Steven Dutch, Professor Emeritus, Natural and Applied Sciences,University
of Wisconsin - Green Bay
Measurements of gravity are less near the equator than at the poles for tworeasons:
The latitude correction for gravity is:
g = 9.7803185(1 + 0.005278895 sin^{2}L + 0.000023462 sin^{4}L) m/sec^{2}where L is latitude.
Note that at latitude 90 the correction factors amount to .0053grather than the .003 due solely to centripetal acceleration. The extra mass ofthe earth's bulge at the equator is not enough to make up for the increaseddistance from the earth's center. That latitudinal variation of 0.5 % is fargreater than variations due to density differences within the earth. Anuncorrected gravity map would be little more than a map of latitude.
On a perfectly smooth ellipsoidal earth, once latitude wascorrected for, there would be nothing left. On the real earth, there are stillsignificant departures from ideal gravity, called anomalies.
The map above is the only published map of raw, or observed, gravity that I know of. Blues denote lower values, reds denote higher values. Total variation across the map is about 2000 milligals, far more than the variations on any non-corrected map. The map is dominated by latitudinal variations, but with some interesting differences. Note that colors between Hudson's Bay and the Labrador Sea are shifted a bit north relative to the sea. That's because of the higher elevations and thicker continental crust on land, and consequent lower gravity readings. The difference is especially dramatic in the Canadian Rockies, where a deep root of lighter crust extends into the mantle. On the other hand, colors are shifted south off the coast of Labrador because of gravity highs caused by dense mafic rocks added to the continental margin during the initial stages of rifting.
The gravitational pull on a mass m at the earth's surface is gm= GMm/r^{2}, where G is the Gravitational Constant (6.67 x 10^{-11}m^{3}/kg-sec^{2}).Therefore g = GM/r^{2}. Now the vertical change in g as r increases isdg/dr = -2GMm/r^{3} = -2g/r. Since g = 9.8 m/sec^{2}, and r =6,371,000 m (global average), dg/dr = 3.08 x 10^{-6} (m/sec^{2})/m.If dr is 1 km (1000m) then dg is .00308 m/sec^{2}, nearly the amount ofthe centripetal acceleration at the equator!
At this point it is useful to introduce a new unit, the gal(short for Galileo), which is one cm/sec^{2} or 0.01 m/sec^{2}.Gravity measurements on the earth are typically expressed in milligals,.001 cm/sec^{2} or 10^{-5} m/sec^{2}. Thus the verticalchange in gravity is about 0.3 mgal/m. This change is easily detectable bymodern gravimeters. The change from the basement to the roof of a small buildingwould be quite obvious. In Green Bay, elevation about 200 meters, the gravitycorrection due to altitude would be 60 mgal, a significant fraction of thegravity variations observed in Wisconsin. Since dg/dr is negative, gravity at anelevated station is less than at sea level; we have to add 0.3 mgal/m tothe observed gravity value to get the sea level value.
Correction for altitude alone is called a free-aircorrection (it assumes there's nothing but air between you and sea level).Variations in gravity after latitude and altitude are removed are called free-airanomalies. When we calculate free-air gravity for regions of high elevation,we get values that are too high, because we have neglected the mass of thetopography beneath us.
Except for hang-gliders, usually the assumption that there'snothing but air between you and sea level is false. There is mass between you and sea level that partially compensates for altitude. An infinite sheet of material has gravitational attraction 2(pi)GDh, where D is density and h is thickness. That works out to 4.19 x 10^{-10} Dh. For unit density (1000kg/m^{3}) and h in meters, that works out to 0.0419 mgal/m. For normal crustal density (2700kg/m^{3}) the correction is about 0.113 mgal/m. Compare this to the free-air correction of 0.3 mgal/m. The extra mass is not enough to offset the greater distance from the center of the earth.
Other Bouguer corrections may be necessary. Over water (like the Great Lakes) you obviously can't assume everything is rock, so you'd correct for the water and the underlying rock separately. On a high mountain peak, you can't assume there's rock all around, so a terrain correction must be applied.Terrain corrections are generally small.
When we calculate Bouguer gravity for regions of high elevation,we get values that are too low. Apparently, we have overcorrected. The Bouguer gravity map of Montana, below, shows this well. Note how values systematically decrease toward the southwest where elevations are highest.
