Chitchat # 14

       It is interesting that Warren Li remembers all the numerical data in the movie. Although I'm not one of the uncles that he gave the assignment of checking out the height of the tidal wave; S.S. wants me to put in my two cents worth of comments. Here comes my analysis:

      I prefer simple and direct approaches instead of elegant but complex ones. One might say that I'm a true believer of "OCCAM'S RAZOR" (i.e., among all hypotheses consistent with the facts, choose the simplest). Firstly, the weight of the asteroid/comet, and its size are incompatible. The weight density of a one billion ton (2000LB X 10⁹) sphere of 7 miles diameter has a r = 2000 X 10⁹ / (4P /3)*(7X5280)³ = 0.0095 LB/ ft³. (Water is 62.4 LB/ft³). However, this is not fatal because one does not need the weight of the asteroid at all, the volume of the sphere is an important input. My estimate gives an upper bound of the tidal wave height. S.S.’s suggestion will furnish a lower bound if the time dependent volume change during re-entry with realistic thermal ablation, drag coefficient change ...etc. are available a prior.

      In lieu of the missing inputs, the following assumptions are used: (1) The earth’s surface is ¾ ocean. (2) The Atlantic is 1/3 of all oceans. (3) The average ocean depth is 2 miles. This is not unreasonable, remembering Snake King Lee's geography classes that Emden Deep is 20,000 ft. (4) The whole non-porous asteroid sinks on impact. From assumptions (1) through (3), the volume of Atlantic waters is as follows: V_Atlantic = (1/3)X(3/4)X(4P /3)X (4002³ -4000³) = 100,581,240.9mi³. The total volume after impact is the sum of V_Atlantic and V_Asteroid , where V_Asteroid =(4P /3)X7³ = 1436.7550mi³. The total volume is V_total = VAtlantic + VAsteroid = 100,581,240.9 + 1436.7550 =100,582,677.7mi³ = Surface area of Atlantic X D R, where D R = tidal wave height in miles or ft.

      This gives D R = 100,582,677.7/4P X(4002)²X(3/4)X(1/3) = 1.9990mi = 10,554.8745ft. Obviously, it is much larger (8.4439 times, to be exact) than the 1250ft tidal wave in the flick!

      To be practical, all oceans are connected, and the surface area of Atlantic cannot be isolated from the rest. This makes the removal of the 1/3 factor admissible, or D R = 10554.8745/3 = 3,518.2915ft. This corrected height is a mere 2.8146 times the 1250ft from the movie. At this point, S.S.'s details can be invoked to knock down the effective asteroid volume just before the impact, and a smaller D R will result.