Intro Geophysics

GRAVITY AND THE FIGURE OF THE EARTH

Consider two point masses, m1 and m2, separated by a distance r. Let r1 be a unit vector pointing from m1 to m2, then we can write the force F on mass m2 due to mass m1 as:

€

rF = −G

m1m2

r 2

r r 1

Newton's Law of Universal Gravitation

Sir Isaac Newton 1643-1727

CGS MKS F dynes newtons r centimeters meters m1, m2 grams kilograms G 6.67 x 10-8 dynes-cm2/g2 6.67 x 10-11 N-m2/kg2

Notes on Units

We will mostly be interested in the gravity field of the Earth, so we take m1 = Me (mass of the Earth). If we are at the surface of the Earth, then often r = Re (radius of the Earth).

Gravitational Acceleration

Since F = ma, we can define the gravitational acceleration of the Earth as a = g or

€

rg =

r F

m= −G

M e

Re2

r r 1

Technically we should use m/s2 as the unit for any acceleration, but a traditional unit often used in gravity work is the gal (after Galileo). 1 gal = 1 cm/sec2. The gravitational acceleration at the surface of the Earth is about 980 gals. Typically we are interested in variations in gravity on the order of a few thousandths of a gal, and so a common unit is the milligal or mgal. For geotechnical work we might even look for signals as small as 10-6 gal or a microgal or gal.

This means that instruments that measure gravity must be accurate to about 1 part in 109!

A bit about GG was first measured by Henry Cavendish (1731-1810) in 1797 using a torsion balance.

"[Cavendish] probably uttered fewer words in the course of his life than any man who lived to fourscore years."   -Lord Brougham

Why?At the surface, Me = gRe

2/G (and we already know g and Re).

Note that current measurements of G are done in almost the same way!

Cavendish’s experiment is often called “the Weighing of the Earth”

A question: is G a constant, or does G = G(r,t)?

Is G changing with time?

Note that G is a very weak force. For a proton:

Electromagnetic Force/Gravitational Force = 1042

Now 1042 is a universal constant, not an arbitrary number. It is also very big. The physicist Paul Dirac noted that this number also is represented by the following ratio:

Age of the Universe/Time for light to traverse a Proton

More than a coincidence? Because the Universe is getting older, Dirac postulated that gravity is getting weaker with time. So far, no one has found any evidence that this is that case.

Is G a function of distance?

Some scientists have proposed the existence of a “fifth force” that causes departures from Newton’s gravitational law at certain distances because of “exchange interaction” terms. In this case, we would have

€

G(r ) = G∞ 1+α 1+r

λ

⎡

⎣ ⎢ ⎤

⎦ ⎥e−r / λ

⎡

⎣ ⎢

⎤

⎦ ⎥

were G∞ is G measured at large distances and a is the magnitude of the exchange interactions, and is a characteristic distance. Some estimate that = -.007 and = 200 meters, but so far experiments to nail this down have not been conclusive.

Potential Fields

Recall that a force moving an object a distance dr in the direction of the force does an amount of work dE equal to the product of the force times the distance:

dE = Fdr

The gravitational potential U represents the potential energy of a unit mass in a field of gravitational attraction. The total energy of a mass m is then

E = mU

The change in energy due to work being done in the field is

dE = mdU = -Fdr = -mgdr

Thus

€

g = −∂U

∂r

To generalize for all space, the we introduce the grad operator:

€

r∇ =

∂∂x

,∂

∂y,∂

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

or

€

rg = −

r ∇U = −

∂U

∂x,∂U

∂y,∂U

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

2r

MG

r

Ug −=−=

∂∂

For a point mass M:

r

GMU −=or (U -> zero at large r)

Some facts to know about potential fields:

Many different mass distributions can generate identical potential fields, so there is an inherent nonuniqueness to gravity work.

A surface along with U is constant is called an equipotential surface. No work against gravity is required to move a mass from one point to another on such a surface. An example would be any water surface.

Note that g is not necessarily constant along an equipotential surface. This is because g depends on the gradient of U, not on U itself. However, (and for the same reason) g is always perpendicular to an equipotential surface.

Because U is a scalar, it is often easier to compute gravity fields by first determining U and then taking a gradient rather than summing up all the g vectors.

We use U all the time in day-to-day life. We’ll come back to that one in a bit.

