2. Gravity, the shape of the Earth, isostasy, moment of inertia
Gravity is one of the four fundamental forces (the others are the electromagnetic, the
weak force and the strong force).
Kepler’s Laws
1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one
focus.
2. The Law of Areas: A line that connects a planet to the sun sweeps out equal
areas in equal times.
3. The Law of Periods: The square of the period of any planet is proportional to
the cube of the semi major axis of its orbit.
Basic Relations
Newton's Law of Gravitation:- F = G m1 m2 / r2 Newtons
F = force acting between two point masses
m1, m2 = the masses
r = separation of the two masses
G = Universal gravitational constant = 6.67 x 10-11 Nm2kg-2
Newton also found that on Earth: F=mg
Combining gives:
g = G.M / R2
Numerically, on Earth:
g is the acceleration due to gravity.
m.g= G.m.M/ R2
g ~9.81 ms-2 .
Geologically, density important. If ρ is the average density of the Earth, then
ρ = mass/volume = M / [(4/3)πR3] = 3M / 4πR3
We can substitute for M using the relationship between it and g, i.e. M = R2g / G.
Therefore:
ρ = 3g / 4πRG
Thus if we know g, R and G, we can calculate ρ. With current values:ρ = 5.52 x 103 kgm-3
Since most surface rocks have densities in the range 2-3 x 103kg.m-3, density must
increase with depth in Earth.
How do we Measure Gravity?
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Pendulum
Mass Dropping
Gravimeter
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Pendulum
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Period of swing = T
T  2 l / g
g  4 2l / T 2

Mass Dropping
Distance traveled in time t’ = L
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L  gt 2 / 2
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

g  2 L / t 2
Both absolute measurements
Both methods independent of mass, m.
Both methods should give g ~ 9.81 ms-2
Gravimeters
In field use spring-balance gravity meters estimate changes in g.
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g on a mass is balanced by a force exerted on spring
m.g = k.x
x = stretch in spring, k is spring compliance
Linking such measurements to places where absolute values are
known allows us to determine absolute values with gravimeters.
Gravity and the Shape of the Earth
The variation of g over the surface >> shape and internal structure of the Earth.
If we rearrange F = G mM/ R2 substituting for M via ρ = mass/volume = M / [(4/3)πR3] =
3M / 4πR3, so M = 4πR3 ρ/3 we obtain:
g = 4π ρ RG / 3
If the Earth were a perfect sphere of uniform density, g constant
If the Earth deviates from spherical (i.e. if R varies or if there is a local density anomaly,
g will vary.
The Earth is not spherical, but an ellipsoid of revolution i.e. it is flattened at the Poles
- this is a rotational effect. NB figure wrong: flattened sphere should have shorter polar R
Satellite studies have provided a very accurate measure of ellipticity:equatorial radius = 6378 km.
polar radius = 6356.6 km.
Flattening = (6378 - 6356.6 )/ 6378 = 1 / 298.26
Since g = GM / R2, g will be larger where R is smaller. Therefore g at the poles is larger
than g at the Equator.
g is also affected by the fact that the Earth rotates and an observer on its surface
therefore experiences a centrifugal force.
We can summarise by saying:
(1) If the Earth were a non-rotating perfect sphere, g would be constant.
(2) Because of rotation, the Earth is flattened at poles. This affects g in two ways:(a) g at the poles > g at the equator because R at the poles < R at the equator.
(b) rotational force at the Earth surface is at right angles to the axis of rotation
and proportional to the distance from that axis. It is therefore zero at the poles
and a maximum at the Equator. It acts outwards, reducing g.
The net g force at the surface is equal to the resultant of the forces due to internal mass
and the centrifugal action. If the gravitational force due to M is a = GM / R2 and the
centrifugal force is c, then the total effective gravitational force is b the vector sum of a
and c. However, b does not act towards the centre of Earth, but at right angles to the
surface of the elliptical Earth:
Thus a perfectly homogeneous plastic body will deform until the combination of a and c
meets this criterion. In mathematical jargon, the surface of the ellipsoid is an
equipotential surface.
