GRAVITY AND GRAVITY ANOMALIES Newtonian Gravitation

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Berenice Powers
5 years ago
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1 Gravity Exploration

2 GRAVITY AND GRAVITY ANOMALIES Newtonian Gravitation Gravity: force of attraction between objects with mass Consider two objects with mass m 1 and m 2 : m 1 m 2 F g F g distance (r) Newton s theory of gravitation: F = g Gm m r 1 2 G is the universal gravitational constant ( big G ) 2 G = m 3 kg -1 s -2 (or N m 2 kg -2 )

3 Variation in F g with distance For m 1 = m 2 = 1,000,000 kg F = Gm1m 2 r 2 inverse square law

Weighing the Earth ----- Lord Cavendish Measurement of Gravitational Constant/ Mass of the Earth (first performed in 1797-1798 by Lord Cavendish, performed using lead balls of 1.

4 Weighing the Earth Lord Cavendish Measurement of Gravitational Constant/ Mass of the Earth (first performed in by Lord Cavendish, performed using lead balls of 1.6 pounds and 348 pounds, the former are suspended on 6 ft bar and free to rotate, accurate to within 1%) kθ = LF = torque θ = angle of rotation k = rotational stiffness Coef. L = length of wood bar F = gravitational force Why called Weighing the Earth?

5 Consider the Earth (mass M E ) and a small object (mass m) M E F g F g m For a spherical, non-rotating, homogeneous Earth à the force of gravity (outside the Earth) will be the same as if all mass were at the centre. M E F g r F g m Consider mass m. Gravity is a force. This causes an acceleration according to Newton s third law of motion: F g = ma rearranging a = F g m

6 Gravitational Acceleration Acceleration of small object: Newton s Law of Gravitation: Therefore: Note that: GM r m 1 m a g = F = GM a E = = 2 2 r F g m GM E r E 2 = m g GRAVITATIONAL ACCELERATION (units are m/s 2 ) g decreases with distance from the distance from the centre of M E (inverse square law) g does not depend on mass of the small object (m)

the same acceleration (actually, big one beat the small one by 2 inches,

7 g does not depend on mass of object Galileo allegedly dropped masses from the leaning Tower of Pisa in Italy found that a small mass and a large mass fall with the same acceleration (actually, big one beat the small one by 2 inches, according to his daughter) Galileo Galilei ( ) Why does this work? What assumptions?

Galileo Galilei was often credited with the first discovery of gravitation accelaration at the Earth s surface g through a kinematic experiment involving an

8 Galileo Galilei was often credited with the first discovery of gravitation accelaration at the Earth s surface g through a kinematic experiment involving an inclined plane (to slow down the motion so that he could make a measurement, which is not easy to do in the leaning tower experiment)

9 Average g for the Earth Calculate the average gravitational acceleration at the surface of the Earth: Mass (M E ) = x kg (~Venus, 1000% of Mars, a lot more of 2000% of Mercury, 600% of Moon) Avg. radius (r) = 6371 km G = 6.67 x m 3 kg -1 s -2 g = GM r 2 E à g = m/s 2 (near surface)

10 Units for gravity (milligals) average g at the Earth s surface is m/s 2 in gravity exploration, we are interested in tiny variations in g: à convenient to use a smaller unit: the milligal (mgal) 1 mgal = 10-5 m s m s -2 = cm s -2 = Gal (after Galileo) = 981,700 milligals (mgal)

11 Gravity variations over Earth s surface à g is not constant over the surface of the Earth Variations are due to: (A) shape (B) rotation of the Earth (latitude) (C) topography (elevation) (D) heterogeneities within the Earth (local geology)

12 (A) Variations in gravity with latitude The Earth is rotating à causes Earth to become distorted into an oblate spheroid Equatorial radius = 6378 km Polar radius = 6357 km (21 km less) Both the non-spherical shape and rotation affect surface g

1. Shape of the Earth Value of g depends on the distance to the centre of the Earth (inverse square law): g = GM r à since the equatorial radius is greater than the polar radius,

13 1. Shape of the Earth Value of g depends on the distance to the centre of the Earth (inverse square law): g = GM r à since the equatorial radius is greater than the polar radius, gravity should be weaker at the equator. 2 E (M E = x kg) Equatorial radius = 6378 km à g E = 979,540 mgal Polar radius = 6357 km à g P = 986,022 mgal g E < g P g P g E = 6482 mgal

14 2. Variations in mass distribution The Earth has an equatorial bulge à there is more mass between the Earth s surface and its centre at the equator than at the poles This causes an increase in gravity at the equator. g E > g P g P g E = mgal

15 3. Rotation of the Earth a person on the Earth s surface travels around the axis of rotation once a day highest rotation velocity at equator - travel a circle of radius 6378 km every 24 hrs (1670 km/hr) objects in circular motion experience centrifugal force (virtual) - acts in opposite direction to g v 2 a = R _ earth - effect decreases with latitude - no centrifugal force at poles à causes an decrease in gravity at the equator. g E < g P g P g E = 3370 mgal 1,674.4 km/h (465.1 m/s

16 Combining all three effects: shape: g P g E = mgal mass: g P g E = mgal rotation: g P g E = mgal TOTAL: g P g E = mgal à g is less at equator Observations: Gravity at equator = 978,032 mgal Gravity at poles = 983,218 mgal Gravity difference: g P -g E = 5186 mgal

17 GRS67 equation The theoretical variation in gravity with latitude given by the Geodetic Reference System for 1967 (GRS67) equation: g(θ) = * ( sin 2 θ sin 2 2θ) Gravity (m/s2) Latitude (degrees) For CCIS building: latitude = ºN FROM GRS67, g = 981, mgal = m/s 2