Rotational Accelerations

Gravity is all about acceleration, and so we have to be able to understand the effects other sources of acceleration

We live in rotating reference frames (rotating earth, rotating earth-moon and earth-sun systems) that cause accelerations (anything that causes a change in magnitude or direction of velocity is an acceleration).

Centripetal (center seeking) forces – these are the forces that are applied to keep an object going in a circle (e.g., stone on a string). Directed from the object to the center.

Gravity examples: Moon about Earth, Earth about Sun.

An observer in a non-inertial (accelerating) frame will experience a “Centrifugal” force; e.g. you push against a door as you go around a corner. It’s not “real” – just a noninertial figment. Still, it can be useful to talk about if you are in this kind of system, like when you live on a rotating planet.

From an article on Tractor Safety

€

Uc = −1

2ω 2x2

We describe a rotation by its angular velocity (radians/unit time) about an axis (we often use as a vector oriented along this axis using the right hand rule). The velocity at any distance x from the axis of rotation is v = x. As a vector, it is tangent to the rotation. The centripetal acceleration is a = 2x = v2/x, and is directed towards the axis of rotation.

A mass on a rotating earth experiences a centrifugal acceleration of 2x. The change to gravitational potential is found by integrating:

Gravitational Effects of the Sun and Moon on the Earth: The Tides

The gravitational acceleration of the sun and moon deform the Earth, and modify its rotation.

Rotation about the Center of massOrbiting bodies rotate about their center of mass. When one body is much more massive than the other (e.g. an artificial satellite around the Earth) the center of mass is virtually the same as that of the larger body.This is pretty much the case for the Earth-Sun system. BUT this is not the case for the Earth-Moon, because the moon is both massive AND close.

Let d be the distance of the mass center from the center of the Earth, E is the mass of the Earth, m is the mass of the moon, rL is the distance between the centers of the Earth and moon. The center of mass (also known as the barycenter) is determined by setting the rotational moments of the Earth and moon equal:

( )drmEd L −=

LrmE

md

+=

or

Plugging in known values (rL = 382,000 km, m/E = 0.0123) gives d = 4600 km.

Note that Re = 6371 km, so this is more than half way to the surface of the Earth.NB: We can take advantage of this bit of knowledge and look for large planets in other solar systems by detecting this type of tiny wobble in the star's position.

The result is a “revolution without rotation” as shown to the left.

Points on the surface of the Earth traverse the small circles as shown in this figure (radii = 4600 km).This rotation causes a centrifugal acceleration.

Note that this acceleration:1. Is the same for all points on the surface (all the circles are the same).2. Is directed away from the moon along the line of the Earth-Moon system

How does this effect the total acceleration at the surface?

The gravitational acceleration of the moon on the earth at the Earth’s center is

€

aL = Gm

rL2

Since the net acceleration of the center of the Earth towards the moon is zero (in the Earth-moon reference frame: rL is constant), the centrifugal acceleration is exactly equal and opposite to this amount. At a point (B) closest to the moon, the gravitational acceleration is

€

aB = Gm

rL − Re( )2

and this amount is reduced by the centrifugal acceleration, so

€

aB = Gm1

rL − Re( )2 −

1

rL2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟=

Gm

rL2 1−

Re

rL

⎛

⎝ ⎜

⎞

⎠ ⎟

−2

−1 ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥≈

Gm

rL2 2

Re

rL

+ 3Re

rL

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+L ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

At the farthest point (A) the gravitational acceleration is:

Note that the difference in acceleration goes like 1/rL

3.Remember that we are talking about accelerations here – NOT the shape of the Earth per se. We’ll get back to that.

The effect is to deform the Earth into a prolate ellipsoidal shape. This explains two tides per day, and the rotational axis tilt explains difference in elevation of tides.

€

aA = Gm1

rL2 −

1

rL + Re( )2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟≈

Gm

rL2 2

Re

rL

− 3Re

rL

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+L ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

The Sun has a similar effect but although the gravitational attraction of the Sun is about 200 times that of the Moon, the 1/r3 dependence of tides makes the tidal acceleration less than half that of the moon.

Nevertheless, there are spring and neap tides.

Again, we are talking about accelerations here – NOT the shape of the Earth per se. We’ll get to that in a minute.

Note that the height in this figure is exaggerated: tides in the open ocean are on the order of 60 cm (~2 ft) and those on land are about 20 cm.

