Download The geoid on a rotating earth 1 Potentials for Gravity and Magnetism

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GS388 Notes : The geoid on a rotating earth
POTENTIALS FOR GRAVITY AND MAGNETISM....................................................1
GEOID ON A PERFECTLY RIGID ROTATING EARTH..............................................3
THE REAL, DEFORMABLE EARTH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Potentials for Gravity and Magnetism
Magnetic, electrical and gravity fields can all be expressed in terms of potentials, which are
simple scalar fields that are often easier to deal with than the vector fields. We need potentials
especially for gravity, but it is also useful for magnetism, where one can treat the magnetic field as
generated by magnetic "poles". Physically, the most elementary source of the magnetic field is the
dipole formed by a current loop. The concept of poles, however, is quite useful in analysing
magnetization and magnetic anomalies. A brief review follows. The units for the magnetic field
below are "emu" or electromagnetic units.
H=
p
R
R2
g =-G m R
R2
magnetic field vector
gravity field vector
R
R
+
+
pole strength = p
mass = m
The potential energy derives from the work required to move a unit mass from one point to
another. In the case of the gravity field, work is required to move a mass away from the center of
the earth. This increases the potential energy, which is then available to be converted into kinetic
energy if the mass is then released and allowed to fall. Thus for gravity, the force needed to move a
particle to greater R is in the direction of R, i.e. in the direction opposite to the gravitational
acceleration vector, so the potential energy accumulated is
P.E. = Ug =
F dR =
= Gm
G m dR
R2
dR = Gm - 1 = - Gm
R
R
R2
GS388 Notes : The geoid on a rotating earth
while for magnetics the force requried to move the test pole is the opposite direction, in the
direction of -R, so that moving a particle to greater R should decrease its potential energy rather
than increase it. Thus
p
P.E. = UH =
F dR =
dR
R2
p
= - p dR = - p - 1 =
R R
R2
The vector field can be derived from the potential as the spatial derivative of the potential. In vector
notation, using the gradient operator ∇ (which calculates the spatial derivative of the potential in
each of the three coordinate directions to form a vector), this is
g = - ∇Ug = -
∂Ug
R = - Gm R
∂R
R2
where the derivatives in the other two directions are zero because of spherical symmetry, and
∂U H
p
H =- ∇UH = R= 2 R
∂R
R
These simple expressions apply only to the fields of a single point mass for gravity or a single
point pole for magnetics. For more complicated distributions of mass or poles, there will be
components and derivatives in the other coordinate directions as well.
The dipole magnetic field, and the gravitational effect of rotation in "flattening" the earth
can all be expressed in terms of variations of the potential with respect to R and co-latitude, θ,
alone. The fields have symmetry about the dipole direction (the line through the two poles) in the
case of the magnetic field, or the axis of rotation in the case of the gravity field. The spatial
derivative of the potential in spherical coordinates looks like the following for gravity:
g =-
∂U g
1 ∂Ug
R+
θ
∂R
R ∂θ
H =-
∂U H
1 ∂UH
R+
θ
∂R
R ∂θ
and for magnetism:
Potentials are especially useful in analysis of the earth's shape and gravity field. The idea of
an equi-potential surface, a surface on which the gravitational potential is everywhere the
same, is the key idea in defining the global-scale shape of the earth. If the oceans were
homogeneous in composition and temperature and did not have currents, sea level would be a
perfect equipotential surface. It is not a bad one even in the real world. The shape of the surface of
the solid earth is not an equipotential surface. Although gravitational potential is constant on an
equipotential surface, the magnitude of the gravitational acceleration vector is not. On the surface,
however, the acceleration vector is everywhere perpendicular to the surface.
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GS388 Notes : The geoid on a rotating earth
Geoid on a perfectly rigid rotating earth
We first consider the simplest possible model to show how rotation affects the gravity
field. The model has a perfectly spherical earth which rotates but does not deform – it remains
spherical, which implies that it has infinite rigidity. The following figure shows the coordinate
system for our rotating earth. The dashed elliptical (ellipticity greatly exaggerated) shape in red
indicates the "flattened" equipotential surface
Ro
R sin θ
θ
φ
R
φ = latitude
θ = colatitude
Ro = radius of sphere
R = radial distance to
point on equipotential
surface
The next figure shows the balance of forces in a rotating earth. Here g m is the acceleration
due only to the mass of the sphere, and would be the only acceleration if the sphere were not
rotating. g 0 is the gravity observed on the rotating surface of the sphere. It is g m minus the part of
g m used to keep the observer on the rotating sphere. The observer does not sense this part. Hence
the effect of rotation is to reduce the observed gravity field by the amount required to hold the
observer on the sphere. The amount of reduction is given by - (centripetal acceleration) or - g ,
whose magnitude (from elementary physics) is just ω2 R sin θ, i.e. the rate or rotation (radians per
second) squared time the distance from the axis of rotation. This is the acceleration required to
make a body follow a circular path at the angular rate ω.
