## How to Calculate and Solve for Total Anomalous Mass | Gravity

The image above represents total anomalous mass.

To compute for total anomalous mass, one essential parameter is needed and this parameter is Number of Area Segments (n).

The formula for calculating total anomalous mass:

ME = 23.9 Σ(ΔgδA)

Where:

ME = Total Anomalous Mass
Δg = Change in Gravity
δA = Area Segment

Let’s solve an example;
Find the total anomalous mass when the number of area of segments is 2 (change of gravity (Δg1) is 5, area segment (δA1) is 12, change of gravity (Δg2) is 8 and area segment (δA2) is 10).

This implies that;

δA1 = Area Segment = 12
Δg1 = Change in Gravity = 5
δA2 = Area Segment = 10
Δg2 = Change in Gravity = 8

Δg δA ΔgδA
5 12 60
8 10 80

Σ(ΔgδA) = 60 + 80
Σ(ΔgδA) = 140

Therefore,

ME = 23.9 Σ(ΔgδA)
ME = 23.9 (140)
ME = 3346

Therefore, the total anomalous mass is 3346 g.

## How to Calculate and Solve for Actual Mass of Geological Body | Gravity

The image above represents actual mass of geological body.

To compute for actual mass of geological body three essential parameters are needed and these parameters are Total Anomalous Mass (ME), Anomalous Density of the Body (ρ1and Host Rock Density (ρ0).

The formula for calculating actual mass of geological body:

M = ME (ρ1 / ρ1 – ρ0)

Where:

M = Actual Mass of Geological Body
ME = Total Anomalous Mass
ρ1 = Anomalous Density of the Body
ρ0 = Host Rock Density

Let’s solve an example;
Find the actual mass of geological body when the total anomalous mass is 12, anomalous density of the body is 21 and the host rock density is 13.

This implies that;

ME = Total Anomalous Mass = 12
ρ1 = Anomalous Density of the Body = 21
ρ0 = Host Rock Density = 13

M = ME (ρ1 / ρ1 – ρ0)
M = 12 (21 / 21 – 13)
M = 12 (21 / 8)
M = 12 (2.625)
M = 31.5

Therefore, the actual mass of geological body is 31.5 kg.

## How to Calculate and Solve for Total Mass Determination | Gravity

The image above represents total mass determination.

To compute for total mass determination, two essential parameters are needed and these parameters are Maximum Gravity (Δgmaxand Half Width (x1/2).

The formula for calculating total mass determination:

M ≃ 255Δgmax(x1/2

Where:

M = Total Mass
Δgmax = Maximum Gravity
x1/2 = Half Width

Let’s solve an example;
Find the total mass determination when the maximum gravity is 10 and the half width is 6.

This implies that;

Δgmax = Maximum Gravity = 10
x1/2 = Half Width = 6

M ≃ 255Δgmax(x1/2
M ≃ 255(10)(6)²
M ≃ 255(10)(36)
M ≃ 91800

Therefore, the total mass determination is 91800 kg.

## How to Calculate and Solve for Geoid Height Anomaly for an Ocean Basin | Gravity

The image above represents genoid height anomaly for an ocean basin.

To compute for genoid height anomaly for an ocean basin, one essential parameter is needed and this parameter is depth (d).

The formula for calculating genoid height anomaly for an ocean basin:

Δh ≃ 3.85d (0.7 – 0.046d)

Where:

Δh = Geoid Height Anomaly for an Ocean Basin
d = Depth

Let’s solve an example;
Find the genoid height anomaly for an ocean basin when the depth is 20.

This implies that;

d = depth = 20

Δh ≃ 3.85d (0.7 – 0.046d)
Δh ≃ 3.85(20) (0.7 – 0.046(20))
Δh ≃ 77 (0.7 – 0.919)
Δh ≃ 77 (-0.219)
Δh ≃ -16.93

Therefore, the genoid height anomaly for an ocean basin is -16.93 m.

