Astronomy Picture Of The Day : A Gravity Map of Earth

NASA ^ | 11.13.01 | Robert Nemiroff (MTU) & Jerry Bonnell (USRA)

[rogers21774]

On top of Everest you will weigh slighly more because of the increased mass (caused by the mass of Mt. Everest) pulling against you.

Of course that's just speculation on my part and I really don't know squat.

[RightWhale]

Does anyone know if you weigh less on top of MT. Everest compared at sea level?

The force due to earth's gravity decreases as the distance from the center of mass increases. Aside from the lumpiness due to masscons, gravity is less at the top of Everest.

[rogers21774]

This could explain why I weigh more at the Dr. Office than I do at home. :)

[Prince Caspian]

Does anyone know if you weigh less on top of MT. Everest compared at sea level?

Yes, but it has more to do with all the calories you will have burned off by the time you climb to the top!

[Chemist_Geek]

You will weigh more on top of Mt. Everest. You would be right about the inverse square law if the Earth were smooth, but that is more than compensated by the gravity from Everest itself. Look at the big blue-and-white blob over the Himalayas!

Hey, atoms are perfect spheres (as far as I'm professionally concerned). I get lost very quickly beyond spherically symmetric force fields. Thanks for the assist.

[Wm Bach]

Does it change with time? 1600 Pennsylvania Avenue seems to posses much more gravity than it has in years.

[Physicist]

I'm wrong.

Everest is about 10,000 meters high. Its volume is about a trillion cubic meters. If we assume a density of 4 grams per cubic centimeter, we get a mass of about 4x10¹⁵ kilograms. If the center of gravity is a little more than 6 kilometers away from Mr. Hillary, the acceleration due to gravity from Everest itself is somewhat more than 6x10^-3 meters per second squared.

The radius of the Earth is 6400 kilometers. The gravity of a sphere of that size will be reduced by a factor of about 3x10^-3. The acceleration due to gravity at sea level is 10 meters per second squared, so the change in acceleration is about 3x10^-2, or about 5 times the change caused by Everest itself.

When in doubt, work it out.

[Physicist]

The gravity of a sphere of that size will be reduced by a factor of about 3x10^-3.

...from an additional 10,000 meters out, I meant.

[Libertarianize the GOP]

bump

[Physicist]

I've worked out the loss of buoyancy. The density of a human being is about 1 gram per cubic centimeter. The density of air at sea level is .00129 g/cc. The pressure at 10,000 feet is 0.24 times the pressure at sea level, so your effective density increases by about one part in a thousand. So the increase in weight due to loss of buoyancy is about a third as big as the weight lost to the change in radius.

You really do weigh less on Everest.

[RightWhale]

I was eliminating the local variances of earth's gravity as shown on the color chart by assuming the question wanted a comparison of gravity at the base of Everest to gravity at the top of Everest and ignoring the mass of Everest itself. A lot of people seem attracted to Everest.

[Physicist]

At the risk of Clintonesque waffling, I'd like to amend my testimony. Everest isn't standing by itself; it's surrounded by other giant mountains. These will also exert a nontrivial gravitational influence roughly downward. So the answer is that I don't know whether you'd weigh more or less on Everest than at home. It's just one of those things you have to measure. All I can tell you is that the mass of Everest alone won't overcome the change in radius, but there is still an outside chance that my physics intuition was right the first time. :-)

[Chemist_Geek]

So the answer is that I don't know whether you'd weigh more or less on Everest than at home.

But does it depend upon what the meaning of the word "is" is? All the mountains surrounding Everest have mass concentrations below the observer atop Everest, in the same general direction as the center of mass of the (assumed spherical) Earth. Imagine, if you will, a sphere with a (small) torus sitting on one side, and a cone extending through the doughnut hole. The C-O'M of that system is still on the axis from the sphere's center to the tip of the cone. It changes the symmetry from spherical to cylindrical.

Are the mountains around Everest perfectly arranged? Is Everest a cone? Of course not. But I'd bet money that a person would weigh less up there than at sea level.

[r9etb]

Well shoot -- I generated one of those for my class a couple of years ago. They shoulda just asked me....

[xp38]

The assumed perfect sphere of the earth is in error of course. From my high school days it is an oblate spheroid and thus the top of Everest is not the furthest point on the surface from the the center of the earth. I am not sure where that point is but the difference wipes out the 5 to 6 miles above sea level that Everest is.

[Chemist_Geek]

The assumed perfect sphere of the earth is in error of course. From my high school days it is an oblate spheroid and thus the top of Everest is not the furthest point on the surface from the the center of the earth. I am not sure where that point is but the difference wipes out the 5 to 6 miles above sea level that Everest is.

The difference is that the geometric center is no longer the center of gravity? For a sufficiently small body that would make a major difference. What the degree of oblateness of the Earth? How out-of-round is it?

[xp38]

I don't know for certain but I do know you are closer to the center at the poles than at the equator but I also believe the southern hemisphere bulges out more than the northern one. Something to research I suppose.

[RightWhale]

oblate spheroid and thus the top of Everest is not the furthest point on the surface from the the center of the earth

The earth's poles are lower than the equator, so gravity is higher on the surface at the poles. Everest is slightly closer to the equator than to the pole, so it is possible there is a mountain more southerly of Everest that is actually farther from the center of the earth at the peak.

[RightWhale]

physics intuition

And a well-seasoned intuition it is.

[xp38]

I just checked....polar radius according to my small science data book is 6356.8 km....equatorial radius 6378.2 km for a difference of at least 21 km which is 13 miles of difference.