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Geophysical gravity
1 Earth’s orbit in the Solar System
The concept of Gravity was introduced to science by Sir Isaac Newton in his Philosophiae Naturalis Principia Mathematica in 1687. He recognized that he could explain the periodic orbits of the then-known planets about the Sun by intoducing a very sim- ple concept: the planets are attracted toward the Sun by a gravitational force that is inversely proportional to the square of their momentary distance from the Sun. Newton’s Law of Gravitation can be stated in simple mathematical form:
F^ ~gravity = G m^1 ·^ m^2 |~r|^3
· ~r
where ~r is the vector displacement of one mass, taken to be at the coordinate origin, to the other. Newton recognized that it was gravity that provided the force required to accelerate each planet along its elliptical orbit.
Let us assign m 1 = M to the coordinate origin centred on our Sun. The force acting on an orbiting planet, mass m 2 = mp, is simply
mp
d^2 ~r dt^2
which balances the gravitational attraction of the Sun so that
d^2 ~r dt^2
= −G
M
|~r|^3
~r.
While it might not be immediately clear to you, this is a non-linear differential equation that describes the evolution of the orbital position of the planet in time. Non-linear differential equations are not easy to solve by straight-forward analytic manipulations but given initial conditions, we can quite easily model the incrementing position using computers. Digital computers, however, cannot sufficiently resolve the position (due to short numerical word length and round-off error) to obtain an exact solution. Errors accumulate in the recursions.
v t
F cent^ v n
v r
Figure 1 The Sun, sitting at one focus of the ellipse of the orbit provides the gravitational attraction, the centripetal force, to hold the planet into its orbit. Without this gravity-provided centripetal force (green vector), the planet would escape, moving with constant velocity, along its orbital tangent (red vector). Note that the other focus is empty. The velocity of the planet normal to the vector direction of the cen- tripetal force is indicated in black. You might recognize that if the orbit were circular (both foci assembled at the circle’s centre), the normal (black) and tangental (red) velocities would be identical.
While the classical mechanics of stable elliptical orbits^1 is quite complicated, if the orbit is circular, it is not. For a circular orbit, vr = 0, vt = vn and the planet is always in the same gravitational potential field of the Sun. For elliptical orbits, we have to take into account and balance the accelerations due to the planet’s varying distance from the Sun as well as just providing the necessary centripetal acceleration. Let us look at a circular orbit where M is the mass of the Sun and mp is the mass of a planet moving in a circular orbit at radial distance |~r| from the Sun (taken as coordinate origin) with tangential velocity vt. If this orbit is to be stable, the centripetal force,
F^ ~cent = −mp|~vt|
(^2) ~r
|~r|^2 required to maintain the circular orbit is just that provided by gravity
F^ ~gravity = −GM^ mp |~r|^3
~r.
(^1) Prof. Walter Lewin’s lecture on elliptical orbits in his MIT Classical Mechanics course
1.1 Earth’s Moon
Earth’s Moon orbits the Earth in 1 month. What is the length of that month? There are many measures of the month depending upon the perspective one takes in determining the period of orbit. The best measure of the orbital period of the Moon for purposes of determining the mass of the Earth about which it orbits is measured in reference to the “fixed stars”: 1 tropical month. Properly, this orbital period is that about the centre of mass of the Earth-Moon system and not about Earth’s own centre of mass. For elliptical orbits such as that of the Moon about Earth, the appropriate |~r| in the equation above is the “length of the semimajor axis of the orbital ellipse”. This is just half the longest diameter through the ellipse. A fully classical mechanical derivation of the the description above would take us to this conclusion. That’s too involved for here and too much for me. The length of the tropical month is 27. 321582241 days of 86400 seconds. Until the advent of atomic clocks, this orbital period offered the most precise measure of time. The semimajor axis of the lunar orbit, measured from the Earth-Moon system’s centre of mass is 384748 km. The centre of mass of the Earth-Moon system is displaced along a line from the Earth’s centre of mass toward the Moon at perigee (Moon closest to Earth during its elliptical orbit.) by 4428 km and at apogee (Moon farthest from Earth) by 4943 km. Both the Moon and Earth are in elliptical orbits about this centre of mass. The Earth’s orbit is entirely within the body of the Earth whose radius is about 6371 km.
