Geophysical exploration methods

Quick refresher

Each method sees a contrast in one property. Gravity sees density. Magnetic surveys see magnetisation (magnetic susceptibility, which shows how strongly a rock becomes magnetised in a field, and remanent magnetisation, which the rock keeps). Seismic sees elastic properties and acoustic impedance (density times wave velocity). Electrical surveys see electrical resistivity. No contrast, no anomaly.
Gravity anomalies are small. 1 mGal $= 10^{-5}$ m/s², which is about a millionth of the acceleration due to gravity. Before interpretation, we remove three things from the measurements: the normal field of the Earth (which depends on latitude), the effect of the station elevation (0.3086 mGal/m) and the attraction of the intermediate layer (the rock between the station and sea level; 0.0419 mGal/m for each 1 g/cm³).
Over a sphere, the anomaly is greatest above the centre, $\Delta g_{max} = GM/z^2$ , and it is half that at a distance $x_{1/2} \approx 0.766\,z$ . The width of the anomaly gives the depth. The amplitude, together with the depth, gives the mass. We cannot find the size of the body and the density contrast separately.
Sedimentary rocks are weakly magnetised, so in basins the magnetic anomalies come mainly from the basement. The deeper the basement, the wider and weaker the anomalies. This is how we make maps of basement depth and cover thickness, accurate to within tens of per cent.
Seismic surveying measures time, not depth: in each layer, $z = Vt/2$ . Fast salt “pulls up” everything beneath it on a time section. We get depths only from a velocity model tied to wells (checked against well data).
A solution from geophysical fields is non-unique (more than one answer fits the data), so we combine methods. Cheap regional surveys show where to run seismic. Seismic shows where to drill. Wells give better densities and velocities for all the models.

Key relationships

\Delta g(x) = \dfrac{G M z}{(x^2 + z^2)^{3/2}}

gravity anomaly above a sphere whose anomalous mass (extra or missing mass compared with the surrounding rock) is $M = \tfrac43\pi a^3\Delta\rho$ and whose centre is at depth $z$

z \approx 1.305\,x_{1/2}

depth of the centre of a sphere, from the half-width of the anomaly; the top of a body of any shape with the same anomaly is no deeper

\Delta g_\text{slab} = 0.04194\,\rho\,h

attraction of an infinite flat slab, in mGal ( $\rho$ in g/cm³, $h$ in m)

z = \sum_i \dfrac{V_i\,t_i}{2}

depth from the two-way time $t_i$ in each layer, using the interval velocities $V_i$ (the velocity of sound within each layer)

Common pitfalls

Confusing the milligal with the gal: 1 mGal $= 10^{-3}$ Gal $= 10^{-5}$ m/s². A sign error in the exponent ( $10^3$ instead of $10^{-3}$ ) makes the anomaly of a dome a million times too large, which is more than the field of the whole Earth.
Taking the depth from the half-width rule as the depth of the top. It is the depth of the centre of the sphere, and the top of a real body is shallower.
Reading a time section as a depth section. Beneath salt, and where velocity changes sharply along the profile, structures on a time section can be false.
Using the apparent density of salt from the density log (2.03 g/cm³) in a density model, instead of the true density (about 2.16 g/cm³).

After this lesson you can

name, for gravity, magnetic, seismic and electrical surveys, the physical property of rocks that creates the anomaly, and explain why a method is “blind” when that property has no contrast (no difference between neighbouring rocks)
calculate the gravity anomaly of a body approximated by a sphere (the profile, the maximum value in milligals and the half-width), then use the half-width to estimate the depth of the centre of the body and the amplitude to estimate its missing or excess mass
explain why we apply corrections for latitude, station elevation, the intermediate layer and terrain, and why the elevations of gravity stations are needed to an accuracy of about 10 cm
explain how the depth of the basement (the hard rock below the sedimentary cover) is estimated from the width of magnetic anomalies, and why this estimate is only approximate
convert the two-way time of a reflected wave (the time it takes to go down and come back) to depth using layer velocities, and show how fast salt distorts a time section
compare land, marine, airborne and satellite surveys, and explain why one anomaly can have many explanations and why methods are combined

Why a petroleum engineer needs geophysics

A well 20 cm in diameter opens the section at a single point, while a trap occupies tens of square kilometres. Core (rock cylinders cut from the well) and logs (measurements made in the well) give detailed information about the rocks, but only beside the borehole. Between wells we have to infer the structure (Structure maps and contouring). Geophysics measures physical fields over the whole area, both between wells and where there are no wells yet.

It rests on a simple observation: rocks differ in physical properties, and a body with different properties changes the field around it. Light salt slightly weakens the attraction above it. A magnetised intrusion (a body of igneous rock that pushed into the rock around it) adds its field to the magnetic field of the Earth. The boundary between rocks of different elasticity reflects elastic waves. A bed with oil conducts electric current less well than its neighbours. We measure the field wherever we can get to — on the ground, in the air, at sea, in a well — and from its changes we judge what lies below.

Calculating the field of a given body is the forward problem, and it has one solution. Finding the body from the measured field is the inverse problem, and it usually has many solutions: the same field can be produced by bodies of different shape, size and depth. So the interpreter always keeps a model in mind, a geologically reasonable set of bodies, and tests it against all the available data (see the section on ambiguity).

Methods of exploration geophysics

Method: Gravity survey
Rock property: density $\rho$
What is measured: gravity, mGal
Main tasks: salt, reefs, basement relief (the shape of the top of the basement), basin boundaries

Method: Magnetic survey
Rock property: magnetisation: susceptibility $\kappa$ and remanent
What is measured: magnetic field induction, nT
Main tasks: basement depth, faults, volcanic rocks

Method: Seismic survey
Rock property: elastic properties: velocity $V$ and impedance $\rho V$
What is measured: arrival times and amplitudes of elastic waves
Main tasks: shape of horizons, traps, faults, sometimes fluids

Method: Electrical survey
Rock property: resistivity $R$
What is measured: electric and magnetic fields
Main tasks: thickness of the conductive cover, salt, sometimes saturation

Method	Rock property	What is measured	Main tasks
Gravity survey	density $\rho$	gravity, mGal	salt, reefs, basement relief (the shape of the top of the basement), basin boundaries
Magnetic survey	magnetisation: susceptibility $\kappa$ and remanent	magnetic field induction, nT	basement depth, faults, volcanic rocks
Seismic survey	elastic properties: velocity $V$ and impedance $\rho V$	arrival times and amplitudes of elastic waves	shape of horizons, traps, faults, sometimes fluids
Electrical survey	resistivity $R$	electric and magnetic fields	thickness of the conductive cover, salt, sometimes saturation

Seismic is the main method of exploration and appraisal; the others are cheaper per square kilometre, cover large areas faster and help where seismic is difficult.

SchematicEach method responds to its own body (schematic). Salt is lighter than the host sediments, so there is a gravity minimum above it. The intrusion in the basement is magnetised, so there is a magnetic maximum above it, and it is broader because the source is deeper. An elastic wave is reflected from layer boundaries. An accumulation beside a salt wall stands out by its high resistivity. The shapes of the curves follow the formulas of the lesson; the scales are arbitrary.

Regional structure: basin boundaries, depth of the basement and thickness of the sedimentary cover, and large faults. Exploration starts here (Plate tectonics and sedimentary basins, Petroleum system and the play concept).
Traps: crests, salt domes, reefs and blocks between faults (Traps and timing). We place exploration wells on them.
The accumulation: its shape and size, the faults and blocks within it, and how far the water-bearing part extends. These are the input data for the volumetric estimate (the calculation of volumes) of hydrocarbons in place and for the development plan.
Fluids — sometimes: amplitude anomalies of reflections (unusually strong or weak reflections) over gas, and raised resistivity over an accumulation. In development, repeat seismic surveys show how fluids move (4D seismic).