The ideal (ellipsoidal) mean sea level surface is called the Earth ellipsoid or (Earth)
spheroid. The gravitational force over the spheroid varies, with a maximum at the poles
(where c = 0) and a minimum at the equator (where c is a maximum).
The gravitational acceleration on the surface of the spheroid is given by the
International Gravity Formula (IGF).
g = 9.780318 (1 + a sin2(λ) - b sin2 (2λ))
where g = sea-level gravitational acceleration on the spheroid and λ = latitude, and a =
0.0053024 and b = 0.00000587
g at Equator (lat = 0) = 9.780318 m.s-2 g at Pole (lat = 90) = 9.832177 m.s-2
The difference amounts to approximately a half of one percent. The value of the
theoretical or normal gravity varies smoothly between the two extremes, but
inhomogeneities in the Earth produce shorter wavelength perturbations in the smooth
curve.
Variation of gravity with height
At Q (h=0, ie surface) >> g = GM/r2
At P (h=h) >> g = GM/(r+h)2
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g decreases with height
gradient
= dg/dh = -2g/R
= 3.086x10-6 ms-2/m
= 3.086 gu/m
Units of gravitational acceleration
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1 gu = 10-6ms-2
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1 mgal = 10-5 ms-2
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10 gu = 1 mgal
Variation of gravity through the Earth
At surface g = GM/r2
In the Interior, at some radius, r:
g(r) = gMI + gMO
gMI = GMI/r2
gMO = 0
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g(r) = GMI/r2
What is MI?
MI = (4/3)πr3ρ
 g(r) = (4/3)πrρG
• gravity is zero at Earth’s centre
In reality the depth curve varies from simple equation because core is much denser
than mantle.
Effect of Inhomogeneities: the Geoid
Large-scale inhomogeneities produce departures of the measured values of g at sealevel from those predicted by the I.G.F.
g does not vary smoothly from equator to pole >> lateral inhomogeneities in the Earth.
The real sea level equipotential surface is known as the geoid and has "highs" and
"lows" relative to the spheroid.
Values of g can be determined by surface measurements and by satellite studies.
Sea-level is +54 metres higher in the North Atlantic than predicted by the IGF spheroid,
and the maximum departure is -94 m, over India.
+ve features = active magmatic regions:- e.g. Mid-Atlantic Ridge,The Andes, The Philippines
-ve features = old, inactive ocean basins and continents:- e.g. Antarctica, Canada, Siberia, India
Major physical undulations (e.g. mountains) are NOT associated with geoid anomalies,
and so must be balanced by deeper seated mass excesses or deficiencies (Isostasy).
Possible that long wavelength undulations in the geoid reflect the convective system in
the mantle, or some other deep phenomena (undulations on surface of the core).
Crustal Gravity Anomalies
In addition to the global anomalies due to convection, there are smaller scale effects
because of crustal inhomogeneities (sedimentary basins, intrusions, etc.).
Their analysis is important in exploration for natural resources, but they also need to be
taken into account in global scale investigations and surveys
What causes gravity anomalies?
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If ρ1 ≠ ρ2 then there is a local mass excess or mass deficiency in the vicinity of
the geological body causing a local very small variation in the value of g
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positive anomaly if ρ2 > ρ1
negative anomaly if ρ2 < ρ1
So a local survey might give:
Can fit to background, and get the difference to show up ore body:
Gravity Corrections
The measured value of gravity at a field station might vary from the value at the base
station for a variety of reasons, even if there were no crustal or geoid anomalies.
Once the value has been obtained it must be corrected to account for effects such as:(1) Latitude differences
(2) Elevation effects
(3) Topographic effects
Latitude Correction
We know g varies from pole to equator
Thus if stations are at different latitudes, we would expect g to be different.
For small N-S distances (up to a few km) the difference in g latitude λ is approximately:ΔgLAT = 8.1 sin(2 λ) g.u. per km
Free-air Correction
If stations are at different elevations, we would expect g to be different because of the
different distances to the centre of the Earth.
