**
****Impact on the Sea**

**These are the assumptions used for an impact of icy particles of a comet coma impacting the Pacific Ocean at high velocity. **

**The rise in mean Sea Level from Pleistocene to Holocene time of 110 meters was a result of a large water ice laden comet impacting the area of the Pacific Ocean. The total mass of the coma water ice impacting, first the atmosphere and then the Ocean is therefore 3.995 times 10^19 kg. Of this mass, 5.943 times 10^17 kg was sublimated into vapor in the impact with the atmosphere as described above under the heading of Impact with the Atmosphere. The remaining mass available to impact the ocean is therefore 3.935 times 10^19 kg.**

**The average depth of the Pacific Ocean is now approximately 4200 meters. It would have been approximately 4090 meters before the impact described herein.**

**The mass of the Ocean impacted by the comet coma over the area of impact will approximately equal the water ice mass of the coma remaining after the impact with the atmosphere. The kinetic energy (KE) conversion as heat due to the collision friction of the coma ice and ocean will sublimate the coma ice to steam and vaporize an equivalent Ocean mass to a temperature of 100 degrees Celsius. The remaining kinetic energy will be conserved as the velocity of the steam explosion of the combined expanding cloud mass of the water vapor produced.**

**The velocity of the impacting comet is assumed to be 13.5 km/sec relative to the Ocean.**

**The mean temperature of the comet coma water ice is assumed to be approximately -200 degrees Celsius. **

**The mean temperature of the Pacific Ocean water is assumed to be approximately 10 degrees Celsius.**

**The surface area of the Earth is 4 times pi times the radius (6380000 m) squared or 5.115 times 10^14 square meters. **

**The area of impact assumed to be approximately 10 percent of the area of the surface of the Earth; or an area of 5.115 times 10^13 square meters of the Pacific Ocean.**

**The mass of water in the Ocean, in the area of impact to a depth of approximately 769 meters is 3.935 times 10^19 kg (10^22 grams) and is equivalent to the remaining water ice mass of the coma available to impact. This amount of ocean water, at the area of impact, is the mass of ocean assumed to be boiled away by the impact. **

**Water liquid and water ice convert to steam at a volume of approximately 1600 times that of the liquid and ice.**

**Joules convert to calories at the rate of 1 calorie to 4.184 Joules.**

**One calorie is required to raise one gram of water or water ice one degree Celsius.**

600 calories per gram are absorbed with no temperature change in state transformation from liquid water to water vapor. 80 calories per gram are absorbed with no temperature change in state transformation from solid ice to liquid water. Sublimation of solid ice to water vapor will absorb 680 calories per gram.

**The calories and KE sublimating the ice in the coma to water vapor at 100 degrees Celsius from -200 degrees Celsius (3.935 times 10^22 grams times 300) plus the heat of state transformation (680 c times 3.935 times 10^22 grams) is therefore 3.857 times 10^25 calories or 1.614 times 10^26 Joules.**

**The calories and KE required to raise the temperature of the ocean to steam at 100 degrees Celsius from 10 degrees Celsius (3.935 times 10^22 grams times 90) plus the heat of state tranformation (600 times 3.935 times 10^22 grams) is therefore 2.715 times 10^25 calories or 1.137 times 10^26 Joules.**

**The KE of the coma ice (½mv^2) or (.5 times 3.9365 times 10^19 times 13500^2) is therefore 3.583 times 10^27 Joules. By subtracting the energy in Joules used to raise the temperature of the coma ice and ocean water, there remains 3.311 times 10^27 Joules from the mass and velocity of the coma.**

**The combined mass of the coma ice vapor and sea water vapor (3.935 times 10^19 kg times 2) is therefore 7.87 times 10^19 kg. The remaining KE will now be applied to obtain the new velocity of the combined mass to conserve the energy.**

**Since v^2 is equal to KE divided by (mass times .5), then (3.311 times 10^27 J divided by (7.87 times 10^19 kg times .5) equals the velocity squared or 84164380. The square root is 9172 m/s or the velocity of the expanded bubble of coma ice vapor mixed with vaporized ocean. This mass of steam would explode into a ballistic or parabolic vector as a huge plume multiple angles to the horizon. It would proceed along a conic trajectory similar to the incoming impact trajectory and its orbital characteristic would be similar to the atmospheric explosion that preceded it. This vector would quickly take the mass into space where it would condense back into ice, and its latent heat would be lost to space. It would become ice crystals again. The now solid ice mass would return to the surface of the Earth in a few hours repeat the impact process over a much wider area and at a reduced velocity to eventually fall as rain, hail or brittle ice. The secondary impact of double the mass at a lower, but still substantial velocity would absorb heat energy through vaporization and expand to lose it again to space through condensation, until the velocity was dampened to normal atmospheric conditions.**