Physical Geodesy

Definition

Physical geodesy is the study of the gravitational field of the earth and it’s geodetic

applications.

Under Physical Geodesy following subjects are studied.

Potential theory.

The gravity field of the Earth.

Gravimetric methods.

Heights above sea level.

Astro-geodetic methods.

Reference

Physical Geodesy by Weikko A. Heiskanen and Helmut Moritz

Fundamentals of Potential Theory

Newton’s Universal Law of Gravitation.

𝐹 = 𝐾 𝑚1𝑚2

𝑙2

𝐾 = 66.7 × 10−9 𝑐𝑚3𝑔−1𝑠𝑒𝑐−2

= 6.67 × 10−11 𝑚3𝑘𝑔−1𝑠𝑒𝑐−2

Gravitational Acceleration (Gravity)

Gravity is the gravitational force acting on a unit mass. In other word it is the

acceleration of a mass under the influence of a gravitational force. The units are

𝑚/𝑠𝑒𝑐2

Average value 9.8 𝑚 𝑠𝑒𝑐−2

980 𝑐𝑚 𝑠𝑒𝑐−2 (gals). In geodesy unit mgal is often used. 1 mgal = 10−3 gals

𝑚1

Gravity Potential

Potential is work done to bring a unit mass from infinity to the required point.

Potential is scalar therefore it can be added algebraically.

Gravity (Gravitational Acceleration) can be derived from the potential

mathematically.

Acceleration

𝐹𝑋 = −𝐹 cos𝛼 = − 𝑘𝑚

𝑙2(𝑥 − 𝜉)

𝑙 = −

𝑘𝑚 (𝑥 − 𝜉)

𝑙3

𝐹𝑌 = −𝐹 cos𝛽 = − 𝑘𝑚(𝑦 − 𝜂)

𝑙3

𝐹𝑍 = −𝐹 cos 𝛾 = − 𝑘𝑚 (𝑧 − 𝜁)

𝑙3

𝑊ℎ𝑒𝑟𝑒 𝑙 = √(𝑥 − 𝜉)2 + (𝑦 − 𝜂)2 + (𝑧 − 𝜁)2

Potential

𝑉 = ∫𝐹 𝑑𝑙 = ∫𝑘𝑚

𝑙2

𝑙

∞

𝑙

∞

𝑑𝑙 = [𝑘𝑚

𝑙]∞

𝑙

𝑉 = 𝑘𝑚

𝑙 𝑠𝑐𝑎𝑙𝑎𝑟

𝜕𝑉

𝜕𝑥 = −

𝑘𝑚

𝑙2 (𝑥 − 𝜉)

𝑙 = −

𝑘𝑚 (𝑥 − 𝜉)

𝑙3

// y

𝜕𝑉

𝜕𝑦 = −

𝑘𝑚 (𝑦 − 𝜂)

𝑙3

𝜕𝑉

𝜕𝑧 = −

𝑘𝑚 (𝑧 − 𝜁)

𝑙3

∴ 𝐹𝑋 = 𝜕𝑉

𝜕𝑥 𝑖 𝐹𝑦 =

𝜕𝑉

𝜕𝑦 𝑗 𝐹𝑧 =

𝜕𝑉

𝜕𝑧 𝑘

Gravitational force F = grad V

= ∇𝑉

𝑊ℎ𝑒𝑟𝑒 ∇ = 𝜕

𝜕𝑥 𝑖 +

𝜕

𝜕𝑦 𝑗 +

𝜕

𝜕𝑧 𝑘

The force is the gradient vector of the scalar function V

If we have several point masses m1 m2 …………..mn the potential of the system is

𝑉 = 𝑘𝑚1

𝑙1 +

𝑘𝑚2

𝑙2 + …………… +

𝑘𝑚𝑛

𝑙𝑛 𝐷𝑒𝑠𝑐𝑟𝑒𝑡𝑒 𝑚𝑎𝑠𝑠𝑒𝑠

Properties of Gravitational Potential

1. Potential is a continuous function

2. Vanishes at infinity lim𝑙→∞

𝑉 = 0

3. At great distances body acts like point mass

4. ∇𝑉 = 𝐹

5. V(xyz) = a Equipotential surface

∇𝑉 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑒𝑞𝑢𝑖𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

Poission’s Equation

∆𝑉 = −4𝜋𝑘𝜌 𝑤ℎ𝑒𝑟𝑒 ∆ = 𝜕2

𝜕𝑥2 +

𝜕2

𝜕𝑦2 +

𝜕2

𝜕𝑧2

Potential V satisfies Poission’s Equation

∴ 𝜕2𝑉

𝜕𝑥2 +

𝜕2𝑉

𝜕𝑦2 +

𝜕2𝑉

𝜕𝑧2 = −4𝜋𝑘𝜌

∆ − 𝐿𝑎𝑝𝑙𝑎𝑠𝑖𝑎𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟

Laplace Equation

Outside the attracting body, 𝜌 is 0

∴ ∆V = 0

The solutions of Laplace equation are called harmonic functions

The gravity potential is a harmonic function outside the attracting masses. But inside

the masses it satisfies poission’s equation.

Potential of a solid body

Consider element mass dm at (𝜉, 𝜂, 𝜁) inside the solid body

Potential at P due to dm is ,

𝑑𝑉 = 𝑘.𝑑𝑚

𝑙

If the density is 𝜌(𝜉, 𝜂, 𝜁)

𝑑𝑉 = 𝑘.𝜌. 𝑑𝑉

𝑙 =

𝑘. 𝜌

𝑙 𝑑𝜉 𝑑𝜂 𝑑𝜁

Total potential

𝑉 = ∭𝑑𝑉

𝑉 = 𝑘 ∭𝜌(𝜉 𝜂 𝜁)

√(𝑥 − 𝜉)2 + (𝑦 − 𝜂)2 + (𝑧 − 𝜁)2 𝑑𝜉 𝑑𝜂 𝑑𝜁

𝐹𝑥 = 𝜕𝑉

𝜕𝑥 = 𝑘.

