Fairly nconsequential correction follows: The collision velocity of the two initial impactors will be conservatively:
V = (2 G M/(R))^0.5 V = (2 G (4.7x10^20 kg)/(730 km))^0.5 V = 293 m/s So the energy E converted to heat is: E = 2 * .5 m*V^2 = (4.7x10^20 kg)(293 m/s)^2 = 4x10^25 J Thus the heat per gram H is: H = E/(2*m) = (4x10^25 J)/(4.7x10^20 kg) = 43 J/g <==== note correction H = 10 cal/g which is not a lot of heat to dissipate, so this could simply result in increased temperature, or as you noted, be dissipated by ice. Even 4 times that number will not produce much incremental temperature. If it has the heat capacity of water that is only about 40 deg. C., not enough to boil water starting from 0 deg. C ice. Iapetus is so small one has to wonder how enough energy is developed to smush two bodies together to make it one spherical body. Looks like the three body theory is not even necessary, unless I have a computation error. Iapetus is not very dense, or very big. See: <http://www.star.ucl.ac.uk/~idh/solar/eng/iapetus> At 11:23 AM 2/19/5, revtec wrote: >293 m/s is 649 mph! The heat of collision would be intense and localized >for planetary sized bodies. The 8.6 J/g, if correct, is not evenly >distributed. In the area of contact, billions of tons of material would be >heated to incandesence. > >Jeff Well, you shoot a bullet at that speed and it will not warm up itself or the target much due to the collision. It has lots of momentum and destructive power well focused, but not much heat. What you say about the heat being concetrated at the surface is certainly true, and that can account for the ridge, but the bulk of the masses should remain solid. Very strange. Regards, Horace Heffner
