5 earth gravitation

Earth Gravitation

SOLO HERMELIN

Updated: 17.01.13

16.12.14

1

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Table of Content

SOLO

2

Earth Gravitation

Introduction to Gravitation

Gravitation of a Point Mass M (Netwon Gravitation Law)

Gravitation of a Distribution of Mass Defined by the Density ( )Srρ

Gravitation of a Uniform Distribution of Mass in a Spherical Volume

( )( ).ConstrS =ρ

Physical Meaning of the Low Degree and Order SHCs

Reference Earth Model

Clairaut's theoremMac Cullagh’s Approximation

World Geodetic System (WGS 84)

Reference Ellipsoid

Air Vehicle in Ellipsoidal Earth Atmosphere

References

Gravity

Nicolaus Copernicus (1473 – 1543)“De Revolutionibus Orbium Coelestium” (On the Revolutions of the Heavenly Spheres”, Nuremberg 1543)

SOLO

Gravity

Kepler’s Laws of Planetary Motion

From 1601 to 1606 Kepler tried to fit various Geometric Curves to Tycho Brache’s Data on Mars Orbit. The result is“Kepler’s Law of Planetary Motion”

Tycho Brahe(1546-1601)

Johaness Kepler (1571-1630)

SOLO

KEPLER’S LAWS OF PLANETARY MOTION 1609-1619

•FIRST LAW THE ORBIT OF EACH PLANET IS AN ELLIPSE, WITH THE SUN AT A FOCUS.

b

a

•SECOND LAW THE LINE JOINING THE PLANET TO THE SUN SWEEPS OUT EQUAL AREAS IN EQUAL TIME.

a dAh

dt2=

b

•THIRD LAW THE SQUARE OF THE PERIOD OF A PLANET IS PROPORTIONAL TO THE CUBE OF ITS MEAN DISTANCE FROM THE SUN.

a 2/322a

GMGM

abTP

ππ ==

b

GravitySOLO

GALILEO GALILEI (1564-1642( “DISCOURSES AND MATHEMATICAL DEMONSTRATIONS CONCERNING TWO NEW SCIENCES ” 1636

THE FIRST TWO CHAPTERS DEAL WITH STRENGTH OF MATERIALS . THIS IS ESSENTIALLY THE FIRST NEW SCIENCE.

THE THIRD AND FOURTH SUBJECT IS :

• UNIFORM MOTION WITH CONSTANT ACCELERATION AND MOTION OF PROJECTILES

•THE LAW OF INERTIA

•THE COMPOSITION OF MOTIONS ACCORDING TO VECTOR ADDITION (“GALILEAN TRANSFORMATION”) AND

•THE STUDY OF UNIFORMLY ACCELERATED MOTION

GravitySOLO

NEWTON’S LAWS OF MOTION

“THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY”1687

• FIRST LAW EVERY BODY CONTINUES IN ITS STATE OF REST OR OF UNIFORM MOTION IN A STRAIGHT LINE UNLESS IT IS COMPELLED TO CHANGE THAT STATE BY FORCES IMPRESSED UPON IT.

• SECOND LAW THE RATE OF CHANGE OF MOMENTUM IS PROPORTIONAL TO THE FORCE IMPRESSED AND IS IN THE SAME DIRECTION AS THAT FORCE.

• THIRD LAW TO EVERY ACTION THERE IS ALWAYS OPPOSED AN EQUAL REACTION.

GravitySOLO

NEWTON’S LAW OF UNIVERSAL GRAVITATION

ANY TWO BODY ATTRACT ONE ONOTHER WITH A FORCE PROPORTIONAL TO THE PRODUCT OF THE MASSES AND INVERSLY PROPORTIONAL TO THE SQUARE OF THE DISTANCE BETWEEN THEM.

THE UNIVERSAL GRAVITATIONAL CONSTANT INSTANTANEUS PROPAGATION OF THE FORCE ALONG THE DIRECTION BETWEEN THE MASES (“ACTION AT A DISTANCE”).

M m

GravitySOLO

THE LAGEOS SATELLITE MONITORSITS POSITION RELATIVE TO THE EARTHUSING REFLECTED LIGHT.

THE TORSION BALANCE EXPERIMENTS:* HENRY CAVENDISH 1797 * RESEARCH GROUP OF UNIVERSITY OF WASHINGTON, SEATTLE (EOTWASH)

CAVENDISH EXPERIMENTS HAVE NOT BEEN ABLE TO TEST THE GRAVITATIONAL FORCE AT SEPARATIONSSMALLER THAN A MILLIMETER.

IF THERE ARE n EXTRA DIMENSIONS (TO THE 3 SPACE + 1 TIME) CURLED UP WITH DIAMETERS R, AT SCALES

SMALLER THAN R THE GENERALIZED NEWTON POTENTIAL WILL BE: Rrr

MGrV

nn <<=+1)(

Gravity

Henry Cavendish(1731 – 1810)

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SOLO

10

SOLOEarth Gravitation


SFSF

SF

SF

SFFM rrr

rr

rr

rr

MmGgmF

−=

−−

−−== :2

According to Newton the Gravitation Force of Point Mass M on the Point mass mF is given by

MF

The acceleration of Mass m due to Gravity is

( ) ( ) rrfr

r

r

MGrg

=

−=

2

( ) ( ) ( )r

r

r

rfr

r

rfrf

∂∂=∇

∂∂=∇

Any function of the form has the property that( ) rrf

Because of this 1.we can write2.we have

( ) ( )rUrg −∇=

( )( )

( )( )

( )

( )( )

( )

( )( )

( )

( )( ) ( )[ ]1

1111

rUrUmrUdmrdrUmrdrgmrdrF F

rB

rA

F

rB

rA

rB

rA

rB

rA

−=−=⋅∇−=⋅=⋅ ∫∫∫∫

The Work necessary to move the Mass mF in the Gravity Field of Point Mass M,from point to point , is not a function of the trajectory chosen but onthe values of the scalar function U (called Potential) at those two points.

( )rA ( )1rB

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Earth Gravitation


( ) ( ) ( )( )

( )

( )

( )

( )

( )

r

MG

r

MGrd

r

MGrd

r

r

r

MGrdrgrUrU

f

rB

rA

rB

rA

rB

rA

f

fff

−=−=⋅

−=⋅=− ∫∫∫ 22

The Gravitation Potential U of a Point Mass M is given by

The Gravitation Equi-Potential Surface of a Point Mass M is Spherical Surface centered at M location.


The Gravitation Potential of the Distribution is obtaining byIntegrating the Gravitation Potentials of the Point Masses dmover the Volume V

( ) ( ) ( )∫∫ −

−=−

−=V

SS

S

M S

S Vdrr

rG

rr

rmdGrU

ρ

If we take the reference rf → ∞ we get 0. ==fr

MGConst

( )r

MGrU

−=

( ) .Constr

MGrU +−=


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12

Earth Gravitation

( ) S

SSS

rrMd

r

r

r

rr

G

rr

MdGrUd >

+

−

−=−

−=2

cos21

1

γ

( ) ( ) SSSSSSSS

onDistributiDensity

SSS ddrdrrVdrMd ϕθθθϕρθϕρ sin,,,, 2==

( ) ( )S

r SS

SSSSSSSS rr

r

r

r

r

ddrdrr

r

GrU

S S S

>

+

−

−= ∫ ∫ ∫ϕ θ γ

ϕθθθϕρ2

2

cos21

sin,,

( ) Sn

n

n

SSS rrPr

r

r

r

r

r >

=

+

− ∑

∞

=

−

0

2/12

coscos21 γγ

Pn (u) are Lagrange Polynomials, of order n, given explicitly by Rodrigues’ Formula:

( ) ( )n

nn

kn ud

ud

nuP

1

!2

1 2 −=


( ) ( ) ( ) ( ) 2/13&,1 2210 −=== uuPuuPuPExamples:

Using Spherical Coordinates (rS, φS, θS)

The Potential Differential of any distribution of mass is given by:

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13

Earth Gravitation

( ) ( ) ( ) S

r Md

SSSSSSSSn

n

n

S rrddrdrrPr

r

r

GrU

S S S

>

−= ∫ ∫ ∫ ∑

∞

=ϕ θ

ϕθθθϕργ

sin,,cos 2

0

From Spherical Trigonometry:

( )SSS ϕϕθθθθγ −+= cossinsincoscoscos

According to Addition Theorem for Spherical Harmonics:

( ) ( ) ( ) ( )( ) ( )[ ] ( ) ( )∑

=

−+−+=

n

mS

mn

mnSSnnn PPm

mn

mnPPP

1

coscoscos!