𝜕

𝜕𝑥 ∭

𝜌 (𝜉 𝜂 𝜁)

𝑙𝑑𝜉 𝑑𝜂 𝑑𝜁

= 𝑘 ∭𝜌(𝜉 𝜂 𝜁) 𝜕

𝜕𝑥 (

1

𝑙) 𝑑𝜉 𝑑𝜂 𝑑𝜁

= 𝑘 ∭−(𝑥 − 𝜉)

𝑙3 𝜌. 𝑑𝑉

The Potential of Spherical shell

𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑎𝑡 𝐴 = 𝑘 𝛿𝑚

𝑙

If the thickness of the shell is Δ𝑅 and the density is 𝜌

𝛿𝑚 = 𝑅. 𝑑𝜃 𝑅 𝑆𝑖𝑛 𝜃 𝑑𝜆 . Δ𝑅. 𝜌

𝛿𝑉 =𝑘 𝜌 𝑅2 𝑆𝑖𝑛 𝜃 𝑑𝜃. 𝑑𝜆. Δ𝑅

𝑙

𝑉 = ∭𝑘 𝜌 𝑅2 Δ𝑅 𝑆𝑖𝑛 𝜃 𝑑𝜃. 𝑑𝜆

𝑙

If 𝜌 and Δ𝑅 are constants

𝑉 = 𝑘 𝜌 𝑅2 Δ𝑅 ∬𝑆𝑖𝑛 𝜃 𝑑𝜃 𝑑𝜆

𝑙

Due to spherical symmetricity 𝑙 is not a function of 𝜆 lim𝜆

0 → 2𝜋

𝑉 = 2𝜋𝑘 𝜌𝑅2 ∆𝑅∫𝑆𝑖𝑛 𝜃

𝑙

𝜋

0

𝑑𝜃

𝑙2 = 𝑟2 + 𝑅2 − 2𝑟𝑅 cos 𝜃

2𝑙 𝑑𝑙 = 2𝑟𝑅 𝑆𝑖𝑛 𝜃 𝑑𝜃

∴ 𝑆𝑖𝑛 𝜃

𝑙 𝑑𝜃 =

𝑑𝑙

𝑟𝑅

𝑉 = 2𝜋𝑘𝜌𝑅2 ∆𝑅 ∫𝑑𝑙

𝑟𝑅

Now by considering following cases

Case I

When A lies outside the sphere

𝑙1 = 𝑟 − 𝑅

𝑙2 = 𝑟 + 𝑅

𝑉 = 2𝜋𝑘𝑅2 ∆𝑅 𝜌∫𝑑𝑙

𝑟𝑅

𝑙2

𝑙1

𝑉 =2𝜋𝑘𝑅2 ∆𝑅 𝜌

𝑟. 𝑅 [𝑟 + 𝑅 − (𝑟 − 𝑅)]

=4𝜋𝑘𝑅2 ∆𝑅 𝜌

𝑟

But 4𝜋𝑅2 Δ𝑅 𝜌 = 𝑚 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒

𝑉 =𝑘𝑚

𝑟

Case II

When A lies inside the sphere

𝑉 =2𝜋𝑘 𝑅2 Δ𝑅 𝜌

𝑟𝑅 [𝑅 + 𝑟 − (𝑅 − 𝑟)]

𝑉 =2𝜋𝑘 𝑅2 Δ𝑅 𝜌 2𝑟

𝑟𝑅

= 4𝜋𝑘 𝑅 ∆𝑅 𝜌

𝑉 = 𝑘𝑚

𝑅

The potential inside the sphere is constant

Let us consider the case I above

𝑉 = 𝑘𝑚

𝑟

𝑑𝑉

𝑑𝑟 = −

𝑘𝑚

𝑟2

𝑑2𝑉

𝑑𝑟2=

2𝑘𝑚

𝑟3

But 𝑟 = √(𝑥2 + 𝑦2 + 𝑧2)

𝑑𝑉

𝑑𝑥 =

𝑑𝑉

𝑑𝑟 .𝑑𝑟

𝑑𝑥 = −

𝑘𝑚𝑥

𝑟3

𝑑2𝑉

𝑑𝑥2= −𝑘𝑚 [

𝑟3 − 𝑥 3 𝑟2 𝑥𝑟

𝑟6]

= 𝑘𝑚 [3𝑥2

𝑟5−

1

𝑟3]

// y

𝑑2𝑉

𝑑𝑦2= 𝑘𝑚 [

3𝑦2

𝑟5−

1

𝑟3]

𝑑2𝑉

𝑑𝑧2= 𝑘𝑚 [

3𝑧2

𝑟5−

1

𝑟3]

𝑑2𝑉

𝑑𝑥2 +

𝑑2𝑉

𝑑𝑦2 +

𝑑2𝑉

𝑑𝑧2= 𝑘𝑚 [

3(𝑥2 + 𝑦2 + 𝑧2)

𝑟5−

3

𝑟3]

= 𝑘𝑚 [3𝑟2

𝑟5−

3

𝑟3] = 0

∴ 𝑑2𝑉

𝑑𝑥2 +

𝑑2𝑉

𝑑𝑦2 +

𝑑2𝑉

𝑑𝑧2= 0

Potential of a spherical shell satisfies Laplace Equation

Case II

𝑉 =𝑘𝑚

𝑅→ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝜕𝑉

𝜕𝑥= 0

𝑑2𝑉

𝑑𝑥2= 0

∴ 𝑑2𝑉

𝑑𝑥2 +

𝑑2𝑉

𝑑𝑦2 +

𝑑2𝑉

𝑑𝑧2= 0 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛

Potential at a point due to a spherical shell with finite thickness

There are three cases to be considered

i.) r > a

ii.) b < r <a

iii.) r < b

Case I

The potential at A due to thin spherical shell

𝛿𝑉 = 4𝜋𝑘 𝜌 𝑅2 𝑑𝑅

𝑟

𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑎𝑡 𝐴 = 𝑉 = ∫4𝜋𝑘 𝜌 𝑅2 𝑑𝑅

𝑟

𝑎

𝑏

𝑉 = 4𝜋𝑘𝜌

3 [𝑎3 − 𝑏3

𝑟]

𝑉 = 𝑘𝑚

𝑟

Even sphere can be considered as point mass and Laplace equation is

satisfied

Case II (b<r<a)

The potential to be calculated in two components [ie. shell inside and

shell outside.]

