SEG_Class11_31.08.15

Solid Earth Geophysics(ES655)(ES655)

Earth’s Gravity and Isostasy -- 1

Dr. Animesh MandalAssistant Professor, Department of Earth Sciences, p

Indian Institute of Technology Kanpur

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What is gravity?• Gravity or gravitation is the force of attraction between two

masses– Including stars, planets, galaxies and even light and sub-atomic particles

• Gravity has played a big part in making the universe the way itGravity has played a big part in making the universe the way it is – Gravity is what makes pieces of matter clump together into planets, moons, and y p p g p , ,

stars

– Gravity is what makes the planets orbit the stars--like Earth orbits our star, the SunSun

– Gravity is what makes the stars clump together in huge, swirling galaxies

Nobody truly knows what gravity is or cause of gravity 

.........thanks to scientists throughout the history we know

2

.........thanks to scientists throughout the history we know 

how it behaves and it’s effects.....

Galileo Galilei (1564 1642):History of gravity

Galileo Galilei (1564 – 1642):• Realized that gravity causes bodies to fall towards Earth

– All bodies fall at the same rate– He told us how gravity works, but not quantified

Sir Isaac Newtown (1643 – 1727):U i l l f it ti• Universal law of gravitation

Pierre Bouguer (1698 1758):

forceforce forceforcerrmm11 mm22

221

rmmGF

Pierre Bouguer (1698 – 1758):• Then these experimental and mathematical analyses were applied to

several areas of geophysics:

r

– Earth’s shape, density, and– Gravity field variation with latitude, elevation, attraction of mountain

• Gravity was the first geophysical technique used in oil and gas• Gravity was the first geophysical technique used in oil and gas exploration Torsion balance by Lorand Eötvös

Albert Einstein (1879 – 1955):• Gravity is what happens when space itself

is curved or warped around a mass 3

Gravity-Gravimetry• The gravitational force depends on the rocks below us

• Small changes in gravitational forces inferences about the rocks below us• This is the essence of gravity surveying

• Gravimetry: Study (measurement) of spatial as well as temporal variation of Earth’s gravitational field and their relationship to distribution of mass within EarthEarth s gravitational field and their relationship to distribution of mass within Earth

• Closely related to Geodesy: Shape and dimensions of the Earth

•    Applications of gravimetry– Internal structure of the Earth (from surface to core)

– Exploration for ore, oil, water

– Isostasy and mechanical properties of the lithosphere

– Earth tides

– Transfer of geophysical fluids between reservoirs: water, magma, ice => temporal variations of gravityg , p g y

– Artificial satellites: orbitography

– Planetary gravimetry4/15

Gravitation• Newton’s second law:

– Force (in Newtons) acting on mass m (in kg), responsible for its acceleration a (in m s-2)

amF

its acceleration a (in m.s-2)

• Newton’s law of gravitation:– Two masses m and m attract each other

F Frm1 m2

– Two masses m1 and m2 attract each other

– This attraction results in a force:

– G = 6.673 x 10-11 m3.kg-1.s-2

rrmmGF ˆ2

21

• In the absence of any other force besides the one generated by M, one can write

rmGarmmGam ˆˆ 22

221

1

mrr

Garr

Gam 221 r

rmGa ˆ2

1'

• a = gravitational acceleration of mass m1 due to the attraction of mass m2 5/15

Gravitational potential• Gravitational potential (V), due to a point mass m, at a distance r from m,

is the work done by the gravitational force in moving a unit mass from infinity to a iti fposition r from m.

– Potential energy of mass m is mV

• Change in gravitational potential energy mdVrdF

.Change in gravitational potential energy mdVrdF.

drdVadVadrmdVrdam

. 22,rdrGmdV

drdV

rmGor

drVar

drdVa

ˆrmGV VmFVa

Gravitational acceleration is equal to the rate of change in the potential field

• The gravitational potential energy of a mass m2 at distance r from mass m1 is −Gm1m2/r

• If mass m2 moves away from mass m1 to a new position, r’, the gravitational energy that is released is

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Gravitational potential• If, instead of just one mass m1, we imagine a distribution of

masses,

• We can now use the last integral form to calculate the potential of g pa spherical shell

Spherical shell ofSpherical shell of mass M, mass density ρ, inner y ρ,radius b & outer radius a;Find potential at P at a distance R from centerfrom center

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Gravitational potential• Worked on the Board!• Outside the shell, R > a,

– V = -(GM)/R ------------------------(1)

– M = (4π)ρ(a3 - b3)

The same as if M were a point mass at the– The same as if M were a point mass at the origin!

• Completely inside shell, R < b, V 2 G( 2 b2) (2)– V = -2πρG(a2 - b2) ------------------------ (2)

– V = constant, independent of position

• Within the shell, b R a, , ,– V = -4πρG[a2- (b3/R) - R2] ---------(3)

If R  a, (1) & (3) are the same!