!2coscoscos θθϕϕθθγ

( ) ( ) ( )uPud

duuP nm

mmm

n

2/21: −=

Where Pnm(u) is the Associated Lagrange Function of the First Kind of Degree n

and Order m, given by:


( ) ( )uPuP nn =0 ( ) ( ) ( ) ( )222

2

121

2 13,13 ttPtttP −=−=They satisfy:

The Potential of any distribution of mass is given by:

See Presentation on“Legendre Functions”

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Earth Gravitation


( ) ( ) ( )

( ) ( ) Sn r Md

SSSSSSSSn

n

S

k

S

r Md

SSSSSSSSk

n

n

S

rrddrdrrPa

r

r

a

r

G

rrddrdrrPr

r

r

GrU

S S S

S S S

>

−=

>

−=

∑ ∫ ∫ ∫

∫ ∫ ∫ ∑∞

=

∞

=

0

2

2

0

sin,,cos

sin,,cos

ϕ θ

ϕ θ

ϕθθθϕργ

ϕθθθϕργ

According to Addition Theorem for Spherical Harmonics:

( ) ( ) ( ) ( )( ) ( )[ ] ( ) ( )∑

=

−+−+=

n

mS

mn

mnSSnnn PPm

mn

mnPPP

1

coscoscos!

!2coscoscos θθϕϕθθγ

Using those results we obtain:

( ) 1cos0 =γP

( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( ) ( ) ( )∑ ∫ ∫ ∫

∑ ∑ ∫ ∫ ∫

∑ ∫ ∫ ∫∫ ∫ ∫

∞

=

∞

= =

∞

=

+−−

+−−

−

−=

1

2

1 0

2

1

22

1

0

0

sin,,sincossincos!

!2

sin,,coscoscoscos!

!2

sin,,coscossin,,cos

n r

SSSSSSSSSSn

n

Sn

n

n

n

m r

SSSSSSSSSSm

n

n

Smn

n

n r

SSSSSSSSSn

n

Sn

n

M

r Md

SSSSSSSSS

S S S

S S S

S S SS S S

ddrdrrmPa

rmP

r

a

mn

mn

r

G

ddrdrrmPa

rmP

r

a

mn

mn

r

G

ddrdrrPa

rP

r

a

r

GddrdrrP

a

r

r

GrU

ϕ θ

ϕ θ

ϕ θϕ θ

ϕθθθϕρϕθϕθ

ϕθθθϕρϕθϕθ

ϕθθθϕρθθϕθθθϕργ


a – any reference length

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Earth Gravitation

( ) ( ) ( ) ( ) ( )( )

+

+

+−= ∑∑∑

∞

= =

∞

= 1 110 sincoscoscos1

n

n

mnmnm

mn

n

nnn

n

mSmCPr

aCP

r

a

r

MGrU ϕϕθθ

( ) ( )∫ ∫∫

==

S S Sr Md

SSSSSSSSSn

n

Snn ddrdrrP

a

r

MCC

ϕ θ

ϕθθθϕρθ

sin,,cos1

: 20

( )( ) ( ) ( )

( ) ( )∫ ∫ ∫

+−=

S S Sr Md

SSSSSSSSS

SS

mn

n

S

nm

nm ddrdrrm

mP

a

r

mn

mn

S

C

ϕ θ

ϕθθθϕρϕϕ

θ

sin,,sin

coscos

!

!2: 2



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16

Earth Gravitation

(1) A tremendous simplification results if the mass distribution is symmetric about the z axis, i.e. ρ is a function only of rS and θS. Since

( ) ( ) ,2,10cossin2

0

2

0

=== ∫∫ jdjdj SSSS

ππ

ϕϕϕϕ

the coefficients Cmn and Smn vanish identically.

(2) In addition to axial symmetry, the origin of coordinates coincides with the center of mass, the constant C1 is identically zero.

( ) ( ) 0cos1

sin,cos1

: 2

cos

11 =

=

= ∫∫ ∫ ∫

m

SS

r md

SSSSSSSSS md

a

r

MddrdrrP

a

r

MC

S S S S

θϕθθθρθϕ θ θ

C1 is proportional to the first moment of the mass M with respect to the xy plane.


In this case

( ) ( ) 01

:cos1 nnn

nn

n

CJJPr

a

r

MGrU −=

−−= ∑

∞

=

θ

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17

Earth Gravitation

(3) Finally, if the mass is distributed in homogeneous concentric layers, i.e. ρ is a function only of rS , then Cn vanishes identically for all n

( ) ( ) 0sincos:2

0

0

00

22

=

= ∫∫∫

+ ππ

ϕθθθρ SSSSk

R

SS

n

Sn ddPrdr

a

r

M

aC


( )r

MGrU =In this case


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18

Earth Gravitation

Gravitation of a Uniform Distribution of Mass in a Spherical Volume( )( ).ConstrS =ρ

For a Static Spherical Volume with an Uniform DistributedMass the Gravitation Potential at a distance r > R is given by

( )r

MG

r

RGVd

rrGrU S

V

SSS

−=−=−

−= ∫3/41 3πρρ

The Gravitation Potential for a Static Spherical Volume with an Uniform Distributed Mass is equivalent to a Point MassGravitation Potential concentrated at the Center of the Sphere. The Equi-Potential Surfaces will be Concentric Spherical Surfaces outside the Spherical Mass.

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19

Earth Gravitation

The Potential of the Reference Ellipsoid is given by:

where:UG – Gravitational Potential (m2/s2)GM – Earth’s Gravitational Constantr - Distance from the Earth’s Center of MassrS - Distance from the Earth’s Center of Mass to d m

Gravitation of a Uniform Distribution of Mass in a Ellipsoid Volume( )( ).Constr =ρ

( )Srr

mdGrUd

−=

Typically the Potential is expanded in a series. This can be done in two ways, which lead to the same result:1.By expanding the term and integrate the result term by term.2. By writing the Potential as solution of Lalace’s Equation using Spherical Harmonics.

Srr −

1

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20

Earth Gravitation

The Potential Equation of Gravity Field

where:UG – Gravitational Potential (m2/s2)GM – Earth’s Gravitational Constantr - Distance from the Earth’s Center of MassrS - Distance from the Earth’s Center of Mass to dm

( ) ( )

( ) ( )

( )rG

VdrrrG

Vdrr

rGrU

S

S

V

SSS

V

SS

SS

ρπ

δπρ

ρ

4

4

122

−=

−−=

−

∇−=∇

∫∫∫

∫∫∫

( ) ( )∫∫∫∫ −−=

−−=

SV S

SS

M S rr

VdrG

rr

mdGrU

ρStart with:

Inside the mass M we have:

( ) ( )rGrU ρπ42 −=∇Poisson Equation:

( ) 02 =∇ rULaplace Equation:

Outside the mass M we have:

Siméon Denis Poisson

1781-1840

Pierre-Simon Laplace(1749-1827)


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Earth Gravitation



As pointed out in the previous slides the first term of the Spherical Harmonic Coefficients (SHC) of Gravity Potential is equal to the Potential for Spherical Mass. The remaining terms then represent the Gravitational Potential due to Non-spherical, Non-uniform Mass.

( ) ( ) ( ) ( ) ( )( )

+

+

+−= ∑∑∑

∞

= =

∞


n

n

mnmnm

mn

n

nnn

n

mSmCPr

aCP

r

a

r

MGrU ϕϕθθ

We have (see Figure)

=

=

S

SS

SS

S

s

s

s

S r

z

y

x

r

θϕθϕθ

cos

sinsin

cossin

( ) 0&1

cos1

: 10

cos

1

1

110 ===

== ∫∫ S

a

zMdz

MaMdP

a

r

MCC G

M

S

M

SS

S

θ

θ

( )

=

=

=

=

∫∫ ∫ ay

axMd

y

x

MaMd

a

r

MMdP

a

r

MS

C

G

G

M S

S

M M S

SS

S

S

SS

S

/

/2

2

sin

cossin

2

sin

coscos

21

11

1

11

11

ϕϕ

θϕϕ

θ

( ) ( ) 2

121

1 1 ttP −=where we used Legendre Polynomials. GGG zyx

,, are the coordinates of the Mass Center of Gravity

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Earth Gravitation



( ) ( ) ( ) ( ) ( )( )

+

+

+−= ∑∑∑

∞

= =

∞


n

n

mnmnm

mn

n

nnn

n

mSmCPr

aCP

r

a

r

MGrU ϕϕθθ

=

=

S

SS

SS

S

s

s

s

S r

z

y

x

r

θϕθϕθ

cos

sinsin

cossin

+−−=

−=

2

1

2

312220

SSSS

SSSSSS

yyxxzzzzrr

III

MaII

MaC

( )∫

=

M

SS MdPa

r

MS

C

0

1cos

2 02

2

20

20 θ

( ) ( )

+−++=

+−=−−−=−==2

31

2

1

2

3

2

1cos3coscos

2222

2

222

22

22222

2

2

22

2022

2SS

SSSSS

SSSSSSS

SS

SS yx

zyxa

yxz

aa

zyxz

a

rP

a

rP

a

r θθθ

( ) ∫∫∫

=

=

=

=

M zy

zx

SS

SS

M S

SSSS

M S

SS

S

MaI

MaIMd

zy

zx

MaMdr

MaMdP

a

r

MS

C

SS

SS

2

2

2

2

2

12

2

21

21

/

/1

sin

coscossin3

3

1

sin

coscos

!3

12

ϕϕ

θθϕϕ

θ

( )( ) ( )