𝛿𝑉 = 4𝜋𝑘 𝜌 𝑅1

2

𝑟 𝑑𝑅1 + 4𝜋𝑘𝜌 𝑅2𝑑𝑅2

𝑉 = ∫4𝜋𝑘 𝜌 𝑅1

2

𝑟 𝑑𝑅1 + ∫ 4𝜋𝑘𝜌 𝑅2 𝑑𝑅2

𝑎

𝑟

𝑟

𝑏

𝑉 = 4𝜋𝑘𝜌

3[𝑟3 − 𝑏3

𝑟] +

4𝜋𝑘𝜌

2 [𝑎2 − 𝑟2]

𝑉 = 4𝜋𝑘𝜌 [𝑎2

2−

𝑏3

3𝑟−

𝑟2

6]

Case III (r<b)

𝛿𝑉 = 4𝜋𝑘𝜌 𝑅 𝑑𝑅

𝑉 = ∫ 4𝜋𝑘𝜌 𝑅 𝑑𝑅 = 4𝜋𝑘𝜌

2 [𝑎2 − 𝑏2]

𝑎

𝑏

𝑉 = 2𝜋𝑘𝜌 [𝑎2 − 𝑏2]

The potential inside the sphere is constant .Therefore satisfies Laplace

Equation

Will consider case II

𝑊ℎ𝑒𝑟𝑒 𝑉 = 4𝜋𝑘𝜌 [𝑎2

2−

𝑏3

3𝑟−

𝑟2

6]

𝑟 = √(𝑥2 + 𝑦2 + 𝑧2)

𝑑𝑉

𝑑𝑟= 4𝜋𝑘 [

𝑏3

3𝑟2−

𝑟

3]

𝑑𝑉

𝑑𝑥= 4𝜋𝑘𝜌 [

𝑏3

3𝑟2−

𝑟

3] .

𝑥

𝑟

𝑑𝑉

𝑑𝑥= 4𝜋𝑘𝜌 [

𝑏3𝑥

3𝑟3−

𝑥

3]

𝑑2𝑉

𝑑𝑥2= 4𝜋𝑘𝜌 [

𝑏3

3[𝑟3 − 𝑥 3𝑟2 𝑥

𝑟𝑟6

] −1

3]

𝑑2𝑉

𝑑𝑥2= 4𝜋𝑘𝜌 [

𝑏3

3[1

𝑟3−

3𝑥2

𝑟5] −

1

3]

// y

𝑑2𝑉

𝑑𝑦2= 4𝜋𝑘𝜌 [

𝑏3

3[1

𝑟3−

3𝑦2

𝑟5] −

1

3]

𝑑2𝑉

𝑑𝑧2= 4𝜋𝑘𝜌 [

𝑏3

3[1

𝑟3−

3𝑧2

𝑟5] −

1

3]

𝑑2𝑉

𝑑𝑥2+

𝑑2𝑉

𝑑𝑦2+

𝑑2𝑉

𝑑𝑧2= 4𝜋𝑘𝜌 [

𝑏3

3[3

𝑟3−

3(𝑥2 + 𝑦2 + 𝑧2)

𝑟5] − 1]

= 4𝜋𝑘𝜌 [𝑏3

3[3

𝑟3−

3

𝑟3] − 1]

𝑑2𝑉

𝑑𝑥2+

𝑑2𝑉

𝑑𝑦2+

𝑑2𝑉

𝑑𝑧2= −4𝜋𝑘𝜌 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛

Potential due to a Ring

Assume the thickness is ∆𝑅 and the height is ∆ℎ

Potential at A due to small segment 𝛿𝑉

𝛿𝑉 = 𝑘 Δ𝑅 ΔΗ 𝑑𝑙

𝑎 𝜌

𝑇𝑜𝑡𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑉 = ∫𝑘 Δ𝑅 ΔΗ 𝑑𝑙

𝑎 𝜌

2𝜋𝑅

0

= 2𝜋𝑅 𝑘 Δ𝑅 ΔΗ 𝜌

√(𝑅2 + 𝑟2)

𝐵𝑢𝑡 2𝜋𝑅 Δ𝑅 ΔΗ 𝜌 = 𝑚

∴ 𝑉 = 𝑘𝑚

𝑎

Potential due to Disk

Potential at A due to ring 𝛿𝑉

𝛿𝑉 = 2𝜋𝑅 𝑘 Δ𝑅 ΔΗ 𝜌

√(𝑅2 + 𝑟2)

Potential at A due to Disk

𝑉 = ∫2𝜋𝑅𝑘 ΔΗ 𝜌

√(𝑅2 + 𝑟2)

𝑎

0

𝑑𝑅 𝑅 = 𝑟 tan𝜃

= 2𝜋𝑘𝜌 ΔΗ ∫𝑅

√(𝑅2 + 𝑟2)

𝑎

0

𝑑𝑅 𝑑𝑅 = 𝑟𝑠𝑒𝑐2𝜃 𝑑𝜃

𝑉 = 2𝜋𝑘𝜌 ΔΗ ∫𝑟 tan𝜃 𝑟 𝑠𝑒𝑐2 𝜃

𝑟 sec 𝜃

𝜃1

0

𝑑𝜃 lim𝑅 = 0 𝑅 = 𝑎

𝜃 = 0 𝜃 = 𝑡𝑎𝑛−1 (𝑎

𝑟)

𝑉 = 2𝜋𝑘𝜌 ΔΗ ∫ 𝑟 tan𝜃 sec 𝜃𝜃1

0

𝑑𝜃

𝑉 = 2𝜋𝑘𝜌 ΔΗ [𝑟 sec 𝜃]0𝜃1

𝑉 = 2𝜋𝑘𝜌 ΔΗ 𝑟 [√(𝑎2 + 𝑟2)

𝑟− 1]

𝑉 = 2𝜋𝑘𝜌 ΔΗ [√(𝑎2 + 𝑟2) − 𝑟]

Potential due to a Cylinder

Potential at A due to Disk 𝛿𝑉

𝛿𝑉 = 2𝜋𝑘𝜌 𝑑𝐻 [√𝑅2 + (𝑟 − 𝐻)2 − (𝑟 − 𝐻)]

Total potential at A

𝑉 = ∫ 2𝜋𝑘𝜌 𝑑𝐻 [√𝑅2 + (𝑟 − 𝐻)2 − (𝑟 − 𝐻)]𝐻

−𝐻

= 2𝜋𝑘𝜌 ∫ √𝑅2 + (𝑟 − 𝐻)2 𝑑𝐻 −𝐻

−𝐻

2𝜋𝑘𝜌 ∫ (𝑟 − 𝐻) 𝑑𝐻𝐻

−𝐻

= 𝐼1 − 𝐼2

𝐼2 = 2𝜋𝑘𝜌 ∫ (𝑟 − 𝐻) 𝑑𝐻 𝐻

−𝐻= 2𝜋𝑘𝜌 [𝑟𝐻 −

𝐻2

2]−𝐻

𝐻

= 2𝜋𝑘𝜌 [𝑟𝐻 −𝐻2

2+ 𝑟𝐻 +

𝐻2

2]

𝐼2 = 4𝜋𝑘𝜌𝑟𝐻

𝐼1 = 2𝜋𝑘𝜌 ∫ √𝑅2 + (𝑟 − 𝐻)2𝐻

−𝐻

𝑑𝐻

𝐼1 = 2𝜋𝑘𝜌 [(𝑟 + 𝐻)√(𝑅2 + (𝑟 + 𝐻)2 − (𝑟 − 𝐻)√(𝑅2 + (𝑟 − 𝐻)2

2𝑅2]

