__________P
= ρRT/M___________(1)
where P is the
atmospheric pressure, ρ is the atmospheric
density, R is the Universal Gas
Constant (= 8.31432
Nm/Mole), T is the temperature and M is the molar
mass of
air (= 2.89644 x 10-2).
If we assume that in the vertical direction (z) the
atmosphere is hydrostatic equilibrium then:
_________dP
= −ρ
g dz___________(2)
where dP is the
change in pressure with change in
height dz and g (=9.80 m/s2) is
the acceleration due
to gravity (N.B. g = 9.80665 m/sec2 at 0 m,
g =
9.797 m/sec2 at 3000 m, and g =
9.776 m/sec2
at 10000 m).
Hence, from
equations (1) and (2) we find that:
________dP/P
= −ρ
g dz M / ρRT
= −gMdz
/ RT___(3)
which gives the
barometric formula as its solution:
________P
= PO exp [−gMdz/RTO]_____________(4)
where PO
(in Pascals) and TO (in Kelvin) are the pressure
and temperature at
the Earth’s surface and dz = (h – hO)
where h is the height in
metres above some reference height hO.
Now Newton’s
Second Law tells us that:
________P
= MAT g / A_____________________(5)
where MAT
is the mass of atmosphere above a given area
A of the Earth’s surface, above a
given height h.
Combining equations (4) and (5) we get:
_____MAT
= [PG A /g] x {1 − exp [−gMh/RTG]}__(6)
where [PG
A /g] is the total mass of the atmosphere above
a given area A of the Earth’s
surface, PG and TG are the
pressure and temperature at
ground level
(i.e. TG =
288.15 K (15 CO) and PG = 101,325 Pascals),
respectively,
and h is the height above ground level in metres.
If we take A to be
the surface area of the whole Earth = 4πRR,
R = 6.371 x 106
m, we get:
Area
of the Earth’s Surface
(1014 m2)
|
Height
(m)
|
Pressure
(hPa)
|
TO
(K)
|
Mass
of atmosphere above height h
(1018 Kg)
|
Mass
Faction
|
5.1006
|
0
|
1013.25
|
288.15
|
5.270
|
1.000
|
5.1006
|
3000
|
709.98
|
268.66
|
1.577
|
0.299
|
5.1006
|
3000
|
701.20
|
268.66
|
1.623
|
0.308
|
where the data in
the second last row uses the pressure
value derived from the barometric formula
while the data
in the last row uses the value for the US Standard Atmosphere.
All
of the values for TO, are those for the US standard atmosphere.
The actual NCAR value for the total mass
of the atmosphere
that is derived from sophisticated models is 5.1480 x 1018
kg for
wet air (+/- 1.2 to 1.5 x 1015 kg due to water vapour) and
5.1352 x 1018 kg for dry air. Our simple model gives a value
of
5.270 x 1018 kg most of the
difference explained by the
reduction of volume due to mountain ranges.
The surface area
of the Earth’s southern summer sub-tropical
high pressure ridges can be obtained
by assuming that bottom
surface area this body of air is a zone of a sphere with
the radius
of the Earth, centred at 40O ± 10O South of the
Equator.
__________Area
= 2 π
R h_______________(7)
where h = height of
the zone = R (sin LU – sin LL) and LU
and LL
are the latitudes of the bottom and top arcs of the zone.
_________Area
= 2 π
R2 (sin 50O – sin 30O)__(8)
which gives and area
of 6.7850 x 1013 m2 (approximately
13.3024 % of the area
of the whole Earth’s surface).
Area
Earth’s Sub-Tropical high pressure ridge.
(1013 m2)
|
Height
(m)
|
Pressure
(hPa)
|
TO
(K)
|
Mass
of atmosphere above height h
(1017 Kg)
|
Mass
Faction
|
6.7850
|
0
|
1013.25
|
288.15
|
7.010
|
1.000
|
6.7850
|
3000
|
709.98
|
268.66
|
2.098
|
0.299
|
6.7850
|
3000
|
701.20
|
268.66
|
2.159
|
0.308
|
B. Change in Angular
Momentum of the Earth
CASE A: The Summer Southern
Sub-Tropical
High-Pressure Ridge shifts south by 1O.
Let X = the radius of circle of latitude of the Earth at
latitude L.
____________X
= R cos L___________(9)
___________ΔX
= -R sin L ΔL________(10)
for ΔL
= 1O = 0.017453293 radians and L = 40O
ΔX = 71.48 km
(N.B. In
winter time the corresponding numbers would be:
for ΔL
= 1O = 0.017453293 radians and L = 30O
ΔX = 55.60 km)
Now the Total Angular
Momentum of the Earth (L) is
given by:
___________L = I ω_______________(11)
where I =
MAT X2, ω = VT/R is the angular frequency
of
rotation of the Earth and VT
is the tangential velocity
of rotation of the Earth’s surface (in m s-1) then:
__________L
= MAT X2 ω__________(12)
____________= MAT X VT___________(13)
The
movement of the Summer Southern Sub-Tropical
High-Pressure Ridge 1O
south means that there is
decrease in the total angular momentum of the Earth’s
atmosphere since part its mass (i.e. the sub-tropical ridge)
is now rotating
about the Earth’s axis of rotation at a
radius that is 71.48 km smaller. This
decrease in
atmospheric angular momentum is compensated
for by an increase in
the angular momentum of the
solid-Earth so that the total change in the Earth’s
overall angular momentum (ΔL) remains unchanged
(i.e. ΔL = 0):
_____MAT
VT ΔRAT = -
ME RE ΔVT___(14)
or_ΔVT
(Earth)
___= - (MAT/ME) (ΔRAT /RE)
VT(Atmosphere)__(15)
Now_______VT
= (2 π R cos L)/LOD______(16)
where the
nominal length-of-day (LOD) is 86400 seconds.