## How to Calculate and Solve for Geoid Height Anomaly for Crustal and Mantle Densities | Gravity

The image above represents genoid height anomaly for crustal and mantle densities.

To compute for genoid height anomaly for crustal and mantle densities, one essential parameter is needed and this parameter is height (h1).

The formula for calculating genoid height anomaly for crustal and mantle densities:

Δh ≃ 6h1 (0.7 + 0.066h1)

Where:

Δh = Geoid Height Anomaly for Crustal and Mantle Densities
h1 = Height

Let’s solve an example;
Find the genoid height anomaly for crustal and mantle densities when the height is 12.

This implies that;

h1 = Height = 12

Δh ≃ 6h1 (0.7 + 0.066h1)
Δh ≃ 6(12) (0.7 + 0.066(12))
Δh ≃ (72) (0.7 + (0.792))
Δh ≃ (72) (1.492)
Δh ≃ 107.424

Therefore, the genoid height anomaly for crustal and mantle densities is 107.424 m.

## How to Calculate and Solve for Bouguer Anomaly | Gravity

The image above represents bourger anomaly.

To compute for bourger anomaly, three essential parameters are needed and these parameters are Free Air Correction (gF), Bouguer Correction (δgB) and Terrain Correction (δgT).

The formula for calculating bourger anomaly:

gB = gF – δgB + δgT

Where:

gB = Bouguer Anomaly
gF = Free Air Correction
δgB = Bouguer Correction
δgT = Terrain Correction

Let’s solve an example;
Find the bourger anomaly when the free air correction is 14, the bourger anomaly is 18 and the terrian correction is 20.

This implies that;

gF = Free Air Correction = 14
δgB = Bouguer Correction = 18
δgT = Terrain Correction = 20

gB = gF – δgB + δgT
gB = 14 – 18 + 20
gB = 14 – 38
gB = -24

Therefore, the bourger anomaly is -24 mGal.

## How to Calculate and Solve for Elevation Correction | Gravity

The image above represents elevation correction.

To compute for elevation correction, two essential parameters are needed and these parameters are Free Air Correction (δgFand Bouguer Correction (δgB).

The formula for calculating elevation correction:

δgE = δgF – δgB

Where:

δgE = Elevation Correction
δgF = Free Air Correction
δgB = Bouguer Correction

Let’s solve an example;
Find the elevation correction when the free air correction is 12 and the bourger correction is 17.

This implies that;

δgF = Free Air Correction = 12
δgB = Bouguer Correction = 17

δgE = δgF – δgB
δgE = 12 – 17
δgE = -5

Therefore, the elevation correction is -5 mGal.

Calculating the Free Air Correction when the Elevation Correction and the Bourger Correction is Given.

δgF = δgE + δgB

Where;

δgF = Free Air Correction
δgE = Elevation Correction
δgB = Bouguer Correction

Let’s solve an example;
Find the free air correction when the elevation correction is 34 and the bourger correction is 21.

This implies that;

δgE = Elevation Correction = 34
δgB = Bouguer Correction = 21

δgF = δgE + δgB
δgF = 34 + 21
δgF = 55

Therefore, the free air correction is 55.

## How to Calculate and Solve for Gravity Anomaly for a Semi-Infinite Horizontal Sheet | Gravity

The image above represents gravity anomaly for a semi-infinite horizontal sheet.

To compute for gravity anomaly for a semi-infinite horizontal sheet, five essential parameters are needed and these parameters are Gravitational Constant (G), Anomalous Density (Δρ), Thickness of the Sheet (t), Horizontal Distance (x) and Depth Buried (d).

The formula for calculating gravity anomaly for a semi-infinite horizontal sheet:

δgz = 2GΔρt [π/2 + tan-1(x/d)]

Where:

δgz = Gravity Anomaly for a Semi-Infinite Horizontal Sheet
G = Gravitational Constant
Δρ = Anomalous Density
t = Thickness of the Sheet
x = Horizontal Distance
d = Depth Buried

Let’s solve an example;
Find the gravity anomaly for a semi-infinite horizontal sheet when the gravitational constant is 6.67E-11, the anomalous density is 12, the thickness of the sheet is 14, the horizontal distance is 18 and the depth buried is 5.