+1

2 𝑙𝑛 |

√(𝑅2 + (𝑟 + 𝐻)2 + (𝑟 + 𝐻)

√(𝑅2 + (𝑟 + 𝐻)2 + (𝑟 − 𝐻)|

∴ 𝑉 = 𝐼1 − 𝐼2

Gauss Formula

∭𝑑𝑖𝑣 𝐹 𝑑𝑉 = ∬𝐹 𝑑𝑆

𝑆𝑉

𝑉 − 𝑉𝑜𝑙𝑢𝑚𝑒 𝑆 − 𝑆𝑢𝑟𝑓𝑎𝑐𝑒

𝑑𝑖𝑣 = 𝜕𝑋

𝜕𝑥+

𝜕𝑌

𝜕𝑦+

𝜕𝑍

𝜕𝑧

Where X,Yand Z are the components of vector F in x.y,z directions.

F = X+Y+Z

Physical meaning

Assume that an incompressible fluid is

generated or absorbed by the jets (sources)

inside a surface. Vector F is the velocity.

Then div F is the strength of the jets.

∭𝑑𝑖𝑣 𝐹 𝑑𝑉

𝑣

is the amount of fluid generated in unit time

∬𝐹 𝑑𝑆

𝑠

is the amount of fluid flowing across the surface S in unit time.

In physical geodesy F is the vector of the gravitational force

F = X +Y+Z

𝑋 =𝜕𝑉

𝜕𝑥 𝑖 𝑌 =

𝜕𝑉

𝜕𝑦 𝑗 𝑍 =

𝜕𝑉

𝜕𝑧 𝑘

𝑑𝑖𝑣 𝐹 = 𝜕2𝑉

𝜕𝑥2+

𝜕2𝑉

𝜕𝑦2+

𝜕2𝑉

𝜕𝑧2= ∆𝑉

By poisson’s equation 𝑑𝑖𝑣 𝐹 = −4𝜋𝑘𝜌

Masses are the sources of gravitational field. Div F is proportional to

mass density.

F dS is the gravitational flux of Force.

∴ ∭∆𝑉 𝑑𝑣 = ∬𝜕𝑉

𝜕𝑛𝑆𝑣

𝑑𝑆

The above equation is Gauss integral formula for the potential.

Harmonic Functions

As the Gravitational Potential in space satisfies ∆𝑉 = 0 and it is continuous function

and it has continuous First derivatives.

Gravitational Potential in space is harmonic function.

The simplest Harmonic Function is Reciprocal Distance:

1

𝑙=

1

√(𝑥 − 𝜉)2 + (𝑦 − 𝜂)2 + (𝑧 − 𝜁)2

𝜕 (1𝑙)

𝜕𝑥= −

(𝑥 − 𝜉)

𝑙3

𝜕 (1𝑙)

𝜕𝑦= −

(𝑦 − 𝜂)

𝑙3

𝜕 (1𝑙)

𝜕𝑧= −

(𝑧 − 𝜁)

𝑙3

𝜕2 (1𝑙)

𝜕𝑥2= −

𝑙3 − (𝑥 − 𝜉) 3𝑙2

𝑙6(𝑥 − 𝜉)

𝑙 =

−𝑙2 + 3(𝑥 − 𝜉)2

𝑙5

Similarly,

𝜕2 (1𝑙)

𝜕𝑦2=

−𝑙2 + 3(𝑦 − 𝜂)2

𝑙5

𝜕2 (1𝑙)

𝜕𝑧2=

−𝑙2 + 3(𝑧 − 𝜁)2

𝑙5

𝜕2 (1𝑙)

𝜕𝑥2+

𝜕2 (1𝑙)

𝜕𝑦2+

𝜕2 (1𝑙)

𝜕𝑧2=

−3𝑙2 + 3[(𝑥 − 𝜉)2 + (𝑦 − 𝜂)2 + (𝑧 − 𝜁)2]

𝑙5

=−3𝑙2 + 3𝑙2

𝑙5= 0

∴1

𝑙 𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 ∆𝑉 = 0

Reciprocal Distance is a solution of ∆𝑉 = 0

Reciprocal Distance is a harmonic function.

Potential of a solid body is given by,

𝑉 = 𝐾 ∭𝜌

𝑙𝑣

𝑑𝑣

Δ𝑉 = 𝐾 ∭𝜌Δ(1

𝑙)

𝑣

𝑑𝑣 = 0

The potential of a solid body is also harmonic at any point outside the masses.

Strokes Theorem

A function V harmonic outside the surface S can be uniquely determined by its values

on S.

Potential at A can be determined by the potentials on the surface S.

The interior mass distribution is not important.

Direct Problem of Potential Theory

Determination of the potential from the masses

𝑉 = ∭𝑑𝑚

𝑙

Inverse Problem of Potential Theory

Determination of the masses from the potential. There is no unique solution for inverse

problem.

i.e. Same potential can be generated from different size of spheres with equal

masses.

Dirichlet’s Problem

Problem of computing harmonic function inside or outside the surface from the

boundary values on the surface. This is also known as the 1st boundary value problem

of potential theory.

