GRAVITY EXPLORATION. subsurface density. (material property) Gravity anomalies of some simple shapes

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1 GRAVITY EXPLORATION g at surface (observation) subsurface density (material property) subsurface geology Gravity anomalies of some simple shapes Reminder: we are working with values about mgal g!!! N

2 Applications of Gravity Surveys Exploration of fossil fuels (oil, gas, coal) Exploration of bulk mineral deposit (mineral, sand, gravel) Exploration of underground water supplies Engineering/construction site investigation Cavity detection Glaciology Regional and global tectonics Geology, volcanology Shape of the Earth, isostasy Military (e.g., detection of underground bunkers)

3 Gravity and gravity anomalies Gravity forward problems: Density contrast (variation in subsurface mass distribution relative to some average Earth) Gravity anomaly (difference between observed gravity and expected gravity) We computed the gravity anomaly of a spherical ore body at X (immediately on top of the ore body). à How does the gravity anomaly vary laterally?

4 Calculating gravity anomalies

5 Total gravity g + Δg Vertical gravity g + Δg Instruments can only measure vertical component of gravity. Also, Δg is usually small θ is small g+δg θ g g Δg x is small à consider only anomalies in vertical gravity Δg Δg Δg x Δg (for exploration studies for tectonic studies, the horiontal component can be significant, e.g., in mountainous areas)

6 Vertical gravity anomaly for a point mass For a point mass at distance r: Δg = GΔm h 2 Δm is mass difference w.r.t. surrounding material Vertical component is: Δg GΔm = cosθ = 2 h GΔm h 3 Note: cosθ = Δg Δg = h

7 Gravity profile for a point mass

8 Gravity anomaly of any body divide body into point masses at each position calculate gravity anomaly due to each point mass Δg = GΔm 3 h add up all anomalies Δg 1 Δg 2 Δg 3 Δg 4 Mathematically (in 3D): Where: ( ' ) Δg = GΔρ 3 h h = ( x' x) 2 + ( y' y) 2 + ( ' ) 2 dx'dy'd' & (x, y, ) is the observation point

9 Gravity anomaly of a buried sphere Sphere is more dense than surrounding material therefore has a mass excess (M S ) Can show that the vertical component of gravity will vary as: g (x) = GM S 3 ( x ) 2 (just like for a point mass) NOTE: x is the horiontal distance from the center of sphere to the point of measurement

10 g (x) = SOME VALUES = 100 m r = 50 m Δρ =2000 kg m -3 M S ~ 10 9 kg GM S 3 ( x ) 2 Gravity anomaly, g (mgal) Distance (m) Where: M S = 4 3 πr 3 ( Δρ) g (x) =

11 Gravity anomaly of a buried sphere maximum g at point above the centre of the sphere (x=0 m) g decreases rapidly away from the sphere maximum value of g is: GM max S g = Distance (m) 2 Gravity anomaly, g (mgal)

12 Gravity anomaly of a buried sphere maximum g at point above the centre of the sphere (x=0 m) g max g decreases rapidly away from the sphere maximum value of g is: GM max S g = Distance (m) 2 half-width of the curve is where g = g max / 2 can show that x 1/2 =0.766 equivalently: = x 1/2 Gravity anomaly, g (mgal) X 1/2

13 Gravity anomaly of a buried sphere Forward problem: Given, r, and Δρ can calculate g (x) Inverse problem: Observe g (x) à find g max, x 1/2 can now determine depth: = x 1/2 Gravity anomaly, g (mgal) X 1/2 g max Distance (m) and M S : M S = g max G 2

14 Changing the depth of the sphere max g = GM 2 S x 1/2 = Deeper sphere à broader, smaller amplitude gravity anomaly (Lowrie, 2007)

15 Half-width technique (half maximum) The width (wavelength) of the anomaly is related to the depth of the body that causes it. Deeper body = wider anomaly à measure the half-width (x 1/2 ) of an anomaly to get the approximate depth of the body. 2D: cylinder = x 1/2 3D: sphere = x 1/2 where is the center. - for other shapes, represents the maximum depth of the top of the body (Lowrie, 2007)

16 Point Mass From ETH

17 Buried horiontal cylinder Cylinder with radius a is at depth. The density contrast between the cylinder and surrounding material is Δρ. Can show that the vertical component of gravity will vary as: g (x) = 2Gπr2 Δρ x ( )