What each method sees: physical properties of rocks

The density ( $\rho$ ) of a rock depends on the density of its minerals and on the fluid in its pores (mineral densities). As sediments are buried, they compact, lose water and become heavier. Their density rises from 2.0 g/cm³ at the surface to 2.4–2.6 g/cm³ at a depth of several kilometres (compaction trend). Rock salt does not compact, so its density is about 2.16 g/cm³ at any depth. In the upper hundreds of metres it is heavier than young sediments; deeper it is lighter (why salt flows). A gravity survey senses not the density itself but its contrast $\Delta\rho$ , the difference between the body and the host rocks (the rocks around it).

The magnetisation of a rock has two parts. The induced magnetisation arises in the present magnetic field of the Earth and is proportional to it. The coefficient of proportionality is the magnetic susceptibility $\kappa$ , which has no units in SI. Almost all of it comes from the content of magnetite (a strongly magnetic iron oxide, Fe₃O₄). The remanent magnetisation was fixed in the rock when the rock formed — for example, when basalt cooled — and it may point in a different direction from the present field (remanent magnetisation). The ratio of remanent to induced magnetisation is called the Koenigsberger ratio $Q$ . When $Q > 1$ , the remanent part dominates. Above the Curie point (about 580 °C) magnetite loses its magnetisation. So the sources of magnetic anomalies lie above the depth where the temperature reaches this value (this isotherm), in the crust.

The elastic properties determine the velocity of elastic waves. The velocity rises with the stiffness of the rock and falls with its density (velocity formulas). In porous rocks it also depends on the porosity and the fluid (Sonic log). Reflections arise where the product of density and velocity changes. This product is the acoustic impedance $\rho V$ . How the strength of a reflection depends on the contrast of impedances is covered in the topic Seismic waves, acoustic impedance and reflections.

The resistivity $R$ of a rock (how strongly it resists electric current) is set by the water in the pores. The rock matrix (the solid grains) and hydrocarbons hardly conduct current, while brine (salty water) conducts well (resistivity of rocks, Archie's equation and saturation of clean reservoirs). So shales and water-bearing beds are conductors. Salt, anhydrite, tight carbonates (carbonate rocks with almost no pore space), crystalline basement and beds with oil and gas are insulators. Here, as in the lessons on logging, we write resistivity as $R$ and keep the letter $\rho$ for density.

Approximate physical properties of rocks

Rock: Shales, mudstones
$\rho$ , g/cm³: 2.0–2.6
$V_p$ , km/s: 2.0–4.3
$R$ , Ω·m: 0.5–10
$\kappa$ , $10^{-3}$ SI: 0.01–15

Rock: Sandstones
$\rho$ , g/cm³: 2.0–2.6
$V_p$ , km/s: 2–5
$R$ , Ω·m: 0.2–10; higher with oil
$\kappa$ , $10^{-3}$ SI: 0–20

Rock: Limestones, dolomites
$\rho$ , g/cm³: 2.3–2.87
$V_p$ , km/s: 3.5–7
$R$ , Ω·m: 0.2–10; tight ones from $10^3$
$\kappa$ , $10^{-3}$ SI: 0–3

Rock: Rock salt
$\rho$ , g/cm³: 2.16
$V_p$ , km/s: 4.5
$R$ , Ω·m: more than $10^3$
$\kappa$ , $10^{-3}$ SI: about −0.01

Rock: Anhydrite
$\rho$ , g/cm³: 2.96–2.98
$V_p$ , km/s: 6.1
$R$ , Ω·m: more than $10^3$
$\kappa$ , $10^{-3}$ SI: about 0

Rock: Granites, gneisses
$\rho$ , g/cm³: 2.6–2.7
$V_p$ , km/s: 5.8–6.1
$R$ , Ω·m: $10^3$ – $10^4$
$\kappa$ , $10^{-3}$ SI: 0–50

Rock	$\rho$ , g/cm³	$V_p$ , km/s	$R$ , Ω·m	$\kappa$ , $10^{-3}$ SI
Shales, mudstones	2.0–2.6	2.0–4.3	0.5–10	0.01–15
Sandstones	2.0–2.6	2–5	0.2–10; higher with oil	0–20
Limestones, dolomites	2.3–2.87	3.5–7	0.2–10; tight ones from $10^3$	0–3
Rock salt	2.16	4.5	more than $10^3$	about −0.01
Anhydrite	2.96–2.98	6.1	more than $10^3$	about 0
Granites, gneisses	2.6–2.7	5.8–6.1	$10^3$ – $10^4$	0–50

These are orders of magnitude, for orientation. $V_p$ is the velocity of compressional waves. The mineral densities and the velocities in tight rocks are as in the lessons on minerals and on sonic logging; the rest comes from reference compilations of rock properties. Porous rocks and shales at depth are denser and faster than at the surface. The susceptibility of sedimentary rocks is usually less than $10^{-3}$ SI units, and large values come from grains of magnetite. Rock salt is weakly diamagnetic (it has a negative susceptibility, so it slightly weakens a magnetic field). The most strongly magnetised rocks are mafic and ultramafic igneous rocks (dark rocks rich in iron and magnesium, such as basalt, gabbro and peridotite), up to 0.1–0.2 SI units.

Gravity surveying: the anomaly above a buried body

The acceleration due to gravity at the Earth's surface is about 9.78 m/s² at the equator and 9.83 m/s² at the poles. Geological bodies change it by tiny fractions, so gravity surveying uses a small unit: 1 gal (Gal, named after Galileo) $= 1$ cm/s², and 1 milligal (mGal) $= 10^{-3}$ Gal $= 10^{-5}$ m/s² $= 10$ µm/s². A milligal is about a millionth of the full acceleration. The anomalies that interest exploration are usually a few to tens of milligals.

A gravimeter is in essence a very sensitive spring balance: a mass hangs on a spring, and we measure how much the stretch of the spring changes from station to station. Such an instrument is relative: it shows the difference in gravity between stations, and its reading repeats to within thousandths of a milligal. The readings slowly “drift” (instrument drift), and the tides caused by the Moon and the Sun change gravity by about 0.3 mGal. So we run the survey in loops, returning to a base station, and we subtract the drift and the tides. The base network is tied to absolute gravimeters. In these instruments a reflector falls freely in a vacuum chamber.

A body with an excess or a deficit of mass $M$ adds its own attraction to that of the Earth, positive or negative. The gravimeter is levelled by a plumb line, so it measures the vertical component of gravity. The gravity anomaly is therefore the vertical component of the attraction of the anomalous mass. It is simplest to calculate for a sphere: outside, a uniform sphere attracts as if all its mass were collected at the centre.

Gravity anomaly above a sphere (a point mass)

\Delta g(x) = \frac{G M z}{(x^2 + z^2)^{3/2}}, \qquad M = \tfrac{4}{3}\pi a^3\,\Delta\rho

where:

$\Delta g(x)$ anomaly at a distance $x$ from the point above the centre of the sphere; the value in m/s² multiplied by $10^5$ gives milligals, m/s²
$G$ gravitational constant, $6.6743\cdot10^{-11}$ , m³/(kg·s²)
$M$ anomalous mass of the body — excess (positive) or deficit (negative), kg
$z$ depth of the centre of the sphere below the line of observation, m
$x$ horizontal distance from the point above the centre, m
$a$ radius of the sphere, m
$\Delta\rho$ density contrast between the body and the host rocks; for salt among compacted sediments it is negative, kg/m³

Units: SI; 1 mGal $= 10^{-5}$ m/s²

Holds when

The body is a uniform sphere or any spherically symmetric body: outside it, its field equals that of a point mass at the centre.
The host rocks are uniform horizontally. The effect of everything else (deep sequences, basement relief) has already been subtracted as a regional background.
Corrections for elevation and terrain have moved the observations to a horizontal surface (next section).
We measure the vertical component; an ordinary gravimeter does not sense the horizontal components.