( )∫∫∫

−=

−=

−⋅

=

=

M yx

xxyy

SS

SS

M SS

SSSS

M S

SS

S

MaI

MaIIMd

yx

yx

MaMdr

MaMdP

a

r

MS

C

SS

SSSS

2

222

2

2222

2

22

2

22

22

2/

4/

24

1

cossin2

sincossin3

34

1

2sin

2coscos

!4

!12

ϕϕϕϕ

θϕϕ

θ

( ) ( ) ( ) ( )222

2

121

2 13,13 ttPtttP −=−=where we used

( ) ( ) ( ) ( ) ( )SSSSSSSSSSSSSS zzyyxx

M

SSSrr

M

SSzz

M

SSyy

M

SSxx IIIMdzyxIMdyxIMdzxIMdzyI ++=++=+=+=+= ∫∫∫∫ 2

1::,:,: 222222222

Define Moments of Inertia

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Earth Gravitation



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24

Earth Gravitation




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25

Earth Gravitation

Earth is a Rotating Non-Spherical Body, with a slightly non-uniform mass distribution. The force acting on an external mass is due to Gravitation and Centrifugal Accelerations. To find a Model for the Gravitation Acceleration the Earth is approximated by an Ellipsoid

The flattering of the Earth was already discovered by the end of the 18th century.It was noticed that the distance between a degree of Latitude as measured, for instance with a sextant, differs from that expected from a Sphere:RE (θ1 – θ2) ≠ RE dθ, with RE the radius of the Earth, θ1 and θ2 two different Latitudes

f - the flattening of a meridian section of the Earth, defined as:

a

baf

−=:

R = 6,371.000 kma = 6,378.136 kmb = 6,356.751 km



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26

Reference Ellipsoid

Meridian Ellipse Equation: 12

2

2

2

=+b

z

a

p

Slope of the Normal to Ellipse:

2

2

tanbp

az

zd

pd =−=φ

The Slope of the Geocentric Line to the same point

p

zO =φtan

OO RzRp φφ sincos ==

Deviation Angle between Geographic and GeodeticAt Ellipsoid Surface

Ob

a φφ tantan2

2

=

= −

Ob

a φφ tantan2

21 ( ) ( ) φφφφ tan1tan1tantan 22

2

2

fea

bO −=−==

a

baf

−=:

22

222 2: ff

a

bae −=−=

Earth Gravitation

SOLO

27

Reference Ellipsoid

Meridian Ellipse Equation:

Oφφδ −=

12

2

2

2

=+b

z

a

p

Slope of the Normal to Ellipse:

2

2

tanbp

az

zd

pd =−=φ

The Slope of the Geocentric Line to the same point

p

zO =φtan

−=

−+

−

=+

−=

+−= 1

11

1

1tantan1

tantantan

2

2

2

2

2

2

2

2

2

22

22

2

2

b

a

a

zp

ap

pa

ba

pz

bpazpz

bpaz

O

O

φφφφδ

OO RzRp φφ sincos ==

( )

( ) ( )OO

f

O fba

R

b

ba

a

baR

ba

ba φφφδ 2sin2sin2

tan2sin2

tan

1

2

11

1222

221 ≈

+

−=

−=

≈≈<<

−−

Deviation Angle between Geographic and GeodeticAt Ellipsoid Surface

Earth Gravitation

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28

Reference Ellipsoid

For a point at a Height h near the Ellipsoid the value of δ must be corrected:

u−= 1δδ

From the Law of Sine we have:

Deviation Angle between Geographic and GeodeticAt Altitude h from Ellipsoid Surface

( ) R

h

hR

huu ≈+

≈=− 11 sin

sin

sin

sin

δδπ

Since u and δ1 are small: 1δR

hu ≈

The corrected value of δ is:

( )OfR

h

R

hu φδδδ 2sin11 11

−=

−=−=

Therefore:

( )OO fR

h φλδφφ 2sin1

−+=+=

Earth Gravitation

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29


where λ – Longitude e – Eccentricity = 0.08181919


In Earth Center Earth Fixed Coordinate –ECEF-System (E)the Vehicle Position is given by:

( )( )( )( )

+++

=

=

φφλφλ

sin

cossin

coscos

HR

HR

HR

z

y

x

P

M

N

N

E

E

E

E

( )NhH

e

aRN

+=−

= 2/12 sin1 φ

Another variable, used frequently, is the radius of the Ellipsoid referred as the Meridian Radius

( )( ) 2/32

2

sin1

1

φe

eaRM

−−=

Earth Gravitation

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30

Reference Ellipsoid

b

a

Oφ

?O – Geographic Latitude? – Geodetic Latitude

Equator

North Pole

Tangentto Ellipsoid

pNR

zP

a

r

φ

β

Meridian Ellipse

Ellipsoid Equation:

( ) 2222

2

2

22

11 aebb

z

a

yx −==++

Define: 22: yxp +=

12

2

2

2

=+b

z

a

p

Differentiate: 022

=+b

dzz

a

dpp

z

p

a

b

dp

dz2

2

cot =−=φFrom the Figure:

φtan2

2

pa

bz =

( )φφ 2224

222

4

2

2

2

tantan1 baa

pp

a

b

a

p +=+=

φφφ

φφφ

2222

2

2222

2

sincos

sin,

sincos

cos

ba

bz

ba

ap

+=

+=

Earth Gravitation

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31

Reference Ellipsoid

( )φφ

φφφ

φφ

φφφ

22

2

2222

2

222222

2

sin1

cos1

sincos

sin

sin1

cos

sincos

cos

e

ea

ba

bz

e

a

ba

ap

−−=

+=

−=

+=

( ) 222 1 aeb −=

Earth Gravitation

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32

Reference Ellipsoid

( ) 222 1 aeb −=

12

2

2

2

=+b

z

a

p

0

0

sin

cos

φφ

rz

rp

==

1sincos

20

2

20

22 =

+

bar

φφ

( )2/1

20

2

20

2

0

sincos−

+=

bar

φφφ

( )

( ) ( )022/1

02

2/1

02

2

2/1

02

2

222/1

02

2

2

02

0

sin1sin21

sin1sin1sincos

φφ

φφφφφ

fafa

b

ba

b

a

a

baa

b

baa

b

aar

f

−≈+≈

+−+=

−+=

+=

−

−

≈

−−

Earth Gravitation

SOLO

33

Reference Ellipsoid

The Meridians and Parallels are the Lines of Curvature of the Ellipsoid. The principal Radii of Curvature are therefore in the Plane of Meridian (Meridian Radius of Curvature RM) and in the Plane of Prime Vertical, perpendicular to Meridian Plane (Radius of Curvature in the Prime Vertical RN)

Radiuses of Curvature of the Ellipsoid

Meridian Ellipse Equation: ( ) 2222

2

2

2

11 aebb

z

a

p −==+

From this Equation, at any point (x,y) on the Ellipse, we have:

φtan

12

2

−=−=az

bp

pd

zd

32

4

32

2222

2

2

2

2

22

2

22

2

2

2 111

za

b

za

pbza

a

b

z

p

a

b

z

p

za

b

pd

zd

z

p

za

b

pd

zd −=+−=

+−=

−−=

From the Ellipse Equation:

( )φ

φφ2

22

2

2

2222

2

2

2

2

2

2

2

2

cos

sin1

1

1tan1111

e

a

p

ee

a

p

b

a

p

z

a

p −=

−−+=

+=

( )( )

( ) 2/122

2

2

2

2/122 sin1

sin1tan

sin1

cos

φφφ

φφ

e

eap

a

bz

e

ap

−−==→

−=

From the Figure on right: ( ) 2/122 sin1cos φφ e

apRN

−==

Earth Gravitation

SOLO

34

Reference Ellipsoid

Let develop the RN and RM (continue):

we have at any point (p,z) on the Ellipse:

φtan

12

2

−=−=az

bp

pd

zd ( )3

22

32

4

2

2 11

z

ea

za

b

pd

zd −−=−=

The Radius of Curvature of the Ellipse at the point (p,z) is:

( )( )

( )( )

( ) 2/322

2

2/322

3323

22

2/3

2

2

2

2/32

sin1

1

sin1

sin1

1

tan1

11

:φφ

φφe

ea

e

ea

ea

pdzd

pdzd

RM−

−=−

−−

+

=

+

=

( )( )

( ) 2/122

2

2/122 sin1

sin1

sin1

cos

φφ

φφ

e

eaz

e

ap

−−=

−=

( )( ) 2/322

2

sin1

1:

φe

eaRM

−−=


Earth Gravitation

SOLO

35

Reference Ellipsoid

( )( ) 2/322

2

sin1

1:

φe

eaRM

−−=

( ) 2/122 sin1cos φφ e

apRN

−==

a

baf

−=: 22

222 2: ff

a

bae −=−=Using

( )( )[ ] ( ) ( ) [ ] ++−≈

+−++−≈

−−−= φφ

φ2222

2/322

2

sin321sin22

3121

sin21

1: ffaffffa

ff

faRM

( )[ ] ( ) [ ]φφφ

2222/122

sin31sin22

31

sin21faffa

ff

aRN +≈

+−+≈

−−=

[ ]φ2sin321 ffaRM +−≈

[ ]φ2sin31 faRN +≈

We used and we neglect f2 terms( )( )

+−++=− !2

11

1

1 nnxn

x n



Earth Gravitation

SOLO

36


Clairaut's theorem

Clairaut's theorem, published in 1743 by Alexis Claude Clairaut in his “Théorie de la figure de la terre, tirée des principes de l'hydrostatique”, synthesized physical and geodetic evidence that the Earth is an oblate rotational ellipsoid. It is a general mathematical law applying to spheroids of revolution. It was initially used to relate the gravity at any point on the Earth's surface to the position of that point, allowing the ellipticity of the Earth to be calculated from measurements of gravity at different latitudes.

