– The surface layer may be uniform over the nucleus but the uppermost part might
be desiccated with gas emission from below the surface at a depth where the heat
available for sublimation is substantially below that available at the surface
(Fig. 2.41 right).
A combination of these three end-members is also conceivable.
Because the reflectance of dust-laden ice can appear to be similar to ice-free
surfaces at optical wavelengths, these three models may look similar in imaging
observations. One might think that at thermal wavelengths, the temperature contrast
between dusty and icy surfaces could help distinguish between these models.
However, there are significant ambiguities.
We have seen the black-body emission from a surface at a temperature, T, in
Eq. (2.9). When the integrated flux from a surface is normalised to 1, i.e.
Z 1
0
F N λ
ð Þ dλ ¼ 1
ð2:124Þ
Then
F N λ
ð Þ ¼
h
4 c
4
π 4 k
4 T
4
15
λ
5 e hc=λkT À 1
ð2:125Þ
The surfaces in Fig. 2.41 have two components that will have different temperatures. At 1 AU the dark, dusty, non-volatile component when at the surface and
illuminated may have a temperature exceeding 350 K, whereas the water ice surfaces
will be at 200 K because of vigorous sublimation. One can then compute the emitted
(thermal) flux from a two-component surface
F N λ
ð Þ ¼ F N1 λ
ð ÞσT
4
1 A 1 þ F N2 λ
ð ÞσT
4
2 1 À A 1
ð
Þ
ð2:126Þ
where A 1 is the fractional area of the surface at T 1 and (1ÀA 1 ) is the fractional area of
the surface at a temperature T 2 . F N (λ) is in units of [W m
À2 nm
À1 ]. The radiated
fluxes for the two terms of this equation using T 1 ¼ 350 K, A 1 ¼ 0.9 and T 2 ¼ 200 K,
are shown in Fig. 2.42. One can see immediately that the contribution to the total flux
from the cold surface is small. This is similar to the effect of surface roughness
discussed in relation to Fig. 2.1. The brightness temperature of the total surface area
can then be calculated. In the case shown, the integral over all wavelengths corresponds to 341.81 K—which should be a detectable difference compared to T 1 .
However, it is important to note that the thermal emissivity, ε, has been set to 1 in
Eq. (2.126). This quantity is not well enough known and simply changing the
assumed emissivity of the dusty material by a factor equal to the reciprocal of the
fractional area would produce an almost identical change in the observed brightness
temperature. It is this combination of the low fractional area of the subliming
material and the lack of knowledge of ε (even ignoring the further complication
arising from the beaming parameter, η th , discussed below) that limits the usefulness
108
2 The Nucleus
be desiccated with gas emission from below the surface at a depth where the heat
available for sublimation is substantially below that available at the surface
(Fig. 2.41 right).
A combination of these three end-members is also conceivable.
Because the reflectance of dust-laden ice can appear to be similar to ice-free
surfaces at optical wavelengths, these three models may look similar in imaging
observations. One might think that at thermal wavelengths, the temperature contrast
between dusty and icy surfaces could help distinguish between these models.
However, there are significant ambiguities.
We have seen the black-body emission from a surface at a temperature, T, in
Eq. (2.9). When the integrated flux from a surface is normalised to 1, i.e.
Z 1
0
F N λ
ð Þ dλ ¼ 1
ð2:124Þ
Then
F N λ
ð Þ ¼
h
4 c
4
π 4 k
4 T
4
15
λ
5 e hc=λkT À 1
ð2:125Þ
The surfaces in Fig. 2.41 have two components that will have different temperatures. At 1 AU the dark, dusty, non-volatile component when at the surface and
illuminated may have a temperature exceeding 350 K, whereas the water ice surfaces
will be at 200 K because of vigorous sublimation. One can then compute the emitted
(thermal) flux from a two-component surface
F N λ
ð Þ ¼ F N1 λ
ð ÞσT
4
1 A 1 þ F N2 λ
ð ÞσT
4
2 1 À A 1
ð
Þ
ð2:126Þ
where A 1 is the fractional area of the surface at T 1 and (1ÀA 1 ) is the fractional area of
the surface at a temperature T 2 . F N (λ) is in units of [W m
À2 nm
À1 ]. The radiated
fluxes for the two terms of this equation using T 1 ¼ 350 K, A 1 ¼ 0.9 and T 2 ¼ 200 K,
are shown in Fig. 2.42. One can see immediately that the contribution to the total flux
from the cold surface is small. This is similar to the effect of surface roughness
discussed in relation to Fig. 2.1. The brightness temperature of the total surface area
can then be calculated. In the case shown, the integral over all wavelengths corresponds to 341.81 K—which should be a detectable difference compared to T 1 .
However, it is important to note that the thermal emissivity, ε, has been set to 1 in
Eq. (2.126). This quantity is not well enough known and simply changing the
assumed emissivity of the dusty material by a factor equal to the reciprocal of the
fractional area would produce an almost identical change in the observed brightness
temperature. It is this combination of the low fractional area of the subliming
material and the lack of knowledge of ε (even ignoring the further complication
arising from the beaming parameter, η th , discussed below) that limits the usefulness
108
2 The Nucleus