Normal gravity on the reference ellipsoid is smaller at the equator than at the poles. What is the main reason?

Centrifugal acceleration is largest at the equator, and the equatorial bulge places it farther from the centre

The mantle beneath the poles is much denser than beneath the equator, so a larger mass attracts a body placed at the pole

Gravity is observed at sea level at the poles but on high ground near the equator, where g is lower with height

Centrifugal acceleration is largest at the equator, and the equatorial bulge places it farther from the centre

The Earth's magnetic field adds to the gravitational pull at the poles and subtracts from it at the equator

Which of the following instruments measures the value of g itself, rather than the difference in g between stations?

A free-fall meter that times a falling mass in vacuum with a laser interferometer

A LaCoste and Romberg zero-length-spring meter read at a base station and at a field station

A fused-quartz spring meter whose dial is read in a loop through several field stations

A gradiometer that measures the vertical gradient of g in Eötvös units over a fixed baseline

A gravity station stands 100 m above the datum. Using 0.3086 mGal per metre, what is the free-air correction?

+30.86 mGal, added to the observed gravity, because g falls with height

−30.86 mGal, subtracted, because the station is above the datum

+3.086 mGal, added, because the factor applies to every ten metres of height

+0.3086 mGal, added, because the correction depends only on the sign of the height

+30.86 mGal, added to the observed gravity, because g falls with height

Why is the terrain correction always added to the observed gravity, whether the relief near the station is a hill or a valley?

Because a hill pulls the meter upward, lowering g, while a valley lacks mass the slab correction assumed

Because all terrain lies below the station level, so its mass always adds to the Bouguer slab being removed

Because a hill pulls the meter upward, lowering g, while a valley lacks mass the slab correction assumed

Because the terrain correction simply undoes the free-air correction, which is always subtracted first

Because rock above the station pulls the instrument downward and so always raises the observed g value

Over a broad mountain range in local Airy isostatic equilibrium, which pattern of anomalies is expected?

Free-air anomaly near zero, Bouguer anomaly strongly negative, isostatic anomaly near zero

Free-air anomaly strongly positive, Bouguer anomaly near zero, which is the pattern of an uncompensated range

Free-air anomaly strongly negative, Bouguer anomaly near zero, isostatic anomaly strongly positive

Free-air anomaly near zero, Bouguer anomaly strongly negative, isostatic anomaly near zero

All three anomalies strongly positive, because the extra mass of the mountains attracts the meter

Why is downward continuation of gravity data a numerically unstable operation?

It multiplies each wavelength by exp(+2πz/λ), which magnifies short-wavelength noise most

It multiplies each wavelength by exp(−2πz/λ), which removes too much of the true signal as well

It needs more stations than upward continuation, so a sparse survey makes the operation fail

It changes the mean value of the field, which cannot be restored afterwards by any later step

Which statement about the principle of equivalent stratum is correct?

Any anomaly on a plane can be reproduced by a thin sheet at depth, so it alone cannot fix its source

It shows that a gravity anomaly determines the true depth and the density of its source layer uniquely

It states that every regional anomaly is produced by one stratum lying at the depth of the Moho boundary

It states that gravity anomalies can be produced only by horizontal layers and never by compact bodies

Any anomaly on a plane can be reproduced by a thin sheet at depth, so it alone cannot fix its source

For an infinitely long horizontal cylinder whose axis is perpendicular to the profile, how is the depth z of the axis related to the half-width x½ of the gravity anomaly?

z = x½, because Δg ∝ 1/(x² + z²) falls to half its peak where x = z

z = 1.305 x½, the rule that applies to a buried sphere

z = x½, because Δg ∝ 1/(x² + z²) falls to half its peak where x = z

z = 0.766 x½, the ratio of half-width to depth for a sphere

z = 2 x½, because the cylinder's attraction extends twice as far as a sphere's

A residual gravity anomaly Δg(x, y) is integrated over the whole plane, and the result is divided by 2πG. What does this quantity give?

The excess mass, that is density contrast times volume, whatever the shape and depth of the body

The total mass, density times volume, of the body, because the host rock contributes nothing to the integral

The mass of the body only if it is a sphere, since the result depends on the assumed shape

The mass lying above the datum plane only, because gravity measures nothing below the surface

Which of the following statements about the corrections applied in gravity data reduction are correct?

The Bouguer slab correction is 2πGρh, which equals 0.04193ρh mGal for ρ in g/cm³ and h in metres

The terrain correction is always positive, that is added, for a hill as well as for a valley

The Bouguer slab correction is 2πGρh, which equals 0.04193ρh mGal for ρ in g/cm³ and h in metres

The terrain correction is always positive, that is added, for a hill as well as for a valley

The free-air correction is subtracted from the observed gravity for a station above the datum

The Bouguer correction adds the attraction of the rock slab between the datum and the station

Which of the following statements about continuation of potential-field data are correct?