Clairaut's formula for the acceleration due to gravity g on the surface of a spheroid at latitude , was:ϕ

where G is the value of the acceleration of gravity at the equator, m the ratio of the centrifugal force to gravity at the equator, and f the flattening of a meridian section of the earth, defined as:

a

baf

−=:

Alexis Claude Clairaut )1713 – 1765(

−+= φ2sin2

51 fmGg

Earth Gravitation


SOLO

37


Mac Cullagh’s Approximation (1845)

James MacCullagh (1809 – 1847)

Mac Cullagh’s used the Approximation:

( )( )

( )

( ) ( )

−

−

+

+−≈ ∫∫

≈

MP

SS

P

ra

M

M P

SS

P

Mda

r

Mr

aMd

a

r

Mr

aM

r

GrU

S

CM

S

θθθθ

θθθθ

cos

22

cos

22

cos

1

cos

1

22

0

11

2

1cos31

2

1cos3cos

1cos1

∫∫∫

+−++=

+−=−=M

SSSSS

M

SSS

M

SS dMyx

zyxMa

dMyx

zMa

dMa

r

MC

23

1

2

1

2

1cos31:

2222

2

222

2

2

2

2

20

θ

222220 2

1

2

31J

Ma

ACIII

MaII

MaC SSSS

SSSSSS

yyxxzzzzrr −=−−=

+−−=

−=

( ) ( ) ( ) ( ) ( )SSSSSSSSSSSSSS zzyyxx

M

SSSrr

M

SSzz

M

SSyy

M

SSxx IIIMdzyxIMdyxIMdzxIMdzyI ++=++=+=+=+= ∫∫∫∫ 2

1::,:,: 222222222

Define Moments of Inertia

2:,: SSSS

SS

yyxxzz

IIAIC

+==

( )( )

Ma

ACJJ

r

aMGM

r

GrU

P

22

2

23

2

:2

1cos3

2

−=−+−≈

θ

θ

Earth Gravitation


SOLO

38


The definition of geodetic latitude (φ) and longitude (λ) on an ellipsoid. The normal to the surface does not pass through the centre

Reference Ellipsoid

Geodetic latitude: the angle between the normal and the equatorial plane. The standard notation in English publications is ϕ

Geocentric latitude: the equatorial plane and the radius from the centre to a point on the surface. The relation between the geocentric latitude (ψ) and the geodetic latitude ( ) is ϕderived in the above references as

The definition of geodetic (or geographic) and geocentric latitudes

( ) ( )[ ]φφψ tan1tan 21 e−= −

To use previous results, where we used spherical coordinates, rS, φS, θS we will use

θπψϕλ −==2

&

Earth Gravitation

SOLO

39

Earth Gravitation

Centrifugal Potential

Since Earth is a Rotating Body, a Centrifugal Acceleration is exerted on a Mass m, at a (x,y,z) Position, given by

( ) Cyx

yxVyx

C Uyxyx

yxVa ∇=+Ω=

+

+Ω=

+Ω=

1111 2

22

22

where:Ω = 7292115.1467x10-11 rad/s – Earth Angular Velocity

The Centrifugal Earth Potential UC at (x,y,z) Position is given by

( ) ( )( ) .

.1111..

2

2

Constydyxdx

ConstydxdyxConstrdUConstUdU yxyxCCC

++Ω=

++⋅+Ω=+⋅∇=+=

∫∫∫∫∫

Centrifugal Potential

Earth Gravitation Model

( )222

2

1yxUC +Ω=

By choosing Const. = 0, we obtain

SOLO

40

Earth Gravitation

Potential of the Rotating Reference Ellipsoid

The Potential of the Rotating Reference Ellipsoid is the sum of the Gravitational and Centrifugal Potentials:

CG UUW +=

( ) .2

1 222 ConstyxUC ++Ω=

The Gravitational Potential of the Reference Ellipsoid (assuming Uniform Density Distribution) is:

( ) ( )

+

+−= ∑∑

= =

max

2 0

sincossin1n

n

n

mnmnmnm

n

G mSmCPr

a

r

MGU λλψ

where:UG – Gravitational Potential (m2/s2)GM – Earth’s Gravitational Constantr - Distance from the Earth’s Center of Massa - Semi-Major Axis of the WG 84 Ellipsoidn,m – Degree and Order respectivelyψ– Geocentric Latitudeλ – Geocentric Longitude = Geodetic Longitude


SOLO

41

Earth Gravitation

The WGS 84 Gravity Model is defined in terms of normalized coefficients:

( ) ( ) ( ) ( ) ( )( )

+

+

+−= ∑∑∑

∞

= =

∞

= 1 110 sincossinsin1

n

n

mnmnm

mn

n

nnn

n

mSmCPr

aCP

r

a

r

MGrU λλψψ

( )∫

==

M

Sn

n

Snn MdP

a

r

MCC ψsin

1:0

( )( ) ( ) ( )

( )∫

+−=

M S

SS

mn

n

S

nm

nm Mdm

mP

a

r

mn

mn

S

C

λλ

ψsin

cossin

!

!2:

( ) ( ) ( ) ( ) ( )( )

+

+

+−= ∑∑∑

∞

= =

∞

= 1 110 sincossinsin1

n

n

mnmnm

mn

n

nnn

n

mSmCPr

aCP

r

a

r

MGrU λλψψ

( ) ( ) ( )( ) ( )

≠=

=

+−+=

02

01sin

!

!12sin

m

mkP

mn

mnknP m

nm

n ψψ

( )( )( )

≠=

=

−+

+=

02

01

!

!

12

1

m

mk

S

C

mn

mn

knS

C

nm

nm

nm

nm

Cnm and Snm are called Spherical Harmonic Coefficients (SHC).

The Gravitation Potential of the Earth Model is given by:Earth Gravitation Model


SOLO

42



Spherical Harmonics

Visual representations of the first few spherical harmonics. Red portions represent regions where the function is positive, and green portions represent regions where the function is negative.

Earth Gravitation

SOLO

43



Carlo Somigliana (1860 –1955)

The Theoretical Gravity on the surface of the Ellipsoidis given by the Somigliana Formula (1929)

84

22

2

2222

22

sin1

sin1

sincos

sincos

WGS

epe

e

k

ba

ba

φφγ

φφ

φγφγγ

−+=

+

+=

where

1: −=e

p

a

bk

γγ

2

22

:a

bae

−= - Ellipsoid Eccentricity

a - Ellipsoid Semi-major Axis = 6378137.0 m

b - Ellipsoid Semi-minor Axis = 6356752.314 m

γp – Gravity at the Poles = 983.21849378 cm/s2

γe – Gravity at the Equator = 978.03267714 cm/s2

– Geodetic Latitudeϕ

The Theory of the Equipotential Ellipsoid was first given byP. Pizzetti (1894)

Earth Gravitation

SOLO

44



The coordinate origin of WGS 84 is meant to be located at the Earth's center of mass; the error is believed to be less than 2 cm. The WGS 84 meridian of zero longitude is the IERS Reference Meridian. 5.31 arc seconds or 102.5 meters (336.3 ft) east of the Greenwich meridian at the latitude of the Royal Observatory.The WGS 84 datum surface is an oblate spheroid (ellipsoid) with major (transverse) radius a = 6378137 m at the equator and flattening f = 1/298.257223563. The polar semi-minor (conjugate) radius b then equals a times (1−f), or b = 6356752.3142 m. Presently WGS 84 uses the EGM96 (Earth Gravitational Model 1996) Geoid, revised in 2004. This Geoid defines the nominal sea level surface by means of a spherical harmonics series of degree 360 (which provides about 100 km horizontal resolution).[7] The deviations of the EGM96 Geoid from the WGS 84 Reference Ellipsoid range from about −105 m to about +85 m.[8] EGM96 differs from the original WGS 84 Geoid, referred to as EGM84.

