(LOD) associated with the 18.6 year Draconic lunar tides,
is a direct result of a systematic one degree shift to the
South/North by the Southern/Northern Summer SubTropical
High Pressure Ridge that is produced by lunar atmospheric
tides.
A. The Mass of Summer SubTropical
_____LOD^{/} = 2 π R_{E} cos L /V_{T}^{ /} (Earth)_____(17)
is a direct result of a systematic one degree shift to the
South/North by the Southern/Northern Summer SubTropical
High Pressure Ridge that is produced by lunar atmospheric
tides.
A. The Mass of Summer SubTropical
_______HighPressure Ridge
The ideal gas equation states that:
__________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/s^{2}) is
the acceleration due
to gravity (N.B. g = 9.80665 m/sec^{2} at 0 m,
g =
9.797 m/sec^{2} at 3000 m, and g =
9.776 m/sec^{2}
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
= P_{O} exp [−gMdz/RT_{O}]_____________(4)
where P_{O}
(in Pascals) and T_{O} (in Kelvin) are the pressure
and temperature at
the Earth’s surface and dz = (h – h_{O})
where h is the height in
metres above some reference height h_{O}.
Now Newton’s
Second Law tells us that:
________P
= M_{AT} g / A_____________________(5)
where M_{AT}
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:
_____M_{AT}
= [P_{G} A /g] x {1 − exp [−gMh/RT_{G}]}__(6)
where [P_{G}
A /g] is the total mass of the atmosphere above
a given area A of the Earth’s
surface, P_{G} and T_{G} are the
pressure and temperature at
ground level
(i.e. T_{G} =
288.15 K (15 C^{O}) and P_{G} = 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 = 4πR^{R},
R = 6.371 x 10^{6}
m, we get:
Area
of the Earth’s Surface
(10^{14 }m^{2})

Height
(m)

Pressure
(hPa)

T_{O}
(K)

Mass
of atmosphere above height h
(10^{18 }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 T_{O,} 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 10^{18}
kg for
wet air (+/ 1.2 to 1.5 x 10^{15} kg due to water vapour) and
5.1352 x 10^{18} kg for dry air. Our simple model gives a value
of
5.270 x 10^{18} kg most of the
difference explained by the
reduction of volume due to mountain ranges.
The surface area
of the Earth’s southern summer subtropical
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 40^{O} ± 10^{O} South of the
Equator.
__________Area
= 2 π
R h_______________(7)
where h = height of
the zone = R (sin L_{U} – sin L_{L}) and L_{U}
and L_{L}
are the latitudes of the bottom and top arcs of the zone.
_________Area
= 2 π
R^{2} (sin 50^{O} – sin 30^{O})__(8)
which gives and area
of 6.7850 x 10^{13} m^{2} (approximately
13.3024 % of the area
of the whole Earth’s surface).
Area
Earth’s SubTropical high pressure ridge.
(10^{13 }m^{2})

Height
(m)

Pressure
(hPa)

T_{O}
(K)

Mass
of atmosphere above height h
(10^{17 }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
SubTropical
HighPressure Ridge shifts south by 1^{O}.
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
= 1^{O} = 0.017453293 radians and L = 40^{O}
ΔX = 71.48 km
(N.B. In
winter time the corresponding numbers would be:
for ΔL
= 1^{O} = 0.017453293 radians and L = 30^{O}
ΔX = 55.60 km)
Now the Total Angular
Momentum of the Earth (L) is
given by:
___________L = I ω_______________(11)
where I =
M_{AT} X^{2}, ω = V_{T}/R is the angular frequency
of
rotation of the Earth and V_{T}
is the tangential velocity
of rotation of the Earth’s surface (in m s^{1}) then:
__________L
= M_{AT} X^{2} ω__________(12)
____________= M_{AT} X V_{T___________}(13)
The
movement of the Summer Southern SubTropical
HighPressure Ridge 1^{O}
south means that there is
decrease in the total angular momentum of the Earth’s
atmosphere since part its mass (i.e. the subtropical 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
solidEarth so that the total change in the Earth’s
overall angular momentum (ΔL) remains unchanged
(i.e. ΔL = 0):
_____M_{AT}
V_{T } ΔR_{AT} = 
M_{E} R_{E} ΔV_{T___}(14)
or_ΔV_{T}
(Earth)
___=  (M_{AT}/M_{E}) (ΔR_{AT} /R_{E})
V_{T}(Atmosphere)__(15)
Now_______V_{T}
= (2 π R cos L)/LOD______(16)
where the
nominal lengthofday (LOD) is 86400 seconds.
So for L = 40^{O} V_{T}
= 354.91773191286 m s^{1}.
Then if M_{AT} = 2.098 x 10^{17}
kg, M_{E} = 5.9722 x 10^{24} kg,
R_{E} = 6.371 x 10^{6}
m, ΔR_{AT}
= 7.148 x 10^{4} m:
____ΔV_{T}
(Earth) = 1.398865 x 10^{7} m s^{1}
^{}
and since the new
lengthofday (LOD ^{/}) is given by:
_____LOD^{/} = 2 π R_{E} cos L /V_{T}^{ /} (Earth)_____(17)
where the new lower
velocity V_{T} ^{/} (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 SubTropical
HighPressure Ridge shifts south by 1^{O} at the
same time.
HighPressure Ridge shifts south by 1^{O} at the
same time.
At this stage we have
not taken into account a comparable
shift (i.e. ~ 1^{O}) towards the
equator by the northern
subtropical high pressure ridge, which is likely to
normally be located at its winter position 30^{O} North of
the
Equator. The movement of the Winter Northern
SubTropical
HighPressure Ridge 1^{O} south means
that there is increase in the
total angular momentum
of the Earth’s atmosphere since part its mass (i.e.
the
subtropical 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 solidEarth 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 30^{O} North is:
___________V_{T}
= 401.24013019223 m s^{1}
So for L =
30^{O} V_{T} = 401.2401301922 m s^{1},
if M_{AT}
= 2.098 x 10^{17} kg, M_{E} = 5.9722 x 10^{24} kg,
R_{E}
= 6.371 x 10^{6} m, ΔR_{AT} = 5.560 x 10^{4}
m, then
the Earth’s rotation speed slows down by:
__________ΔV_{T}
(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 HighPressure ridges will
produce
a change in the Earth’s rotation rate at the 9.3/18.6
year time scale of:
year time scale of:
_____ΔLOD_{TOT}
= (0.0341 + 0.0264) ms
______________= 0.0077 ms
______________= 7.7 μsec
Note that ΔLOD_{TOT}
would be zero if both the northern
and southern subtropical 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 nonzero
during these
seasons. This compares with
the 12.57 μsec
change in the lengthofday that is associated with the
effect of 18.6 years
lunar tides upon the earth’s rotation.
Remember that this is only a
backoftheenvelope
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^{ })_(10^{20} 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 subtropical
high
pressure ridges in both hemispheres by ~ 1^{O} every
9.3/18.6
years, which is what we observed.
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