Upward continuation attenuates short wavelengths more strongly than long wavelengths

Upward continuation is a stable, smoothing operation and never increases the amplitude of any wavelength

Upward continuation attenuates short wavelengths more strongly than long wavelengths

Upward continuation is a stable, smoothing operation and never increases the amplitude of any wavelength

Downward continuation suppresses noise because it multiplies each wavelength by exp(−2πz/λ)

Upward continuation sharpens the anomalies of shallow sources, which makes them easier to separate

Which of the following statements about gravity anomalies of simple bodies are correct?

The peak anomaly of a buried sphere is GM/z², with M the excess mass and z the depth of its centre

For a horizontal cylinder the half-width of the anomaly equals the depth to its axis

The vertical attraction of an infinite horizontal slab is independent of the depth at which the slab lies

The peak anomaly of a buried sphere is GM/z², with M the excess mass and z the depth of its centre

For a horizontal cylinder the half-width of the anomaly equals the depth to its axis

The vertical attraction of an infinite horizontal slab is independent of the depth at which the slab lies

Moving a buried sphere nearer the surface broadens its anomaly while lowering its peak value

Gravity: The Earth as a Planet, Measurement, Data Reduction, Filtering and Interpretation of Anomalies

This chapter answers the Gravity heading of Part B2 of the Geology and Geophysics (GG) paper: “The earth as a planet; Absolute and relative gravity measurements; The various corrections for gravity data reduction – free air, Bouguer and isostatic anomalies; density estimates of rocks; Regional and residual gravity separation; Principle of equivalent stratum; Upward and downward continuation; Wavelength filtering; Gravity anomalies and their interpretation – anomalies due to geometrical and irregular shaped bodies, depth rules, calculation of mass.” It teaches the figure and gravity field of the Earth with the units used, how g is measured, the reductions that turn a reading into an anomaly, isostasy and density estimates, regional-residual separation, continuation and filtering with the equivalent stratum, and the anomalies of simple and irregular bodies with depth rules and mass. The numericals are drift-corrected differences, latitude, free-air and Bouguer corrections, Airy roots, continuation factors and the depth and mass of a buried sphere.

1. The Earth as a planet and the units of gravity

The Earth is close to an oblate spheroid, flattened by rotation. The geoid, the equipotential surface that coincides with mean sea level, departs from the smooth reference ellipsoid by undulations of about ±100 m at most. The mass follows from g = GM/R²: M = gR²/G ≈ 5.97 × 10²⁴ kg. The mean density, about 5.5 g/cm³, is twice that of surface rocks (about 2.7), so the interior must be dense. The moment-of-inertia factor C/MR² = 0.33, less than 0.4 for a uniform sphere, shows that density rises with depth, towards the core.

Earth constants worth remembering
Quantity	Value
Equatorial and polar radius	about 6378 km and 6357 km (flattening about 1/298)
Mean radius	6371 km
Normal gravity, equator and pole	about 9.780 and 9.832 m/s²
Free-air gradient	−0.3086 mGal per metre of height

Normal gravity γ(φ), the value on the reference ellipsoid, rises from equator to pole by about 5.2 × 10³ mGal. Two effects cooperate: the centrifugal acceleration ω²R cos φ opposes gravity and is largest at the equator (about 3.4 × 10³ mGal), and the equatorial bulge of about 21 km puts the equator farther from the centre. Units: 1 Gal = 1 cm/s², 1 mGal = 10⁻⁵ m/s², 1 μGal = 10⁻⁸ m/s², and the gravity unit g.u. = 10⁻⁶ m/s² = 0.1 mGal. Anomalies of 0.1 to 100 mGal ride on g ≈ 980 000 mGal, so a survey instrument must resolve about one part in 10⁸. One Eötvös, the unit of gradient, is 10⁻⁹ s⁻² = 0.1 mGal/km.

⚠️ Gal, mGal and m/s²

A common slip is to treat 1 mGal as 10⁻³ m/s². It is 10⁻⁵ m/s². Also remember that the equator-to-pole difference is a difference in normal gravity, which is why a latitude correction is needed even when the ground is level.