Earth Gravitation

SOLO

45



Earth Gravitation

SOLO

46

World Geodetic System (WGS 84)Reference Earth Model

Geoid product, the 15-minute, worldwide Geoid Height for EGM96 The difference between the Geoid and the Reference Ellipsoid exhibit the following statistics: Mean = - 0.57 m, Standard Deviation = 30.56 mMinimum = -106.99 m, Maximum = 85.39 m

Earth Gravitation

SOLO

47

World Geodetic System (WGS – 84)


Parameters Notation Value

Ellipsoid Semi-major Axis a 6.378.137 m

Ellipsoid Flattening (Ellipticity) f 1/298.257223563(0.00335281066474)

Second Degree Zonal Harmonic Coefficient of the Geopotential C2,0 -484.16685x10-6

Angular Velocity of the Earth Ω 7.292115x10-5 rad/s

The Earth’s Gravitational Constant (Mass of Earth includes Atmosphere)

GM 3.986005x1014 m3/s2

Mass of Earth (Includes Atmosphere) M 5.9733328x1024 kg

Theoretical (Normal) Gravity at the Equator (on the Ellipsoid) γe 9.7803267714 m/s2

Theoretical (Normal) Gravity at the Poles (on the Ellipsoid) γp 9.8321863685 m/s2

Mean Value of Theoretical (Normal) Gravity γ 9.7976446561 m/s2

Geodetic and Geophysical Parameters of the WGS-84 Ellipsoid

Earth Gravitation

SOLO

48

World Geodetic System (WGS 84)Reference Earth Model

a

baf

−=:f - Ellipsoid Flattening (Ellipticity)

a - Ellipsoid Semi-major Axis

b - Ellipsoid Semi-minor Axis

e - Ellipsoid Eccentricity 22

222 2: ff

a

bae −=−=

( ) 211 eafab −=−=

Reference Ellipsoid

Earth Gravitation

SOLO

49


where λ – Longitude e – Eccentricity = 0.08181919


In Earth Center Earth Fixed Coordinate –ECEF-System (E)the Vehicle Position is given by:

( )( )( )( )

+++

=

=

φφλφλ

sin

cossin

coscos

HR

HR

HR

z

y

x

P

M

N

N

E

E

E

E

( )NhH

e

aRN

+=−

= 2/12 sin1 φ

Another variable, used frequently, is the radius of the Ellipsoid referred as the Meridian Radius

( )( ) 2/32

2

sin1

1

φe

eaRM

−−=

Radius of Curvature in the Prime Vertical


Earth Gravitation

50

SOLO Air Vehicle in Ellipsoidal Earth AtmosphereSIMULATION EQUATIONS

51

SOLO

7. Forces Acting on the Vehicle (continue – 4)

Gravitation Acceleration

( ) ( )

−

−

−==

zgygxg

gg100

0

0

0

010

0

0

0

001

χχχχ

γγ

γγ

σσσσ cs

sc

cs

sc

cs

scC EWE

W

( ) gg

−=

γσγσ

γ

cc

cs

sW

2sec/174.322sec/81.90

2

0

00gg ftmg

HR

R==

+=

The derivation of Gravitation Acceleration assumes an Ellipsoidal Symmetrical Earth.The Gravitational Potential U (R, ) is given byϕ

( ) ( )

( ) ( )φ

φµφ

,

sin1,2

RUg

PR

aJ

RRU

EE

n n

n

n

∇=

−⋅−= ∑∞

=

μ – The Earth Gravitational Constanta – Mean Equatorial Radius of the EarthR=[xE

2+yE2+zE

2]]/2 is the magnitude of the Geocentric Position Vector

– Geocentric Latitude (sin =zϕ ϕ E/R)Jn – Coefficients of Zonal Harmonics of the Earth Potential FunctionPn (sin ) – Associated Legendre Polynomialsϕ


52

SOLO

7. Forces Acting on the Vehicle (continue – 5)

Gravitation Acceleration

Retaining only the first three terms of theGravitational Potential U (R, ) we obtain:ϕ

R

z

R

z

R

z

R

aJ

R

z

R

aJ

Rg

R

y

R

z

R

z

R

aJ

R

z

R

aJ

Rg

R

x

R

z

R

z

R

aJ

R

z

R

aJ

Rg

EEEEz

EEEEy

EEEEx

E

E

E

⋅

+⋅−⋅

⋅−

−⋅

⋅−⋅−=

⋅

+⋅−⋅

⋅−

−⋅

⋅−⋅−=

⋅

+⋅−⋅

⋅−

−⋅

⋅−⋅−=

342638

515

2

31

342638

515

2

31

342638

515

2

31

2

2

4

44

42

22

22

2

2

4

44

42

22

22

2

2

4

44

42

22

22

µ

µ

µ

φ

φλ

φλ

sin

cossin

coscos

=

⋅=

⋅=

R

zR

yR

x

E

E

E

( ) 2/1222EEE zyxR ++=


53

SOLO

23. Local Level Local North (LLLN) Computations for an Ellipsoidal Earth Model

( )( )( )

( )( ) 2

2210

20

20

20

5

21

20

60

sin

sin1

sin321

sin1

sec/10292116557.7

sec/051646.0

sec/780333.9

26.298/.1

10378135.6

Ae

e

p

m

e

HR

RLatggg

LatfRR

LatffRR

LatfRR

rad

mg

mg

f

mR

++=

+=

+−=

−=

⋅=Ω

=

==

⋅=

−

LatHR

V

HR

V

HR

V

Ap

EastDown

Am

NorthEast

Ap

EastNorth

tan+

−=

+−=

+=

ρ

ρ

ρ

Lat

Lat

Down

East

North

sin

0

cos

Ω−=Ω=Ω

Ω=Ω

DownDownDown

EastEast

NorthNorthNorth

Ω+==

Ω+=

ρφρφρφ

East

North

Lat

LatLong

ρ

ρ

−=

=

•

•

cos

( )

( ) ∫

∫•

•

+=

+=

t

t

dtLatLattLat

dtLongLongtLong

0

0

0

0

SIMULATION EQUATIONS



SOLO

54

References


R.,H.,Battin, “Astronautical Guidance”, McGraw-Hill, 1964

George M. Siouris, “Aerospace Avionics System”, Academic Press, 1993, Appendix B

S. Hermelin, “Legendre Functions”

Broxmeyer, C,. “Inertial Navigation Systems”, McGraw-Hill, 1964

Averil B. Chatfield, “Fundamentals of High Accuracy Inertial Navigation”, Progress in Astronautics and Aeronautics 174, 1997

X. Li, H-J. Götze, “Tutorial: Ellipsoid, Geoid, Gravity, Geodesy, and Geophysics”, Geophysics, Vol. 66, No. 6, Nov-Dec. 2001

http://en.wikipedia.org/wiki/

Department of Defense, World Geodetic System 84, NIMA (National Imagery and Mapping Agency) TR8350.2, Third Edition

http://earth-info.nga.mil/GandG/images/ww15mgh2.gif

http://earth-info.nga.mil/GandG/publications/tr8350.2/wgs84fin.pdf

Earth Gravitation

55

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

56

Vector AnalysisSOLO

Vector Operations in Various Coordinate Systems

Laplacian 2∇=∇⋅∇• Cartesian:

2

2

2

2

2

22

z

U

y

U

x

UU

∂∂+

∂∂+

∂∂=∇

• Cylindrical:

2

2

2

2

22

2

2

2

2

2

22 1111

z

UUUU

z

UUUU

∂∂+

∂∂+

∂∂+

∂∂=

∂∂+

∂∂+

∂∂

∂∂=∇

θρρρρθρρρ

ρρ

• Spherical:

( )φθθ

θθθ

φθ2

2

2222

22

sin

1sin

sin

11,,

∂∂+

∂∂

∂∂+

∂∂

∂∂=∇ U

r

U

rr

Ur

rrrU

57

SOLO

Laplace Differential Equation in Spherical Coordinates

0sin

1sin

sin

112

2

2222

22 =

∂∂+

∂∂

∂∂+

∂∂

∂∂=∇

φθθθ

θθU

r

U

rr

Ur

rrU

Let solve this equation by the method of Separation of Variables, by assuming a solution of the form :

( ) ( ) ( )φθφθ ,,, SrRrU =

Spherical Coordinates:

θϕθϕθ

cos

sinsin

cossin

rz

ry

rx

===

In Spherical Coordinates the Laplace equation becomes:

Substituting in the Laplace Equation and dividing by U gives:

0sinsinsin

112

2

222

2=

∂∂+

∂∂

∂∂+

φθ

θθ

θθ

SS

Srrd

Rdr

rd

d

Rr

The first term is a function of r only, and the second of angular coordinates. For the sum to be zero each must be a constant, therefore:

λφθ

θθ

θθ

λ

−=

∂∂+

∂∂

∂∂

=

2

2

2

2

sinsinsin

1

1

SS

S

rd

Rdr

rd

d

R

58

SOLOLaplace Differential Equation in Spherical Coordinates

λ=

rd

Rdr

rd

d

R21

( ) ( )ϕθ ,SrR=Φ

We get:

022

22 =−+ R

rd

Rdr

rd

Rdr λor:

Assume a solution of the form: R = C rα, where α is a constant to be defined and C is determined by the boundary conditions. Substituting in the Differential Equation gives

( )[ ] 021 =−−− αλααα rC

or:( ) 01 =−+ λαα

Let define l as l (l+1):=λ

( ) ( ) ( ) 0111 2 =+−+=+−+ llll αααα

( ) ( )