The effect for a positive height (h) above sea-level is approximately equal to -3.086
g.u./metre, an increase in height produces a decrease in gravity.
So gground-level > ghighup
ΔgELEV = -3.086h g.u.
The correction, known as the free-air correction (because material between the station
and the reference surface is air), must therefore be positive. So Δg is added to ghigh-up to
bring it into line with the reference gground-level.
Gravity corrections require accurate elevations, and getting these is often the most
expensive part of a gravity survey.
Bouguer Correction
In reality, gravity station will be on rockwhich exerts a +ve (downwards) gravitational pull.
The Bouguer correction adds the effect of rock to the free-air correction.
Assume g of real topography = g uniform flat plate, thickness h, extending to infinity (fig).
ΔgROCK = 2π G ρ h = 41.91 x 10-5ρ h g.u.
The effect is positive (ie it increases the gravity field) and therefore the correction for the
presence of rock must be negative.
For granite ρ is approximately 2670 kg m-3, and this has been adopted as a “standard”
density for the upper crust, giving a correction of -1.118 g.u./metre.
The free-air correction is 3.086 g.u./metre, but when the effect of intervening rock is
considered the net correction is reduced, so that the net elevation correction, the
Bouguer Correction, is about 1.968 g.u./meter.
So in this case, again, gground-level > ghigh-up, but much less than in the free-air case due to
the intervening rock.
Terrain Correction
If the station is next to a mountain or valley, topography matters:
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A mountain will attract upwards, and so reduce the value of gravity measured.
______g1______________________________________________g2_∧
g1 > g2, so Δg is added to g2 to bring it into line with the reference g1
A valley will not attract as much as it should if it were filled with rock and so will also give
rise to a gravity value which is smaller than would be expected.
______g1______________________________________________g2_
________
g1 > g2, so Δg is added to g2 to bring it into line with the reference g1
Thus terrain correction must be positive to give corrected gravity differences.
Once all the corrections have been made, the reduced gravity records variations in
gravity field due solely to subsurface density variations (see web notes for examples).
If only the latitude and free-air corrections have been applied, the quantity calculated is
known as the free-air gravity (free-air anomaly).
If, in addition, the Bouguer correction has been applied, the quantity is known as the
(simple) Bouger gravity (or anomaly).
If, in addition, terrain corrections have been made, the quantity is known as the
extended Bouger gravity or complete Bouger gravity (or anomaly).
Isostasy
Early gravity measurements showed that the mountains did not deflect a plumb bob as
much as expected from their exposed mass.
Mountains have low-density roots beneath them which supply buoyancy that supports
the additional mass exposed above mean sea level; that is, variations in surface
elevation are hydrostatically supported.
This is the principle of isostasy: above some depth in the Earth (called the level of
compensation), all columns of rock exert the same pressure. The level of compensation
is the depth below which hydrostatic pressure in the Earth is independent of location
(latitude and longitude)
Isostasy applies on a broad scale – mountain ranges, mid-ocean ridges. The basic idea
is that of flotation. Large-scale gravity anomalies indicate that the lithosphere is
hydrostatically supported, i.e., the rock column “floats” above the level of compensation.
Consequently large scale gravity anomalies reflect the structures of the lithosphere.
NB: Some areas of the Earth, though, are not in isostatic balance (see later).
A gravity survey across a mountain range will show a negative Bouguer gravity, because
mountains have low density roots. This isostatic balance is responsible for their
elevation.
Examples of Bouger gravity profiles:
a) density of sedimentary rock < basement >> gravity low
b) density of granite less than rock >> gravity low
c) density of ore > rock >> gravity high
The mountain range and mid-ocean ridge are in isostatic equilibrium, so the freeair gravity profiles are virtually flat. The Bouger profiles show gravity lows due to
the low-density mountain roots in (a) and from the low density magma chamber
in (b) (of the order of -300 g.u. and -1000 g.u respectively).