2. Absolute and relative gravity measurements

Absolute methods measure g itself. A free-fall meter drops a test mass in vacuum and tracks it with a laser interferometer: g = 2h/t² for a fall from rest, with a precision of a few μGal. A pendulum gives g from its period: T = 2π√(L/g) for a simple pendulum, and for the reversible (Kater) pendulum, two knife edges a distance L apart about which the periods are equal, g = 4π²L/T², with no need to know the centre of mass or the moment of inertia. Example: L = 0.995 m and T = 2.000 s give g = 4π² × 0.995/4 = 9.82 m/s².

Relative meters measure the difference in g between stations; they are portable and precise to about 0.01 mGal or better. Most are spring-type: g changes the extension of a spring supporting a mass. The zero-length spring (LaCoste & Romberg) gives a large linear response, and fused-quartz meters are compact. The reading creeps with instrument drift and also changes with the Earth tide of the Sun and Moon (up to a few tenths of a mGal over hours). A survey is therefore run in loops: read at a base, occupy the stations, and return to the base. The difference of the two base readings is the drift, assumed linear in time, and is removed in proportion to elapsed time. A superconducting gravimeter records continuously with μGal stability, and a gradiometer measures the vertical gradient directly, in Eötvös units.

Drift example. The base reads 2450.20 at 0 h and 2450.60 at 4 h, so the drift is 0.10 per hour. A station read at 2 h gives 2461.00; its drift-corrected reading is 2461.00 − 0.20 = 2460.80, so g there exceeds the base by 10.60 (in mGal, if the readings are already calibrated).

⚠️ Absolute versus relative

A relative meter never gives g by itself: it gives differences from the base, and the absolute value comes from tying the base to an absolute station. Calibration of the dial against mGal is also needed, and the drift correction assumes the drift is smooth between base visits.

3. Gravity data reduction: from reading to anomaly

Reduction removes the effects of what is known, so that only the anomaly from unknown density variations remains. From g = GM/R², dg/dh = −2g/R ≈ −0.31 mGal/m, which is the origin of the standard free-air value 0.3086 mGal/m: gravity falls with height, so the free-air correction is added for a station above the datum. The Bouguer slab is an infinite horizontal slab of thickness h whose attraction is 2πGρh.

Corrections applied to the observed gravity
Correction	Size	Sign and reason
Drift and tide	From base loops and tide tables	Removed first
Latitude (normal gravity)	Subtract γ(φ); north–south gradient about 0.81 sin 2φ mGal/km	g increases towards the pole
Free-air	+0.3086 mGal per m above datum	Add: g falls with height
Bouguer	−0.04193 ρ h mGal (ρ in g/cm³, h in m)	Subtract: remove the slab between station and datum
Terrain (Hammer)	Computed from the relief around the station	Always add: hills pull up, valleys lack the slab mass

Free-air anomaly Δg_FA = g_obs − γ + 0.3086 h. Bouguer anomaly Δg_B = g_obs − γ + 0.3086 h − 0.04193 ρ h, plus the terrain correction for the complete Bouguer anomaly. For ρ = 2.67 g/cm³ the slab term is 0.112 mGal/m, so the combined elevation correction is 0.3086 − 0.1119 = 0.197 mGal/m. Example: g_obs − γ = −12.0 mGal at h = 250 m with ρ = 2.67 and terrain correction 0.8 mGal gives Δg_B = −12.0 + 77.15 − 27.99 + 0.8 = 38.0 mGal.

⚠️ Each correction has its own sign

The free-air term is added and the Bouguer term subtracted, so they work against each other. The terrain correction is added whether the relief is a hill or a valley. Over the oceans the Bouguer step replaces water by rock, using the density contrast ρ − ρ_water.

4. Isostasy, isostatic anomalies and density estimates

Isostasy is the hydrostatic balance of lithospheric columns above a compensation depth. In the Airy model a crust of constant density ρc floats on mantle of density ρm, so topography h is supported by a root t = hρc/(ρm − ρc); for ρc = 2.7 and ρm = 3.3 g/cm³, t = 4.5h. In the Pratt model the compensation depth D is fixed and density varies, ρ(D + h) = ρ0 D, so higher columns are lighter. Vening Meinesz compensation is regional: the lithosphere has flexural strength and spreads the load.

An isostatic anomaly is the Bouguer anomaly plus the isostatic correction, the effect of the compensating root or density variation, which is a mass deficit and so is a positive correction. For compensated topography the pattern is: free-air anomaly near zero over broad features, Bouguer anomaly strongly negative over mountains (roughly −100 mGal per km of elevation) and strongly positive over oceans, and isostatic anomaly near zero. A large isostatic anomaly marks a region out of balance, as under a recently melted ice sheet.