. A

Laplace Equation in Spherical Coordinates

𝑥 = 𝑟 sin 𝜃 cos 𝜆

𝑦 = 𝑟 sin 𝜃 sin 𝜆

𝑧 = 𝑟 cos 𝜃

𝑑𝑥 = 𝜕𝑥

𝜕𝑟𝑑𝑟 +

𝜕𝑥

𝜕𝜃𝑑𝜃 +

𝜕𝑥

𝜕𝜆𝑑𝜆

𝑑𝑦 = 𝜕𝑦

𝜕𝑟𝑑𝑟 +

𝜕𝑦

𝜕𝜃𝑑𝜃 +

𝜕𝑦

𝜕𝜆𝑑𝜆

𝑑𝑧 = 𝜕𝑧

𝜕𝑟𝑑𝑟 +

𝜕𝑧

𝜕𝜃𝑑𝜃 +

𝜕𝑧

𝜕𝜆𝑑𝜆

𝜕𝑥

𝜕𝑟= sin 𝜃 cos 𝜆

𝜕𝑥

𝜕𝜃= 𝑟 cos 𝜃 cos 𝜆

𝜕𝑥

𝜕𝜆= −r sin 𝜃 sin 𝜆

𝜕𝑦

𝜕𝑟= sin 𝜃 sin λ

𝜕𝑦

𝜕𝜃= 𝑟 cos 𝜃 sin 𝜆

𝜕𝑦

𝜕𝜆= r sin 𝜃 cos 𝜆

𝜕𝑧

𝜕𝑟= cos 𝜃

𝜕𝑧

𝜕𝜃= −𝑟 sin 𝜃

𝜕𝑧

𝜕𝜆= 0

𝑑𝑥 = sin 𝜃 cos 𝜆 𝑑𝑟 + 𝑟 cos 𝜃 cos 𝜆 𝑑𝜃 − 𝑟 sin 𝜃 sin 𝜆 𝑑𝜆

𝑑𝑦 = sin 𝜃 sin 𝜆 𝑑𝑟 + 𝑟 cos 𝜃 sin 𝜆 𝑑𝜃 + 𝑟 sin 𝜃 cos 𝜆 𝑑𝜆

𝑑𝑧 = cos 𝜃 𝑑𝑟 − 𝑟 sin 𝜃 𝑑𝜃

𝑑𝑠2 = 𝑑𝑥2 + 𝑑𝑦2 + 𝑑𝑧2 = 𝑑𝑟2 + 𝑟2𝑑𝜃2 + 𝑟2𝑠𝑖𝑛2𝜃𝑑𝜆2

The general form of the arc in orthogonal coordinate 𝑞1 , 𝑞2 ,𝑞3 is

𝑑𝑠2 = ℎ1 2𝑑𝑞1

2 + ℎ22𝑑𝑞2

2 + ℎ32𝑑𝑞3

2

When,

𝑑𝑠2 = ℎ1 2𝑑𝑞1

2 + ℎ22𝑑𝑞2

2 + ℎ32𝑑𝑞3

2

It can be shown that Laplace operator is,

Δ𝑉 = 1

ℎ1ℎ2ℎ3[

𝜕

𝜕𝑞1(ℎ2ℎ3

ℎ1

𝜕𝑉

𝜕𝑞1) +

𝜕

𝜕𝑞2(ℎ3ℎ1

ℎ2

𝜕𝑉

𝜕𝑞2) +

𝜕

𝜕𝑞3(ℎ1ℎ2

ℎ3

𝜕𝑉

𝜕𝑞3)] ⇒ (1)

In our case ℎ1 = 1 ℎ2 = 𝑟 ℎ3 = 𝑟 sin 𝜃

∴ Δ𝑉 = 1

𝑟2 sin 𝜃[𝜕

𝜕𝑟(𝑟2 sin 𝜃

𝜕𝑉

𝜕𝑟) +

𝜕

𝜕𝜃(sin 𝜃

𝜕𝑟

𝜕𝜃) +

𝜕

𝜕𝜆(

1

sin 𝜃

𝜕𝑉

𝜕𝜆)]

=1

𝑟2 sin 𝜃[2𝑟 sin 𝜃

𝜕𝑉

𝜕𝑟+ 𝑟2 sin 𝜃

𝜕2𝑉

𝜕𝑟+ cos 𝜃

𝜕𝑉

𝜕𝜃+ sin 𝜃

𝜕2𝑉

𝜕𝜃2+

1

sin 𝜃

𝜕2𝑟

𝜕𝜆2]

Δ𝑉 = 𝜕2𝑉

𝜕𝑟2+

2

𝑟

𝜕𝑉

𝜕𝑟+

1

𝑟2

𝜕2𝑉

𝜕𝜃2+

cot 𝜃

𝑟2

𝜕𝑉

𝜕𝜃+

1

𝑟2𝑆𝑖𝑛2𝜃

𝜕2𝑉

𝜕𝜆2

Laplace Equation in Spherical Coordinates

𝜕2𝑉

𝜕𝑟2+

2

𝑟

𝜕𝑉

𝜕𝑟+

1

𝑟2

𝜕2𝑉

𝜕𝜃2+

cot 𝜃

𝑟2

𝜕𝑉

𝜕𝜃+

1

𝑟2𝑆𝑖𝑛2𝜃

𝜕2𝑉

𝜕𝜆2= 0 ⇒ (2)

X 𝑟2

𝑟2𝜕2𝑉

𝜕𝑟2+ 2𝑟

𝜕𝑉

𝜕𝑟+

𝜕2𝑉

𝜕𝜃2+ cot θ

∂V

∂θ+

1

𝑆𝑖𝑛2𝜃

𝜕2𝑉

∂λ2= 0 ⇒ (3)

𝑉(𝑟, 𝜃, 𝜆) = 𝑓(𝑟). 𝑌(𝜃, 𝜆)

𝜕𝑉

𝜕𝑟= 𝑓′(𝑟). 𝑌(𝜃, 𝜆)

𝜕2𝑉

𝜕𝑟2= 𝑓′′(𝑟). 𝑌(𝜃, 𝜆)

𝜕𝑉

𝜕𝜃= 𝑓(𝑟)

𝜕𝑌

𝜕𝜃

𝜕2𝑉

𝜕𝜃2= 𝑓(𝑟)

𝜕2𝑌

𝜕𝜃2

𝜕𝑉

𝜕𝜆= 𝑓(𝑟)

𝜕𝑌

𝜕𝜆

𝜕2𝑉

𝜕𝜃2= 𝑓(𝑟)

𝜕2𝑌

𝜕𝜆2

By substituting above values to Laplace Equation

𝑟2𝑓′′𝑌 + 2𝑟𝑓′𝑌 + 𝑓𝜕2𝑌

𝜕𝜃2+ cot 𝜃 𝑓

𝜕𝑌

𝜕𝜃+

1

𝑆𝑖𝑛2𝜃

𝜕2𝑌

𝜕𝜆2= 0

1

𝑓[𝑟2𝑓′′ + 2𝑟𝑓′] = −

1

𝑌[𝜕2𝑌

𝜕𝜃2+ cot 𝜃

𝜕𝑌

𝜕𝜃+

1

𝑆𝑖𝑛2𝜃

𝜕2𝑌

𝜕𝜆2]

L.H.S. of the above equation is a function of 𝑟 and R.H.S. is a function of 𝜃, 𝜆.

Therefore both sides must be equal to a constant.