The anomaly is greatest above the centre, $\Delta g_{max} = GM/z^2$ , and it decreases with distance: at $x = z$ it is already $2^{3/2} \approx 2.83$ times smaller. The shape of the curve depends only on the depth. The mass merely stretches the curve vertically. The sign of the anomaly matches the sign of the contrast: above salt among denser rocks there is a minimum, and above a dense body there is a maximum.

Half-width of the anomaly and the depth of the centre of the sphere

x_{1/2} = z\sqrt{2^{2/3} - 1} \approx 0.7664\,z, \qquad z \approx 1.305\,x_{1/2}

where:

$x_{1/2}$ half-width of the anomaly: the distance from the extremum (the highest or lowest value) to the point where the anomaly is half as large, m
$z$ depth of the centre of the sphere, m

Units: any consistent units of length

Holds when

The body is a sphere (a point mass), and the anomaly is isolated: the regional background has been subtracted correctly.
For other shapes the factor is different. For a long horizontal cylinder (a body elongated along strike), $x_{1/2} = z$ . For a thin vertical column extending to a great depth, $x_{1/2} = \sqrt3\,z$ , where $z$ is the depth of its top.

The width of the anomaly tells us the depth, not the size of the body: the deeper the source, the wider and weaker the anomaly. Any body can be represented as a set of point masses. Mass spread out horizontally gives a wider anomaly than mass gathered at a single point. So for the anomaly to stay as narrow as it is, at least part of the mass must lie no deeper than the centre of the sphere. This means that the top of a body of any shape that creates such an anomaly lies no deeper than the centre of the sphere (if the density contrast has the same sign throughout the body). We call the depth from this rule the limiting depth: the greatest depth that the top can have.

CalculatedCalculated anomaly above the salt dome of the example below, a sphere 2 km in radius with its centre at a depth of 3 km (solid line), and above the same dome lowered to twice the depth (dashed line). The dots mark the half-width. At twice the depth the anomaly is twice as wide and four times weaker.

Worked exampleGravity anomaly above a salt dome

Given (assumed data)

The dome is approximated by a sphere: radius $a = 2.0$ km, centre at a depth of $z = 3.0$ km: top at 1 km, bottom at 5 km
Density of salt: 2.16 g/cm³, as in the lesson on folds and diapirs
Host sediments: they compact with depth following the trend of the same lesson: at 1 km the salt is 0.10 g/cm³ lighter than they are, and at 5 km it is 0.44 g/cm³ lighter
Contrast $\Delta\rho$ averaged over the volume of the sphere: −0.33 g/cm³, that is, −330 kg/m³

Solution

Volume of the sphere: $V = \tfrac43\pi a^3 = 3.351\cdot10^{10}$ m³ at $a = 2000$ m.
Missing mass $M = V\Delta\rho$ : $3.351\cdot10^{10}\cdot(-330)$ , that is, $-1.106\cdot10^{13}$ kg, which is about 11.1 billion t.
Anomaly above the centre $\Delta g_{max} = GM/z^2$ : multiply $6.6743\cdot10^{-11}$ by $-1.106\cdot10^{13}$ and divide by $3000^2$ . This gives $-8.20\cdot10^{-5}$ m/s², that is, −8.20 mGal.
Half-width: $x_{1/2} = 0.7664\cdot 3.0 = 2.30$ km. At $x = 3$ km, which is $x = z$ , the anomaly is $2^{3/2}$ times smaller: −2.90 mGal. At 6 km it is −0.73 mGal, and at 10 km it is −0.19 mGal.
Reverse step: the depth of the centre from the half-width is $z = 1.305\cdot 2.30 = 3.00$ km.

Answer: The dome gives a gravity minimum of about −8.20 mGal. The anomaly falls to half at 2.30 km from the centre and almost vanishes at 10 km.

What it means. An anomaly of several milligals is clearly visible. A gravimeter notices thousandths of a milligal, and the accuracy of a survey after corrections is hundredths. But this is only millionths of gravity, and the corrections for elevation and latitude easily reach tens of milligals, so we apply them very carefully. The width of the anomaly, about 4.6 km between the half-value points, is set by the depth and not by the size of the dome. A sphere of any radius with its centre at 3 km would give the same width. The same dome twice as deep would give an anomaly twice as wide and four times weaker (figure above). The model is simplified. In reality the contrast grows with depth, from −0.10 g/cm³ at the top to −0.44 g/cm³ at the bottom. So most of the missing mass lies in the lower half of the sphere. The real anomaly will be somewhat weaker and wider than calculated, and the depth from its half-width will be greater than 3 km. The stem of the dome and the source layer of salt (the deep layer the salt rose from) add a broad minimum, which in processing is assigned to the regional background. The data in this example are assumed for illustration.

In exploration the problem is the inverse one: we have measured the anomaly, and the body is unknown. The half-width gives the depth of the centre for the shape we assume. The amplitude, together with the depth, gives the anomalous mass $M = \Delta g_{max} z^2/G$ . But the radius and the density contrast enter the mass only as the product $a^3\Delta\rho$ , so from one anomaly we cannot separate them.

Worked exampleDepth and mass of a dome from a measured anomaly

Given (assumed data)

Local anomaly above another dome (regional background subtracted): minimum −11.5 mGal
Half-width of the anomaly $x_{1/2}$: 2.9 km
Density contrast between salt and host rocks from a neighbouring well: −0.30 g/cm³; as a check, −0.20 g/cm³

Solution

Depth of the centre, if the dome is a sphere: $z = x_{1/2}/0.7664$ , that is, $2900/0.7664 = 3784$ m. The top of any body with such an anomaly is no deeper than 3.78 km.
Missing mass $M = \Delta g_{max} z^2/G$ : we divide the product $11.5\cdot10^{-5}\cdot 3784^2$ by $6.6743\cdot10^{-11}$ , which gives 2.467·10¹³ kg.
Radius of the sphere at $\Delta\rho = -0.30$ g/cm³, that is, $-300$ kg/m³: $a =\sqrt[3]{3|M|/(4\pi|\Delta\rho|)} = 2698$ m. The top is at a depth of $3784 - 2698 = 1086$ m.
At $\Delta\rho = -0.20$ g/cm³ ( $-200$ kg/m³) the sphere is larger: $a = 3088$ m, and the top is at 696 m.
Now suppose the dome is not a sphere but a stock (a column of salt) that extends down to the source salt. The model of a thin vertical column with the same half-width gives the depth of its top, $z = x_{1/2}/\sqrt3 = 1674$ m. But the amplitude rejects this model. Such a stock needs a missing mass $\lambda = |\Delta g_{max}|\,z/G$ , that is, $2.88\cdot10^9$ kg for each metre of length. At a contrast of −0.30 g/cm³ (−300 kg/m³) its radius $r =\sqrt{\lambda/(\pi|\Delta\rho|)}$ is about 1750 m, which is greater than the depth of the top. This is no longer a thin column, and the anomaly of such a thick stock would be markedly wider than the one measured.

Answer: The top of the salt is no deeper than 3.78 km (the depth where the centre of the sphere would be). The missing mass is about 2.467·10¹³ kg. If the dome is close to a sphere, the top lies at 0.70 km at a contrast of −0.20 g/cm³ and at 1.09 km at −0.30 g/cm³. The model of a thin stock does not agree with the amplitude.