If R  b, (2) & (3) are the same!  8/15

Gravitational potential• These results are very important• For, R > a, V = -(GM)/RFor, R a, V (GM)/R

– The gravitational potential at any point outside a spherically symmetric mass distribution is independent of the size of the distribution, & is the same as th t f it t b t t d t th t f th di t ib ti ithat for its mass to be concentrated at the centre of the mass distribution, i.e., equivalent point mass

• Shell or solid; a solid is composed of infinitesimally thick shells!

• For, R < b, V = -2πρG(a2 - b2)– The potential is constant anywhere inside a spherical shellThe potential is constant anywhere inside a spherical shell

• Given the results for the potential V, we can compute the acceleration due to gravity or gravitational field asacceleration due to gravity or gravitational field as

F h i l h ll lid b d V d d R l

VardrdVa

ˆ

• For spherical shell or solid body V depends on R only acceleration is radially directed 9/15

Gravitational potential

• a = a er = - (dV/dR)er

O id h h ll R• Outside the shell, R > a, – a = - (GM)/R2, [M (4π)ρ(a3 -

b3)]b3)]

– The same as if M were a point mass at the origin! V = constant

• Completely inside shell, R < b,

V = constant

‐V

V ∞ 1/R V = ‐(GM)/R

– a = 0

– Since V = constant, independent of position

• Within the shell, b R a, ‐a

a=0

a ∞ 1/R2 a = ‐ (GM)/R2

– a = (4π)ρG[(b3/R2) - R]a = 0 10/15

Equipotential Surface• On which the potential is constant

• For a sphere or spherical mass distribution the gravitational

V0 V1

V2

For a sphere or spherical mass distribution the gravitational potential depends on– Total mass inside the sphere m mGVp

– The distance r from its center

• The equipotential surfaces of the original spherical mass form

rGV

• The equipotential surfaces of the original spherical mass form a set of concentric spheres– One of which (e g V0) coincides with the surface of the spherical massOne of which (e.g., V0) coincides with the surface of the spherical mass

• No change in potential No Work done in movingF d l ti f th it ti l fi ld t t

0. rdF

– Force and acceleration of the gravitational field must act perpendicular to the surface

– This normal to the equipotential surface defines the vertical, direction q p ,

– The plane tangential to the equipotential surface at a point defines the horizontal at that point 11/15

Earth’s gravitational acceleration (g)

• Earth isn’t a point mass

• Earth is approximately spherically• Earth is approximately spherically

symmetrical Earth as a point source

of mass located at the center of the Earth!

• To consider acceleration…– Insert Newton’s second law to solve for the force on the small body…

EE MmM

Note that g is independent on the mass of the object being accelerated toward

22E

E

E

sEs R

MGgR

mMGgmF

– Note that g is independent on the mass of the object being accelerated toward the center of the Earth

– This relationship implies that all bodies on earth’s surface should experience the same falling acceleration

– g on the surface of the Earth ~9.81 m.s-2 12/15

Gravity Units• Lets have an example:

kgrmwheremGg 63

101574

• As 6.673×10−11 N·(m/kg)2 . Thus the gravity

kgmwhered

Gg 2 101573

, 3/300 mkg

As 6.673 10 N (m/kg) . Thus the gravityanomaly in g is ~1.048x10-6 m/s2.– gravity anomalies in m/s2 is not convenientg y

• Instead we adopt a new unit: Gal (named in honor of Galileo)– 1 Gal = 0.01 m/s2 = 1 cm/s2

• But this is still too large, so we typically use milliGals (mGal) or gravity units (g.u.)

– 1 mGal = 0 001 Gal = 10 g u = 1x10-5 m/s2– 1 mGal = 0.001 Gal = 10 g.u. = 1x10 5 m/s

• So our buried sphere produced an anomaly of:– 1.048x10-6 m/s2 = ~ 0.1 mGal

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Gravitational acceleration (g) on other planets

Wh t ill b• What will be your body weight on surface of moon or mars?

• g on Moon ~1/6 of that on Earthof that on Earth

• g on mars ~1/3 of that on Earthof that on Earth

• Weight = mg

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Earth’s gravitational acceleration (g)• Is the g value is constant on Earth Surface?

• is true only for attraction of Earth & stationary objects2E

RMGg is true only for attraction of Earth & stationary objects

– Earth is not a stationary sphere

• Earth also rotates about its own axis

ER Earth

• Earth also rotates about its own axis– Creates centrifugal force depends on:

• Distance from axis

Equator::9.78 m/s2

• Distance from axis

• Rate of rotation

• Linear velocity at equator is much faster than at poles

– Deformation of the shape of Earth: polar flattening• Earth is an oblate spheroid

Poles: 9.83 Poles: 9.83 m/sm/s22

• Radius of Earth is greatest at equator, least at poles

• Effects of other celestial bodies mainly effects of Moon and Sun– Tides 15/15