This implies that;

G = Gravitational Constant = 6.67E-11
Δρ = Anomalous Density = 12
t = Thickness of the Sheet = 14
x = Horizontal Distance = 18
d = Depth Buried = 5

δgz = 2GΔρt [π/2 + tan-1(x/d)]
δgz = 2(6.67e-11)(12)(14) [π/2 + tan-1(18/5)]
δgz = 2.2411200000000002e-8 [1.570 + tan-1(3.6)]
δgz = 2.2411200000000002e-8 [1.570 + 74.475]
δgz = 2.2411200000000002e-8 [76.046]
δgz = 0.000001704

Therefore, the gravity anomaly for a semi-infinite horizontal sheet is 0.000001704 mGal.

## How to Calculate and Solve for Gravity Anomaly for an Infinitely Long Cylinder | Gravity

The image above represents gravity anomaly for an infinitely long cylinder.

To compute for gravity anomaly for an infinitely long cylinder, five essential parameters are needed and these parameters are Gravitational Constant (G), Anomalous Density (Δρ), Radius of Cylinder (b), Depth Buried (d) and Horizontal Distance (x).

The formula for calculating gravity anomaly for an infinitely long cylinder:

δgx = GΔρ2πb²d / x² + d²

Where:

δgx = Gravity Anomaly for an Infinitely Long Cylinder
G = Gravitational Constant
Δρ = Anomalous Density
d = Depth Buried
x = Horizontal Distance

Let’s solve an example;
Find the gravity anomaly for an infinitely long cylinder when the gravitational constant is 6.67E-11, the anomalous density is 9, the radius of cylinder is 11, the depth buried is 8 and the horizontal distance is 10.

This implies that;

G = Gravitational Constant = 6.67E-11
Δρ = Anomalous Density = 9
b = Radius of Cylinder = 11
d = Depth Buried = 8
x = Horizontal Distance = 10

δgx = GΔρ2πb²d / x² + d²
δgx = 6.67e-11(9)2π(11)²(8) / (10)² + (8)²
δgx = 6.67e-11(9)2π(121)(8) / (100) + (64)
δgx = 6.67e-11(9)(6082.123) / 164
δgx = 0.000003651098 / 164
δgx = 2.226e-8

Therefore, the gravity anomaly for an infinitely long cylinder is 2.226e-8 mGal.

## How to Calculate and Solve for Free Air Anomaly | Gravity

The image above represents free air anomaly.

To compute for free air anomaly, four essential parameters are needed and these parameters are Observed or Measured Gravity Value (gobs), Reference Value of Gravity (g(λ)), Elevation (h) and Radius of the Earth (R).

The formula for calculating free air anomaly:

δgF = gobs – g(λ) [1 – 2h/R]

Where:
δgF = Free Air Anomaly
gobs = Observed or Measured Gravity Value
g(λ) = Reference Value of Gravity
h = Elevation
R = Radius of the Earth

Let’s solve an example;
Find the free air anomaly when the observed or measured gravity value is 9, reference value of gravity is 10, elevation is 16 and radius of the earth is 4.

This implies that;

gobs = Observed or Measured Gravity Value = 9
g(λ) = Reference Value of Gravity = 10
h = Elevation = 16
R = Radius of the Earth = 4

δgF = gobs – g(λ) [1 – 2h/R]
δgF = 9 – 10 [1 – 2(16)/4]
δgF = 9 – 10 [1 – 32/4]
δgF = 9 – 10 [1 – 8]
δgF = 9 – 10 [-7]
δgF = 9 – -70
δgF = 79

Therefore, the free air anomaly is 79 Gal.