What it means. Gravity surveying tells us with confidence that a dome exists, where it lies in plan view (on the map) and how much mass it lacks. But it knows the depth of the top only roughly. Even for a sphere, the depth of the top shifts by 390 m between −0.20 and −0.30 g/cm³, and from one anomaly we cannot tell the contrast and the shape of the body apart. A trial model must explain both the width and the amplitude of the anomaly at a realistic density contrast. This is how the thin-stock model was rejected. So we constrain the gravity model with other data. We take the contrast from the density logs of the nearest wells, the shape from seismic sections, and the top from a well if one has passed nearby. The reverse is useful too. If a salt body built from seismic data gives a calculated anomaly that differs markedly from the measured one, we review the interpretation. The data in this example are assumed for illustration.

From a gravimeter reading to the Bouguer anomaly

A gravimeter shows the total gravity at a station. The geology in it is a small addition to three large parts, which tell us nothing for exploration (they are “empty”). The first is the normal field of the Earth. Because the planet spins and is slightly flattened at the poles, gravity grows from the equator to the poles. The second is the elevation of the station: the farther from the centre of the Earth, the weaker the attraction. The third is the rock between the station and the reference level, usually sea level: a hill adds mass under the instrument. We subtract these three parts, and what remains is the anomaly, the deviation from what the Earth would give without variations in density.

Normal gravity on the ellipsoid

g_\text{n}(\varphi) = 978{,}032.7\,\bigl(1 + 0.0053024\sin^2\varphi - 0.0000058\sin^2 2\varphi\bigr)

where:

$g_\text{n}$ normal gravity on the surface of the Earth's ellipsoid, mGal
$\varphi$ geographic latitude of the station, °

Units: mGal

Holds when

An approximate form of the normal field for the ellipsoid (the smooth, slightly flattened model of the Earth's shape) of the Geodetic Reference System 1980 (GRS80) adopted in geodesy.
We calculate the correction from the latitude of each station.

From the equator (978,032.7 mGal) to the pole (983,218.6 mGal), normal gravity increases by almost 5.2 Gal. At the latitudes of Kazakhstan, from 41 to 55° N, it increases by about 0.77–0.81 mGal for every kilometre to the north. Over a profile a few kilometres long this adds up to milligals, which is as much as a small structure gives.

Bouguer anomaly

\Delta g_\text{B} = g - g_\text{n}(\varphi) + 0.3086\,h - 0.04194\,\rho\,h + \delta g_\text{T}

where:

$\Delta g_\text{B}$ Bouguer anomaly at the station, mGal
$g$ observed gravity, corrected for tides and instrument drift, mGal
$g_\text{n}(\varphi)$ normal gravity at the latitude of the station, mGal
$h$ elevation of the station above the reference level (usually above sea level), m
$\rho$ density of the intermediate layer — the rocks between the station and the reference level, g/cm³
$\delta g_\text{T}$ terrain correction (the effect of the hills and hollows around the station), mGal

Units: mGal; $h$ in m, $\rho$ in g/cm³

Holds when

0.3086 mGal/m is the mean normal vertical gradient: gravity decreases by this amount for every metre of rise. This is the free-air correction (also called the Faye correction).
$0.04194\,\rho h$ is the attraction of an infinite flat slab of thickness $h$ , that is, $2\pi G\rho h$ in milligals for a density in g/cm³. This is the correction for the intermediate layer (also called the Bouguer correction).
The terrain correction accounts for how the ground around the station differs from a flat slab. On land it is positive both for hills above the station and for hollows below it.
For a station below the reference level, $h$ is negative. At sea and in the air the corrections are different, and on a moving ship or aircraft we add the Eötvös correction for the motion.

If we apply only the elevation correction, we get the free-air anomaly (the Faye anomaly). If we also apply the corrections for the intermediate layer and terrain, we get the Bouguer anomaly (in full, the gravity anomaly in the Bouguer reduction). It keeps only the influence of the density variations that lie below the reference level, so on land exploration is based on it.

Worked exampleElevation corrections on a hilly profile

Given (assumed data)

Stations A and B on the same latitude (a profile along a parallel): elevations 112.0 and 168.5 m
Observed gravity difference $g_B - g_A$: −10.20 mGal after corrections for tides and drift
Density of the intermediate layer: 2.30 g/cm³ (near-surface sediments); for comparison, 2.67 g/cm³
Terrain around the stations: gentle, so we neglect the terrain correction

Solution

Difference of elevations: $\Delta h = 168.5 - 112.0 = 56.5$ m. The normal field is the same at both stations, because they are on the same latitude.
Free-air correction: $0.3086\cdot 56.5 = 17.44$ mGal. This is how much less the gravimeter recorded at B because of the height. We add it.
Correction for the intermediate layer: $0.04194\cdot 2.30\cdot 56.5 = 5.45$ mGal. This is the attraction of the extra layer of rock under B. We subtract it.
Difference of Bouguer anomalies: to −10.20 mGal we add 17.44 and subtract 5.45, which gives 1.79 mGal.
With a density of 2.67 g/cm³ the slab gives 6.33 mGal, and the difference of anomalies is 0.91 mGal.
Sensitivity to elevation: the slab gives $0.04194\cdot 2.30 = 0.0965$ mGal per metre. Together with the free-air correction, the net effect is $0.3086 - 0.0965 = 0.212$ mGal per metre.

Answer: At station B the gravimeter showed 10.20 mGal less than at A, but the Bouguer anomaly at B is 1.79 mGal greater than at A. The difference of elevations explains the whole observed drop, and more.

What it means. The corrections here are larger than the anomaly itself, so everything depends on them. An elevation error of 1 m gives 0.212 mGal. To keep the error within 0.02 mGal, we need the elevations of stations to an accuracy of about 10 cm. We measure them with satellite receivers and by levelling (surveying heights with a level). The density of the intermediate layer also has a marked effect: at 2.67 g/cm³ the difference of anomalies is half as large. We choose the density so that the anomaly does not copy the relief (Nettleton's method). If maxima appear on all the hills and minima in the valleys, the density we adopted is too low. The data in this example are assumed for illustration.

The Bouguer anomaly contains the influence of all masses below the reference level, from salt domes to basement relief and the Moho (the boundary between the crust and the mantle). Broad, smooth changes from deep sources are called the regional background (regional). Narrow changes from shallow bodies are called local anomalies (residual). We remove the background by smoothing, by subtracting a smooth surface or by filtering the map. The amplitude of the local anomaly depends markedly on how this is done, so the interpretation states the method of separation. Over large areas we also see anomalies that come from departures from isostatic equilibrium (the balance in which the crust floats on the mantle) (isostasy).

Magnetic surveying: the depth of the basement

The magnetic field of the Earth is created mainly by currents in the liquid outer core (the Earth's core). The induction of the field (its strength) at the surface is from about 30 thousand nanotesla (nT; in old reports gammas, γ; 1 γ = 1 nT) at the equator to 60 thousand nT and more at the poles. In Kazakhstan, according to the International Geomagnetic Reference Field (IGRF) for 2025, it is about 51–59 thousand nT at an inclination of 61–74° (the angle of the field below the horizontal): the field points steeply downwards. Anomalies from rocks are usually from a few to hundreds of nanotesla.

Modern magnetometers, proton and quantum (for example caesium), measure the magnitude of the total induction vector. They sense its changes down to fractions of a nanotesla. Soviet airborne magnetometers of the 1950s gave an accuracy of 15–20 nT. The field also changes in time (daily variations and magnetic storms). So during a survey a magnetic variation station works continuously at a base point, and we subtract its record. We also subtract the normal field from the IGRF model, and what remains is the anomaly $\Delta T$ .