1

𝑓[𝑟2𝑓′′ + 2𝑟𝑓′] = −

1

𝑌[𝜕2𝑌

𝜕𝜃2+ cot 𝜃

𝜕𝑌

𝜕𝜃+

1

𝑆𝑖𝑛2 𝜃

𝜕2𝑌

𝜕𝜆2] = 𝑛(𝑛 + 1) ⇒ (4)

∴ 𝑟2𝑓′′ + 2𝑟𝑓′ − 𝑛(𝑛 + 1)𝑓 = 0 ⟹ (5)

𝜕2𝑌

𝜕𝜃2+ cot 𝜃

𝜕𝑌

𝜕𝜃+

1

𝑆𝑖𝑛2𝜃

𝜕2𝑌

𝜕𝜆2+ 𝑛(𝑛 + 1)𝑌 = 0 ⟹ (6)

Solutions of (5) ⟹ 𝑓(𝑟) = 𝑟𝑛 𝑜𝑟 𝑓(𝑟) =1

𝑟𝑛+1 ⟹ (7)

∴ 𝑉 = 𝑟𝑛𝑌𝑛 (𝜃 𝜆) 𝑜𝑟 1

𝑟𝑛+1𝑌𝑛(𝜃 𝜆) 𝑆𝑜𝑙𝑖𝑑 𝑠𝑝ℎ. ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐

𝑌𝑛(𝜃 𝜆) is Laplace Surface Spherical Harmonics

𝑌(𝜃 𝜆) = 𝑔(𝜃) ℎ(𝜆)

𝜕𝑌

𝜕𝜃= 𝑔′ℎ

𝜕2𝑌

𝜕𝜃2= 𝑔′′ℎ

𝜕2𝑌

𝜕𝜆2= 𝑔ℎ′′

By substituting to equation (6)

𝑔′′ℎ + cot 𝜃 𝑔′ℎ +1

𝑆𝑖𝑛2𝜃𝑔ℎ′′ + 𝑛(𝑛 + 1)𝑔ℎ = 0

sin 𝜃

𝑔[sin 𝜃𝑔′′ + cos 𝜃𝑔′ + sin 𝜃 𝑛(𝑛 + 1)𝑔] = −

ℎ′′

ℎ

L.H.S. of the above equation is a function of 𝜃 and R.H.S. is a function of 𝜆.

Both sides must be equal to a constant.

sin 𝜃

𝑔[sin 𝜃 𝑔′′ + cos 𝜃𝑔′ + sin 𝜃 𝑛(𝑛 + 1)𝑔] = 𝑚2

− ℎ′′

ℎ= 𝑚2

sin 𝜃𝑔′′ + cos 𝜃𝑔′ + [𝑛(𝑛 + 1) sin 𝜃 −𝑚2

sin 𝜃] 𝑔 = 0

ℎ′′ + ℎ𝑚2 = 0

Solution ℎ = sin𝑚𝜆 𝑜𝑟 cos𝑚𝜆

𝑔 = 𝑃𝑛𝑚(cos 𝜃) Legendre function

Where 𝑛 ≥ 𝑚 𝑛 − 𝑑𝑒𝑔𝑟𝑒𝑒;𝑚 − 𝑜𝑟𝑑𝑒𝑟

𝑌𝑛(𝜃, 𝜆) = 𝑃𝑛𝑚(cos 𝜃) cos𝑚𝜆 and 𝑃𝑛𝑚(cos 𝜃) sin𝑚𝜆

The most general solution for Laplace Surface Spherical Harmonics is,

𝑌(𝜃, 𝜆) = ∑[𝑎𝑛𝑚 𝑃𝑛𝑚 (cos𝜃) cos𝑚𝜆 + 𝑏𝑛𝑚𝑃𝑛𝑚(cos θ) sin𝑚𝜆]

𝑛

𝑚=0

∴ 𝑉(𝑟, 𝜃, 𝜆) = ∑ 𝑟𝑛 ∑[𝑎𝑛𝑚𝑃𝑛𝑚(cos𝜃) cos𝑚𝜆 + 𝑏𝑛𝑚𝑃𝑛𝑚(cos𝜃) sin𝑚𝜆]

𝑛

𝑚=0

∞

𝑛=0

Function Harmonic inside a sphere

OR

𝑉(𝑟, 𝜃, 𝜆) = ∑1

𝑟𝑛+1∑[𝑎𝑛𝑚𝑃𝑛𝑚(cos 𝜃) cos𝑚𝜆 + 𝑏𝑛𝑚𝑃𝑛𝑚(cos 𝜃) sin𝑚𝜆]

𝑛

𝑚=0

∞

𝑛=0

Function Harmonic outside

𝑡 = cos 𝜃

𝑃𝑛𝑚(𝑡) =1

2𝑛 𝑛! (1 − 𝑡2)

𝑚2

𝑑𝑛+𝑚(𝑡2 − 1)𝑛

𝑑𝑡𝑛+𝑚

When 𝑚 = 0 , 𝑃𝑛 0(𝑡) is called as Legendre Polynomials.

𝑃𝑛(𝑡) =1

2𝑛 𝑛!

𝑑𝑛(𝑡2 − 1)𝑛

𝑑𝑡𝑛

𝑃𝑛𝑚 (𝑡) = (1 − 𝑡2)𝑚2

𝑑𝑚𝑃𝑛(𝑡)

𝑑𝑡𝑚

Rodrigus Formula

Recursion Formula

𝑃𝑛(𝑡) = −𝑛 − 1

𝑛 𝑃𝑛−2(𝑡) +

2𝑛 − 1

𝑛+ 𝑃𝑛−1(𝑡)

Another explicit formula for Legendre function is

𝑃𝑛𝑚(𝑡) = 2−𝑛(1 − 𝑡2)𝑚2 ∑(−1)𝐾

(2𝑛 − 2𝐾)!

𝐾! (𝑛 − 𝐾)! (𝑛 − 𝑚 − 2𝐾)!𝑡𝑛−𝑚−2𝐾

𝑟

𝐾=0

This formula can be used for programming

Where 𝑟 is the greatest integer ≤ (𝑛 − 𝑚)/2

𝑃0 (𝑡) = 1 𝑃1 (𝑡) = 𝑡 𝑃2 (𝑡) =3

2 𝑡2 −

1

2

𝑃3 (𝑡) =5

2 𝑡3 −

3

2 𝑡 𝑃4 (𝑡) =

35

8 𝑡4 −

15

4 𝑡2 +

3

8

𝑃11 (cos𝜃) = sin 𝜃 𝑃21 (cos𝜃) = 3sin 𝜃 cos 𝜃

𝑃22 (cos 𝜃) = 3 𝑆𝑖𝑛2𝜃 𝑃31 (cos 𝜃) = sin 𝜃 (15

2𝐶𝑜𝑠2 𝜃 −

3

2)

𝑃32(cos𝜃) = 15 𝑆𝑖𝑛2𝜃 cos 𝜃 𝑃33 (cos𝜃) = 15 𝑆𝑖𝑛3𝜃

Legendre’s polynomials (𝑚 = 0) have 𝑛 zeros in the interval 0 ≤ 𝜃 ≤ 𝜋 (1 ≤ 𝑡 ≤ −1)

As 𝑚 = 0 they don’t change with 𝜆. They divide the sphere into 𝑛 + 1 zones

(Change the sign 𝑛 times) and are called as zonal harmonics.

