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Weighting function Transmission to space

Figure 6.8. Calculated transmission weighting functions (nadir) for Jupiter at 600 cm-1 (solid line) and 1,300 cm-1 (dotted line).

Figure 6.8. Calculated transmission weighting functions (nadir) for Jupiter at 600 cm-1 (solid line) and 1,300 cm-1 (dotted line).

Figure 6.9. Variation of peak of calculated transmission weighting function with wavelength for Jupiter. Below 4.5 ^m, the pressure level where the transmission to andfrom space is 0.5 has been plotted.

6.4.2 Net flux and disk averaging

The radiative transfer equation (Equation 6.38) very successfully models the IR observations of the giant planets made by passing spacecraft instruments such as Voyager IRIS which have a small field of view and thus sample only a small range of positions and zenith angles. Ground-based observations of the planets historically have poorer spatial resolution and thus often (especially for planets such as Uranus and Neptune which have a small angular size) the spectrum measured is that averaged over the whole planet and thus samples all zenith angles between 0° and 90°. If we assume that the temperature and composition profiles are the same all over the planet, then the disk-averaged spectrum is directly proportional to the averaged spectrum emitted by a single location on the planet into a hemisphere. Hence, to calculate the disk-averaged spectrum, all we need to do is integrate Equation (6.38) over all zenith angles and then scale this appropriately to get the disk-averaged spectral radiance.

The spectral radiance emitted upwards at the top of the atmosphere at zenith angle d (Equation 6.38) may be rewritten as

where the vertical coordinate is now the optical depth of the atmosphere r. This expression is valid for 0 < p < 1 and gives us the spectral radiance emitted in any given direction. To calculate the total spectral irradiance (or flux) leaving the atmosphere, we need to integrate Equation (6.41) over the solid angle of a hemisphere. The total upward spectral flux (W m-2(cm-1)-1) is then given by

I- p d Q = 2k where dQ = 2k sin d dd = —2k dp is the element of solid angle and where the extra factor of p arises since it is assumed that the emitting surfaces are horizontal. Substituting for I-1 (p) in Equation (6.42) we find

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