Now the vector sum of g m and - g could be obtained from plane trigonometry for the
triangle shown, but a simpler form of the vector addition can be derived for the situation where we
are adding (or subtracting) a large vector and a much smaller vector. The complete formula would
be
g02 = gm2 + gω 2 - 2 gm gω sin θ = gm2 [ 1 + (gω /gm)2 - 2 gω /gm sin θ)]
In the actual case where gω is very much smaller than gm, we can greatly simplify by throwing
away the squared term, (gω /gm)2, since it will be very tiny. We then get
g0 = gm[ 1 - 2 gω /gm sin θ ]1/2
GS388 Notes : The geoid on a rotating earth
which can be again simplified by expanding the square root with a power series. Remember the
expansion from calculus:
[ 1 + x ]n = 1 + nx + (terms with x2, x 3 etc)
If x is much less than 1, than the terms with powers of 2 and higher are tiny and can be neglected,
leaving just the first two terms 1 + nx. Applying this to the square root term of the preceding
equation, we have
g0 = gm [ 1 - (1/2) 2 gω /gm sin θ ] = gm [ 1 - gω /gm sin θ ]
or for R = R0, (also remember gω = ω2 R sin θ)
g0 = gm [ 1 - m sin2 θ ]
The blue line is parallel
to the surface of the
rigid sphere
the red line is the
equipotential surface
θ
gw
centripetal acceleration
perpendicular to axis of rotation
- gw
φ
+ gm
gm
gravitational
acceleration
due to
spherical mass
go
net observed acceleration:
go = gm - gw
where m is defined as the ratio of centripetal acceleration at the equator to the acceleration due to
the mass of the earth. We are considering gravity on the surface of the planet. g 0 given above is
the magnitude of the gravitational acceleration observed on the surface of our ideal planet. The
magnitude of the mass attraction by itself, g m , is simply GMe/R02. To represent the earth, Me and
R0 are taken as 5.973 x 10 24 kg and 6371 km, respectively, for a sphere of volume equal to the
actual spheroidal earth. m is therefore equal to [(2π radians/day)2 x R0] / [GMe/R02] or m = 3.4 x
10-3 = 1/291 for values representing a spherical earth. Thus the reduction of gravity at the equator
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GS388 Notes : The geoid on a rotating earth
compared to gravity at the pole due to the effect of rotation on our simple non-deforming earth is
just a part in 291.
The potential can be obtained by adding together the potentials for the mass attraction and
the effect of rotation. The potential for the mass attraction, as we found in the previous handout, is
simply U m = -GMe/R. The potential for the rotational part can be shown to be
U
= (1/2) ω2 R2 sin2 θ
The test of this is to take derivatives according to the defining equation. The equation is written in
terms of the orthogonal unit vectors in the R and θ directions; the "θ direction" is horizontal and
parallel to lines of longitude. These are parts of the standard spherical coordinate system.
∂U
∂Uω
gω = – ∇Uω = – ∂Rω R– 1R
θ
∂θ
If the expression for Uω is substituted in the above, you will get the appropriate representation of
gω . I will leave this to you as an optional exercise. Now, back to the main story. The combined
potential, U0, is then
U0 = - GMe/R + (1/2) ω2 R2 sin2 θ
U0 = - GMe/R [ 1 + (m/2) sin2 θ ]
Note that we have made another approximation here by representing
m = (ω2R0)/(GMe/R02) ≈ (ω2R)/(GMe/R2)
for R - R0 « R0.
If we pick a particular equipotential surface given by U0 = constant = - GMe/R0, we can
then think of the resulting equation as an implicit equation for R as a function of θ to give us the
shape of the equipotential surface:
- GMe/R0 = - GMe/R [ 1 + (m/2) sin2 θ]
or
R/R0 = [ 1 + (m/2) sin2 θ ]
which is the approximate equation for an ellipsoid of revolution, with ellipticity or flattening, f,
given by
f = (Requator - Rpole)/Rpole = m/2 ≈ 1/580
The actual flattening of the earth's equipotential surface is about twice this due to the effects
of the earth mass itself flattening into a spheroidal shape and thereby redistributing mass relative to
a spherically symmetric shape, as discussed in the next section.
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GS388 Notes : The geoid on a rotating earth
The real, deformable earth
The real earth flattens as a result of rotation, and so the gravitational potential due to the
earth's mass alone (without the effects of rotation), Um, is no longer simply -GM/R as for a
sphere, but is something more complicated. In general, Um can be represented as an infinite series
of simpler functions that are designed to work nicely on a sphere, called spherical harmonics. The
main advantage of this approach is that only the first few terms of the series need be used, since
each succeeding term is very much smaller than the preceding. Because of the overwhelming effect
of rotation, the shape of the earth has symmetry about the axis of rotation and is thus mainly a
function of latitude, φ, rather than longitude. The standard base model for the earth's shape and
gravity field has only this variation. The first two terms of the spherical harmonic expansion of Um
are as follows, where we use Req = Requator and co-latitude, θ (θ = 90° - latitude).