−−=+±−=

++±−=

12

121

2

14112,1 l

llllα

( )

=+1

1l

l

r

rrR

59

SOLO

( )1sinsinsin

12

2

2+−=

∂∂+

∂∂

∂∂

llSS

S φθθ

θθ

θ

( ) ( )ϕθ ,SrR=ΦWe obtain:

Multiply this by S sin2θ and put to get:( ) ( ) ( )φθφθ ΦΘ=,S

01

sinsinsin1

2

22 =Φ

Φ+

+

Θ∂∂

Θ φθλ

θθ

θθ

d

d

d

d

Again, the first term, in the square bracket, and the last term must be equal and opposite constants, which we write m2, -m2. Thus:

( )

Φ−=Φ

=Θ

−++

Θ∂∂

22

2

2

2

0sin

1sinsin

1

md

d

mll

d

d

φ

θθθ

θθ

The Φ ( ) must be periodical in (a period of 2 π) and because ϕ ϕthis we choose the constant m2, with m an integer. Thus:

( ) φφφ mbma sincos +=Φ


( ) φφφ mjmj ee +−=Φ ,or

With m integer, we have the Orthogonality Condition

21

21,

2

0

2 mmmjmj dee δπϕ

πφφ =⋅∫ +−

60

SOLO

( ) ( ) ( ) ( )φθφθ ΦΘ= rRrU ,,

We get:

or: ( ) Θ+−=Θ−Θ+Θ1

sincot

2

2

2

2

llm

d

d

d

d

θθθ

θ

( ) 0sin

1sinsin

12

2

=Θ

−++

Θ∂∂

θθθ

θθm

lld

d


Change of variables: t = cos θ

θθ dtd sin−=

td

d

d

d Θ−=Θ θθ

sin

( )

td

dt

td

dt

td

d

d

d

td

d

td

d

td

d

td

d

d

d

d

d

d

d Θ−Θ−=Θ+Θ=

Θ−−=

Θ=Θ

−

2

22

cossin/1

2

22

2

2

1sin

sinsinsinsinθθ

θθθθθθθ

θθθ

( ) ( ) Θ

−

−++Θ−Θ−Θ=Θ

−++Θ+Θ=

=

2

2

2

2cos

2

2

2

2

11

sin1cot0

t

mll

td

dt

td

dt

td

dmll

d

d

d

d t θ

θθθ

θ

We obtain:( )

1cos

,2,1,001

122

2

2

2

≤⇒=

==Θ

−

−++Θ−Θ

tt

mt

mll

td

dt

td

d

θ

Associate Legendre Differential Equation

61

SOLO

( ) ( ) ( ) ( )φθφθ ΦΘ= rRrU ,,


We obtain:

( )

1cos

,2,1,001

122

2

2

2

≤⇒=

==Θ

−

−++Θ−Θ

tt

mt

mll

td

dt

td

d

θ

Associate Legendre Differential Equation

Let start with m = 0 with:

( )

1cos

0122

2

≤⇒=

=Θ++Θ−Θ

tt

lltd

dt

td

d

θLegendre Differential Equation

They are named after Adrien-Marie Legendre. This ordinary differential equation is frequently encountered in physics and other technical fields. In particular, it occurs when solving Laplace's equation (and related partial differential equations) in spherical coordinates.

The Legendre polynomials were first introduced in 1785 by Adrien-Marie Legendre, in “Recherches sur l’attraction des sphéroides homogènes”, as the coefficients in the expansion of the Newtonian potential

Adrien-Marie Legendre(1752 –1833(

SOLO

62

Legendre Polynomials

Olinde Rodrigues (1794-1851)

Start from the function: ( ) .12 constktkyn

=−=

( ) 12 12:'−−== n

ttkntd

ydy

( ) ( ) ( ) 222122

2

11412:''−− −−+−== nn

ttnkntkntd

ydy

Let compute:

( ) ( ) ( ) ( ) ( ) '12211412''112222 ytntnttnkntknyt

nn −+=−−+−=− −

or: ( ) ( ) 02'12''12 =−−+− ynytnyt

Let differentiate the last equation n times with respect to t:

( )[ ] ( ) ( ) ( ) ( )

( ) ( ) ''1''2''1

00''13

''12

''11

''1''1

2

2

1

12

3

3

0

23

3

2

22

2

2

1

1222

ytd

dnny

td

dtny

td

dt

ytd

dt

td

dny

td

dt

td

dny

td

dt

xd

dny

td

dtyt

td

d

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

−

−

−

−

−

−

−

−

−

−

−++−=

+−

+−

+−

+−=−

( ) ( ) ( ) ( )

+−=

+−=− −

−

−

−

''12'1

'12'121

1

1

1

ytd

dny

td

dtny

td

dt

td

dny

td

dtnyt

td

dn

n

n

n

n

n

n

n

n

n

n

Derivation of Legendre Polynomials via Rodrigues’ Formula


SOLO

63



Start from the function: ( ) .12 constktkyn

=−=

( ) ( ) 02'12''12 =−−+− ynytnytDifferentiate n times with respect to t:

( ) ( ) ''1''2''12

2

1

12 y

td

dnny

td

dtny

td

dt

n

n

n

n

n

n

−

−

−

−

−++−

( ) 02''121

1

=−

+−+ −

−

ytd

dny

td

dny

td

dtn

n

n

n

n

n

n

Define: a Polynomial( ) ( )[ ]n

n

n

n

n

ttd

dk

td

ydtw 1: 2 −==

( ) ( ) ( ) [ ] 02'121'2''12 =−+−+−++− wnwnwtnwnnwtnwt

( ) ( ) ( ) ( )[ ] 02121'2''12 =−−+−+−++− wnnnnnwtnttnwt

( ) ( ) 01'2''12 =+−+− wnnwtwt

This is Legendre’s Differential Equation. We proved that one of the solutions are Polynomials. We can rewrite this equation in a Sturm-Liouville Form:

( ) ( ) 0112 =+−

− wnnw

td

dt

td

d



SOLO

64



Let find k such that:

by choosing Pn (1) = 1

( ) ( )[ ]n

n

n

n

n

n ttd

dk

td

ydtP 12 −==

( ) ( )[ ] ( ) ( ) ( ) ( )

−+=

+−=−= ∑

>0

22 1!2111i

inn

v

n

u

n

n

nn

n

n

n ttatnktttd

dktk

td

dtP

!2

1

nk

n=

We obtain the Rodrigues Formula:( ) ( )[ ]n

n

n

nn ttd

d

ntP 1

!2

1 2 −=

Let use Leibnitz’s Rule (Binomial Expansion for the n Derivative of a Product - with u:=(t-1)n and v:=(t+1)n ):

( ) ( )( )

udvudvdnvddunn

vddunvdu

vdudmnm

nvud

nnnnn

n

m

mnmn

+++−++=

−=⋅

−−−

=

−∑1221

0

!2

1

!!

!

We have:

( ) ( )

1!2!2

11

1!20

12

0

21

00

==

+++−++==

=

−−− nkudvudvdnvddunn

vddunvdukxP n

xn

nnnnnn

n

We can see from this Formula that Pn (t) is indeed a Polynomial of Order n in t.



SOLO

65

Legendre PolynomialsThe first few Legendre polynomials are:

( )

( )( )

( )( )

( )( )

( )( )

( ) 256/63346530030900901093954618910

128/31546201801825740121559

128/35126069301201264358

16/353156934297

16/51053152316

8/1570635

8/330354

2/353

2/132

1

10

246810

3579

2468

357

246

35

24

3

2

−+++−+−+−

+−+−−+−

−+−

+−+−

−−

xxxxx

xxxxx

xxxx

xxxx

xxx

xxx

xx

xx

x

x

xPn n


66

SOLO

Orthogonality of Legendre Polynomials

Define ( ) ( )tPwtPv nm == :&:

We use Legendre’s Differential Equations:

( ) ( ) 011 2 =++

− vmm

td

vdt

td

d

( ) ( ) 011 2 =++

− wnn

td

wdt

td

d

Multiply first equation by w and integrate from t = -1 to t = +1.

