## Wednesday, June 20, 2012

### A SIMPLE MODEL FOR THE 18.6 YEAR ATMOSPHERIC TIDAL OSILLATION

Hypothesis: The -12.57 μsec change in the length-of-day
(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 Sub-Tropical
High Pressure Ridge that is produced by lunar atmospheric
tides.

A. The Mass of Summer Sub-Tropical
_______High-Pressure 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/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.