Geoscience Reference
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Finally, the problem is reduced to the differentiation of the volume coefficients
of the molecular scattering and absorption. The first coefficient is equal to the
sum of the coefficients of absorbing gases (all, including O
2
) by (1.22). The
corresponding sum is inferred for the derivatives too. Specifying the concrete
gas with subscript
k
, with accounting for (5.18) we get:
=
κ
m
,
k
(
P
i
)
−
.
∂κ
m
,
k
(
P
i
)
∂
∂
1
T
(
P
i
)
+
1
C
a
,
k
C
a
,
k
(5.44)
∂
T
(
P
i
)
T
(
P
i
)
The absorption cross-sections of gases NO
2
,NO
3
,O
3
within the range 426-
848 nm don't depend on temperature, hence, equality
∂
|
∂
=
0is
correct. The following is obtained from (5.7) for O
3
within the range 330-
356 nm:
∂
C
a
,
k
(
T
(
P
i
))
λ
C
a
,
k
(
,
T
(
P
i
))
=
λ
λ
C
1
(
)+2
C
2
(
)
T
(
P
i
) .
(5.45)
∂
T
(
P
i
)
Equation (5.8) yields the following expression with taking into account the
linear interpolation of cross-sections over wavelength:
λ
j
)
T
(
P
i
)
λ
j
+1
−
λ
∂
λ
λ
j
,
P
,
T
(
P
i
))
C
2
(
C
a
,
k
(
,
P
i
,
T
(
P
i
))
=
−
C
a
,
k
(
∂
λ
j
+1
−
λ
j
T
(
P
i
)
(5.46)
λ
j
+1
)
T
(
P
i
)
λ
j
λ
j
+1
−
λ
λ
j
+1
,
P
,
T
(
P
i
))
C
2
(
−
−
C
a
,
k
(
.
λ
j
The following is obtained for the derivative of the volume coefficient of the
molecular scattering with (1.25) and (1.26):
=
σ
z
,
m
(
P
i
)
4
m
m
2
−1
,
∂σ
z
,
m
(
P
i
)
∂
∂
m
1
T
(
P
i
)
T
(
P
i
)
+
(5.47)
∂
T
(
P
i
)
and expression (1.27) yields the following:
b
(
)
2.178
∂
10
−6
1 + 0.003661
T
(
P
i
)
m
T
(
P
i
)
=
λ
10
−11
P
i
×
∂
10
−6
P
i
(1 + 10
−6
P
i
(1.049 − 0.0157
T
(
P
i
)))
1 + 0.003661
T
(
P
i
)
− 5.079
×
(5.48)
λ
−2
+10
−4
P
w
2.284 − 0.0249
1 + 0.003661
T
(
P
i
)
After analyzing the obtainedderivativeswith themethods described in Sect. 4.4
the concrete sets of altitudes and wavelengths are selected for the retrieval of
the atmospheric parameters, namely:
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