( ) ( ) 0111

1

1

1

2 =++

− ∫∫

+

−

+

−dtwvmmdtw

td

vdt

td

d

Integrate the first integral by parts we get

( ) ( ) ( ) 01111

1

1

1

2

0

1

1

2 =++−−− ∫∫+

−

+

−

+=

−=

dtwvmmdttd

wd

td

vdtw

td

vdt

t

t

In the same way, multiply second equation by v and integrate from t = -1 to t = +1.( ) ( ) 011

1

1

1

1

2 =++−− ∫∫+

−

+

−dtwvnndt

td

wd

td

vdt


67

SOLO


( ) ( ) 0111

1

1

1

2 =++−− ∫∫+

−

+

−dtwvmmdt

td

wd

td

vdt

Subtracting those two equations we obtain

( ) ( ) 0111

1

1

1

2 =++−− ∫∫+

−

+

−dtwvnndt

td

wd

td

vdt

( ) ( )[ ] ( ) ( )[ ] ( ) ( ) 011111

1

1

1=+−+=+−+ ∫∫

+

−

+

−dttPtPnnmmdtwvnnmm nm

This gives the Orthogonality Condition for m ≠ n

( ) ( ) nmdttPtP nm ≠=∫+

−0

1

1

To find let square the relation and integrate between t = -1 to t = +1. Due to orthogonality only the integrals of terms having Pn

2(t) survive on the right-hand side. So we get

( )∫+

−

1

1

2 dttPn( )∑

∞

=

=+− 0

221

1

nn

n tPuutu

( )∑ ∫∫∞

=

+

−

+

−=

+− 0

1

1

221

1 221

1

nn

n dttPudtutu


68

SOLO


( )∑ ∫∫∞

=

+

−

+

−=

+− 0

1

1

221

1 221

1

nn

n dttPudtutu

( ) ( )( ) 1

1

1ln1

1

1ln

2

121ln

2

1

21

12

21

1

21

1 2<

−+=

+−

−=−+

−=

−+

+=

−=

+

−∫ uu

u

uu

u

utuu

udt

tuu

t

t

( ) ( ) ( ) ( ) ( ) ( ) ( )∑∑∑∞ ++∞ +∞ +

+−−−=

+−−−

+−=−−+

0

11

0

1

0

1

11

1

11

1

11

11ln

11ln

1

n

uu

un

u

un

u

uu

uu

u

nnn

nn

nn

( ) ( ) ( )( ) ( ) ( )

( ) ∑∑∑∑∞∞ +∞ ++

+∞ ++

+=

+=

+−−−+

+−−−=

0

2

0

12

0

0

121212

0

12122

12

2

12

12

121

1

121

1 nnnn

nnn

n unn

u

un

uu

un

uu

u

Let compute first

Therefore

( )∑ ∫∑∫∞ +

−

∞+

−=

+=

+− 0

1

1

22

0

21

1 2 12

2

21

1dttPuu

ndt

utu nnn

Comparing the coefficients of u2n we get ( )12

21

1

2

+=∫

+

− ndttPn


( ) ( ) nmmn ndttPtP δ

12

21

1 +=∫

+

−Hence

SOLO

69

Associated Legendre Functions

Let Differentiate this equation m times with respect to t, and use Leibnitz Rule of Product Differentiation: ( ) ( )[ ] ( )

( ) ( )im

im

i

im

im

m

td

tgd

td

tsd

imi

mtgts

td

d−

−

=∑ −

=⋅0 !!

!

Start withLegendre DifferentialEquation:

( ) ( ) ( ) ( ) 1011 2 ≤=++

− ttwnntw

td

dt

td

dnn

or: ( ) ( ) ( ) ( ) ( ) 101212

22 ≤=++−− ttwnntw

td

dttw

td

dt nnn

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )twmmtwtmtwttwtd

dt

td

d mn

mn

mnnm

m

1211 1222

22 −−−−=

− ++

( ) ( ) ( ) ( ) ( )twmtwttwtd

dt

td

d mn

mnnm

m

+=

+1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )twnntwmtwttwmmtwtmtwt mn

mn

mn

mn

mn

mn 122121 1122 ++−−−−−− +++

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) 011121 122 =+−+++−−= ++ twmmnntwtmtwt mn

mn

mn


SOLO

70


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) 011121 122 =+−+++−− ++ twmmnntwtmtwt mn

mn

mn

Define: ( ) ( ) ( )twty mn=:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) 011121 122 =+−+++−− tymmnntytmtyt

Now define: ( ) ( ) ( )tyttum

221: −=Let compute:

( ) ( ) ( )122122 11 ytyttm

td

ud mm

−+−−= −

( ) ( ) ( ) ( )1122222 111 ytyttm

td

udt

mm +−+−−=−

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )21221221221

2222222 1121111 ytyttmyttmyttmytmtd

udt

td

d mmmmm +− −+−+−−−−+−−=

−

( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) yt

tmmnnmmtymmnnytmytt

mm

−

+−+−+−++−+++−−−=2

22222

0

12222

111111211

We get:( ) ( ) 0

111

2

22 =

−

−++

− u

t

mnn

td

udt

td

d Associate Legendre Differential Equation


SOLO

71


Define: ( ) ( ) ( )twtd

dttu nm

mm

221: −=

We get: ( ) ( ) 01

112

22 =

−

−++

− u

t

mnn

td

udt

td

d

Start with Legendre Differential Equation:

( ) ( ) ( ) ( ) 1011 2 ≤=++

− ttwnntw

td

dt

td

dnn

Summarize

But this is the Differential Equation of Θ (θ) obtained by solving Laplace’s Equation

by Separation of Variables in Spherical Coordinates .

02 =Φ∇( ) ( ) ( ) ( )φθφθ ΦΘ=Φ rRr ,,

The Solutions Pnm(t) of this Differential Equation are called Associated Legendre

Functions, because they are derived from the Legendre Polynomials

( ) ( ) ( )tPtd

dttP nm

mmm

n221: −=


SOLO

72


Examples

( ) ( ) ( )tPtd

dttP nm

mmm

n221: −=

( ) ( ) 10 000 === tPtPn

( ) ( )( )

( )

( ) ( )

( ) ( ) ( ) θ

θ

θ

θ

θ

θ

sin1

cos:

sin111

cos2

121

11

1

cos

10

1

cos2

122

121

1

1

−=−−=−=

===

=−=−==

=−

=

=

t

t

t

tP

ttPtP

ttPtP

tttd

dttPn

( ) ( ) ( )( ) ( )tP

mn

mntP m

nmm

n !

!1

+−−=−

( ) ( )( )

( )

( ) ( )( )

( )

( ) ( )

( ) ( ) ( )( )

( ) ( ) ( ) ( ) θ

θθ

θ

θθ

θ

θ

θ

θ

θ

θ

2cos

222

222

cos2

1211

2

2cos2

202

cos2

12

2

2

121

2

2cos

22

2

2

2

222

2

sin8

11

8

1

!4

!01

2

cossin13

!3

11

2

1cos3

2

13

cossin3132

131

sin3132

1312

12

2

1

=−

=−

=

=

=

=−=−=

−=−−=

−=−==

=−=

−−=

=−=

−−==

t

t

tP

t

t

tP

t

tP

ttPtP

tttP

ttPtP

ttt

td

dttP

tt

td

dttPn

( ) 10 =tP ( )2

1

2

3 22 −= ttP( ) ttP =1


SOLO

73


Examples

( ) ( ) ( )tPtd

dttP nm

mmm

n221: −= ( ) ( ) ( )

( ) ( )tPmn

mntP m

nmm

n !

!1

+−−=−


SOLO

74


Orthogonality of Associated Legendre Functions

( ) ( ) ( )[ ]n

mn

mnm

nm

n ttd

dt

ntP 11

!2

1: 222 −−= +

+

Let Compute

( ) ( ) ( ) ( )[ ] ( )[ ]∫∫+

−+

+

+

+

+

+

−

−−−=1

1

2221

1

111!!2

1dtt

td

dt

td

dt

qpdttPtP

q

mq

mqp

mp

mpm

qpm

qm

p

Define X := x2 -1

( ) ( ) ( ) [ ] [ ]∫∫+

−+

+

+

+

+

+

−

−=1

1

1

1 !!2

1dtX

td

dX

td

dX

qpdttPtP q

mq

mqp

mp

mpm

qp

mm

qm

p

If p ≠ q, assume q > p and integrate by parts q + m times

[ ] [ ] mqidtXtd

dvdX

td

dX

td

du q

imq

imqp

mp

mpm

i

i

+==

= −+

−+

+

+

,,1,0

All the integrated parts will vanish at the boundaries t = ± 1 as long as there is a factorX = x2-1. We have, after integrating m + q times

( ) ( ) ( ) ( ) [ ]∫∫+

−+

+

+

+

+

++

−

−−=1

1

1

1 !!2

11dtX

td

dX

td

dX

qpdttPtP p

mp

mpm

mq

mqq

qp

mqmm

qm

p

SOLO

75


( ) ( ) ( ) ( ) [ ] 1:!!2

11 21

1

1

1

−=

−−= ∫∫+

−+

+

+

+

+

++

−

xXdtXtd

dX

td

dX

qpdttPtP p

mp

mpm

mq

mqq

qp

mqmm

qm

p

Because the term Xm contains no power greater than x2m, we must haveq + m – i ≤ 2 m

or the derivative will vanish. Similarly,p + m + i ≤ 2 p

Adding both inequalities yieldsq ≤ p

which contradicts the assumption that q > p, therefore is no solution for i and the integral vanishes.

Let expand the integrand on the right-side using Leibniz’s formula

( )( )∑

+

=++

++

−+

−+

+

+

+

+

−++=

mq

i

pimp

impm

imq

imqqp

mp

mpm

mq

mqq X

td

dX

td

d

imqi

mqXX

td

dX

td

dX

0 !!

!

( ) ( ) qpdttPtP mq

mp ≠=∫

+

−

01

1

This proves that the Associated Legendre Functions are Orthogonal (for the same m).