Large tidal effects are seen only along the coast. Why is this?

Changes in the Earth’s Rotation: Length of Day

There is a tidal lag of about 12 minutes (difference of time between maximum tide and shortest distance to moon) or about 3 degrees, due to anelasticity of the Earth.

This results in an asymmetric torque on the Earth by the Moon, acting in a direction opposite to the rotation of the Earth.

The result is to slow the Earth down at a rate of 2.4 ms per century. Observations are more like 1.4 ms/century, so other factors, like the redistribution of the mass in the earth (and hence it’s angular momentum) or just redistribution of momentum itself (e.g. core mantle coupling) could be responsible.

Eventually, the earth will stop rotating relative to the moon – the length of an earth day will be equal to the rotational period of the moon (if we ignore the sun).

This has already happened to the moon – which is why we only see one side of it.

Angular momentum of the Earth-Moon system:

E Earth’s mass Earth’s rate of rotationC Earth Moment of inertiam Moon’s mass Moon’s rotation = rotation about EarthCl Moon’s moment of inertia

The total angular momentum is

C Due to Earth’s rotationEd2 Due to rotation about center of massm(rL-d)2 Due to Moon’s rotation about center of massCl Due to Moon’s rotation about its axis – this is a small number relative to the others and we will ignore it

€

Cω + EΩd2 + mΩ rL − d( )2

= Cω +E

E + mmΩrL

2 = const.

€

d = rL

m

E + m

Summing all the contributions to angular momentum gives:

Recalling that

The centripetal acceleration of the Moon about the center of rotation is

€

2 rL − d( ) and this equals the gravitational acceleration of the moon by the Earth:

€

GE

rL2 = Ω2 rL − d( ) = Ω2rL

E

E + m

⎛

⎝ ⎜

⎞

⎠ ⎟

€

rL2 = G E + m( )rLAnd so

Substitute this into the equation at the top of this slide and we get:

€

Cω +EM

E + m( )GrL = const.

Note the Earth will not stop rotating, it will just rotate at a rate equal to the rotation of the moon about the earth (actually, after a while the Sun’s tidal forces will be greater than the moon’s and so will eventually make the length of the day = 1 year).

€

rL2 = G E + m( )rL*Recall from before that

Tidal friction reduces , which means the second term must increase to compensate. This means rL should increase. And if rL increases, must decrease*, meaning that months are getting longer.

The Earth-Moon distance has been measured with lasers since the 1970s. The measured rate of increase is about 3.7 cm/yr.

Actual Measurements of the Length of Day

It turns out that the Earth is a terrible clock. In addition to the long term effects, the Earth speeds up and slows down by a few ms because of seasonal exchanges in momentum with the atmosphere and bimonthly solid Earth Tides.

Using VLBI, we have a very good record of LOD.

Chandler Wobble

The stable rotation of a body should be around the axis about which the moment of inertia is the greatest. This is called the axis of figure.

But, in reality, the real (instantaneous) axis of rotation is off a bit (by a few meters). For the most part this is because of uneven mass distribution within the body. The result is the while the angular momentum is constant, the direction of the rotation axis meanders.

This type of motion is called a nutation, and in this case is a free nutation because is occurs in the absence of an externally applied torque.

Such a wobble (free nutation) would result in periodic variations of the astronomic latitudes of all points on Earth. The expected period of the variation of latitude was 305 days.

Several astronomers attempted to detect the phenomenon during the succeeding century, without success.

Leonhardt Euler

1707-1783

About 1765 Leonhardt Euler, a Swiss mathematician studying the dynamics of rotating fluid bodies, developed equations that suggested the Earth might wobble slightly about its axis of rotation.

In 1888 German astronomer Friedrich Küstner published an analysis indicating that the latitude of the Berlin Observatory had changed during 1 year of observation.

Küstner's observations were not continuous enough to detect any periodicity in the variation of latitude. However, he argued strongly that the apparent change in latitude was real and his evidence was sufficiently convincing that the International Geodetic Association (now the International Association of Geodesy) organized a special observational campaign to verify his discovery.

Küstner subsequently refined his analysis, finding a total variation in latitude of 0.5 seconds of arc, but gave no value of the period or direction of the motion of the pole (Küstner, 1890).