Sedimentary rocks are weakly magnetised, and salt is even slightly diamagnetic. The crystalline basement (granites, gneisses and mafic intrusions) has more magnetite. So in a sedimentary basin the magnetic anomalies come mainly from the basement: from the relief of its surface and from blocks and intrusions of different composition. The exceptions are volcanic rocks in the cover and, in detailed surveys, interbeds with magnetite grains in the sediments themselves. The deeper the basement, the wider and weaker its anomalies. This is the same law as in gravity surveying. Hence the main use of magnetic surveying in petroleum geology: a map of the depth of the magnetic basement, and so of the thickness of the sedimentary cover. A second use is a map of faults and block boundaries, which show up as chains of anomalies and steps in the field.

At mid-latitudes the anomaly above a body is asymmetric. The field and the magnetisation are inclined, so the maximum is displaced towards the equator, and a minimum appears on the polar side. To bring the anomalies over their sources, we recalculate the map to the form it would have at the magnetic pole, where the field and the magnetisation are vertical. This is reduction to the pole. The recalculation assumes that the magnetisation is induced. Strong remanent magnetisation distorts the result.

Anomaly of a magnetic pole — the top of a long vertical body

\Delta T(x) = \frac{C\,z}{(x^2 + z^2)^{3/2}}, \qquad C = \frac{\kappa\,B_0\,A}{4\pi}

where:

$\Delta T(x)$ anomaly at a distance $x$ from the point above the body (the map is reduced to the pole), nT
$z$ depth of the top of the body, m
$\kappa$ magnetic susceptibility of the body relative to the host rocks, SI units (dimensionless)
$B_0$ induction of the Earth's magnetic field, nT
$A$ area of the horizontal cross-section of the body, m²
$C$ strength of the “pole”, nT·m²

Units: $x$ , $z$ in m; $A$ in m²; $B_0$ and $\Delta T$ in nT

Holds when

The body is a thin vertical column, magnetised along its axis, that extends far deeper than its top. The field of the lower end is negligible, so only the field of the upper “pole” remains.
The magnetisation is induced and the field is vertical, or else the map has been reduced to the pole.
The susceptibility is small ( $\kappa \ll 1$ ), and we do not take the demagnetisation of the body into account (the weakening of its magnetisation by its own field).

The curve has the same shape as the gravity anomaly above a sphere, so the half-width rule is the same: $z \approx 1.305\,x_{1/2}$ . The amplitude grows with the susceptibility, the field and the cross-section of the body, and it falls as the square of the depth. For a compact body magnetised vertically (a dipole, with a north and a south pole close together) the curve is sharper: $x_{1/2} \approx 0.50\,z$ , that is, $z \approx 2.0\,x_{1/2}$ .

CalculatedCalculated anomalies of the same source from the example below (the pole model, with the map reduced to the pole). The top of the intrusion is at a depth of 4 km (solid line) and at 12 km (dashed line). At three times the depth the anomaly is three times wider and 9 times weaker.

Worked exampleAnomaly of an intrusion on a basement high and in a basement depression

Given (assumed data)

Intrusion in the basement — a vertical body extending to depth: cross-section $A = 6$ km², susceptibility $\kappa = 0.05$ SI units
Field of the Earth: $B_0 = 55{,}000$ nT; the map is reduced to the pole
Top of the intrusion: 4 km beneath a basement high (a raised part of the basement); for comparison, 12 km in a depression (a low part)
Another anomaly on the same map: half-width 6.0 km

Solution

Strength of the pole $C = \kappa B_0 A/(4\pi)$ : $0.05\cdot 55{,}000 = 2750$ nT, multiplied by $6\cdot10^{6}$ m² gives $1.65\cdot10^{10}$ , divided by $4\pi$ : $C = 1.313\cdot10^{9}$ nT·m².
Anomaly above the top at a depth of 4 km: $\Delta T_{max} = C/z^2 = 82$ nT. The half-width is $0.7664\cdot 4 = 3.07$ km.
The same source in a depression, at 12 km: $\Delta T_{max} = 9.1$ nT, which is 9 times weaker. The half-width is 9.20 km, three times wider.
The inverse problem for an anomaly with a half-width of 6.0 km. By the pole model, $z = 6.0/0.7664 = 7.83$ km. If the source is compact (a dipole), $z = 6.0/0.50 = 12.0$ km.

Answer: Above a basement high the intrusion gives an anomaly of about 82 nT, a few kilometres wide. In a depression it gives about 9.1 nT, smeared over tens of kilometres. The source of an anomaly with a half-width of 6 km lies at a depth of 7.83 to 12.0 km, depending on its shape.

What it means. The depth of the basement from magnetic surveying is an estimate with an accuracy of tens of per cent. The factor in the half-width rule depends on the shape of the source, and the shape is unknown. Practical methods give many estimates over the whole area. They include Peters' half-slope method, Euler and Werner deconvolution (computer methods that find the depth of the source from the field), and spectral analysis of maps. In Peters' method the depth is the distance between the points where the steepness of the curve is half its greatest value, divided by 1.2–2.0, most often by 1.6. We combine the estimates into a map and tie it to wells and seismic data. Even for Peters' method the uncertainty of the factor alone is about 30%. A body with a cross-section of 6 km² at a depth of 4 km is no longer “thin”. Its anomaly is wider than that of a pole, and for it the half-width rule slightly overestimates the depth. The data in this example are assumed for illustration.

Seismic surveying: time, velocity and depth

Reflection seismic is the main method of exploring for oil and gas. A source at the surface sends an elastic wave downwards. The source can be the explosion of a small charge in a shallow hole, a vibrator on land, or an air gun at sea (it releases a burst of compressed air under water). At each boundary where the acoustic impedance changes, part of the energy is reflected and returns to the surface. There, seismic receivers record it (geophones on land, hydrophones at sea). The record of one receiver is a seismic trace: the oscillations that have arrived from all the boundaries, in order of arrival time.

The arrival time of a reflection is the two-way time (TWT; in Russian-language practice, the time $t_0$ ). It is the time for the path down to the boundary and back, reduced to the vertical. Traces laid out along a profile form a time section. Its vertical axis is not depth but time. To convert to depth, we need the velocity of the waves in each layer.

Depth from two-way time

z = \sum_{i=1}^{n} \frac{V_i\,t_i}{2}, \qquad t = \sum_{i=1}^{n} t_i = \sum_{i=1}^{n} \frac{2 h_i}{V_i}

where:

$z$ depth of the reflecting boundary, m
$t$ two-way time to it, s
$t_i$ two-way time in the $i$ -th layer, s
$h_i$ vertical thickness of the $i$ -th layer, m
$V_i$ interval velocity of the compressional wave in the $i$ -th layer, m/s
$n$ number of layers above the boundary, —

Units: SI: m, s, m/s

Holds when

The rays are vertical. After processing, the time section is reduced to the case where the source and the receiver are at the same point.
The layers are horizontal, and the velocity in each layer is constant.
The velocities are known: from check-shot surveys and vertical seismic profiling (VSP) in wells (measurements of wave travel time from the surface down the hole), from sonic logs (travel time from the sonic log) or, less accurately, from the seismic data themselves.

For a single layer, $z = Vt/2$ . The ratio of depth to half the two-way time is the average velocity down to the boundary. Times add, so the average velocity is not the plain mean of the layer velocities. Do not confuse the two-way time with the interval transit time $\Delta t$ of the sonic log. The interval transit time is the travel time per unit length, the reciprocal of the velocity.

As long as velocities change only with depth, a time section looks like a depth section that is merely stretched unevenly in the vertical. If the velocity changes along the profile, the picture is distorted. Beneath a fast body (salt, a reef or tight carbonates), reflections arrive earlier and “rise” on a time section (velocity pull-up). Beneath a slow body (a gas accumulation or a thick sequence of young shales), they “sink” (push-down). A false rise beneath salt can be taken for a trap, and a real trap can be missed.