SOLO

76


( )[ ] ( ) ( )1:

!!2

11 21

12

21

1

2 −=

−−= ∫∫+

−+

+

+

++

−

xXdtXtd

dX

td

dX

ppdttP p

mp

mpm

mp

mpp

p

mpm

p

Let expand the integrand on the right-side using Leibniz’s formula

( )( )∑

+

=++

++

−+

−+

+

+

+

+

−++=

mp

i

pimp

impm

imp

imppp

mp

mpm

mp

mpp X

td

dX

td

d

impi

mpXX

td

dX

td

dX

0 !!

!

For the case p = q we have

Because X = x2 – 1 the only non-zero term is for i = p - m

( )[ ] ( ) ( )( ) ( ) ( )

( ) ( )

1:!2!!2

!1 21

1

!2

2

2

!2

2

2

22

1

1

2 −=⋅−+−= ∫∫

+

−

+

−

xXdtXtd

dX

td

dX

mmpp

mpdttP

p

pp

p

m

mm

mp

p

pm

p

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )( )

( )( )!

!

12

2

sin1!!2

!2!11

!!2

!2!1

!12

!21

0

1222

cos1

1

222

212

mp

mp

p

dmpp

pmpdtx

mpp

pmp

p

p

pp

p

ptp

X

p

p

pp

−+⋅

+=

−−

+−=−−

+−=

+−

+=+

−+

∫∫

θθ

πθ


SOLO

77


( ) ( ) ( ) ( ) ( )( ) qp

mq

mp

tm

qm

p mp

mp

pdPPdttPtP ,

0

cos1

1 !

!

12

2sincoscos δθθθθ

πθ

−+

+== ∫∫

=+

−

Therefore the Orthonormal Associated Legendre Functions is

( ) ( )( ) ( ) mnmP

mn

mnnΘ m

nm

n ≤≤−+−+= θθ cos

!

!

2

12cos


78

SOLO

Absolute Angular Momentum Relative to a Reference Point O

The Absolute Momentum Relative to a Reference Point O, of the particle of mass dmi at time t is defined as:

( ) ( ) iiOiiiOiiOiO dmVrdmVRRPdRRHd

×=×−=×−= ,, :

The Absolute Momentum Relative to a Reference Point O, of the mass m (t) is defined as:

( ) ( ) ∑∑∑===

×=×−=×−=N

ii

I

iOi

N

iiiOi

N

iiOiO dm

td

RdrdmVRRPdRRH

1,

11, :

By taking a very large number N of particles, we go from discrete to continuous

∫⇒∑∞→

=

NN

i 1

( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

The Absolute Momentum Relative to a Reference Point O, of the system (including the mass entering (+)/leaving (-) through surface S), at time t + Δt is given by:

( ) ( )∑∑ ∆

∆+×∆++

∆+×∆+=∆+

= openingsiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOiOO m

td

Rd

td

Rdrrdm

td

Rd

td

RdrrHH

,,1

,,,,

EQUATIONS OF MOTION OF A VARIABLE MASS SYSTEMSIMPLIFIED PARTICLE APPROACH

79

SOLO

Absolute Angular Momentum Relative to a Reference Point O (continue – 1)

By subtracting

I

O

tI

O

t

H

td

Hd

∆∆

=→∆

,

0

, lim

( ) ( )

t

dmtd

Rdrm

td

Rd

td

Rdrrdm

td

Rd

td

Rdrr

openings

N

ii

I

iiOiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOi

t ∆

×−∆

∆+×∆++

∆+×∆+

=

∑ ∑∑==

→∆

1,,

1,,

0lim

∑∑∑ ×+×+×=== openings

iflow

I

iflowOiflow

N

ii

I

iOiN

ii

I

iOi m

td

Rdrdm

td

Rd

td

rddm

td

Rdr

,1

,

12

2

,

Now let add the constraint that at time t the flow at the opening is such that

iopenS

( ) ( ) ( ) ( )trtrtRtR OiflowOiopeniflowiopen ,,

=→=

to obtain (next page)


( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

dividing by Δt, and taking the limit, we get

from ( ) ( )∑∑ ∆

∆+×∆++

∆+×∆+=∆+

= openingsiflow

I

iflow

I

iflowOiflowOiflow

N

ii

I

i

I

iOiOiOO m

td

Rd

td

Rdrrdm

td

Rd

td

RdrrHH

,,1

,,,,

80

SOLO


∑∑∑ ×+×+×=== openings

iflow

I

iflowOiopen

N

ii

I

i

I

OiN

ii

I

iOi

I

O mtd

Rdrdm

td

Rd

td

rddm

td

Rdr

td

Hd

,1

,

12

2

,,

( )∑ ×−+∑ ×

−+∑ ×=

== openingsiflow

I

iflowOiopen

N

ii

I

i

I

O

I

iN

ii

I

iOi m

td

RdRRdm

td

Rd

td

Rd

td

Rddm

td

Rdr

112

2

,

( )∑ ×−+∑×−∑ ×=== openings

iflow

I

iflowOiopen

N

ii

I

i

I

ON

ii

I

iOi m

td

RdRRdm

td

Rd

td

Rddm

td

Rdr

112

2

,

By taking a very large number N of particles, we go from discrete to continuous

∫⇒∑∞→

=

NN

i 1

( )( )∑ ×−+×−∫ ×=

openingsiflowiflowOiopenO

tmI

O

I

O mVRRPVdmtd

Rdr

td

Hd

2

2

,,

( ) ( ) ∑∑ −+=−−+=openings

OiopeniflowO

I

O

openingsOiopeniflowO

I

O rmVmtd

cdRRmVm

td

cdtP ,

,,

Substitute to obtain

( )( ) ( )∑ ×−+×

∑ −−++∫ ×=

openingsiflowiflowOiopenO

openingsiflowOiopenO

I

O

tmI

O

I

O mVRRVmRRVmtd

cddm

td

Rdr

td

Hd

,

2

2

,,

or

( )( )∑ −×+×+∫ ×=

openingsiflowOiflowOiopenO

I

O

tmI

O

I

O mVVrVtd

cddm

td

Rdr

td

Hd

,,

2

2

,,


81

SOLO


We obtained

( )( )

( )( ) ( )

∫ ×=∫ ×−=∫ ×−=tm

Otm

Otv

OO dmVrdmVRRdvVRRH

,, ρ

Substitute in the previous equation

OIO

O

OO

I

O

I

O

I

OO rtd

rdV

td

rd

td

Rd

td

RdVrRR ,

,,, :&

×++=+==+= ←ω

( )( ) ( )

∫

×++×=∫ ×−= ←

tmOIO

O

OOO

tmOO dmr

td

rdVrdmVRRH ,

,,,

ω

( )( )

( ) ( )∫

×+∫ ××+×

∫= ←tm O

OO

tmOIOOO

tmO dm

td

rdrdmrrVdmr ,,,,,

ω

We obtain

(a) (b) (c)

Let develop those three expressions (a), (b) and (c).

where is the angular velocity vector from I to O.IO←ω


82

SOLO


(a)( ) ( ) ( ) ( ) ( )

( ) OOC

tm

OC

tm

OC

tm

OC

tm

C

tm

O cmRRdmrdmrdmrdmrdmr ,,,,,,

=−===+= ∫∫∫∫∫

Where we used because C is the Center of Mass (Centroid) of the system.( )

0, =∫tm

C dmr

( )OOOOCO

tm

O VcVrmVdmr ×=×=×

∫ ,,,

( )( )

( )[ ]( )

IOOIOtm

OOOOtm

OIBO Idmrrrrdmrr ←←← ⋅=⋅∫ −⋅=∫ ×× ωωω ,,,,,,, 1(b)

where ( )[ ]( )∫ −⋅=

tm

OOOOO dmrrrrI ,,,,, 1:

2nd Moment of Inertia Dyadic of all the mass m(t) relative to O

We obtain (a) + (b) + (c)

( )( ) ( )

( )( ) ( )

∫

×+∫ ××+×

∫=∫ ×−= ←tm O

OO

tmOIOOO

tmO

tvOO dm

td

rdrdmrrVdmrvdVRRH ,,,,,, :

ωρ

( )∫

×+⋅+×= ←

tm O

OOIOOOO dm

td

rdrIVc ,,,,

ω


SOLO


( )[ ]( )∫ −⋅=

tm

OOOOO dmrrrrI ,,,,, 1:

2nd Moment of Inertia Dyadic of all the mass m(t) relative to O


=

O

O

O

O

z

y

x

r

,

,

,

,

( )[ ] [ ] [ ]OOO

O

O

O

O

O

O

OOOOOOO zyx

z

y

x

z

y

x

zyxrrrr ,,,

,

,

,

,

,

,

,,,,,,,

100

010

001

1

−

=−⋅

( )∫

+−−

−+−

−−+

=tm

OOOOOO

OOOOOO

OOOOOO

O dm

yxzyzx

zyzxyx

zxyxzy

I2

,2

,,,,,

,,2

,2

,,,

,,,,2

,2

,

, :

Science

5 earth gravitation