Using the Almucantar results, along with observations made in Berlin, Prague, Potsdam, and Pulkova, Chandler found a periodic variation of latitude, with a total range of about 0.7 seconds of arc and a period of 427 days, approximately 14 months (Chandler, 1891a and 1891b). The 40 percent discrepancy between the 305-day period predicted by theory and the 427-day observed period was explained by Simon Newcomb as the consequence of the "fluidity of the oceans" and the "elasticity of the Earth" (Newcomb, 1891).

The Royal Astronomical Society of London awarded Chandler the Gold Medal, but many scientists (particularly European scientists) continue today to credit Küstner with the discovery of the wobble. However, the 14-month wobble of the pole is universally referred to as the Chandler wobble, and there is no argument that Chandler illuminated the complex nature of the phenomenon, dominating the subject for decades.

Seth Carlo Chandler, born in Boston in 1846, finished his education at English High School in Boston in 1861.

In the 1880’s Chandler invented a “floating telescope” called the Almucantar which he installed at Harvard and made detailed observations of latitude in 1884-1885.

Chandler discovered two components of nutation in 1891:1. 0.1 seconds of arc, annual period – due to changing of seasons 2. 0.15 seconds of arc, period of 435 days – the free nutation but

longer than 305 days due to elasticity (non-rigidity) of the Earth

The “beating” of two frequencies is very clear in this record of the Chandler Wobble from the IERS.

Precession and nutation of rotation axis

The earth is tilted by about 23.5 degrees to the ecliptic. This means that the flattening of the earth caused by its rotation will make for asymmetric torques similar to Tidal friction by sun and moon and this causes a precession of the axis. Minimum at equinoxes, maximum at solstices.

The sense of the precession is retrograde – opposite direction of earth’s rotation. Period of precession is 25,700 years. Effect is to change the location of the equinoxes by 50.4 arcsec/year. Polaris is now at north pole, but in 3000 BC it was Alpha Draconis.

The variation in the torque due to position of sun and moon produces forced nutation from both (period of 1/2 year from sun and 1/2 month from moon). There are others due to the precession of the lunar orbit with a period of 18.6 years

Hipparchus calculated the length of the year to within 6.5 minutes and discovered the precession of the equinoxes. Hipparchus's value of 46" for the annual precession is good compared with the modern value of 50.26" and much better than the figure of 36" that Ptolemy was to obtain nearly 300 years later.

Precession was first observed by Hipparchus in 120 BC.

QuickTime™ and aSorenson Video 3 decompressorare needed to see this picture.

NOTE: Chandler is the motion of the rotation within the earth (relative to inertial frame) but others are the motion of the rotation axis w.r.t. the solar system.

Why is this important?

1. Precision navigation depends on knowing the reference frame, which in this case is changing all the time. That’s why we have an IERS now.

2. Effect on climate: Milankovich cycles. The position of the earth in space changes the amount of solar radiation and the periods of the seasons. This can create ice ages.

Variations in the Earth's orbital eccentricity—the shape of the orbit around the sun.

Changes in obliquity—changes in the angle that Earth's axis makes with the plane of Earth's orbit.

Precession—the change in the direction of the Earth's axis of rotation, i.e., the axis of rotation behaves like the spin axis of a top that is winding down; hence it traces a circle on the celestial sphere over a period of time.

Orbital Variations Changes in orbital eccentricity affect the Earth-sun distance. Currently, a difference of only 3 percent (5 million kilometers) exists between closest approach (perihelion), which occurs on or about January 3, and furthest departure (aphelion), which occurs on or about July 4.This difference in distance amounts to about a 6 percent increase in incoming solar radiation (insolation) from July to January. The shape of the Earth’s orbit changes from being elliptical (high eccentricity) to being nearly circular (low eccentricity) in a cycle that takes between 90,000 and 100,000 years. When the orbit is highly elliptical, the amount of insolation received at perihelion would be on the order of 20 to 30 percent greater than at aphelion, resulting in a substantially different climate from what we experience today.

