Geoscience Reference
In-Depth Information
(positive +
t
direction). So why doesn't all the air leave the planet? The answer
is that gravity exerts a compensating downward force.
Hydrostatic balance
occurs when the downward force of gravity balances the upward force of the
vertical pressure gradient:
2
−= =−
1
2
p
p
g
ρ
g
.
(6.35)
&
ρ
2
z
2
z
Equation 6.35 is an excellent approximation for the vertical balance of forces
in both the atmosphere and the oceans on large scales of motion, away from
the surface where frictional accelerations are strong.
Combining the hydrostatic and adiabatic conditions (section 5.2) for an
ideal gas allows us to calculate the atmospheric
adiabatic lapse rate
. Taking
the natural log (ln) of Eq. 5.6 and the Lagrangian derivative of the resulting
equation, we obtain
dp
dp
1
d
Θ
=−=
1
dT
R
1
dT
R
.
0
=
(6.36)
&
Θ
T
dt
cp
T
dt
cp
dt
dt
dt
p
p
To focus on the height dependence, we consider
T
T
(
z
) and
p
p
(
z
) alone (
T
and
p
depend only on elevation). Then,
/
22
(see Appendix B) and Eq.
ddtw z
( /)
6.36 becomes
2
=− =− ==
2
T
RT
p
g
Γ
98K/km
.
(6.37)
ADIABATIC
2
z
cp
2
z
c
p
p
for an ideal gas under adiabatic, hydrostatic conditions. In other words, as a
parcel of air rises adiabatically in a hydrostatic atmosphere, it will cool 9.8 K
for every 1 km increase in elevation, and we would observe a lapse rate of
9.8 K/km.
Figure 2.8
shows that the observed lapse rate in the troposphere is
roughly half the adiabatic lapse rate, which suggests that diabatic heating (such
as the radiative heating processes and the turbulent heat fluxes discussed in
chapter 5)
is important.
The hydrostatic relation (Eq. 6.35) can be integrated vertically to derive an
equation for
p
as a function of
z
in the atmosphere. A very simple analytical
form of this relationship can easily be derived if two assumptions are made.
The first is that the atmosphere behaves as an ideal gas (section 5.1). This is
an excellent assumption. The second assumption is that the atmosphere is iso-
thermal. This is a terrible assumption, as
Figures 2.8
and
2.9
demonstrate, but
it is useful for a irst-order derivation of how pressure depends on elevation in
the atmosphere.
Using the ideal gas law (Eq. 5.1) to eliminate density from the hydrostatic
relation (Eq. 6.35) and rearranging, we have
dp
g
=−
dz
.
(6.38)
p
RT
Integrating from the surface, where
z
0 and
p
p
0
, to some level (
z, p
) in an
isothermal atmosphere we get
ppe
0
=
−
zH
,
(6.39)