The reality of isostasy is confirmed by the measurable uplift of Fennoscandia during the
last two hundred years as a consequence of the unloading accompanying the melting of
the ice sheets. Below the figure shows a) the crust in isostatic equilibrium before the iceage, b) loading of the crust by an ice-cap, and c) the rebounding crust after the ice has
melted. In b) the crust sags, forming a root that supports the ice cap; the mantle material
flows away from the depression. The root causes a negative Bouger anomaly. When the
ice melts the crust starts to rebound and the mantle material flows back into the region.
The viscosity of the mantle is the controlling factor in the rate of rebound. Mantle
viscosities can be estimated from rates of glacial rebound
But not all structures are in isostatic equilibrium. The Hawaiian chain is such an
example where the free-air gravity map shows highs in line with the topography,
which would not be expected from an isostatically compensated structure.
Free air gravity map of the northern Pacific showing Emperor-Hawaii seamount chain
The sketch below shows the free-air and Bouguer anomaly associated with
Hawaii both as derived from the observations, and also what the gravity
observation would show if Hawaii were in isostatic equilibrium.
It is the gravity observation that tells us the coarse sub-surface structure of the
shallow Earth.
Pratt and Airy Hypotheses
Gravity observations cannot tell us what the structures are like within the Earth, only
whether or not there is a density excess or defecit. Two hypotheses exist which attempt
to explain the gravity observations. That of Airy (below left) assumes that the rigid upper
layer has a constant density that is lower than the substratum beneath. The mountain
“floats” with deep roots like an iceberg. Conversely, Pratt (below right) assumes that the
base of the upper layer is level, and it is the density within the upper layer that changes.
Determining which of these hypotheses operates in the Earth is far from straightforward;
however, in combination with seismology, detailed structures can be observed and
understood.
Moment of Inertia
Circular motion
Just as linear velocity is distance travelled per unit time, so too angular velocity of a
rotating body is the angle rotated per unit time. The angular velocity, ω, of a body
rotating in a circular of radius r, with linear speed v, is: ω = v/r
where ω must be in radians per unit time, normally, radians/s or rads/s.
Centripetal acceleration
Velocity is a vector and therefore is defined by both magnitude (speed) and direction. A
rotating body is therefore changing its velocity continuously as it is always changing
direction. It therefore has an acceleration. A body rotation on a circle of radius r, with
linear speed v is being continuously accelerated towards the centre of the circle with a
magnitude: a = r ω2 = v ω = v2/r where a is in units of m/s2
Mass of the Earth from an orbiting satellite (e.g., the Moon)
From Newton’s Law, a body moving with angular velocity, ω, in circular orbit of radius R
about the Earth will have a centripetal acceleration towards the Earth: F = ma = mR ω2
The force provided by the gravitational attraction between the Earth and the satellite is:
F = GMm/R2 Equating the two gives: M=R3 ω2/G
Moment of Inertia
The mechanics of the Earth’s rotation avout its axis introduces a quantity called the
moment of inertia, which is the rotational mechanical analogue of mass. The moment
of inertia of a point mass, m, rotating at a distance, r, about an axis is mr 2. The moment
of inertia of a body rotating about an axis is the sum of all the point contributions of the
moments of inertia of the single point masses, mi, within the body, each at a distance ri
from the rotation axis: I = ∑miri2
The value for the moment of inertia therefore depends on the mass distribution within the
body. eg, a bicycle wheel with all the mass concentrated on the rim would have a I= mr2;
if the mass was all in the axle, I=0; if the mass was evenly distributed, I = 0.5mr2.
In general, the moment of inertia is given by: I = kmr2
Where k is a dimensionless constant that is object/material dependent.
k = 1 a ring or thin walled cylinder rotating about its centre
k = 0.4 a solid sphere rotating about its centre
k = 0.5 a solid cylinder or disk rotating about its centre
For a sphere made up of homogeneous layers, the moment of inertia can be determined
additively; for example, an Earth of radius R with a metal core of radius r and a silicate
mantle: IE = IM(R) – IM(r) + IC(r)