Associated Legendre functions change their signs 𝑛 − 𝑚 times in the interval

0 ≤ 𝜃 ≤ 𝜋. Mean time function cos𝑚𝜆 and sin𝑚𝜆 have 2𝑚 zeros in the interval

0 ≤ 𝜆 ≤ 2𝜋. ∴The sphere get divided into 𝑛 − 𝑚 zones and 2𝑚 sectors to form

tesseral harmonics.

When 𝑛 = 𝑚 there are no sectors, but the sphere get divide into 2𝑚 sectors. They

are called Sectorial harmonics.

𝑃55(+) cos5𝜆

When 𝑛 increasing number of zones also increasing 𝑚 is also increasing. Therefore

tessarals getting smaller, to represent harmonic function accurately.

𝑅𝑒𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 180

𝑛

The Gravity Field of the Earth

The gravity on a body resting on the earth has two components.

1. Gravitational force

2. Centrifugal force

Centrifugal force on a unit mass = 𝜔2𝑟

∴ 𝑓 = 𝜔2(𝑥2 + 𝑦2)1/2

Centrifugal potential 𝜙 is

𝜙 = 1

2𝜔2(𝑥2 + 𝑦2)

𝑔𝑟𝑎𝑑 𝜙 = 𝜔2 (𝑥𝑖 + 𝑦𝑗)

= 𝜔2𝑝

The gravity potential

𝑊 = 𝑉 + 𝜙

= 𝐾 ∭𝜌

𝑙𝑑𝑉 +

1

2𝜔2(𝑥2 + 𝑦2)

𝑉

Δ𝑊 = −4𝜋𝑘𝜌 + 2𝜔2

This does not satisfies the Laplace equation

When q = 0 Δ𝑊 = 2𝜔2(𝑥2 + 𝑦2)

𝜕𝑊

𝜕𝑥= −𝐾 ∭𝜌

(𝑥 − 𝜉)

𝑙3 𝑑𝑉 + 𝜔2𝑥 = 𝑔𝑥

𝜕𝑊

𝜕𝑦= −𝐾 ∭𝜌

(𝑦 − 𝜂)

𝑙3 𝑑𝑉 + 𝜔2𝑦 = 𝑔𝑦

𝜕𝑊

𝜕𝑧= −𝐾 ∭𝜌

(𝑧 − 𝜁)

𝑙3 𝑑𝑉 + 𝜔2𝑧 = 𝑔𝑧

Gravity Vector 𝑔 = 𝑔𝑥𝑖 + 𝑔𝑦𝑗 + 𝑔𝑧 𝑘 = 𝑔𝑟𝑎𝑑 𝑊 = ∇𝑊⃗⃗⃗⃗⃗⃗ ⃗

The magnitude of 𝑔 is called gravity. In physical Geodesy units of gravity is gals (1gal

= 1cms-2). The unit being named in honour of Galileo Galilei.

𝑔equator= 978 gals 𝑔pole = 983 gals

The direction of the gravity vector is known as the direction of the plumb line.

In addition to the centrifugal force, Coriolis force act on a moving body with respect to

the earth.

Level Surface and Plumb Lines

A surface which is having constant potential is known as equipotential surface of Level

surface.

tant W = W0

Assume 𝑑𝑥⃗⃗ ⃗⃗ is a vector along the equipotential surface

𝑑𝑥 = 𝑑𝑥 𝑖 + 𝑑𝑦𝑗 + 𝑑𝑧𝑘

𝑔. 𝑑𝑥 = ∇𝑊⃗⃗⃗⃗⃗⃗ ⃗. 𝑑𝑥

=𝜕𝑊

𝜕𝑥𝑑𝑥 +

𝜕𝑊

𝜕𝑦𝑑𝑦 +

𝜕𝑊

𝜕𝑧𝑑𝑧

= 𝑑𝑊

But on the E.P. surface 𝑑𝑊 = 0

∴ 𝑔. 𝑑𝑥 = 0

This is a proof for gravity ⊥ E.P. surface.

Now will take 𝑑𝑥 along the plumb line,

𝑑𝑊 = 𝑔. 𝑑𝑋⃗⃗⃗⃗ ⃗

= 𝑔. 𝑑�⃗⃗�

= 𝑔. 𝑑𝐻. cos(1800)

𝑑𝑊 = 𝑔. 𝑑𝐻

∴ 𝑔 = −𝑑𝑊

𝑑𝐻

Plumb lines are slightly curved and always ⊥ to level surfaces.

The height H is measured along the plumb line. Height of a point from Geoid along the

plumb line is called as Orthometric Height

By taking vector 𝑑𝑥 along the plumb line

|𝑑𝑥 | = 𝑑𝐻

𝑑𝑊 = 𝑔. 𝑑𝑥 = 𝑔. 𝑑𝐻 cos 180 = −𝑔𝑑𝐻

𝑔 = −𝜕𝑊

𝜕𝐻

Geodetic measurements (Leveling) are almost referred to the system of level surfaces.

Natural Coordinates

Geographical Latitude, Longitude and Orthometric height are known as natural

coordinates.

PN // rotation axis

GPF ⊥ PN

NPF is the meridional plane

NPG // meridione plane of Greenwich

𝑔𝑥 = −𝑔 cos𝜙 cos λ

𝑔𝑦 = −𝑔 cos𝜙 sin λ

𝑔𝑧 = −𝑔 sin𝜙

𝜙 = tan−1−𝑔𝑥

√𝑔𝑥2 + 𝑔𝑦

2

𝜆 = tan−1 (𝑔𝑦

𝑔𝑥)

The third coordinate is orthometric height.

𝑊 = 𝑊0 − ∫ 𝑔 𝑑𝐻𝐻

0

𝐶 = ∫ 𝑔 𝑑𝐻

𝐻

0

𝐶 = 𝑊 − 𝑊0

∴ 𝐻 = − ∫𝑑𝑊

𝑔

𝑊

𝑊0

= ∫𝑑𝐶

𝑔

𝐶

0

𝐶 − 𝐺𝑒𝑜𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟

The Gravity Field of the Level Ellipsoid

Level ellipsoid is an equipotential surface of a normal gravity field.