CalculatedThe salt dome of the example below on a depth section (above) and a time section (below), with vertical rays. The top of the subsalt rocks, the reflecting horizon P1, lies horizontally at a depth of 4.35 km. But beneath the dome the wave travels through fast salt, so the reflection arrives 0.37 s earlier. It draws an uplift on the time section that does not exist.

Worked exampleA false uplift beneath a salt dome

Given (assumed data)

Upper layer (everywhere): 0–1350 m, $V = 2250$ m/s
Point A — minibasin between domes (a low area between two salt domes): sediments 1350–3900 m, 3400 m/s; salt 3900–4350 m, 4500 m/s
Point B — above the dome: salt 1350–4350 m, 4500 m/s
Top of the subsalt rocks (horizon P1): 4350 m at both points

Solution

Upper layer: $2\cdot1350/2250 = 1.200$ s.
Point A: sediments $2\cdot2550/3400 = 1.500$ s, salt $2\cdot450/4500 = 0.200$ s. So the time to P1 is $t_A = 2.900$ s.
Point B: salt $2\cdot3000/4500 = 1.333$ s, so $t_B = 2.533$ s. That is 0.367 s less than at A.
Now convert the time at B using the velocities of point A, as if there were no salt. After the 1.200 s of the upper layer, $2.533 - 1.200 = 1.333$ s remains. This is $1.333/2\cdot 3400 \approx 2270$ m of sediments, so P1 would come out at a depth of about 3620 m. That is 730 m above the true depth.
Average velocities to P1: at A $4350/(2.900/2) = 3000$ m/s, at B about 3430 m/s.

Answer: P1 is horizontal in reality, but on the time section it looks like an uplift of 0.367 s. If we ignored the salt, it would “rise” on the depth map too, by about 730 m.

What it means. Such distortions are the main difficulty of seismic surveying in salt-dome basins, including the Pre-Caspian Basin. The thickness of salt above a point changes every kilometre along the profile, and so does the travel time. So we build subsalt structures only from a velocity model. In it, the salt bodies are picked out from seismic data and checked against wells and gravity modelling. How time is converted to depth is covered in the topic Seismic interpretation. The calculation is simplified: the rays are vertical and the velocity in each layer is constant. In reality, waves are bent (refracted) at the steep walls of salt, and the image beneath the dome is distorted still more. The data in this example are assumed for illustration.

Seismic surveying does not see everything. Layers thinner than about a quarter of the wavelength cannot be told apart one by one. At a velocity of 3000 m/s and a dominant frequency of 30 Hz this limit is about 25 m (Seismic waves, acoustic impedance and reflections). 2D profiles give sections along lines, and 3D surveys give a continuous volume. How they are acquired and processed is covered in the topic Seismic acquisition and processing. How horizons and faults are tracked, and how direct hydrocarbon indicators (seismic signs of oil or gas) are sought, is covered in the topic Seismic interpretation. Besides reflected waves, we use refracted waves. They give the velocities and depths of extended boundaries; this is how the boundary between the crust and the mantle was discovered (head waves). We also use them to correct for the uneven near-surface layer. Repeat surveys over a producing accumulation (4D, or time-lapse, seismic) show where some fluids have replaced others.

Electrical surveying and electromagnetic methods

Electrical surveying studies the resistivity of rocks. In direct-current methods, we pass current into the ground through two grounded electrodes and measure the voltage (the potential difference) between two other electrodes. Well logging does the same with electrode sondes, except that there the distances are metres, and here they are hundreds of metres and kilometres (electrode sondes). By moving the electrodes farther apart, we make the current penetrate deeper. This is vertical electrical sounding (VES). To reach depths of kilometres, we need lines many kilometres long, so electromagnetic methods are more convenient for deep tasks.

Electromagnetic methods work with an alternating field. It induces eddy currents in conducting rocks. It decays with depth faster when the frequency is higher and the rocks conduct better. This is the same skin effect as in induction logging (skin depth). The skin depth (how deep the field penetrates) is $\delta \approx 503\sqrt{R/f}$ m, where $R$ is resistivity in Ω·m and $f$ is frequency in hertz. For an average resistivity of the section of 4 Ω·m and a frequency of 1 Hz, it is about 1 km ( $503\cdot\sqrt4 \approx 1006$ m). At 0.01 Hz it is about 10 km. By changing the frequency, or the time after the current is switched off, we “illuminate” the section to different depths.

Magnetotellurics (MT): A passive method. We record at the surface the natural variations of the electric and magnetic fields of the Earth, with periods from fractions of a second to hours. They are generated by currents in the ionosphere and magnetosphere (the upper atmosphere and the space around the Earth) and by distant thunderstorms. From the ratio of the electric component to the magnetic one at each period, we find the apparent resistivity (the resistivity that uniform ground would show), and from it how resistivity changes with depth, from hundreds of metres to tens of kilometres. MT is used where seismic is difficult: beneath salt and basalts, and in thrust zones. Conductive sediments above a high-resistivity basement show up well.
Transient electromagnetic sounding in the near zone (TEM): We send a current into an ungrounded loop of wire at the surface and switch it off abruptly. The eddy currents then go ever deeper, and their field decays. A receiving loop records this decay. Early times tell us about the upper part of the section, and late times tell us about the deep part.
Marine controlled-source electromagnetics (CSEM): A ship tows a powerful electric source above the seabed, and receivers stand on the seabed. A bed with oil or gas is a thin insulator among sediments that hold salty water, and it distorts the field that passes beneath the seabed. Commercial surveys have been run since the early 2000s. But other insulators respond in the same way: salt, tight carbonates and volcanic rocks. So we check the result against seismic data and drilling.

Where surveys are made: on land, at sea, from the air and from space

Land surveys: All methods are possible. This has the highest accuracy: a gravimeter repeats its reading to thousandths of a milligal, and 2D and 3D seismic gives the most detailed image. It is slow and expensive per unit area, and it needs access to the terrain. The accuracy of gravimetry is limited by station elevations and relief.
Marine surveys: Seismic, and gravity and magnetic measurements from the same ship; also marine electrical surveying. Large areas can be covered without problems of access. A gravimeter on a moving ship is less accurate, and we apply the Eötvös correction (for the motion of the ship) to it. Shallow water, ice and shipping make the work harder.
Airborne surveys: Aeromagnetic survey, airborne gravimetry and electromagnetic methods. They quickly cover large and hard-to-reach territories: deserts, swamps, mountains and shallow water. A detailed aeromagnetic survey is flown at a height of 80–150 m along lines 250–500 m apart. A regional one has lines farther apart. Airborne gravimetry is coarser than land gravimetry: the best systems give tenths of a milligal at a resolution of 1–5 km (they show nothing smaller than that), and others give 1–2 mGal. There is no seismic surveying from the air.
Satellite surveys: They measure the gravity and magnetic fields of the whole Earth. Gravity satellites give about 1 mGal at a resolution of the order of 100 km. That shows the regional background and gives a basis for compiled maps, but not individual structures.
Borehole surveys: Check-shot surveys and VSP, borehole gravimetry and logging (Open-hole logging: why, how and when). They give velocities and densities at depth, which we use to calibrate surface surveys.

Ambiguity and the combination of methods

Any geophysical field can be explained in many ways. In gravity surveying this is a strict statement: from the field outside a body, we cannot reconstruct uniquely the distribution of masses inside it. Only the total anomalous mass is determined uniquely (see the advanced section). The size, shape and contrast of the body trade off against each other. A sphere 2.0 km in radius with a contrast of −0.33 g/cm³ and a sphere 2.5 km in radius with a contrast of about −0.17 g/cm³ give practically the same anomaly if their centres are at the same depth. A magnetic anomaly is likewise created by different combinations of magnetisation and shape. Seismic is more accurate in geometry, but it is ambiguous too: time depends on velocity, and beneath salt and in complex structures the image is distorted. Electrical surveying cannot tell a thin conductive layer from a thicker but less conductive one.