The Moon’s orbital eccentricity, presently e = 0. 055 , the inclination of its orbit relative to the ecliptic plane i = 5. 15 ◦, the tilt of its rotational axis relative to its orbital axis, 6. 69 ◦, and the inclination of Earth’s rotational axis, 23. 5 ◦, conspire to give us varying views of the near-side face of the Moon: Lunar libration. On long time scales, all of these factors vary cyclically.
An exercise (not for grading): I have given you, here, enough theory and data to obtain quite accurate measures of the masses of Earth and Moon. Try to do it!
It is interesting to note that the tropical month is getting longer by 0. 000000001506 days ( 1. 301184 × 10 −^4 seconds) every year. We shall come to this story later in this section.
Knowing the orbital period of the Moon, the displacement of the Earth-Moon system’s centre of mass from the Earth’s centre, we can determine the mass of the Earth and the Moon. It is by these means that we have determined the masses of all the planets of the Solar System and of those satellites of planets that probes have encountered. Neither Mercury nor Venus has an orbiting natural satellite, so it wasn’t until we had probes pass by these planets that we had accurate measures of their masses. Until we could resolve the 3 major satellites of Pluto and their orbital periods, we had seriously overestimated the size and mass of Pluto. Pluto is smaller and much less
massive than Earth’s Moon and many of the satellites or moons of the larger planets.
1.2 Lunar tides
The 1 /r^2 nature of gravitational forces has important implication for the momentary shape of bodies like Earth and Moon. From an Earth perspective, the differential gravity across the diameter of the Earth due to the 1 /r^2 nature of the gravitational attraction of our Moon lifts and depresses tides across the body of the Earth. That hemisphere of the Earth closest to the Moon is lifted toward the Moon and that opposite the Moon is relaxed relative to the centre of mass of the Earth-Moon system. The gravitational force gradient across the diameter of the Earth amounts to about
- 695 × 10 −^7 m · s−^2 or to about 3. 77 × 10 −^8 of the gravitational attraction of a mass on the surface toward the Earth’s centre. 3. 77 parts in 108 may seem like a very small anomaly in acceleration but the Earth as it spins under the Moon is constantly and locally adjusting to the varying attraction of its internal and surface materials toward the Earth’s centre of mass. That side of the Earth closest to the Moon faces a gravity reduction of about 1. 8 parts in 108 compared to the Earth’s centre of mass while the opposite side faces a gravity reduction as well. That is, the side closest to the Moon is being pulled toward the Moon; the side opposite the Moon is relaxed away from the Moon. These are the Earth tides or body tides of the Earth. There are a multiple infinity of body tidal periods; in our own gravimetry work, we used a theoretical tidal model that predicted the 3500 periods with largest amplitudes.
If the Earth were perfectly elastic, it would adjust immediately (actually with the speed of sound within its materials) to these variations in gravity. The Earth, though, is somewhat plastic in rheology and so the adjustment lags the gravitational pertur- bation along the Earth-Moon line by about 12 minutes.
out, this mechanical energy loss also contributes to a slowing of the rotation of the Earth. While energy is only conserved in the transferrance of form from mechanical to other forms in this case, momentum is necessarily conserved only in the motions. The Earth slows in its rotation and the Moon gains angular momentum.
The angular momentum, L~, of a point mass, m, moving in a circular path at radius r from the center of revolution is obtained as
~L = m · r^2 · ~ω
where ~ω is the angular velocity of the motion.
|~ω| =
2 π T
where T is the period of one rotation.
An exercise (not for grading): Above, I noted by how much the lunar orbital period increases each year. That means that ~ω is decreasing year by year for the Moon’s orbit. The Moon, though is also retreating from Earth as a consequence of the transferrance of angular momentum from the Earth to the Moon. Over the past 40 years, this retreat has been measured to be about 3. 85 cm/y. Calculate the annual increase in angular momentum of the Moon’s orbit. The consequence is that the Earth is losing just this amount of angular momentum annually and so its rotation slows. The moment of inertia of Earth referenced to its rotation axis is known to be Iz = 0. 3308 mEarth r Earth^24. The angular momentum of the Earth is Lz = Iz · ωz. What is the increase in the length of a day (i.e. 1 rotation of the Earth) in one year that corresponds to the angular momentum transfer to the retreating Moon?