𝑈 = 𝑉 +1

2𝜔2(𝑥2 + 𝑦2)

𝑈 − 𝑁𝑜𝑟𝑚𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙

𝑉 − 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙

𝑥2 + 𝑦2

𝑎2+

𝑧2

𝑏2= 1

Normal Gravity

𝛾 = 𝑔𝑟𝑎𝑑 𝑈

Assuming the Curvilinear coordinates of point as (𝑈, 𝜃, 𝜆) it can be shown that,

𝛾 =𝐾𝑀

𝑎√𝑎2𝑆𝑖𝑛2 𝛽 − 𝑏2𝐶𝑜𝑠2𝛽[(1 +

𝑚

3

𝑒′𝑞0′

𝑞0) 𝑆𝑖𝑛2𝛽 + (1 − 𝑚 −

𝑚𝑒′𝑞0′

6𝑞0)𝐶𝑜𝑠2𝛽]

𝑚 =𝜔2𝑎2𝑏

𝐾𝑀 𝑒′ =

√𝑎2 − 𝑏2

𝑏2 𝐸 = √𝑎2 − 𝑏2

𝑞0 =1

2[(1 +

3𝑏2

𝐸2) tan−1

𝐸

𝑏− 3

𝑏

𝐸]

𝑞0′ = 3(1 +

𝑏2

𝐸2)(−1 −

𝑏2

𝐸2) (1 −

𝑏

𝐸tan−1

𝐸

𝑏) − 1

𝛽 = 90 − 𝜃

∴ At the equator,

𝛾𝑎 =𝐾𝑀

𝑎𝑏(1 − 𝑚 −

𝑚𝑒′𝑞0′

6𝑞0)

At the poles,

𝛾𝑏 =𝐾𝑀

𝑎2(1 +

𝑚𝑒′𝑞0′

3𝑞0)

The following equation will be satisfied by 𝛾

Clairaut Equation

𝛾𝑏 − 𝛾𝑎

𝛾𝑎=

𝑎 − 𝑏

𝑎=

𝜔2𝑏

𝛾𝑎(1 +

𝑒′𝑞0′

2𝑞0)

By using above formulas

𝛾 =𝑎 𝛾𝑏 𝑠𝑖𝑛

2 𝛽 + 𝑏 𝛾𝑎 𝑐𝑜𝑠2 𝛽

√𝑎2𝑠𝑖𝑛2 𝛽 + 𝑏2𝑐𝑜𝑠2𝛽

tan𝛽 = 𝑏

𝑎tan 𝜙 𝜙 − 𝑔𝑒𝑜𝑔𝑟𝑎𝑝ℎ𝑖𝑐𝑎𝑙 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒

𝛾 = 𝑎 𝛾𝑎 𝑐𝑜𝑠

2 𝜙 + 𝑏 𝛾𝑏 𝑠𝑖𝑛2 𝜙

√𝑎2𝑐𝑜𝑠2 𝜙 + 𝑏2𝑠𝑖𝑛2 𝜙

Burns formula

𝜕𝛾

𝜕ℎ= −2𝑔𝐽 + 2𝜋𝑔𝜌 − 2 𝜔2

𝐽 = 1

2 (

1

𝑀+

1

𝑁) 𝑀 =

𝑎 (1 − 𝑒2)

(1 − 𝑒2 𝑠𝑖𝑛2 𝜙)32

𝑁 = 𝑎

(1 − 𝑒2 𝑠𝑖𝑛2 𝜙)12

Outside the ellipsoid 𝜌 = 0

𝜕𝛾

𝜕ℎ= −2 𝛾𝐽 − 2𝜔2

𝛾ℎ = 𝛾 [1 −2

𝑎 (1 + 𝑓 + 𝑚 − 2𝑓 𝑠𝑖𝑛2 𝜙)ℎ +

3

𝑎2 ℎ2]

𝛾ℎ = 𝛾 − (0.30877 − 0.00045 𝑠𝑖𝑛2 𝜙)ℎ − 0.000072 ℎ2

𝛾, 𝛾ℎ − 𝑖𝑛 𝑔𝑎𝑙𝑠 ℎ − 𝑖𝑛 𝑘𝑚

Measurement of Gravity

1. Measurement of absolute gravity

2. Measurement of relative gravity

1. Measurement of absolute gravity

a. free falling body method

𝑥1 − 𝑥0 = 𝑣0 𝑡1 +1

2 𝑔𝑡1

2

𝑥2 − 𝑥0 = 𝑣0 𝑡2 +1

2 𝑔𝑡2

2

𝑥3 − 𝑥0 = 𝑣0 𝑡3 +1

2 𝑔𝑡3

2

𝑥2 − 𝑥1 = 𝑣0 (𝑡2 − 𝑡1) +1

2 𝑔(𝑡2

2 − 𝑡1 2)

𝑥2 − 𝑥1

𝑡2 − 𝑡1= 𝑣0 +

1

2 𝑔 (𝑡2 + 𝑡1)

// y

𝑥3 − 𝑥1

𝑡3 − 𝑡1= 𝑣0 +

1

2 𝑔 (𝑡3 + 𝑡1)

𝑔 = 2 [(𝑥3 − 𝑥1)(𝑡2 − 𝑡1) − (𝑥2 − 𝑥1)(𝑡3 − 𝑡1)

(𝑡3 − 𝑡1)(𝑡2 − 𝑡1)(𝑡3 − 𝑡2)]

Free falling body operates uses a laser beam interferometer to measure distance and

atomic clock to measure time.

S. D. for 25 drops = ±8 μgals

1. Measurement of relative gravity

a. Pendulum method

𝑇 = 2𝜋√𝑙

𝑔

𝑇1

2

𝑇22 =

𝑔2

𝑔1

𝑔2 − 𝑔1

𝑔1=

𝑇12 − 𝑇2

2

𝑇22

𝐴𝑐𝑐𝑢𝑟𝑎𝑐𝑦 = ± 10−5 𝑔𝑎𝑙𝑠

Pendulum gravimeters are not portable.

b. LaCoste and Romberg Gravity meter

𝑆𝑝𝑟𝑖𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ ∞ 𝑔

The spring is returned to the original position by rotation of micrometer.

Micrometer reading to be converted to the gravity measurement by calibration table. There are two models of L and R meters

Model G

Range 700 m gal

Precision 0.005 m gal

Accuracy ± 0.04 m gal

Model D

Range 200 m gal (Extended 300 m gal)

Precision 5 𝜇 gal

Accuracy ± 0.01 m gal

𝑔2 − 𝑔1 = 𝑐 (𝑑2 − 𝑑1) 𝑑 − 𝑑𝑖𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 , 𝑔 − 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑣𝑎𝑙𝑢𝑒𝑠