How methods are combined on a new area

1
Regional stage
Satellite and airborne gravimetry and aeromagnetic survey show the outlines of the basin, the depth of the basement and thickness of the cover, and large faults and salt massifs. They are cheap per square kilometre, but coarse.
2
Regional seismic
A sparse network of 2D profiles across the basin, tied to deep wells. It shows the stratigraphy, the main reflecting horizons and the first leads (possible traps at an early stage) (Petroleum system and the play concept).
3
Detailed seismic
A dense 2D network or a 3D survey over the leads gives structure maps, faults and trap volumes. A lead turns into a prospect (a trap mapped well enough to drill).
4
Joint model
Densities come from density logs, velocities from sonic logs and VSP, and resistivities from logs. Gravity and magnetic modelling checks the seismic interpretation, especially the shape of salt and the basement relief, which seismic sees worst of all.
5
Drilling and updating
Each well gives better depths, velocities and densities, and we recalculate the models. In development, repeat surveys follow the fluids.

Geophysics in Kazakhstan: from variometers on the Emba to the subsalt fields

The Pre-Caspian Basin is one of the most difficult basins in the world for geophysics. Kungurian salt (salt of the Kungurian stage of the Permian period) is gathered here into more than a thousand domes (salt of the Pre-Caspian Basin). The subsalt rocks lie at depths from 4–5 km at the margins to 10 km in the centre, and the sedimentary cover is more than 20 km thick. The history of geophysics in Kazakhstan is in large part the history of how people looked here first for salt, and then for what lies beneath it.

1925. Gravity surveys began on the salt domes of the Ural–Emba region, using gravity variometers (torsion balances, which measure the horizontal change of gravity) and pendulum instruments. This was the start of geophysical work in the petroleum industry of the USSR. Detailed variometer surveys of 1935–1941 proved very effective for finding salt-dome structures in the Ural–Emba region. By the end of the 1950s gravimetry had covered practically the whole of the northern and north-western Pre-Caspian and had revealed most of the domes there.
From the mid-1930s. In 1935 the first Soviet five-channel seismic stations appeared, and reflection seismic surveying spread quickly, including to the Ural–Emba region. From 1937 the correlation method of refracted waves developed.
1950s. Aeromagnetic surveys at scales from 1:2,000,000 to 1:200,000 covered large territories, including Kazakhstan. In Western Kazakhstan the stripe-like magnetic anomalies proved to be elongated along the structures of the sedimentary cover. By 1957 reflection seismic accounted for more than three quarters of the spending on geophysical work in the petroleum industry of the USSR.
1970s – early 1980s. The first subsalt accumulation was found at the northern margin of the basin in 1967, and after that exploration went under the salt. We trace the top of the subsalt rocks on seismic sections by the reference reflecting horizon P1, and carbonate platforms and reefs show on it as uplifts. The late 1970s and early 1980s were the time when the largest subsalt fields on land were discovered. In 1979 Tengiz (the first well obtained oil from subsalt deposits at a depth of 4045–4095 m) and Karachaganak were discovered.
1990s – 2000. Marine seismic in the Kazakhstan sector of the Northern Caspian revealed subsalt carbonate buildups (reef-like carbonate structures) similar to Tengiz. A well on one of them, Kashagan, produced oil in 2000.
2009–2013. Two sector projects (industry-wide programmes) were carried out: “Comprehensive study of the sedimentary basins of the Republic of Kazakhstan” and “High-precision innovative aeromagnetic survey of the Pre-Caspian Basin”. Seismic, magnetic and gravity data refined the structure of the Devonian sequences in the south of the basin, among others.

Why is combining methods so important here? Beneath salt, seismic sees worst of all. Fast salt bodies of varying thickness distort the image (see the example with the false uplift above), and the resolution at great depths falls. Gravity and magnetic surveys help to outline salt and the basement relief, and wells give velocities and densities for all the models. Reservoir pressures and fluids of the subsalt accumulations are a separate topic (Abnormal formation pressures: origin, prediction and recognition).

Terms and symbols

Gravity anomaly $\Delta g$: The deviation of the measured gravity from the normal value after the corrections. It is measured in milligals.
Milligal (mGal): A unit of acceleration: $10^{-3}$ Gal $= 10^{-5}$ m/s², about a millionth of the acceleration due to gravity.
Bouguer anomaly: The anomaly after corrections for latitude, station elevation, the intermediate layer and terrain. It is the main map of gravity surveying on land.
Half-width $x_{1/2}$: The distance from the extremum of the anomaly to the half-value point. For a sphere, $z \approx 1.305\,x_{1/2}$ .
Limiting depth: The greatest depth at which the top of the body that creates a given anomaly can lie. The half-width rule for a sphere gives an estimate of it.
Regional background and local anomaly: The regional background is the broad part of the field, from deep sources. The local anomaly is the narrow part, from shallow bodies. Separating them is subjective.
Magnetic susceptibility $\kappa$: The ratio of induced magnetisation to the magnetising field. It has no units in SI and is determined mainly by magnetite.
Remanent magnetisation: Magnetisation that was fixed in the rock when it formed. Its ratio to the induced magnetisation is the Koenigsberger ratio $Q$ .
Reduction to the pole: Recalculation of a magnetic map to vertical field and magnetisation, so that the anomalies lie over their sources.
Two-way time $t$: The travel time of a reflected wave to the boundary and back, reduced to the vertical. It is the vertical axis of a time section.
Interval and average velocity: The interval velocity is the velocity in an individual layer. The average velocity is the ratio of depth to half the two-way time to the boundary.
Forward and inverse problem: The forward problem is calculating the field from a known body, and it has one solution. The inverse problem is finding the body from a measured field, and it has many solutions.
Combination of methods: The joint use of several methods and well data in one model of the subsurface.

Common mistakes and misconceptions

Wrong

“The width of the anomaly shows the size of the body.”

In fact

The width shows the depth of the source: a sphere of any radius with its centre at a depth of 3 km gives an anomaly with a half-width of 2.3 km. The size and the density contrast cannot be separated from one anomaly.

Wrong

“There is an uplift beneath the dome on the time section — so there is a trap.”

In fact

Fast salt shortens the travel time, so the horizons beneath it “rise” on the time section. We confirm a structure on a depth section built from a velocity model that has been checked.

Wrong

“Since the gravimeter is accurate to thousandths of a milligal, station elevations can be known roughly.”

In fact

Each metre of elevation is about 0.2 mGal after the correction for the intermediate layer. An elevation error of a few metres is as large as the anomaly of a small structure. We need station elevations to an accuracy of about 10 cm.

Wrong

“A magnetic anomaly over a structure means that the reservoir rocks are magnetised.”

In fact

Sedimentary rocks are weakly magnetised. In basins the anomalies come mainly from the basement, and the structures of the cover often merely repeat its relief.

Wrong

“Gravity and magnetic surveying are obsolete — we have 3D seismic.”

In fact

They are much cheaper per square kilometre, they quickly cover huge areas, and they help where seismic is difficult: beneath salt and basalts, and in thrust zones. We use them together with seismic, not instead of it.

In depth: advanced detail

Total anomalous mass from the area of the anomaly

There is one quantity that the gravity field determines uniquely: the total anomalous mass. If we integrate (add up) the anomaly over the whole horizontal plane, the result is $2\pi GM$ , whatever the shape, size and depth of the body. This follows from Gauss's theorem on the flux of the gravitational field.