18 g (x) = 2Gπr2 Δρ x ( ) SOME VALUES axis at x = 0 m = 100 m r = 50 m Δρ =2000 kg m -3 Gravity anomaly, g (mgal) Distance (m) (same values that were used for the sphere)

19 Important observations maximum g is at the axis of the cylinder (x=0 m) maximum value of g is: g max = can show that half-width is related to depth of the centre of the cylinder: x 1/2 = 2Gπr 2 Δρ Gravity anomaly, g (mgal) Distance (m)

20 g max = 2Gπr 2 Δρ Half width: x 1/2 = Changing the depth Changing the radius (see textbook for the parameter values used to generate these plots)

21 Compare the gravity anomaly profiles: 0.8 SPHERE: 2.2 CYLINDER: Gravity anomaly, g (mgal) Gravity anomaly, g (mgal) Distance (m) Distance (m) why is g max larger for the cylinder? if you recorded these profiles in the field, how would you decide if the anomaly was due to a sphere or a cylinder?

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23 Simple code to try on matlab % a comparison between cynlinder and sphere x = linspace(-250,250,501); pi= ; G=6.67e-11; =100; r=50; Vs=4*pi*r*r*r/3.0; drho = 2000; g_sph=g*vs**drho./(*+x.*x).^1.5*100000; g_cyl=2*pi*g*r*r*drho*./(*+x.*x)*100000; g_sph_scaled=g_sph.*2.0953/0.7; plot(x, g_sph) hold on plot(x, g_cyl, 'r') hold on plot(x, g_sph_scaled, 'g') xlabel('x (m)') ylabel('gravity anomaly g (mgal)') legend('sphere', 'cylinder', 'sphere scaled');

24 Gravity profiles vs. maps SPHERE CYLINDER PROFILE Gravity anomaly, g (mgal) Gravity anomaly, g (mgal) Distance (m) Distance (m) MAP Y distance (m) Y distance (m) X distance (m) X distance (m) 0

25 Example Problem Buried spherical iron ore body: 2 Centre of ore body: x = m Gravity anomaly, g (mgal) Distance (m) Is it possible to determine the radius or density contrast of the sphere? Maximum value of g : g max = mgal Half-width: x 1/2 = m Depth of sphere: = m Excess mass of sphere: M S = kg (G = 6.67 x m 3 kg -1 s -2 )

26 Infinite horiontal layer Can show that the gravitational attraction of this slab at P is: g = 2πGρΔ à g does not depend on the x-position of the measurement or on the depth of the layer -- WHY?

27 Infinite horiontal layer à g does not depend on the x-position of the measurement or the depth of the layer WHY? g = 2πGρΔ

28 Infinite Horiontal Layer Δ Can also think about this in terms of gravity anomaly caused by a density contrast: Δ g = 2πGΔρ Δ where Δρ=ρ 0 -ρ

29 Compare the gravity response of these two layers: Gravity value is: g = 2πGρΔ Both slabs have the same g can not tell them apart using only gravity data à non-uniqueness

30 Horiontal layer cut by a fault on the left

31 Computer modeling of gravity anomalies for more complex geometries, gravity anomalies can be calculated using computer modeling programs models can be 2D or 3D Δg 1 Δg 2 Δg 3 Δg 4 for 2D: polygon is assumed to extend infinitely in the strike direction (in and out of page)

32 Verification of modeling program before using any program, must verify that the calculations are accurate 8-sided cylinder HOW? construct a model where the answer is known to see if the program gives you the right answer for example buried horiontal cylinder Why is model g less than expected? line = model result dots = expected answer

33 Verification of modeling program 16-sided cylinder line = model result dots = expected answer

34 Example: A sedimentary basin Density contrast = -400 kg/m 3 Infinite layer

35 BASIN 1 BASIN 2 Δρ = -400 kg/m 3 Δρ = -200 kg/m 3 minimum g is the nearly the same for both basins edge effects extend further into Basin 2

36 Forward and Inverse Problems FORWARD PROBLEM: Known geological structure Predicted geophysical data - used to evaluate simple structures and to design surveys INVERSE PROBLEM: Inferred geological structure Observed geophysical data - most common, remember that structure can be non-unique