Samples: dry bulk density from mass over volume; saturated density (1 − φ)ρ_grain + φρ_water. Typical values in g/cm³: sediments about 1.9 to 2.6, granite about 2.65, basalt about 2.8 to 3.0, average crust 2.67 (the standard Bouguer density), mantle about 3.3.
Nettleton's method: along a profile over relief, compute the Bouguer anomaly for several trial densities; the correct density is the one at which the anomaly shows the least correlation with the topography.
Borehole gravity: in horizontally layered rock, g measured at two depths Δz apart increases downward by Δg = (0.3086 − 0.0839ρ)Δz mGal (ρ in g/cm³), which gives the density of the rock in between. Seismic velocity gives density through the Gardner relation ρ ≈ 0.31 Vp^0.25 (Vp in m/s, ρ in g/cm³).

🎯 It is the contrast that matters

A gravity anomaly comes from the density contrast Δρ = ρ_body − ρ_host, not from the density of the body. A body of density 2.9 g/cm³ in rock of 2.9 gives no anomaly, and the same body in sediment of 2.3 gives a large one. The Bouguer density is an assumption about the rock between station and datum.

5. Regional and residual separation, continuation, equivalent stratum and wavelength filtering

A Bouguer map holds a regional field (long wavelength, deep or broad sources) and residual anomalies (short wavelength, shallow targets). Separation methods: graphical smoothing of profiles; polynomial trend-surface fitting by least squares, residual = observed − fitted; moving-average or ring operators (Griffin: residual = value at the centre − mean of the ring at radius r); the second vertical derivative, which strengthens short wavelengths, ∂²g/∂z² = −(∂²g/∂x² + ∂²g/∂y²) from Laplace's equation; and upward continuation, where the continued field is the regional. The result depends on the method, since no separation is unique.

In the wavenumber domain, with k = 1/λ in cycles per unit length, upward continuation by z multiplies each component by exp(−2πkz) = exp(−2πz/λ). It is stable and smoothing, because short wavelengths are attenuated more. Downward continuation multiplies by exp(+2πz/λ), which amplifies short wavelengths and noise, so it is unstable and works only for small z and smooth data, and it fails below the depth of the sources. Example: a 2 km wavelength continued up 0.5 km keeps exp(−2π × 0.5/2) = 0.21 of its amplitude, a 10 km wavelength keeps 0.73.

Wavelength filtering is multiplication in the wavenumber domain: low-pass keeps the regional, high-pass the residual, band-pass a chosen depth range. For sources at depth z the power spectrum falls as exp(−2|k|z) (|k| angular), so the slope of ln P against |k| is −2z, which gives depth. The principle of equivalent stratum states that any anomaly observed on a plane can be reproduced exactly by a thin layer of surface density σ(x, y) at a depth d below it, with g(k) = 2πGσ(k) exp(−|k|d). The layer is not the real source, since different layers give the same field, but it is a device for continuing the field up or down, interpolating, and converting gravity to its derivatives.

⚠️ Do not continue downward through the source

Downward continuation sharpens anomalies and so seems attractive, but it magnifies noise exponentially and is meaningless below the top of the source, where the field is no longer harmonic. Upward continuation never increases the amplitude of any wavelength.

6. Anomalies of geometric bodies and depth rules

The anomaly of a body is the vertical attraction of its density contrast Δρ, along a profile x across it, with z the depth. The half-width x½ is the horizontal distance from the peak at which the anomaly falls to half the maximum.

Standard anomalies and half-width depth rules
Body	Δg(x)	Peak	Depth rule
Sphere, excess mass M, centre depth z	GMz/(x² + z²)^(3/2)	GM/z²	z = 1.305 x½
Horizontal cylinder, mass λ per length, axis depth z	2Gλz/(x² + z²)	2Gλ/z	z = x½
Infinite horizontal slab, thickness t	2πGΔρt = 0.04193 Δρ t mGal	Constant	None: independent of depth
Semi-infinite horizontal slab (fault)	Rises from 0 to 2πGΔρt	2πGΔρt far from the edge (πGΔρt at the edge)	Edge lies at the half-value

Depth rules come from solving for the half-value. For the sphere, (1 + x²/z²)^(−3/2) = 1/2 gives x² = (2^(2/3) − 1)z² = 0.587z², so x½ = 0.766z and z = 1.305x½. For the cylinder, 1/(1 + x²/z²) = 1/2 gives x½ = z. These give the depth to the centre; the top is shallower by the radius. A rule needs an assumed shape, so a deeper sphere and a shallower cylinder can share one half-width. Smith's rule gives a shape-free bound: for a body of one sign of density contrast the depth to its top is at most 0.86 Δg_max/(dg/dx)_max, with equality for a point mass.

