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 = (ca)/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, 17361737 sent expedition to measure degree of
meridian arc in Lapland (Sweden), another, 17351743, measured 3 degrees of
arc in Peru.
 Concluded Earth indeed oblate
 a = 6378.136 km
 c = 6356.751 km
 R = 6371.000 km
 (ac)/a = 1/298.257
 aR = 7.1 km
 Rc = 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: elevationslope, temperatureheat 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
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

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 (r>R) an infinite cylinder with mass per unit
length σ (or, if it makes
is easier to visualize, radius R and density ρ)

find gravity {g(r)}
inside (r<R) a homogeneous infinite cylinder of
radius R with constant density ρ
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,
is the colatitude, and 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 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 (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 timedependent) 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 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/r^{2} 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(θ))
(remember, this is a function, not a constant times cos(r)) is 1, so C_{0}^{0} 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, C_{1}^{0} = 0. This is because P_{0}^{0}(cos(θ))
= cos(θ), which is positive in the N and negative
in the S (or vice versa, since C_{1}^{0}, 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
λ 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 (2.51, p. 63; Lowrie calls the Earth's mass E, instead of M )
 Values determined by satellite:


 (oblateness)
 (pearshapedness)


 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
 lm 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
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 (g_{0}) 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:
International Gravity Formula "Calculator"