The reason Bouguer anomalies are too low in regions of highelevation is isostasy; topography is high because the crust is thick andfloating in the mantle. Ideally, we'd want to correct for isostasy, too; such acorrection is called an isostatic anomaly. Unfortunately, to do it right,we'd have to have independent knowledge (usually seismic) of the thickness ofthe crust. Lacking that, we might assume isostatic compensation and estimate thethickness of the crust from topography. In effect we would reduce the Bouguercorrection by some factor. In practice, though, if there's some regional patternsuperimposed on the features we want to see, we can mathematically filter outthe regional pattern without making any assumptions as to what causes it. So thenext step after Bouguer anomaly maps, unless we're specifically interested inisostasy, is a filtered map of some kind.
Above is an isostatic gravity map of Montana, correcting for variations in crustal thickness. Note how the linear gravity trends, barely evident in the Bouguer map, are much more obvious.
If we don't have good crustal thickness data, we can still remove regional trends using filters. We could average gravity anomalies over, say, a 100-km radius around each point and subtract that average value from each point to reveal local gravity variations.
Let's apply the free air and Bouguer corrections to the earth'sequatorial bulge. The difference between the earth's polar and equatorial radiiis 22,000 m. Assume crustal density of 2700 kg/m^{3}. The free-aireffect of the bulge is 22,000 x -0.3 = -6600 mgal. This is -6.6. gal or -.066 m/sec^{2}.Thus, being 22 km farther from the earth's center at the equator reduces theeffect of gravity by .066 m/sec^{2}^{ }or about -.007g. This is quite a bit larger than the pole-equator latitudecorrection of -.0053g.
Now let's apply the Bouguer correction of 4.19 x 10^{-10} Dh. We have to be careful here. The immediate impulse is to use D = 2700 kg/m^{3},the typical density of the crust. But the earth has crust everywhere. Adding 22km of rock at crustal density amounts to saying that the crust at the equator istwice as thick as at the poles. Instead, we have normal crust and more mantlebeneath it, so we need to use mantle density in our calculation - 3300 kg/m^{3}.We have h = 22,000 m so the Bouguer correction becomes 4.19 x 10^{-10} x 3.3 x 10^{3} x 2.2 x 10^{4} = 0.0304 m/sec^{2}. Thatis, the mass of the bulge adds 0.0304 m/sec^{2} to the earth's gravity,not enough to offset the effect of greater radius. Thus, the net effect of thebulge is to reduce the gravitational pull of the bulge by 0.066 - .0304 = .0356 m/sec^{2}.
Now the earth's rotation also contributes centripetalacceleration, which cannot be distinguished from gravity. The centripetalacceleration is 0.0337 m/sec^{2}; thus the total variation in gravityfrom pole to equator should be:
- .0337 m/sec^{2} due to centripetal acceleration
- .066 m/sec^{2} due to increased radius
+ .0304 m/sec^{2} due to extra mass in the bulge
= -.0693 m/sec^{2} total change
But we see from the latitude formula above that the expectedchange is only about .053 m/sec^{2} from equator to pole. Why is ourprediction so much less? Partly it's due to the fact that deeper and denserlayers of the mantle also contribute to the bulge as well, partly because of thefact that the approximation of the bulge as a flat infinite sheet is, to put itmildly, incorrect.
Since the bulge wraps around the whole earth, we really need toconsider what happens if we add a layer of mass to the whole earth. We have g = GM/r^{2}.Now what happens if we increase the radius of the earth by adding a shell ofthickness dr and density D? We have to do a partial differentiation here sinceboth r and M change, and we have dg = GdM/r^{2} - 2GM/r^{3}dr.The latter term is the change in gravity due to distance and we've already dealtwith that. The change in mass dM will be 4(pi)r^{2}Ddr - in other words,the area of the shell times density times thickness. So dg = GdM/r^{2} =G(4(pi)r^{2}Ddr)/r^{2} = 4(pi)GDdr. Recall that the attractionof a flat sheet is 2(pi)GDh, where h corresponds to dr. So the attraction of athin shell is twice that of an infinite flat sheet. Why? Because there'snow mass on the far side of the earth also pulling on you. If we double theeffect of the mass of the bulge in our calculation above, we'd end up with toolarge a correction because the bulge is not uniform in thickness (only half asthick at 45 degrees latitude and zero at the poles). To go any further, we'dhave to start dealing with the gravitational pull of a non-spherical earth,which gets complicated.
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Created 6 December 2000, Last Update
12 June 2020