Total anomalous mass from the integral of the anomaly

\iint \Delta g\,\dd A = 2\pi G M

where:

$\Delta g$ gravity anomaly, m/s²
$\dd A$ element of area on the plane of observation, m²
$M$ total anomalous mass of all bodies below the plane, kg
$G$ gravitational constant, m³/(kg·s²)

Units: SI

Holds when

The integral is over the whole plane: the anomaly is measured far beyond the body and is completely separated from the regional background.
The observation surface is flat, and all the anomalous masses lie below it.

We find the mass without knowing the shape or the contrast. The weak point is the “tails” of the anomaly. For a point mass at depth $z$ , only a fraction $1 - z/\sqrt{r^2 + z^2}$ of the integral is gathered inside a circle of radius $r$ . For the dome of the example (centre at 3 km), a circle of radius 10 km captures 71% of the integral, and a circle of radius 30 km captures 90%. So the survey for such an estimate must run far beyond the anomaly, and separating the background becomes the main source of error.

Where the factor 0.766 and the constants of the corrections come from

Half-width. The anomaly of a sphere is half the maximum when $z^3/(x_{1/2}^2 + z^2)^{3/2} = 1/2$ . Raise both sides to the power $2/3$ : $x_{1/2}^2 + z^2 = 2^{2/3} z^2$ , so $x_{1/2} = z\sqrt{2^{2/3} - 1} = 0.7664\,z$ . For a long horizontal cylinder the anomaly is proportional to $z/(x^2 + z^2)$ , and it falls to half at $x_{1/2} = z$ . For a thin vertical column the anomaly is proportional to $1/\sqrt{x^2 + z^2}$ , and $x_{1/2} = \sqrt3\,z$ .

Free-air correction. Outside the Earth, $g = GM_\oplus/r^2$ . When we rise by $\dd r$ , gravity changes by $\dd g = -2g\,\dd r/r$ . At $g = 9.81$ m/s² and $r = 6371$ km this is $3.08\cdot10^{-6}$ s⁻², that is, 0.308 mGal per metre. If we allow for the flattening and the rotation of the Earth, we adopt the mean value 0.3086 mGal/m.

Intermediate-layer correction. An infinite flat slab of thickness $h$ attracts with a force $2\pi G\rho h$ that does not depend on the distance to it. At $\rho = 1$ g/cm³ $= 1000$ kg/m³ and $h = 1$ m: $2\pi\cdot 6.6743\cdot10^{-11} = 4.194\cdot10^{-10}$ . Multiplied by 1000, this gives $4.194\cdot10^{-7}$ m/s², that is, 0.04194 mGal. The values 0.0419 and 0.04193 found in compilations come from rounding and from different adopted values of $G$ .

Connections to other topics

Structure and composition of the Earth — the shells of the Earth, seismic waves, isostasy and refraction head waves.
Folds, salt diapirs and their relationship to faults — salt domes and diapirs — the main target of gravity surveying in the Pre-Caspian Basin.
Density log and photoelectric factor (Pe) — rock density in the well and why salt on the log appears lighter than it really is.
Sonic log — the velocity of elastic waves beside the borehole and the travel time for tying seismic data.
Resistivity logs: electrical and induction tools — rock resistivity and the skin effect — the physics of electrical surveying on the scale of the well.
Seismic waves, acoustic impedance and reflections — compressional and shear waves, acoustic impedance, reflections and resolution.
Seismic acquisition and processing — how a seismic survey is run and a time section obtained.
Seismic interpretation — tracking horizons and faults, converting time to depth.
Petroleum system and the play concept — how geophysics leads from a play to a lead and a prospect.

Check yourself

1
Which method will respond most strongly to a salt dome among compacted shales and sandstones at a depth of 2–4 km?
1. A.None: salt does not differ from the host rocks
2. B.Magnetic survey: salt is strongly magnetised
3. C.Gravity survey: salt is lighter than the host rocks
4. D.Electrical survey: salt conducts current well
Show the answer
Correct answer: C. Gravity survey: salt is lighter than the host rocks
At such depths salt is lighter than compacted sediments by 0.2–0.4 g/cm³, so it gives a gravity minimum. Salt is non-magnetic (it is even weakly diamagnetic) and hardly conducts current, so for electrical surveying it is an insulator. On seismic data the dome stands out by the high velocity in the salt.
2
Above a buried body the gravity anomaly of −6.0 mGal falls to −3.0 mGal at 1.9 km from the centre. Treat the body as a sphere. Estimate the depth of the centre and the missing mass. What can be said about the depth of its top?
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The depth of the centre is $z = 1.9/0.7664 \approx 2.48$ km, that is, 2479 m. The mass is $M = \Delta g_{max} z^2/G$ : $6.0\cdot10^{-5}\cdot 2479^2 \approx 368.7$ , and dividing by $6.6743\cdot10^{-11}$ gives about $5.5\cdot10^{12}$ kg of missing mass. The top of a body of any shape with such an anomaly lies no deeper than 2.5 km. How much shallower it lies depends on the shape and the density contrast, and the anomaly does not give us those.
3
A gravity station stands at an elevation of about 25 m, and the density of the intermediate layer is 2.40 g/cm³. How will the Bouguer anomaly at this station change if the elevation is wrongly taken to be 1 m too low?
1. A.It will not change
2. B.It will increase by about 0.31 mGal
3. C.It will decrease by about 0.21 mGal
4. D.It will decrease by about 0.04 mGal
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Correct answer: C. It will decrease by about 0.21 mGal
The slab gives $0.04194\cdot 2.40 \approx 0.101$ mGal per metre, and the elevation enters the Bouguer anomaly with a factor of $0.3086 - 0.101 \approx 0.208$ mGal/m. An elevation that is 1 m too low reduces the sum of the corrections, and so the anomaly, by about 0.21 mGal. The value 0.31 mGal is the free-air correction alone, and 0.04 mGal is the intermediate layer of unit density.
4
The reflection from the top of a bed arrived at a two-way time of 1.84 s. Down to 1300 m the interval velocity is 2600 m/s, and below that it is 3600 m/s. At what depth is the top of the bed, and what is the average velocity down to it?
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In the upper layer the wave spends $2\cdot1300/2600 = 1.000$ s. This leaves 0.84 s of two-way time, that is, 0.42 s one way: $0.42\cdot 3600 = 1512$ m. The top of the bed is at a depth of $1300 + 1512 = 2812$ m. The average velocity is $2812/0.92 \approx 3060$ m/s. This is less than the velocity of the lower layer, because the wave travelled part of the path slowly.
5
On a magnetic map reduced to the pole there are two anomalies of the same shape: the first has a half-width of 2 km, the second 8 km. What is most likely?
1. A.Nothing can be said from the width of the anomalies
2. B.The second source is 4 times larger in size and lies at the same depth
3. C.The second source is more strongly magnetised
4. D.The second source is about 4 times deeper
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Correct answer: D. The second source is about 4 times deeper
For sources of the same shape, the half-width is proportional to the depth. So an anomaly four times wider means four times the depth. If the sources are identical, the second anomaly is also 16 times weaker (the amplitude of a pole anomaly falls as the square of the depth). A large shallow source also gives a wide anomaly, but of a different shape, with a flat top and steep edges.
6
Why can a gravity model of a salt dome, without other data, not tell us how deep we must drill to reach the top of the salt?
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The anomaly gives the depth of the centre for the shape we assume, and it gives the missing mass. It does not give the shape or the density contrast. A sphere of smaller radius with a greater contrast, a sphere of larger radius with a smaller contrast, and a thick stock extending down to the source salt give similar curves. In the example of the lesson, the top of the sphere came out at 0.70 km at a contrast of −0.20 g/cm³ and at 1.09 km at −0.30 g/cm³, and the amplitude rejected the thin-stock model. Seismic sections give the shape, the density log gives the contrast, and a well finally gives the depth of the top.

Updated: 16 October 2026