## Problems with the Classical Scenario

The classical scenario of Oort cloud formation discussed above meets two problems when confronted with the quantitative constraints provided by the current Solar System. Fig. 24. Fraction of the initial planetesimal population that is in the Oort cloud, in its inner and outer parts and in the Scattered disk, as a function of time. From 

As we have seen in Sect. 2.2, the outer Oort cloud should currently contain 1012 comets with H10 < 11. The estimates of the nuclear size of a H10 = 11 comet range from 1km  to 2.3km . Assuming, as in , that a H10 = 11 comet has D ~ 1.7 km, and assuming also a cumulative size distribution proportional to D-2 and a density of 0.6gcm-3 (as for P/Halley), one obtains a total mass of 3 x 1028 g, namely 3Me. Because the overall efficiency of formation of the outer Oort cloud is small (2.5%), this implies that the original planetesimal disk in the Jupiter-Neptune region was ~100Me. This seems rather high compared with the total mass of solids associated with the minimum mass solar nebula . Also, numerical simulations show that a planetesimal disk more massive than 30-50 Me would have driven Neptune beyond 30 AU and that, in such a disk, Jupiter and Saturn would have passed across their mutual 2:5 mean-motion resonance (see Sects. 4 and 5). The uncertainty in the conversion between H10 magnitude and size, however, allows enough room for us to make consistent estimates. For instance, if the nuclear size of H10 = 11 comets is 1.3 km (instead of the assumed 1.7), the required mass of the planetesimal disk falls to a more reasonable value of 50 Me.

A second, more severe problem concerns the number ratio between the comet populations in the Oort cloud and in the Scattered disk. We have seen in Sect. 2.1 that the Scattered disk, to be a sufficient source of JFCs, has to contain 4 x 108 comets with H10 < 9. The number of comets with H10 < 11 depends on the exponent of the H10 distribution of comets, which is still highly debated. Using the largest value available in the literature (0.7 ), the Scattered disk should have 1010 H10 < 11 comets. Using the exponent for the nuclear magnitude distribution in [111,180] (0.28) and assuming a linear scaling between nuclear magnitude and H10, the number of H10 < 11 comets in the Scattered disk reduces to 1.5 x 109. Because the number of comets in the outer Oort cloud with H10 < 11 is 1012, the comet number ratio inferred from observations between the outer Oort cloud and the Scattered disk is in the range 100-1,000. However, in the simulations in , the final ratio is ~ 10 (see Fig. 24).

The way out of this problem is much more difficult than for the total mass problem. The discrepancy does not depend on assumed relationships between total magnitude and size, nor on density. It cannot be alleviated with any reasonable assumption of the exponent of the H10 distribution. Also, different assumptions of the initial planetesimal distribution in the disk would not help. The point is that most of the Oort cloud is made of planetesimals from the Uranus-Neptune-trans-Neptunian zone, which have to pass through the Scattered disk to reach the cloud. Thus, there is a causal relationship between the final numbers of comets in the Scattered disk and Oort cloud. To change this relationship, it would be necessary that a much larger number of planetes-imals could reach the Oort cloud without passing through the Scattered disk. This requires that Jupiter and Saturn were more effective in the real Oort cloud-building process than in the simulations of . A possible scenario in which this can occur is discussed below.