gravity below sealevel?

Listening in for our scientists to respond. Interesting question.

From physics, F = GMm / r^2, where
F = gravitational force
G = constant, 6.67E-11 Nm^2 / kg^2
M,m = masses of 2 particles
r = distance separating particles' centers.

From the equation, the closer you are to Earth's center (i.e., the smaller r is), the higher the gravitational force between yourself and the Earth. Thus, "gravity" at below sea level is expected to be higher than gravity at sea level at the same latitude.

Lab rat what do you mean "expected" do we not know yet? And what has latitude got to do with it. Remember I am a non scientist. Although I did take Rocks for jocks (Geology) and physics for poets.

I might be able to understand some of what you are able to tell us. Is there any way to make the answer a bit easier to understand.

Gravity varies more by the composition of what is beneath you than by altitude. For all its mountains and valleys, the Earth's surface is about as regular as a cue ball's. You would have to go very deep into the Earth to notice a significant gravitational change, and by then you would start to be influenced at least a little by the mass of the Earth above (and to left and right of,etc) you.

Gravity fluctuates by a tiny amount from place to place, not necessarily related to altitude. I have heard, but cannot confirm with a link, that in the USA, you weigh the least in Key West, FL and the most in Minot, ND.

There is an article in this month's National Geographic about how they have detected the rim of an ancient meteor crater in Yucatan by measuring tiny gravitational fluctuations.

Way cool Equus and thank you.

Shawnee Mission, KS - I never noticed that before. My mother and grandmother were born in Plainville, KS and my grandfather in Dodge, we are almost related, hee hee.

(Hi. Although I'm in the same state I regret I'm not that close to your roots. Shawnee Mission is in the extreme east, in the Kansas City suburbs. It is a 5-6 hour drive to Plainville or Dodge City. By comparison it is only about 4 hours to the Illinois border)

labrat,

That formula is only good for a sphere when you are *outside* of the sphere. When you are inside the sphere there is a different formula. I don't remember the exact formula (it is easy to derive with basic calculus if anyone has the time.)

But you can convince yourself that this formula does not apply for objects inside the sphere by imagining a particle at the center of the Earth. This particle would clearly be perfectly balanced at the center (being pulled in all directions equally) and the net force of gravity would be zero (not the infinite force predicted by your formula.

If you consider the Earth a sphere (and you must if you want to use the simple formula) than an object under sea level is below the surface of the sphere. I would expect the gravity to be a bit less.

Eric Brown-Munoz.

I agree with EBrown regarding the gravity effects in a "perfect" spherical condition.

I agree with Equus regarding the variability of gravity on Earth due to differing densities within the crust. More info on this here: http://antwrp.gsfc.nasa.gov/apod/ap030723.html

Best Regards,

i agree with rosborne979 :-)

we all stand together --> Paul McCartney

I agree with myself also Though there have been times when this was not a wise choice

Thanks all. I specified latitude as the latitude lines are closer to true circles than longitude lines or typical great circle routes. This is partly because Earth has about a 24 mile bulge near the Equator making east west symmetry better than north south symmetry. The gravity is reduced (defacto) by the centrifugal force due to the Equator rotating about 1000 MPH. The arctic region is closer to the Earth's center which partially (or over compensates?) for the lack of centrifugal force. The sun and moon positions also cause approximately twice daily tiny fluxuations in local gravity.
Am I to assume that gravity is stronger typically in mountainous regions as the rock has slightly higher density, than is typical in low lands, which I suppose could be considered slightly below the surface, for calculation purposes? The surface of the ocean is about 100? meters below Earth's average radius? Neil

I doubt elevation would be a good guide per se. The Canadian shield isn't particularly elevated, but there's a lot of dense stuff under there, whereas in, say, the Cascades, you'll be over a lot of low-density igneous rock (just guessing as to my specific examples here; I don't know dick about geology).

I'm interested, but have nothing to add.

The author, Pellegrino, wrote a book about how the Nile cut deeply into the basin of the Mediterranean Sea, can't think of the name, just now.

Anyway, he said that those prehistoric animals who lived in that deep gorge were able to soar in the thicker atmosphere, which I'd think, would be a big help in couteracting any slight change in gravity.

Hmmm. Wonder what kind of thermals you get in a deep narrow gorge? Any canyoneers out there? Anybody from southern Utah?

The density of the atmosphere is not linear, but it halves at about 18,000 feet altitude, so Ill guess about 10,000 feet below sea level the density doubles. I'm all but sure the Mediteranian ocean level was never that low, so the soaring would be improved only a little as increased drag goes with increased density partly off-setting the advantage.
My guess is some canyons have excellent thermals, but more than half have down drafts most of the time. Neil

Actually, isn't the weight of the water greater than the gravitational pull? When we visited the Dead Sea which is over 1,000 feet below sea level, we didn't "feel" any difference in gravitational pull, and when at Cuzco which is almost 11,000 feet above sea level, didn't "feel" less pull.

I gotta get in without spoiling the fun. go to google snd look up the Airy and (together and separately) Pratt principles for a .
You are all right in a fashion, and you are all very close together , in practice.

The " simple" equation is, when expanded, accounts for elevations, rotation, density, and gravity tides.
hey dog-get back to your studies

Sorry, I only ask a question here.
What is the radius of the (hypothetical) Earth where the surface of the sea warter can float everything on the Earth.