Example. A spherical ore body gives a peak of 1.5 mGal and a half-width of 300 m. The depth to the centre is z = 1.305 × 300 = 391.5 m, and the excess mass M = Δg_max z²/G = (1.5 × 10⁻⁵ × 391.5²)/(6.674 × 10⁻¹¹) = 3.4 × 10¹⁰ kg. A sphere of radius 100 m, centre at 300 m, Δρ = 500 kg/m³ has M = (4/3)π(100)³ × 500 = 2.1 × 10⁹ kg and peak GM/z² = 0.155 mGal. Moving the sphere nearer the surface makes the anomaly taller and narrower.

7. Irregular bodies, total mass and interpretation

Irregular bodies have no closed formula, so they are modelled numerically. Talwani's method (Talwani, Worzel and Landisman) replaces a 2-D body by a polygon and sums line integrals along its sides, an idea going back to Hubbert's line-integral method. Talwani and Ewing extend this to 3-D bodies as a stack of thin horizontal polygonal laminas. The usual practice is forward modelling: the body is adjusted until the computed anomaly matches the observed one.

Calculation of mass. By Gauss's theorem the integral of the vertical attraction over the whole plane is 2πG times the excess mass, whatever the shape and depth: M = (1/2πG) ∬ Δg dA, and λ = (1/2πG) ∫ Δg dx per unit length of a 2-D body. With Δg in mGal and area in m², M = 2.38 × 10⁴ ∬ Δg dA kg. Check with the sphere: ∬ GMz/(r² + z²)^(3/2) dA = 2πGM. This is the excess mass Δρ V. For a body of density ρ in host rock of density ρ_h, the total mass is M × ρ/(ρ − ρ_h). Truncating the tails of the anomaly underestimates M.

Positive anomalies mark excess density: dense basement ridges, mafic and ultramafic intrusions, massive sulphide and chromite bodies, and anticlines of dense strata.
Negative anomalies mark mass deficiency: salt domes (salt is about 2.2 g/cm³, lighter than most compacted sediment), sedimentary basins, granitic plutons and cavities.
Steep gradients mark faults and contacts, and the maximum of the horizontal gradient lies over a steep contact. A broad low-amplitude anomaly points to a deep or extensive source, a narrow sharp one to a shallow source.

⚠️ Interpretation is non-unique

The same anomaly can come from a small, dense, shallow body or a larger, lighter, deeper one, and the mass integral fixes only the excess mass, not the volume or density separately. Geology, drilling, seismic data and other methods must constrain depth, shape or density.