The differential gravity across the body of the Earth due to the Sun’s attraction also provides a tide with an average 24-hour cycle. The amplitude of the solar tide is about 1 / 2 the amplitude of the lunar tide. The tides add together, beating one with the other, to produce a complicated cycle that is only truly periodic over very long time scales.
1.3 Earth rotation: wobbles, nutation, precession
The axis of rotation of the Earth is nearly space fixed and aligned with the geograph- ical coordinate system so that ± 90 o^ approximately coincides with the rotation axis. Actually, the Earth’s rotation is a very complex subject. There are hundreds of effects that disturb this simple model of rotation.
(^4) Properly, because the Earth is not exactly rotationally symmetrical, the moment of intertia is only properly and fully expressed as a tensor quantity. Here, Iz is the Izz element of the moment of inertia tensor; under rotational symmetry, Ixx = Iyy , the remaining 6 elements of the tensor I being zero.
The geographical coordinate system was “inscribed” on the body of the Earth so as to align in this manner with the rotation axis position averaged during the period January 0, 1900 to December 32, 1905. The average length of day during that period was assigned to be 86400 seconds, so defining the duration of the second. As the Earth’s rotation is slowing down (See previous subsection.), the length of day is increasing. Adjustments are made by adding a leap second to the year every 1 to 5 years as needed in order to bring the noon of atomic-based time in coincidence with the solar noon.
The IERS, International Earth Rotation (and Reference Systems) Service^5 obtains accurate measurements of the Earth’s rotation period and publishes them on various intervals, typically, averaged over a period of 5 days. In Figure 3, you see a record of their observations and adjustments to the atomic clock time in coordinating it with the Earth’s rotation time so that solar noon is never more than 1 second in variance from the UTC, Universal Coordinated Time, the standard time that we use both civilly and geophysically.
In 1972, it was decided that rather than continually adjusting our clocks, the leap seconds would be added when necessary during the moment between June 30 and July 1 or December 31 and January 1 each year.
(^5) IERS
Inertia Tensor. As angular momentum is conserved in the Earth’s rotation apart from the rotational-slowing effect described previously any changes in the Inertial Tensor caused by mass redistributions produce changes in the rotation vector, both in magnitude and direction. Moreover, any spinning body that is not perfectly spher- ically symmetric throughout can be disturbed into a free wobble. This wobble is like that that you may have seen as a child when playing with a spinning top or dradle. Tap the spinning top and it begins to wobble. So does the Earth whenever it’s Inertia Tensor or its rotation direction is changed. For example, if it were hit by an asteroid, both its Inertia Tensor due to the addition of mass at a point on the surface and its rotation direction due to the assymmetrical push on the surface would change immediately. The Earth’s free wobble is called the Chandler wobble^7 ; it’s period is about 438 days.
The Earth is also affected by forced periodic changes in its Inertia Tensor. On an annual cycle, seasonal meteorology moves mass about the planet. Ice and snow accu- mulates in the high latitude winter and melts away in its summer. Wind directions and strengths change the surface atmospheric forcing on the body of the Earth. These effects cause the Earth to wobble with an annual wobble with a 365. 25 -day period. This forced nutation beats with the Chandler wobble (a damped free nutuation with a period of ∼ 438 days) to produce a modulated cyclical motion of the Pole Path. Since the 1900-05 definition of the geographical latitude-longitude coordinate system, Earth’s geographical north pole has moved almost 12 m from the present rotation axis!
The rotation pole path and and rotation period has been monitored for over a century. The ILS, International Polar Motion Service, was established in 1899 to measure the effective change of latitude with time as measured by six observatories distributed around the Earth at 39◦ 08 ′^ N in order to compute the evolving rotation pole position. This service was updated in the 1960s as new and more accurate geodetic technologies became available, especially VLBI, Very Long Baseline Radio Interferometry. Now the IERS, International Earth Rotation Service, coordinates measurements made by all available techniques to produce a pole position averaged over a 5-day period every 5 days.
