Gravity and the Figure of the Earth

The Size of the Earth

see "Earth as a Planet" notes and section 2.11 of Lowrie

The Figure of the Earth (geodesy)

Jean Richer, 1672: pendulum clock, accurate in Paris, lost a 2 1/2 minutes per day in Cayenne, French Guiana
Newton, 15 years later (1687, Philosophiae Naturalis Principia Mathematica) correctly interpreted as due to oblate Earth (pumpkin) caused by centrifugal force due to Earth's rotation
Newton, assuming homogeneous Earth:
flattening = (c-a)/a = 1:230 (~0.5%); actual 1:298 (~0.3%)
implies length of degree of meridian arc should subtend a longer distance in polar regions than near equator
French, based on traverse in France, believed earth was prolate (rugby ball)!
French Academy of Sciences, 1736-1737 sent expedition to measure degree of meridian arc in Lapland (Sweden), another, 1735-1743, measured 3 degrees of arc in Peru.
Concluded Earth indeed oblate
a = 6378.136 km
c = 6356.751 km
R = 6371.000 km
(a-c)/a = 1/298.257
a-R = 7.1 km
R-c = 14.2 km

Figure of the Earth and its Rheology

A fluid with the Earth's density distribution, rotating once every day, would have a flattening of almost exactly observed 1:298.25 flattening.

Is Earth fluid?
Or was it fluid?

Gravity

Gravity is a vector quantity, with direction and magnitude
Often treated as scalar because we generally can only measure |g|
For 2 point masses, we know

For real body, must divide into infinitesimal mass elements, dm, find gravity due to each, then find vector sum
It is often convenient to use concept of potential.
Potential field is a scalar field from which the vector gravity field can be found; other examples: elevation-slope, temperature-heat flow

If a (vector) force field is conservative, it may be represented by (the gradient of) a scalar potential. If a force field has a scalar potential, it is conservative.

Potential field/scalar field example

change in potential between two points is work done to move from point A to point B
since conservative, work done to move from point B to point A is equal and opposite
therefore, work done to go in a closed path is zero; i.e., conservative
Work done to take a point mass from location r to infinity
gravity is the gradient of potential
the components of gravity, then, are

Escape Velocity

Escape Velocity Escape velocity is the velocity required to launch an object to escape a planet's gravitational force (not just in orbit around the planet). We have seen that gravitational potential at a point is the work required to move a unit mass from that point to infinity. Potential has units of energy/mass. We found that gravitational potential, U, is given by

where M is the mass of the point mass for which we are finding potential (in this case, mass of the planet, insofar as a planet is spherically symmetric) and R is the distant from the point mass (or from the center of the planet).

To escape a planet's gravitational field, it must have kinetic energy equal to the gravitational potential energy. Kinetic energy of an object, with mass m, is given by

or, per unit mass,

Escape velocity is obtained by setting these equal and solving for V:

Using values for an object starting at the Earth's surface, we get

Finding g from U in spherical coordinates

Note on signs: defined this way, g will be negative, because it points in the opposite direction of the unit radial vector. For this reason, you sometimes see g defined as the positive gradient of potential, so that g (and |g|) will be a positive number, for convenience.

A note about g, and G!

Integrating over masses to find total field

Because gravity is linear in mass (dm), we could find the gravitational acceleration due to an extended body by vectorially adding (integrating) the gravity due to the infinite infinitesimal masses that make up that body, but this would be complicated. Because potential also depends linearly on mass (dm), and is scalar, integrating the potential over a body is easier. The potential due to several (even infinite) dm's is the sum (integral) of the potentials due to individual dm's. In Cartesian coordinates, for example,

For an arbitrary mass distribution (Cartesian coordinates)

For an arbitrary mass distribution (spherical coordinates)

Example: what is potential due to sphere of density r?

Poisson's and LaPlace's equations

Deriving Poisson's and LaPlace's equations

Mass inside the volume.

From Gauss's theorem:

Since this holds no matter how the volume is chosen,

If M is outside the volume, total solid angle is 0 (2 ways to look at this: the surface presents just as much of its front as its back, so they cancel, or notice that the flux lines which go in one side of the volume bounded by the surface come out the other side, so the net flux is zero), so

Note that Laplace's equation is just the special case of Poisson's equation (where density is zero.)

Applications of Poisson's Equation in Integral Form

From derivation above, we have:

[Consider the equation above "Poisson's Equation in Integral Form"]

1. Gravity due to spherically symmetric body: put imaginary surface ("Gaussian surface") around the sphere

where M is the mass contained within the Gaussian surface.

