In the absence of 26 Al, other radioactive materials give Q « 4 x 10-7 ergg-1 s-1. The thermal conductivity of crystalline ice is given by (Klinger, 1980) as K(T) = 5.67 x 107/T erg cm-1 s-1 K-1, so that a steady state is reached only for bodies of the order of a few hundred kilometers in radius, consistent with previous calculations. If 26Al is present with a primordial mass fraction of 5 x 10-8, then Q = 2 x 10-4 ergg-1 s-1, and steady state can be achieved for bodies of the order of 10 km. Thus bodies larger than this will retain energy, heat up, and may reach melting.

An additional complication is due to the fact that when water vapor is deposited at low temperatures, the ice formed is amorphous, and has a lower thermal conductivity than ordinary crystalline ice.

At a temperature of 90 K (the significance of which will be explained shortly), K(T) = 2.39 x 104 erg cm-1 s-1 K-1. In this case R is reduced to ~1.8 km. In addition to a lower thermal conductivity, amorphous ice, when heated, transforms, exothermically, to a crystalline phase. This transformation can supply up to 9 x 108 ergg-1, the rate of crystallization being given by Eq. (10.2), above. At a temperature of 90 K, the rate of transformation is fast enough, so that the latent heat released is of the order of that released by 26 Al decay. This additional heating will cause the phase transformation to proceed even faster, resulting in runaway heating. In this simple model, 26 Al decay will bring bodies larger than about 2 km to high central temperatures for a limited period.

Cooling by the gas depends on the gas diffusion coefficient, which is Kgas ~ 10 cm2 g-1 (see, e.g. Prialnik, 1992). From Fig. 10.1, 26Al requires some 105 years to heat the interior to 90 K. For a 1-km radius comet the thermal diffusion time is also approximately 105 years. Thus heating and cooling are roughly balanced. The inclusion of gas cooling, however, will ensure that the temperature stays well below melting. The gas diffusion timescale will be greater than approximately 105 years for comets larger than approximately 30-km radius. This is a good rough estimate of the size needed to produce melting.

In Fig. 10.1, the squares show the time it takes to reach a maximum in the central temperature for a series of the models we computed with the standard abundance of radioactive U, Th, and K, and a variable abundance of 26Al. In all cases the maximum temperature equaled or exceeded 260 K, so that melting was possible. The lower curve is the time required for 26Al to produce enough heat to bring the ice to 90 K. As can be seen, the curve follows the squares very closely until an 26Al mass fraction of approximately 3 x 10~9 is reached. Such a low abundance will never produce enough energy to heat the ice to 90 K, and runaway heating will never occur. If 26Al were the only heat source, this abundance would be too low to bring about melting. Melting can still be achieved, however, if additional radioactive heat sources are considered.

The upper line shows the time required to reach 90 K as a result of heating by 40K. As can be seen from Table 10.1, 40K can provide substantial heating, albeit at a much lower rate. When the 26 Al abundance is too low to provide the required heating 40K can do the job, but it takes much longer. We see that while the sequence of squares appears to be continuous, it is really the result of two distinct heating mechanisms.

If we require the radioactive heat to be retained in order to raise the internal temperature, we must have a cometary body that is sufficiently large so that the thermal diffusion time is longer than the time required to produce the heat. The two curves in Fig. 10.2 show the required size for thermal diffusivities of k = 10~2 and k = 10 3, respectively. These values roughly bracket the actual values of k in the comet. The x's in the figure show the minimum depth at which melting occurred in the detailed models. This is a good approximation to the minimum radius body required to achieve melting. The agreement between the simple estimates and the detailed models shows that this simple picture is useful for estimating the effects of additional heat sources.

Clearly, the timescale arguments presented here are only illustrations of the physical principles. A nucleus with zero porosity, or zero volatile gas content, will not undergo gas cooling, so the gas diffusion timescale would be irrelevant.

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Al26 mass fraction

Fig. 10.2. The two curves show the estimated depth of the melting level as a function of 26 Al mass fraction based on the simple model described in the text, for two values of the thermal diffusivity. The x's show results of detailed models.

Fig. 10.2. The two curves show the estimated depth of the melting level as a function of 26 Al mass fraction based on the simple model described in the text, for two values of the thermal diffusivity. The x's show results of detailed models.

The fine details depend on the porosity and gas content of the nucleus, and on their evolution with time. These must be studied with a full numerical model.

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