(^7) An excellent article on the subject by D.E. Smylie Current orientation parameters IERS
Figure 4: The diagram shows the rotation pole path measured in geo- graphical coordinates starting in January, 2015 and through August,
- 1 mas (milli-arc-second) represents about 3. 09 cm Note that, on July 31, 2016, the rotation pole position was about 15. 5 m south of the CIO (Conventional International Origin) along a longitude line 64 ◦^ W. The approximate center of this pole path is oriented 78 ◦^ W. Pole-path data from EOC-Paris Observatory Current-recent position
Figure 6: The diagram shows a high-resolution “maximum-entropy” power spectrum of the rotation path for the years 1990-2009. Note the rotation direction of the path is largely described by a negative frequency (representing clockwise motion in the coordinate system of the measurements), here measured in cycles/per year. The highest peak is that of the Chandler Wobble whose period is found to be about 438 days; that peak just left of the Chandler represents the residual (following our corrections – see below) forced annual wobble. Data used in this plot from IERS
In order to produce this spectrum from the pole-path data set, many corrections based on good theory and past experience were necessarily applied. A low-frequency background drift was removed, then a “best-fitted” model of the annual component of the pole path which, in this case, left us with a residual time series that retained the Chandler free wobble and some other lower amplitude wobbles and nutations. It is this Chandler wobble that might be excited by earthquakes, asteroid impacts and other geophysical effects. The annual component does dominate the pole path record but is, in principle, less than a factor of 2 greater in amplitude. We remove it as best we can in order to see through to the Chandler component.
1.4 Precession
As well as the body of the Earth moving across the rotation axis of the Earth, the rotation axis itself moves relative to the inertial system of the “fixed stars”. Prop- erly, the inertial reference system is determined by the positions of the most distant
quasars on the Celestial Sphere. These are tied to the various geographical reference systems for the Earth through “very long baseline radio interferometry” (VLBI) in- volving many of the largest radiotelescopes on Earth. Even before this extremely high accuracy measurement technique, the path of the rotation axis of the Earth was recognized and monitored as the apparent movement of the “pole star”, Polaris. Presently, Polaris that bright star most closely aligned with Earth’s rotation axis; in about 13000 years, the rotation axis will have moved to become almost coincident with the bright star Vega.
The Precession of the rotation axis is driven by torques applied to the rotating Earth due to differential gravitational forces imposed by the Moon on Earth’s equatorial bulge.
F (^) far side Fnear side
Figure 7: The difference of the gravitational attractive forces toward the Moon acting on the equatorial bulge on the near-Moon side of the Earth and far-Moon side produce a torque on the rotating Earth. The Earth’s reaction to the torque produces a continuous, 25772 -year precession of Earth’s rotation axis.
1.5 Shape of the Earth and surface gravity
If the Earth were isolated, homogeneous, at least layer by layer, non-rotating and in equilibrium, it would be pulled into a perfectly spherical shape through gravity. It is in fact almost spherical but it is rotating once every 86164. 092 seconds relative to the inertial system of the fixed stars. This is one “siderial day”. One “solar day” corresponds to 86400 seconds because, each day, the Earth has to spin just 1 / 365. 25 of one full rotation just to return to the same perspective toward the Sun – that is, from one noon to the next. The Earth rotates 366. 25 times during one year of 365. 25 solar days. The stars are so far away that our perspective on the stars doesn’t much change during a whole year. Actually, for the nearby stars, it does change just a little
Figure 7: The effect of the centripetal acceleration on the total downward acceleration and the direction of the vertical.