2. Gravity inside a spherically symmetric hollow shell: put imaginary surface ("Gaussian surface") anywhere within the hollow region around the sphere

Since the mass contained within the Gaussian surface is zero,

Optional Assignment: Read Edgar Rice Burrough's At the Earth's Core

3. Gravity due to an infinite slab of thickness h and density r: Bouguer's Formula

Consider "pill box" or cylindrical Gaussian surface
no flux out of sides of cylinder, by symmetry
g through top and bottom must be constant and perpendicular to top and bottom (again, symmetry), so:

Problems to contemplate:

find gravity {g(r)} inside and outside a homogeneous sphere
find gravity {g(r)} outside an infinite cylinder with mass per unit length s (or, if it makes is easier to visualize, radius R and density r, r>R)
find gravity {g(r)} inside (r<R) a homogeneous infinite cylinder or radius R with constant density r

General Solution to LaPlace's Equation in Spherical Harmonics (Spherical Harmonic Analysis)

Elliptical, Parabolic and Hyperbolic PDEs

LaPlace's equation is , and in rectangular (cartesian) coordinates,
In spherical coordinates, where r is distance from the origin of the coordinate system, q is the colatitude, and f is azimuth or longitude:
Solutions to LaPlace's equation are called harmonics
In spherical coordinates, the solutions would be spherical harmonics
Example: show that for point mass ()

Solving LaPlace's Equation

Assume variables are separable: , so
Multiply through by :
Last term on LHS depends only on l, yet first two do not depend on l, so last term must be constant (and first two must add up to negative of that constant).

This is of the form

This an ODE, with solution , where m is an integer
Going back to the first two terms, we have
Multiply through by :
Again, terms must be independent, so both must be constant:

or which has the solution

Finally,
This is known as Legendre's Equation, and has solutions of the form , where are the Associated Legendre Polynomials, are constants

The general solution to LaPlace's Equation, then, is: [a is mean Earth radius]

Like any differential equation, the undetermined coefficients, in this case C_l^m, C'_l^m, S_l^m, S'_l^m (an infinite number of them!), must be determined by boundary conditions. A few of these are "common sense" boundary conditions; the rest have to be determined by best fit of the various harmonics to the Earth's gravitational field. Since we are continually improving our knowledge of gravity, the values of these constants are being refined.

Since a body that is finite in three dimensions (x, y, z) will "look like" a point mass at infinity, the gravity must tend to GM/r² as r goes to infinity, so the potential will go to -GM/r. This eliminates the C'_l^m, S'_l^m terms, because they depend on r^l
For l = 0, m = 0, the legendre polynomial P_l^m(cos(q)) (remember, this is a function, not a constant times cos(r)) is 1, so C'₀⁰ is identically equal to GM/r, where G is the Univ..., M is the mass of the body, and r is the distance from it. This term represents the "sphere" part of the potential.
If we set the origin at the center of mass of the body, there will be as much mass east and west of the center of mass, north and south of the center of mass, and in front and behind the center of mass. Therefore, the l = 1, m = 0 term must be zero, because it is asymmetrical between the northern and souther hemispheres. So, C₁⁰ = 0. This is because P₀⁰(cos(q)) = cos(q), which is positive in the N and negative in the S (or vice versa, since C₁⁰, if it weren't zero, could be negative).

Why Do We Care About Spherical Harmonic Analysis of Earth's Gravity?

Spherical Harmonic Analysis consists of determining values for (and significance of) constants
- for rotating Earth, might neglect l dependence, i.e., allow only m = 0 terms:

where are Legendre polynomials

or, for convenience
if we pick origin to be center of mass
, n odd, if equator is plane of symmetry

From Lowrie (2.52, p. 48):
Values determined by satellite:
- (oblateness)
- (pear-shapedness)
measurements of Earth's gravity field (primarily by satellite tracking) show that the biggest effect is due to Earth's rotation and bulge
Finally, we can get g from U by taking the gradient; Just keeping the first 3 terms leads to the International Gravity Formula (IGF; also sometimes called "normal" gravity), which we will present in the next section.

Zonal, Sectoral, Tesseral Components

zonal harmonics: m = 0, no longitudinal variation
sectoral harmonics: l = m, no latitudinal variation
general case: tesseral components (tessarae: Gr., "tiles")
m gives the number of nodal planes crossed on front side, E to W
l-m gives the number of nodal planes crossed from N pole to S pole

Shape of the Earth

This is the approximate shape of a rotating fluid body, and approximates the shape of the Earth. It is exactly of the form:

where a is the equatorial radius, and c is the polar radius, and b is latitude. Equatorial radius is 6378.1 km. Polar radius is 6356.8. [Volumetric mean radius is 6371.0 km.]

International Gravity Formula (IGF)

The IGF is a best fit to the Earth's gravity as determined from spherical harmonic analysis of the Earth's potential, using satellites. In these equations, q is geographic latitude. The IGF (g₀) is also commonly referred to as theoretical gravity or normal gravity

First internationally accepted IGF was 1930:
This was found to be in error by about 13 mgals; with advent of satellite technology, much improved values were obtained.
The Geodetic Reference System1967 provided the 1967 IGF:
Most recently IAG developed Geodetic Reference System 1980, leading to World Geodetic System 1984 (WGS84); in closed form it is: