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
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.
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.
Right. You have the right formula and are applying it correctly.
For the case of the valley surrounded by mountains -- let us consider the
case in which the valley is really just a very deep hole. In fact, let us
into the hole with a supply of food, water, and air and have a friend fill
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
the earth (below us) is pulling us down. The net result is that we are
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
the sensation of weightlessness. In effect, you are in freefall with the
orbit around the sun.
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
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Update: June 2012