Download Fundamental forces in the atmosphere • Real Forces: Pressure

Prof. S.K. Dash, IIT Delhi
Fundamental forces in the atmosphere
• Real Forces:
Pressure gradient force
Gravitational force
Frictional force
• Apparent Forces:
Centrifugal force
Coriolis force
Prof. S.K. Dash, IIT Delhi
[An Introduction to Dynamic Met,
J.R. Holton
Inertial or Newtonian motion
Newton’s first law of motion
A mass in uniform motion relative to a coordinate system
fixed in space will remain in uniform motion in the absence of
any forces.
Non-Newtonian Motion
An object at rest w. r. t. the rotating earth is not at rest or in
uniform motion relative to the coordinate system fixed in
space.
Apparent forces are the inertial reaction terms which arise
because of the coordinate acceleration.
Centrifugal force
Prof. S.K. Dash, IIT Delhi
For an observer in fixed space, earth rotates with constant speed
but there is change in direction and hence velocity is not constant.
Ω → constant angular velocity
δ v = v δθ
δv
is directed towards the axis of rotation.
dv
dθ ⎛⎜ R ⎞⎟
=
= v
−
dt
dt ⎜⎝ R ⎟⎠
dθ
v = Ω R and
=Ω
dt
δv
Lim
δ t →0 δt
∴
dv
= − Ω 2R
dt
Ω
R
Centripetal acceleration
Viewed from fixed coordinates, the motion is one of uniform acceleration
directed towards the axis of rotation.
Prof. S.K. Dash, IIT Delhi
Coriolis Force:
Consider particle moving in the eastward direction. The particle rotates faster
than earth. Let u be eastward velocity relative to earth.
2
→
Centrifugal force = ⎛⎜ Ω + u ⎞⎟ R
R⎠
⎝
→
Ωu → u 2 →
2
R+ 2R
=Ω R+2
R
R
For synoptic motions (horizontal scale of 1000km or so):
u p p ΩR
Ω = 7.292 x 10 −5 sec −1 , R ≈ a = 6.37 x 10 6 m, u = 10 m / sec
3rd term u 2 →
R
u
R
pp 1
=
×
=
→
2 nd term R 2
2
R
Ω
2Ωu R
Hence 3rd term can be neglected compared to the second.
Centrifugal force of a parcel rotating in the eastward direction with relative
velocity u
→
→
Ωu
= Ω2 R + 2
R
R
= Centrifugal force of stationary parcel due to rotating earth + Coriolis force.
Prof. S.K. Dash, IIT Delhi
Prof. S.K. Dash, IIT Delhi
Gravitational Force:
r
Fg
r
r*
GM r
≡ g =− 2 ,
m
r r
G = 6.673 ×10 −11 Nm 2 kg − 2
r
r*
GM r
g0 = − 2
a r
r*
a2
*
g = g0
(a + z )2
a = 6.37 ×106 m
a
Effective Gravity:
r r r r
g ≡ g +Ω R
*
2
r=a+z
z
Prof. S.K. Dash, IIT Delhi
Special Ideal Conditions of Atmosphere
(1) Homogeneous atmosphere: Assume density constant with height
0
H
p0
ρ RT
ρ
ρ
dp
g
dz
p
gH
i
e
H
=
−
;
=
.
.,
=
=
0
∫p0
∫0
ρg
ρg
0
i.e, H = RT0/g
Thus height of homogeneous atmosphere is a function of surface temperature
only. For dry air R = 2.8704 x 106 cm2 /sec2 K-1, g = 981 cm/sec2, T0 = 288 K
So H = 8.43 x 105 cm = 8.43 km. H is called the Scale Height.
Differentiate p = ρ RT with height to get temperature Lapse rate.
dT
dT
dp
=ρ R
or , − ρg = ρR
dz
dz
dz
dT
g
= − = − 34.1 0C / km
dz
R
dT is called temperature lapse rate.
γ ≡−
dZ
This value is unrealistic.
Prof. S.K. Dash, IIT Delhi
(2) Isothermal atmosphere: Assume atmospheric temperature
constant through out dp = − ρg dz = − pg dz
RT
dp
g
=
−
∫p0 p RT
p
∫
z
0
dz ⇒ ln
p
gz
=−
p0
RT
Scale height H = RT0/g = RT/g, since temperature is constant with height.
ln
p
= −
p 0
p = 0 only
z
H
i . e ., p =
p
0
e
− z / H
at z = ∞
At z = H , p = p0 e −1
By definition lapse rate dT/dz =0 for isothermal atmosphere.
(3) Constant lapse rate atmosphere: dT/dz =
γ
T = T0 − γ z
Again from hydrostatic approximation: dp = -(pg/RT) dz
dp
g z dz
=
−
∫p0 p R ∫0 T0 − γz
p
Let
T0 − γz = z
'
− γ dz = dz
'
Prof. S.K. Dash, IIT Delhi
z = 0, z ' = T0
z = z , z ' = T0 − γz
∫
z
0
dz
=
T0 − γz
∫
T0 − γz
T0
= −
dz '
−
γz '
ln z
γ
' T0 − γz
T0
p g T0 − γz
So ln = ln
p0 γR
T0
⎛ T0 − γz ⎞
⎟⎟
p = p0 ⎜⎜
⎝ T0 ⎠
⎛T ⎞
i.e., p = p0 ⎜⎜ ⎟⎟
⎝ T0 ⎠
g / γR
g / γR