What has been left from this analysis is the fact that the Earth’s body and surface adjusts to the lowered, downward vertical acceleration at the equator as well and so the radius of Earth’s equator is greater than the polar radius by about 1 part in
- 25 , the “flattening parameter”:
f =
re − rp re
In order to take into account the full story of the variation in gravitation acceleration over the surface of the Earth and the flattened ellipsoidal shape of the Earth is, I think, beyond the scope of this U2 course. A proper, but still approximate analysis was first accomplished by Alexis Claude Clairault^8 in 1743. The limited argument I have given is essentially that of Newton, a century earlier. Clairault’s analysis is known as Clairault’s Theorem^9 which is regarded as one of the major accomplishments in geophysical theory. Clairault’s analysis obtains the radius of the Earth as a function of latitude as r(φ) = re(1 − f sin^2 (φ)) where re is the equatorial radius and the flattening
f =
3(Izz − Ixx) 2 r^2 e M♁
m 2
where m is the ratio of the centripetal acceleration at the equator to the Newto- nian gravitational attraction at the equator. The IERS’^10 best current estimate for f = 1 / 298. 25642 ± 0. 00001 corresponding to re = 6 , 378 , 136. 6 m and rp = 6, 356 , 751. 9 m.
(^8) Alexis Claude Clairault (^9) Clairault’s Theorem (^10) Earth Rotation Service’s General Definitions and Numerical Standards
Figure 8: Flattening and the shape of the Earth.
On the reference ellipsoid which most closely corresponds to the geoid, g~T is defined, geodetically, as the acceleration, locally downward: g~T : [gTx gTy gTz ] = [0 0 gTz ]. The International Gravity Formula (1967) – Helmhertz’ equation obtains gTz as a function of latitude in accord with Clairault’s theorem,
gTz = 9.780327(1.0 + 0.0053024 sin^2 (φ) − 0 .0000058 sin^2 (2φ)) [m · s−^2 ],
where φ is the local latitude.
1.5.1 The geoid
The geoid is that equipotential surface that most closely corresponds to the equi- librium, hydrostatic, nearly ellipsoidal shape of Clairault’s Earth. Colloquially, it is the mean sea level datum surface. Being an equipotential surface, there is no locally horizontal acceleration on that surface. The gradient of equipotential defines the ac- celeration on that surface, ∇U = g~T , where g~T assembles the sum of the Newtonian mass effect and the centripetal acceleration effect as well as effects of any internally anomalous mass-density within the Earth.
Figure 9: The geoid shown as departures from the reference ellipsoid. The deepest departure, south of India, is approximately 107 metres below the ellipsoid and the highest departure in New Guinea is about
- 4 metres above the reference ellipsoid.
ρ 0
geoid
ρ 0+ (^) ρ 0−
Figure 10: Geoidal surface above high ρ0+ and low ρ 0 − mass densi- ties. Note the gravitational acceleration vector direction is normal to the geoidal surface. Note, as well, that the gravitational accel- eration vector is not a constant on the geoidal surface but that it’s horizontal components are 0.
Before continuing with the interpretation of gravity measurements, it would be useful to understand isostasy.
- Isostasy: By the mid-1800s, the reasons for topography – differences in ele- vation – on the Earth were not yet understood. George Everest had by 1830 recognized that there was something odd about the mass of the Himalayan Plateau when he was correcting his survey measurements across India. We shall deal with his problem later even though his argument for its solution might have already have provided a prior explanation to the question posed to the members of the Royal Society of London: “What is the explanation for the elevation of the Himalayan Plateau and the Ande’s Mountain chain?”. In 1855, two models were offered to the members of the Society in explanation. Archdeacon J.H. Pratt suggested that the reason for high elevations is that light materials ”float” higher than do dense materials and that the rock of areas of high elevation are of low density. G.B. Airy proposed another model: the high-standing regions are compensated by deeper roots but their densities are similar to those of low stands. Airy’s hypothesis accounts, for example, for the height of icebergs. An iceberg that floats high has great root depth. We now know that over the continental regions of the Earth both models contribute
almost equally to topographic variations. Variations in ocean bottom depth is largely explained by Airy’s model. That is, where the oceanic lithosphere is thickest, and hence oldest, it sinks deeper.
Pratt Model
Darker columns have higher density
Depth of Compensation
Airy Model
All columns have equal density
Depth of Compensation
1.5.3 Measuring gravitational acceleration
Sensitive gravimeters are easily capable of measuring variations in gravity equivalent to 1 part in 107 |~g| on the surface of the Earth. That is, we easily and accurately measure to the 7 th^ decimal place in the surface gravitational acceleration. Geodetic instruments can be more sensitive by a factor about 100 and contemporary stationary observatory instruments by a factor of well more than 1000. At such details of
