Science, Creation & EvolutionHeard of this book?I haven't been able to get find a simple asteroid/earth mass ratio anywhere on the net but as best I can figure the value you cite is reasonable. I think the Kuiper belt mass is estimated to be much larger but, for purposes of this discussion, we'll use your figure. For comets coming in from distances further out, the old earth figure is, of course going to be considerable larger than a young earth estimate. However, I will just note here that there do appear to be comets coming in from beyond the Kuiper belt. If they were originally ejected from earth, they had to have been flung waaaaaaaay out there. If I understand Brown correctly, he postulates a crack in the earth running through the middle of both the Atlantic and Pacific Oceans flinging water and rock into space as the earth turned on its axis. He further postulates that the comets, asteroids and water on Mercury all came from this water and rock spewing into space. I don't have the math and astronomy background to come up with hard figures but I don't think we need them to do at least a crude test of these claims against known fact. Some additional info that ought to be useful is that Mercury is about 80 million km from earth at its closest approach and, of course, much closer to the sun. The main asteroid belt is between Mars and Jupiter averaging about 2.7 AU (astronomical units) from the sun. The Kuiper Belt is approximately 30-50 AU from the sun. Now let's ask if it is reasonable for all that stuff in space to have come from earth as Brown hypothesizes. Let's take the water on Mercury first. Here the question becomes how much water must be flung into space in order for some of it to land on Mercury. Let's do a thought experiment here. Assume you are on a revolving platform with a machine gun. Ten kilometers away there is a 1 square centimeter target on a 200 cm high wall. The machine gun is shooting 1 gram bullets. Those that hit the wall do so through its entire vertical distance. How many bullets will be necessary to provide a 50% chance of one of them hitting the target? Others may correct me if I'm wrong but I think it is equal to half the surface area of a 2 meter high circular wall of 10 km diameter calculated in square centimeters. That surface area would be 31.4 km x 100,000 cm/km x 100 cm or 314,000,000 square centimeters. That would be 314 metric tonnes of bullets. I don't really know how closely my analogy above approximates the probability of water from earth hitting (and being retained by) Mercury but I doubt it is an underestimate. Almost certainly it would take many earth masses of water flung sunward in order for just a tiny amount of it to fall on Mercury. Likewise, it would require many orders of magnitude more rock to be flung out past Mars for just a tiny part of it to assume a an orbit around the sun. Assuming the Brown mechanism, most of the rock would either fall back to earth or into the sun. The Kuiper Belt is even much further away and contains more mass. How many earth mass equivalents would it take to put those comets out there? There are some more problems to consider but it's late so I'll continue this later. Rev |
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