Gravity and the Figure of the Earth

The Size of the Earth

The Figure of the Earth (geodesy)

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.


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

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 ρ?

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 ρ:  Bouguer's Formula

Problems to contemplate:

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

Elliptical, Parabolic and Hyperbolic PDEs

Solving LaPlace's Equation

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

This is of the form sho d.e. (this is probably where the term "harmonics" arises; see below "Simple Harmonic Oscillator problem)

  • 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

General comments on differential equations:

  • All soluble differential equations have a general solution (by definition).
  • The general solution always has undetermined constants (which is why they are general), similar to, if not identical to, constants of integration
  • To apply the general solution to a particular problem, you have to apply boundary conditions and (if time-dependent) initial conditions
    (see Simple Harmonic Oscillator problem)
  • For the general solution to LaPlace's equation, there are a couple boundary conditions that just make physical sense. The rest are determined by measurements of gravity around the Earth.

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


Like any differential equation, the undetermined coefficients, in this case Clm, C'lm, Slm, S'lm (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.

  1. 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'lm, S'lm terms, because they depend on r l
  2. For l = 0, m = 0, the legendre polynomial Plm(cos(θ)) (remember, this is a function, not a constant times cos(r)) is 1, so C00 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.
  3. 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(θ)) = cos(θ), 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?  

where are Legendre polynomials

Zonal, Sectoral, Tesseral Components

single harmonics

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 β 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, θ is geographic latitude. The IGF (g0) is also commonly referred to as theoretical gravity or normal gravity

International Gravity Formula "Calculator"

The Geoid