18 Observed gravity for Canada (

The Antarctic continent itself is shaded in blue depending on the thickness of the ice sheet (blue shades in steps of 1000 m); light blue is shelf ice; gray

19 The Antarctic continent itself is shaded in blue depending on the thickness of the ice sheet (blue shades in steps of 1000 m); light blue is shelf ice; gray lines are the major ice devides; pink spots are parts of the continent which are not covered by ice; gray areas have no data NOAA/NGDC (Marks, McAdoo & Smith)

20 Gravity variations over Earth s surface à g is not constant over the surface of the Earth Variations are due to: (A) shape and rotation of the Earth (latitude) (B) topography (elevation) (C) heterogeneities within the Earth (local geology)

21 (B) Effect of topography on gravity The Earth is not smooth à lots of topography! g = GM Gravity depends on the distance from Earth s center: 2 Differentiating this equation, you will find that the change in gravity with a change in elevation is: Δg = (mgal) Δh (m) How does gravity change from the ground to the roof of CCIS? Δh = 36 m à Δg = mgal *** NOTE WE ALSO NEED TO CONSIDER EFFECT OF MASS DISTRIBUTION *** r E

22 Gravity variations over Earth s surface à g is not constant over the surface of the Earth Variations are due to: (A) shape (B) rotation of the Earth (latitude) (C) topography (elevation) (D) heterogeneities within the Earth (local geology) à local variations in mass distribution à observable changes in surface g à material property is density

23 Rock density The mass of an object is given by: m = ρ x V where V is its volume ρ is its density (mass per unit volume) The units for density are kg m -3 - sometimes given in g cm -3 (1 g cm -3 = 1000 kg m -3 ) Density of Earth rocks range from ~1000 to >7000 kg m -3

24 Igneous and metamorphic rocks density primarily determined by composition à mafic rocks are generally more dense due to a decreased silica content and an increased amount of heavier elements (Fe and Pb) Granite ρ = g cm -3 Basalt ρ = g cm -3 Gneiss ρ = g cm -3

25 Sedimentary rocks lower density à have pore space filled with low density materials (e.g., air, water, hydrocarbons) composition has a secondary effect on density range of density reflects porosity and degree of weathering density tends to increase with: - increasing depth à compaction (reduces pore space) - increasing age à increased cementation Water ρ = g cm -3 Clay ρ = g cm -3 Shale ρ = g cm -3 Limestone ρ = g cm -3 Dolomite ρ = g cm -3 Sandstone Cretaceous ρ = g cm -3 Triassic ρ = g cm -3 Carboniferous ρ = g cm -3

26 Pure minerals high density because atoms are closely packed together density reflects the composition - higher density if minerals contain a significant fraction of heavy elements, such as Fe and Pb Galena (PbS) ρ = g cm -3 Magnetite (Fe ) ρ = g cm -3 Pyrite (FeS 2 ) ρ = g cm -3 Halite (NaCl) ρ = g cm -3 Note low density of salt!

27 Gravity anomalies Subsurface variations in rock density à small changes in gravitational acceleration (g) at the Earth s surface Gravity exploration - measurements of g taken at different locations in area of interest g at surface (observation) subsurface density (material property) subsurface geology à usually use gravity anomalies the difference between observed g and an average background g

28 Magnitude of gravity anomalies Buried spherical ore body with density ρ (density larger than surrounding rock) Ore body has larger density than rock à expect higher gravity over ore body (maximum at point X) What is the gravity anomaly due to the sphere? à need to calculate how much extra gravity there is due to the excess mass of the sphere (specific gravity)

29 For a sphere: g = GM 2 r where M is the mass of the sphere We want the gravity anomaly (difference in gravity due to sphere) Can rewrite as: Δg = GM z 2 S Δg is gravity anomaly due to the sphere M S is the excess mass of the sphere

30 Total mass of ore body = volume density 4 3 = πr 3 ρ Mass excess of the sphere: M S = mass of ore body mass of rock = 4 π r ρ πr ρ0 3 3 = 4 3 πr 3 ( ρ ρ ) 0 ρ-ρ 0 is the density constrast à difference in density between ore body and rock

31 M S = 4 3 πr 3 ( ρ ρ ) 0 Therefore: Δg = GM z 2 S = 4Gπr 3 ( ρ ρ ) 3z 2 0 Some values: r = 50 m z = 100 m ρ = 5000 kg m -3 The calculated gravity anomaly is: Δg = mgal (remember that 1 mgal = 10-5 m/s 2 ) ρ 0 = 3000 kg m -3

32 Magnitude of gravity anomalies Δ 4Gπr 3 g = 2 ( ρ ρ ) 3z 0 = mgal positive anomaly - gravity stronger due to extra mass of ore body average g for Earth = 981,700 mgal àδg is less than % of g! How does Δg change if ore body - is deeper? - has a larger radius? - has a larger density?

33 Gravity profiles Consider a buried ore body that is more dense than surrounding rock. How does gravitational acceleration vary as you walk across the surface? (gravity profile) à measure gravitational acceleration by dropping a ball at a number of locations across ore body

34 Gravity profiles EXPERIMENT 2: - use same geometry but assume that all densities are reduced by subtracting the rock density

35 Gravity profiles

36 Gravity profiles Shape of the gravity profile and magnitude of the gravity anomaly only depend on the density contrast (ρ-ρ 0 ) à DOES NOT depend on absolute densities Density contrast (variation in subsurface mass distribution relative to some average Earth) Gravity anomaly (difference between observed gravity and expected gravity) à Gravity anomaly profile contains all the information needed to determine the geometry and depth of the ore body (or any other variation in subsurface density)

Note that gravity exploration is different to seismic exploration in the following way:

224B3 Other factors that cause changes in g and need to be corrected Note that gravity exploration is different to seismic exploration in the following way: In a seismic survey, the travel time depends