R - Earth and Environmental Sciences

Intro Geophysics
GRAVITY AND THE FIGURE OF THE
EARTH
Newton's Law of Universal Gravitation
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:
r
m1m2 r
F = "G 2 r1
r
!
Sir Isaac Newton 1643-1727
Notes on Units
CGS
F
r
m1 , m 2
G
dynes
centimeters
grams
6.67 x 10-8 dynes-cm2/g2
MKS
newtons
meters
kilograms
6.67 x 10-11 N-m2/kg2
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
r
r F
Me r
g = = "G 2 r1
m
Re
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 µgal.
This means that instruments that measure gravity must be accurate
to about 1 part in 109!
A bit about G
G 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
Cavendish’s experiment is often called
“the Weighing of the Earth”
Why?
At the surface, Me = gRe2/G (and we already know g and Re).
Note that current measurements of G are done in almost the same
way!
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
%
% r ( +r / $ (
G(r ) = G"'1+ #'1+ *e
*
&
)
$
&
)
were G∞ is G measured at large distances and a is the magnitude of
the exchange interactions, and λ is a characteristic distance.
Some!estimate that α = -.007 and λ = 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
#U
g= "
#r
To generalize for all space, the we introduce the grad operator:
r $# # # '
"= & , , )
% #x # y #z (
!
or
!
r
% $ U $U $U (
r
g = "#U = "' , , *
& $ x $y $z )
For a point mass M:
g =!
!
"U
M
= !G 2
"r
r
or
GM
U =!
r
(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 earthmoon 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
We describe a rotation by its angular
velocity ω (radians/unit time) about an
axis (we often use ω 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 = ωx. As a vector,
it is tangent to the rotation.
The centripetal acceleration is a = ω2x =
v2/x, and is directed towards the axis of
rotation.
A mass on a rotating earth experiences a
centrifugal acceleration of ω2x. The
change to gravitational potential is found
by integrating:
1 2 2
Uc = " # x
2
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 mass
Orbiting 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:
Ed = m(rL ! d )
or
m
d=
rL
E+m
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 EarthMoon 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 = G
m
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
! is exactly equal and opposite to this
centrifugal acceleration
amount.
At a point (B) closest to the moon, the gravitational acceleration is
aB = G
m
( rL " Re )
2
and this amount is reduced by the centrifugal acceleration, so
"2
2
)#
,
)
,
#
&
&
#
&
1
1
Gm +
Re
Gm
R
R
e
e
.
+
(
aB = Gm%%
"
=
1"
"1
/
2
+
3
%
(
%
( + L.
!
2
2(
2
2
rL ' rL +*$ rL '
.- rL +* rL
.$ rL '
$ ( rL " Re )
At the farthest point (A) the
gravitational acceleration is:
2
# 1
& Gm * R
#
&
1
R
, e " 3% e ( + L/
(
aA = Gm%% 2 "
2( )
2 2
/.
(rL + Re ) ' rL ,+ rL $ rL '
$ rL
Note that the difference in acceleration
goes like 1/rL3.
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.
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
ω
C
m
Ω
Cl
Earth’s mass
Earth’s rate of rotation
Earth Moment of inertia
Moon’s mass
Moon’s rotation = rotation about Earth
Moon’s moment of inertia
The total angular momentum is
Cω
EΩd2
mΩ(rL-d)2
ClΩ
Due to Earth’s rotation
Due to rotation about center of mass
Due to Moon’s rotation about center of mass
Due to Moon’s rotation about its axis – this is a small
number relative to the others and we will ignore it
Summing all the contributions to angular momentum gives:
E
2
C" + E#d + m#( rL $ d) = C" +
m#rL = const.
E+ m
2
2
m
Recalling that d = rL
E+ m
!
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
E '
2
2 $
= " ( rL # d) = " rL &
)
2
%
E+ m(
rL
!
And so
!
"rL2 = G( E + m) rL
Substitute this into the equation at the top of this slide and we get:
!
C" +
EM
( E + m)
GrL = const.
Tidal friction reduces ω, which means the second term must increase
to compensate. This means rL should increase. And if rL increases,
!
Ω 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.
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).
2
*Recall from before that "rL =
G( E + m) rL
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.
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.
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
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).
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.
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.
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
Precession was first observed by Hipparchus in 120 BC.
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.
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-toyear 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):
"ac = 2#vE
in a direction parallel to the normal of the rotation axis. The vertical
component of this acceleration is Δaccos(λ), horizontal is Δacsin(λ).
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 hollow
mountains. More
about isostacy
later.
Calculating the Reference Spheroid
Basically, you start with an expression for equipotential surface like
U = G"
dm x 2 + y 2 2
+
#
r
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
(
MG %
K
$ 2r 3
2
2
U=
1+
1"
3sin
#
+
cos
#
'
*
r & 2r 2
2MG
)
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
MG !
K m $! ( 3K m + 2 $
r=
1+
+ &#1 ' * 2 + - sin . &
#"
2
Uo
2a
2 %" ) 2a
2,
%
where
!
a" 2
m=
MG / a 2
Note that r is of the form
r = a [1" f sin 2 # ]
Where f is the “flattening” = (equatorial radius - polar radius)/equatorial radius.
The gravitational acceleration at any point r is found from
!
#U MG % 3K
2
2 (
g= "
= 2 '1+ 2 (1" 3sin $ ) + m cos $ *
)
#r
r & 2r
Gravity on the spheroid (go) is found by substituting the equation for the shape of
the spheroid (r) into the above equation:
!
%" )
Uo2 "
K
3K , 2 %
go =
$#1+ 2 + m'&$1+ +* 2m ( 2 .- sin / '
MG
2a
2a
#
&
go = gequator [1+ B1 sin 2 " ]
!
The idea then is to figure out what B1 should be based on actual gravity readings.
!
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 + 0.005302sin2φg - 0.0000059sin22φg) cm/sec2
Where φg 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.
How the “plumb line” is
related to equipotential
(“level”) surfaces.
Cartoon demonstrating the
different kinds of elevations
used, along with the relation of
the Geoid to the Reference
Ellipsoid.
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
Isostacy
Back 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.