So for L = 40O VT
= 354.91773191286 m s-1.
Then if MAT = 2.098 x 1017
kg, ME = 5.9722 x 1024 kg,
RE = 6.371 x 106
m, ΔRAT
= 7.148 x 104 m:
____ΔVT
(Earth) = 1.398865 x 10-7 m s-1
and since the new
length-of-day (LOD /) is given by:
_____LOD/
= 2 π
RE cos L /VT / (Earth)_____(17)
where the new lower
velocity VT / (Earth) is equal to
354.91773191286 m s-1
+ 1.398865 x 10-7 m s-1
then the change in
LOD (ΔLOD)
is:
____ΔLOD
= LOD / - 86400 = - 0.0341 ms____(18)
CASE B: The Winter
Northern Sub-Tropical
High-Pressure Ridge shifts south by 1O at the
same time.
At this stage we have
not taken into account a comparable
shift (i.e. ~ 1O) towards the
equator by the northern
sub-tropical high pressure ridge, which is likely to
normally be located at its winter position 30O North of
the
Equator. The movement of the Winter Northern
Sub-Tropical
High-Pressure Ridge 1O south means
that there is increase in the
total angular momentum
of the Earth’s atmosphere since part its mass (i.e.
the
sub-tropical ridge) is now rotating about the Earth’s
axis of rotation at a
radius that is 55.60 km larger.
This increase in atmospheric angular momentum
is
compensated for by an decrease in the angular
momentum of the solid-Earth so
that the total
change in the Earth’s overall angular momentum
(ΔL)
remains unchanged (i.e. ΔL = 0):
From equation 16 the
nominal speed of rotation
of the Earth at 30O North is:
___________VT
= 401.24013019223 m s-1
So for L =
30O VT = 401.2401301922 m s-1,
if MAT
= 2.098 x 1017 kg, ME = 5.9722 x 1024 kg,
RE
= 6.371 x 106 m, ΔRAT = 5.560 x 104
m, then
the Earth’s rotation speed slows down by:
__________ΔVT
(Earth) = 1.2301064334 x 10-7 m s-1
Therefore from equation
17:
_________LOD
/ = 86400.000026488012 seconds
and so____ΔLOD
= LOD - 86400 = +0.0264 ms
Hence, the combined
movements of both the Northern
and Southern Tropical High-Pressure ridges will
produce
a change in the Earth’s rotation rate at the 9.3/18.6
year time scale of:
_____ΔLODTOT
= (-0.0341 + 0.0264) ms
______________= -0.0077 ms
______________= -7.7 μsec
Note that ΔLODTOT
would be zero if both the northern
and southern sub-tropical ridges where at
the same
absolute latitudes (i.e. in Spring and Autumn/Fall). However,
this is
not the case during the Northern and
Southern
Summers and Winters, so it will probably be non-zero
during these
seasons. This compares with
the -12.57 μsec
change in the length-of-day that is associated with the
effect of 18.6 years
lunar tides upon the earth’s rotation.
Remember that this is only a
back-of-the-envelope
calculations with many assumptions thrown in, so it is
amazing the it come out to within a factor of two.
TIDAL INFLUENCE
THROUGH LOD
VARIATIONS ON THE TEMPORAL
DISTRIBUTION
OF EARTHQUAKE
OCCURRENCES
P. VARGA1, D.
GAMBIS2, Ch. BIZOUARD2, Z. BUS1,
M. KISZELY1
Tidal
wave___Period (days)__Δ_LOD___Erot
______________________(10-4 sec )_(1020 J)
_____________________________________
18.6
year____6798.37______-0.1257__3.100
Sa _____________365.26_______0.0222__0.547
Ssa_____________182.62_______0.1400__3.453
Msm_____________31.81_______0.0304__0.750
Mm______________27.55_______0.1589__3.920
Msf______________14.76_______0.0264__0.651
Mf ______________13.66_______0.3008__7.4200
Mstm______________9.56_______0.0109__0.2689
Mtm_______________9.13_______0.0576__1.4209
Msqm______________7.10_______0.0092__0.2269
Mqm_______________6.86_______0.0076__0.1875
So the 18.6 year
period changes to the Earth’s LOD
could produce movements of the sub-tropical
high
pressure ridges in both hemispheres by ~ 1O every
9.3/18.6
years, which is what we observed.