Key takeaways

The Earth is an oblate spheroid with mean density 5.5 g/cm³ and normal gravity rising about 5.2 × 10³ mGal from equator to pole. 1 mGal = 10⁻⁵ m/s², 1 Eötvös = 0.1 mGal/km. Absolute meters time free fall or a Kater pendulum (g = 4π²L/T²); relative meters read differences and need loop drift correction.
Reduction: subtract γ(φ), add 0.3086 h (free-air), subtract 0.04193 ρ h (Bouguer, ρ in g/cm³, h in m) and add the terrain correction. Elevation correction for ρ = 2.67 is 0.197 mGal/m. Latitude gradient about 0.81 sin 2φ mGal/km.
Airy root t = hρc/(ρm − ρc); compensated mountains have near-zero free-air and isostatic anomalies but strongly negative Bouguer anomalies. Densities come from samples, Nettleton's least-correlation method, borehole gravity and velocity relations.
Regional-residual separation is never unique (polynomial, ring, derivative, upward continuation). Upward continuation multiplies by exp(−2πz/λ) and is stable; downward continuation multiplies by exp(+2πz/λ) and amplifies noise. The equivalent stratum reproduces any anomaly with a sheet at depth, so the anomaly cannot fix the source.
Sphere: peak GM/z², z = 1.305 x½. Horizontal cylinder: peak 2Gλ/z, z = x½. Slab: 2πGΔρt = 0.04193 Δρ t mGal, independent of depth. Smith: depth to top ≤ 0.86 Δg_max/(dg/dx)_max.
Mass: M = (1/2πG) ∬ Δg dA gives the excess mass independent of shape and depth, and total mass is M × ρ/(ρ − ρ_h). Irregular bodies use Talwani polygons and Talwani–Ewing laminas. Interpretation is non-unique and needs constraints.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Normal gravity on the reference ellipsoid is smaller at the equator than at the poles. What is the main reason?
1. The mantle beneath the poles is much denser than beneath the equator, so a larger mass attracts a body placed at the pole
2. Gravity is observed at sea level at the poles but on high ground near the equator, where g is lower with height
3. Centrifugal acceleration is largest at the equator, and the equatorial bulge places it farther from the centre
4. The Earth's magnetic field adds to the gravitational pull at the poles and subtracts from it at the equator
Show answer
Answer: C — Centrifugal acceleration is largest at the equator, and the equatorial bulge places it farther from the centre
Rotation produces a centrifugal acceleration ω²R cos φ directed away from the axis, largest at the equator (about 3.4 × 10³ mGal), and the equatorial bulge of about 21 km places the equator farther from the centre. Both lower g at the equator. Hidden mass differences are not the cause, since normal gravity is a property of the reference ellipsoid alone, and magnetism plays no role.
Which of the following instruments measures the value of g itself, rather than the difference in g between stations?
1. A free-fall meter that times a falling mass in vacuum with a laser interferometer
2. A LaCoste and Romberg zero-length-spring meter read at a base station and at a field station
3. A fused-quartz spring meter whose dial is read in a loop through several field stations
4. A gradiometer that measures the vertical gradient of g in Eötvös units over a fixed baseline
Show answer
Answer: A — A free-fall meter that times a falling mass in vacuum with a laser interferometer
A free-fall meter obtains g from distance and time, g = 2h/t², using only length and time standards, so it is absolute. Spring meters respond to changes of g and are read against a base, and the gradiometer measures a gradient, not g. Relative meters need an absolute station to tie them to the true value of g.
A gravity station stands 100 m above the datum. Using 0.3086 mGal per metre, what is the free-air correction?
1. −30.86 mGal, subtracted, because the station is above the datum
2. +3.086 mGal, added, because the factor applies to every ten metres of height
3. +0.3086 mGal, added, because the correction depends only on the sign of the height
4. +30.86 mGal, added to the observed gravity, because g falls with height
Show answer
Answer: D — +30.86 mGal, added to the observed gravity, because g falls with height
Gravity decreases by 0.3086 mGal for every metre of height, so 100 m of elevation has lowered the reading by 100 × 0.3086 = 30.86 mGal, and this is added back. Subtracting it is the tempting sign error. The other values come from slips in the multiplication.
Why is the terrain correction always added to the observed gravity, whether the relief near the station is a hill or a valley?
1. Because all terrain lies below the station level, so its mass always adds to the Bouguer slab being removed
2. Because a hill pulls the meter upward, lowering g, while a valley lacks mass the slab correction assumed
3. Because the terrain correction simply undoes the free-air correction, which is always subtracted first
4. Because rock above the station pulls the instrument downward and so always raises the observed g value
Show answer
Answer: B — Because a hill pulls the meter upward, lowering g, while a valley lacks mass the slab correction assumed
A hill above the station attracts the meter upward, which reduces g, and a valley below the station lacks the mass that the infinite Bouguer slab assumed, so the slab correction over-subtracts. Both effects lower the observed value relative to the reduced one, and both call for an addition. Rock above the station does not raise g, which is the tempting error.
Over a broad mountain range in local Airy isostatic equilibrium, which pattern of anomalies is expected?
1. Free-air anomaly strongly positive, Bouguer anomaly near zero, which is the pattern of an uncompensated range