Obliquity (change in axial tilt) As the axial tilt increases, the seasonal contrast increases so that winters are colder and summers are warmer in both hemispheres. Today, the Earth's axis is tilted 23.5 degrees from the plane of its orbit around the sun. But this tilt changes. During a cycle that averages about 40,000 years, the tilt of the axis varies between 22.1 and 24.5 degrees. Because this tilt changes, the seasons as we know them can become exaggerated. More tilt means more severe seasons—warmer summers and colder winters; less tilt means less severe seasons—cooler summers and milder winters. It's the cool summers that are thought to allow snow and ice to last from year-to-year in high latitudes, eventually building up into massive ice sheets. There are positive feedbacks in the climate system as well, because an Earth covered with more snow reflects more of the sun's energy into space, causing additional cooling.

Precession Changes in axial precession alter the dates of perihelion and aphelion, and therefore increase the seasonal contrast in one hemisphere and decrease the seasonal contrast in the other hemisphere. If a hemisphere is pointed towards the sun at perihelion, that hemisphere will be pointing away at aphelion, and the difference in seasons will be more extreme. This seasonal effect is reversed for the opposite hemisphere. Currently, northern summer occurs near aphelion.

Coriolis and Eötvös accelerations

These are accelerations caused by moving about on the Earth.

If you move to the east, you are increasing your rate of angular rotation (it’s like the earth is rotating faster beneath you) and so you experience an increase in centrifugal acceleration, which will decrease the total gravitational acceleration that you feel. The change in acceleration is given by (da/dv):

€

Δac = 2ωvE

in a direction parallel to the normal of the rotation axis. The vertical component of this acceleration is Δaccos(), horizontal is Δacsin(). The effect is to decrease any measured gravity by this vertical component amount. This is called the Eötvös effect. Note that if you move west the effect is the opposite.

Eotvos, Lórand, Baron von (1848-1919)

Gaspard-Gustave de Coriolis

The horizontal force due to moving east is directed to the south in the northern hemisphere and to the north in the southern hemisphere. This constitutes the N-S component of the Coriolis acceleration.

There is an east-west component due to N-S velocity. As you travel north or south from the equator, the velocity of the earth beneath you to the east decreases, which will appear to you to be an acceleration to the east. As you travel north or south to the equator, the velocity of the earth beneath you increases to the east, and this will appear to you to be an acceleration to the west. This is due to conservation of angular momentum. Importance is in the direction of the winds and ocean currents – accounting for their circulatory patterns.

Now that we understand all about gravity and accelerations on the earth, we can sensibly talk about:

Earth's Size and Shape

BC 550: Pythagoras proposed Earth is a sphere, but he had no idea how big.

First good estimate by Eratosthenes 275-195 BC) in 230 BC using basic trig and the difference in the angle of Sun's rays. (arc length = radius*angle)

Assume that Sun is so far away that rays are parallel.

But is the Earth a sphere?

First let’s ask a more fundamental question: What do we mean by the “shape of the Earth”? Just the contact between the air and the land?

No – we talk about “elevations”, but with respect to what, and how do we know what these are?

There are a lot of practical reasons for wanting to know this: including which way is up, where is the horizon, which way will water flow, etc.

“Up” is defined by the gravity field, so the horizon is perpendicular to this, or defined by an equipotential surface.

You can appreciate that knowing the gravity field is a very important part of the science of GEODESY, which is devoted to understanding the shape of the Earth.

By “shape” we mean, what does a reference equipotential surface look like? Two concepts to understand:

1. The “theoretical” suface: this is the called the Reference Ellipsoid

2. The actual “observed” surface: this is called the Geoid.

The first step in figuring out the shape of the Earth (beyond a sphere) was made by Newton, who argued for a oblate ellipsoid (a flattened sphere) based on what a rotating fluid would do. (Remember that the surface of a fluid will define an equipotential surface).

The problem was that a father-son team in France (Giovanni and Cassini) did an arc-measurement experiment and claimed that the Earth was a prolate spheroid - fatter at the poles than the equator, meaning that Newton had gotten it completely backwards.

So, the French sent an expedition to Finland and Ecuador to measure the distance along one degree of meridian. This would allow an estimate of flattening.

Pierre Bouguer was in Ecuador. As a side effect of this work he discovered isostacy, although he didn’t know it at the time.

He thought he had discovered hollowmountains. More about isostacy later.

Calculating the Reference SpheroidBasically, you start with an expression for equipotential surface like

€

U = Gdm

r∫ +

x2 + y2

2ω 2

Where the first term is gravity and the second is centrifugal force.