Um (R,θ) = –
2
GM e
GM e Req 3 sin2θ – 1
+J
2
R
R
R
2
The first term is the spherically symmetric part, while the second term gives the flattening or
ellipsoidal effect. The size of this effect relative to the spherical part is given by the coefficient J2 ,
and this number is closely related to the amount of flattening. Note that the second term falls of
with distance from the center of the earth proportional to 1/R3 whereas the first term falls off only
proportional to 1/R. Hence, at great distances from the earth only the first term is significant.
The total potential should include both the effect of the mass distribution and the effect of
rotation, i.e.,
U(R,θ)= U m + Uω
As we did for the rigid earth, we can set U = constant to see what the equipotential surface looks
like. Again, it has an ellipsoidal shape characterized by a flattening, f, where f = (R eq - Rp)/Req),
but the value of f is now increased over what it would be if the earth were not deformable. What
comes out (again, see Stacey for how all this works) is the following relationship between the
flattening, f , J2 and m, (where m = the ratio of centripetal acceleration and mass attraction, as
before):
f = (3/2)J2 + (1/2)m
This simple formula applies when just the two terms shown in the first equation are used in the
spherical harmonic expansion of the potential, but it is a pretty good approximation, with more
accurate equations having additional terms of order f2.
Thus the net flattening of the gravitational potential surface is the combined effect of
centripetal acceleration itself and the change in the distribution of mass from a spherical shape to a
slightly flattened shape that results from the spinning. The two effects add to increase the total
flattening from 1/580 due to centripetal acceleration to about 1/298 for the total.
What comes out of all this? For us, four main things:
(1) First, there is the determination of the earth's moment of inertia, which is crucial to the
determination of the distribution of density within the earth. This important quantity is derived
from two sources:
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GS388 Notes : The geoid on a rotating earth
- the coefficient in the series expansion of the potential, J2 , determined from observations
of the precession of the orbits of satellites about the earth (the careful tracking of the paths
of satellites orbiting the earth has become our best source of information about the long
wavelength parts of the gravity field);
- and measurement of the period of the precession of the equinoxes (about 26,000
years), which yields the fractional difference in polar (Ip ) and equatorial ( Ieq) moments of
inertia of the earth, (I p -I eq)/I p . These moments are taken about axes through the earth's
poles and equator, respectively. The precession of the equinoxes is due to the torques
exerted on the spinning earth by the effects of the gravitational attractions of the sun and
moon on the earth's elliptical shape.
If we take a spherical body and slightly flatten it so that the equatorial radius is greater than the
polar radius, the moment of inertia about a polar axis (Ip) will be greater than the one taken about
an equatorial axis (Iq), since mass is distributed out farther from a polar axis of rotation than from
an equatorial one. It can be shown (see Stacey, for example) that the distribution of mass that
produces the potential field is related to the distribution of mass that makes up the moments of
inertia in a particularly neat way, such that the following relationship holds:
J2 =
Ip – I eq
Me Req 2
Thus if we have both (I p -I eq) from determination of J2 and (I p -I eq)/I p from determination of the
precession of the equinoxes, then we can solve for Ip itself.
(2) With the known density distribution in the earth, the shape of a fluid, rotating body can be
calculated and a value for flattening derived for this body. The result of this calculation is that the
measured f and the value calculated for a fluid model are nearly the same, indicating that the earth is
close to hydrostatic equilibrium. The small difference can be explained as a result of dynamic
effects of convective motions in the mantle and outer core.
(3) The analysis of the effect of rotation described above provides the standard formula used to
calculate the variation of gravity as a function of latitude at sea level. The formula that is actually
used employs the next term in the expansion (the spherical term, the ellipsoidal term, and a third
term making the earth model a slightly non-ellipsoidal or "spheroidal" one). The third term
amounts to a rather tiny correction to the first two terms (of the order of f2 compared to f).
(4) The geoid is the actual equipotential surface that exists on the earth, with subtle undulations.
This surface is approximated by the surface of the world's oceans after the effects of tides and
ocean currents are removed. The surface is very much dominated by the flattening term, so to see
things of interest beside the effects of rotation, the elevation of the geoid is computed relative to the
rotational ellipsoidal or spheroidal models described above. The geoid only has a relief of several
hundred meters relative to the ellipsoid, and so is quite smooth, but the subtle geoidal
topography is of great significance to study of convection and plate tectonics. Note how smooth
the geoid is relative to the actual surface of the solid earth (geoid relief of as much as 200 meters
compared to solid earth relief of 20 km!).
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