2. Free-air anomaly strongly negative, Bouguer anomaly near zero, isostatic anomaly strongly positive
3. Free-air anomaly near zero, Bouguer anomaly strongly negative, isostatic anomaly near zero
4. All three anomalies strongly positive, because the extra mass of the mountains attracts the meter
Show answer
Answer: C — Free-air anomaly near zero, Bouguer anomaly strongly negative, isostatic anomaly near zero
In a compensated range the mass of the topography is cancelled at broad wavelengths by the mass deficit of the root, so the free-air anomaly is near zero. The Bouguer correction removes the topography but not the root, so the Bouguer anomaly is strongly negative, roughly −100 mGal per km of elevation. The isostatic correction restores the root, leaving a near-zero isostatic anomaly. The pattern of a strongly positive free-air anomaly with a near-zero Bouguer anomaly would belong to an uncompensated range.
Why is downward continuation of gravity data a numerically unstable operation?
1. It multiplies each wavelength by exp(+2πz/λ), which magnifies short-wavelength noise most
2. It multiplies each wavelength by exp(−2πz/λ), which removes too much of the true signal as well
3. It needs more stations than upward continuation, so a sparse survey makes the operation fail
4. It changes the mean value of the field, which cannot be restored afterwards by any later step
Show answer
Answer: A — It multiplies each wavelength by exp(+2πz/λ), which magnifies short-wavelength noise most
Downward continuation by z multiplies a component of wavelength λ by exp(+2πz/λ), which is largest for the shortest wavelengths, where noise lives, so noise grows exponentially. The factor exp(−2πz/λ) belongs to upward continuation, which smooths and is stable. The number of stations and the mean value are not the cause.
Which statement about the principle of equivalent stratum is correct?
1. It shows that a gravity anomaly determines the true depth and the density of its source layer uniquely
2. It states that every regional anomaly is produced by one stratum lying at the depth of the Moho boundary
3. It states that gravity anomalies can be produced only by horizontal layers and never by compact bodies
4. Any anomaly on a plane can be reproduced by a thin sheet at depth, so it alone cannot fix its source
Show answer
Answer: D — Any anomaly on a plane can be reproduced by a thin sheet at depth, so it alone cannot fix its source
A thin sheet at depth d with surface density σ gives g(k) = 2πGσ(k) exp(−|k|d), so a suitable σ can reproduce any observed anomaly on the plane. This is why the interpretation is non-unique, and why the layer is used as a device for continuation rather than as the real source. It says nothing about the Moho and does not forbid compact bodies.
For an infinitely long horizontal cylinder whose axis is perpendicular to the profile, how is the depth z of the axis related to the half-width x½ of the gravity anomaly?
1. z = 1.305 x½, the rule that applies to a buried sphere
2. z = x½, because Δg ∝ 1/(x² + z²) falls to half its peak where x = z
3. z = 0.766 x½, the ratio of half-width to depth for a sphere
4. z = 2 x½, because the cylinder's attraction extends twice as far as a sphere's
Show answer
Answer: B — z = x½, because Δg ∝ 1/(x² + z²) falls to half its peak where x = z
For the cylinder Δg = 2Gλz/(x² + z²), which is half of its peak 2Gλ/z when x² = z², so x½ = z. The factor 1.305 is the sphere's rule, and 0.766 is its inverse, the sphere's x½/z, so both are tempting confusions.
A residual gravity anomaly Δg(x, y) is integrated over the whole plane, and the result is divided by 2πG. What does this quantity give?
1. The excess mass, that is density contrast times volume, whatever the shape and depth of the body
2. The total mass, density times volume, of the body, because the host rock contributes nothing to the integral
3. The mass of the body only if it is a sphere, since the result depends on the assumed shape
4. The mass lying above the datum plane only, because gravity measures nothing below the surface
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Answer: A — The excess mass, that is density contrast times volume, whatever the shape and depth of the body
By Gauss's theorem the plane integral of the vertical attraction is 2πG times the enclosed excess mass, independent of shape and depth, so M = (1/2πG) ∬ Δg dA. It is the excess mass Δρ V, not the total mass, since the host rock's own attraction is in the removed background. The total mass is M ρ/(ρ − ρ_host). The result does not depend on the shape.
Which of the following statements about the corrections applied in gravity data reduction are correct?
1. The Bouguer slab correction is 2πGρh, which equals 0.04193ρh mGal for ρ in g/cm³ and h in metres
2. The terrain correction is always positive, that is added, for a hill as well as for a valley
3. The free-air correction is subtracted from the observed gravity for a station above the datum
4. The Bouguer correction adds the attraction of the rock slab between the datum and the station
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Answer: A — The Bouguer slab correction is 2πGρh, which equals 0.04193ρh mGal for ρ in g/cm³ and h in metres; B — The terrain correction is always positive, that is added, for a hill as well as for a valley
The slab term is 2πGρh = 0.04193ρh mGal, and the terrain correction is added in both cases. The free-air correction is added for a station above the datum because g falls with height, and the Bouguer correction subtracts the slab, whose attraction is already in the reading.
Which of the following statements about continuation of potential-field data are correct?