Solving this equation gets a bit involved and needs expansion in Legendre polynomials. We won’t go into details here, but keeping only the first couple

terms we find when the dust settles that

€

U =MG

r1+

K

2r 21− 3sin2φ +

ω 2r 3

2MGcos2φ

⎡

⎣ ⎢

⎤

⎦ ⎥

Where K is related to the principal moments of inertia. Now, if we define a reference Uo as the value of the equipotential surface at a point on the equator (r=a) then the shape of this surface is given by

r =MGUo

1+ K2a2

+ m2

⎡

⎣⎢⎤

⎦⎥1− 3K

2a2+ m2

⎛

⎝⎜

⎞

⎠⎟sin2φ

⎡

⎣⎢

⎤

⎦⎥

€

m =aω 2

MG / a2where

Note that r is of the form

€

r = a 1− f sin2φ[ ]

Where f is the “flattening” = (equatorial radius - polar radius)/equatorial radius.

The gravitational acceleration at any point r is found from

€

g = −∂U

∂r=

MG

r 21+

3K

2r 21− 3sin2φ( ) + mcos2φ

⎡

⎣ ⎢ ⎤

⎦ ⎥

Gravity on the spheroid (go) is found by substituting the equation for the shape of the spheroid (r) into the above equation:

€

go =Uo

2

MG1+

K

2a2+ m

⎡

⎣ ⎢ ⎤

⎦ ⎥1+ 2m−3K

2a2

⎛

⎝ ⎜

⎞

⎠ ⎟sin2φ

⎡

⎣ ⎢

⎤

⎦ ⎥

The idea then is to figure out what B1 should be based on actual gravity readings.

€

go = gequator 1+ B1 sin2φ[ ]

The Gravity on the Reference Ellipsoid is called the

International Gravity Formula

It is updated from time to time as measurements increase, but one that is commonly used is was this one adopted by the IUGG in 1967:

g = 978.0318(1 + 0.005302sin2g - 0.0000059sin22g) cm/sec2

Where g is the geographic (not geocentric!) latitude.

The last term comes from including higher order terms in the analysis shown before.

Note that because we have always assumed rotational symmetry, there is no dependence on longitude.

We observe that in fact the Earth is pretty spherical

a = 6378.136 km

c = 6356.751 km

f = (a-c)/a = 1/298.257 (but flatter than a rotating fluid!)

Note that on an ellipse there is a difference between a geographic latitude (in relation to the stars) and geocentric latitude (in relation to the center of the Earth).

The theoretical gravity on the reference ellipsoid is what we define as “normal”.

Note that GPS locations and elevations are specified in terms of a reference ellipsoid called the WGS84 frame.

Could this be a problem? After all, this ellipsoid is only a theoretical surface!

What we actually observe is the

GEOID

This is determined by gravity observations, satellite orbit perturbations, and satellite laser ranging.

Cartoon demonstrating the different kinds of elevations used, along with the relation of the Geoid to the Reference Ellipsoid.

How the “plumb line” is related to equipotential (“level”) surfaces.

Note difference in definition of “vertical”. This can be a headache for drillers!

Geoid Height (N): The Deviation of the Geoid

from the Reference Ellipsoid

The Earth as Couch Potato

Geoid Height over the US.

What’s that big crater-looking thing out west?

Satellite Altimetry

Ocean Bathymetry

Why do these maps look so much alike?

The Geoid is “warped” by too much or too little mass

IsostacyBack in Ecuador in 1735, Pierre Bouguer tried an experiment to figure out the density of rocks in a mountain by observing the deflection of a plumb line from vertical (by looking at the angle w.r.t. the stars).

To his surprise, he found that the densities were much less than normal rock, and he hypothesized that the mountain was hollow.

This did not go down well at the Academy of Science in Paris, but no one had a better idea.

Then, in 1855, the Royal Survey of Great Briton conducted a triangulation of all of India (the “jewel in the crown”) and ran into the same problem with the Himalaya and Tibet - the land based and astronomically based surveys did not match.

George Biddell Airy proposed that the mountains had low density roots, or in other words that the Earth worked by Archimedes Principle

He called this idea “isostacy” which means “equal standing” in Greek.

About 4 years later (1859) John Henry Pratt proposed that while the isopiestic idea of Airy was correct, differences in elevation were more likely to be caused by differences in density.

We know that the truth is somewhere in the middle, but more often looks like Airy isostacy.