1. Upward continuation attenuates short wavelengths more strongly than long wavelengths
2. Upward continuation is a stable, smoothing operation and never increases the amplitude of any wavelength
3. Downward continuation suppresses noise because it multiplies each wavelength by exp(−2πz/λ)
4. Upward continuation sharpens the anomalies of shallow sources, which makes them easier to separate
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Answer: A — Upward continuation attenuates short wavelengths more strongly than long wavelengths; B — Upward continuation is a stable, smoothing operation and never increases the amplitude of any wavelength
The upward-continuation factor exp(−2πz/λ) is below one for every wavelength and smaller for short ones, so the operation smooths and is stable. Downward continuation uses exp(+2πz/λ) and amplifies noise, and upward continuation blurs shallow anomalies instead of sharpening them.
Which of the following statements about gravity anomalies of simple bodies are correct?
1. The peak anomaly of a buried sphere is GM/z², with M the excess mass and z the depth of its centre
2. For a horizontal cylinder the half-width of the anomaly equals the depth to its axis
3. The vertical attraction of an infinite horizontal slab is independent of the depth at which the slab lies
4. Moving a buried sphere nearer the surface broadens its anomaly while lowering its peak value
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Answer: A — The peak anomaly of a buried sphere is GM/z², with M the excess mass and z the depth of its centre; B — For a horizontal cylinder the half-width of the anomaly equals the depth to its axis; C — The vertical attraction of an infinite horizontal slab is independent of the depth at which the slab lies
The sphere peak is GM/z², the cylinder rule is x½ = z, and a slab gives 2πGΔρt wherever it lies, as long as it is wide enough to be treated as infinite. Bringing a sphere nearer the surface raises the peak GM/z² and narrows the half-width 0.766z, so the claim that the anomaly broadens is wrong.
A relative gravimeter reads 1200.00 mGal at the base at 08:00 and 1200.45 mGal at the base at 11:00. Drift is linear in time. At a station occupied at 10:00 it reads 1215.20 mGal. The drift-corrected gravity of the station relative to the base, in mGal to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 14.9
The drift is (1200.45 − 1200.00)/3 h = 0.15 mGal per hour. At 10:00, two hours after the start, the accumulated drift is 0.30 mGal, so the corrected reading is 1215.20 − 0.30 = 1214.90. Relative to the base reading of 1200.00 the difference is 14.9 mGal. Subtracting only the base reading (15.2) forgets the drift.
Take the north–south gradient of normal gravity to be 0.81 sin 2φ mGal per km, evaluated at the station's latitude φ = 30° N. A station lies 10 km north of the base along the meridian. The magnitude of the latitude correction between the two, in mGal to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 7.0
sin 2φ = sin 60° = 0.8660, so the gradient is 0.81 × 0.8660 = 0.7015 mGal/km. Over 10 km the correction is 7.015, which is 7.0 mGal. Using sin φ = 0.5 in place of sin 2φ would give 4.05, which is wrong. Normal gravity is larger at the northern station, so its normal value would be subtracted.
At a station g_obs − γ = +8.5 mGal (observed gravity minus normal gravity at the station's latitude). The station is 180 m above the datum, the Bouguer density is 2.50 g/cm³ and the terrain correction is +1.2 mGal. Use 0.3086 mGal/m for the free-air gradient and 0.04193ρh mGal for the slab with ρ in g/cm³ and h in metres. The complete Bouguer anomaly, in mGal to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 46.4
Free-air term: 0.3086 × 180 = 55.55 mGal, added. Slab term: 0.04193 × 2.50 × 180 = 18.87 mGal, subtracted. Terrain: +1.2. Δg_B = 8.5 + 55.55 − 18.87 + 1.2 = 46.4 mGal. Subtracting the terrain correction, or adding the slab, are the usual sign slips.
A sinusoidal gravity anomaly of wavelength 4.0 km is continued upward by 0.50 km. By what factor is its amplitude multiplied, using the factor exp(−2πz/λ)? Give the answer to two decimal places: ____.
Numerical answer — type the value.
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Answer: 0.46
The factor is exp(−2π × 0.50/4.0) = exp(−0.7854) = 0.456, which is 0.46. The amplitude has fallen to 46% of its value at the original level. Using z/λ = 0.125 without the 2π would give exp(−0.125) = 0.88, which is wrong.
A buried sphere gives a peak gravity anomaly of 2.0 mGal and a half-width x½ = 400 m. Use z = 1.305 x½ for the depth to the centre and G = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². The excess mass, in units of 10¹⁰ kg to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 8.2
The depth to the centre is z = 1.305 × 400 = 522 m. The peak is GM/z², so M = Δg_max z²/G = (2.0 × 10⁻⁵ m/s² × 522²)/(6.674 × 10⁻¹¹) = 8.17 × 10¹⁰ kg, that is 8.2 in units of 10¹⁰ kg. Converting mGal to m/s² is the step that is usually missed.
A mountain range 3.0 km high is in Airy isostatic equilibrium. The normal crustal thickness at sea level is 35.0 km, the crustal density is 2.70 g/cm³ and the mantle density is 3.30 g/cm³. The total crustal thickness beneath the crest, measured from the top of the mountain to the Moho, in km to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 51.5
The root is t = hρc/(ρm − ρc) = 3.0 × 2.70/(3.30 − 2.70) = 13.5 km. The thickness from the top of the mountain to the Moho is the height plus the normal thickness plus the root: 3.0 + 35.0 + 13.5 = 51.5 km. Giving only the root (13.5) or omitting the 3.0 km of relief are the usual slips.