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What Remains of the Day

About "For All Your Rational Thought Needs" What Remains of the Day Mar. 2nd, 2005 @ 08:17 pm Looking at NASA's Astronomy Picture of the Day for 3-2-05 I see that the big earthquake shortened the Earth's day by about 3 microseconds. Pretty awe inspiring. I did some rough calculations, becuase I do that kind of stuff. I did them in my head so everything is only first order and rounded to simply the math.Given:mass of earth (m_e) =~ 6 x 10^24 kgradius of earth (r_e) =~ 6 x 10^3 km1 microsecond = 1 x 10^-6 secondsCalculations:Circumphrance of earth (c_e) = 2(pi)(r_e) = 2(3)(6 x 10^3 km)c_e = 4 x 10^4 km = 4 x 10^7 mDay in seconds: (24 hour/day)(60 min/hour)(60 sec/min) = 8 x 10^4 sec/dayVelocity of Earth (v_e) = (c_e)/(day in seconds) = (4 x 10^7 m)/(8 x 10^4 sec)= .5 x 10^3 = 5 x 10^2 m/sKinetic Energy of earth pre-earthquake (T_ei) = (1/2)(m_e)(v_e^2)= (1/2)(6 x 10^24 kg)(5 x 10^2 m/s)^2~ 7 x 10^29 JKinetic Energy lost (T_e_lost)= (T_ei)(ratio of second/3microseconds)= (7 x 10^29 J) (3 x 10^-6)~ 2 x 10^24 JMomentum of Earth pre-earthquake (p_ei)= (m_e)(v_e)= (6 x 10^24 kg)(5 x 10^2 m/s)~ 3 x 10^27 kg m/sMomentum lost (p_e_lost)= (p_ei)(ratio of second/3microseconds)= (3 x 10^27)(3 x 10^-6)~ 1 x 10^21 kg m/sConclusions:So the quake stripped from the Earth system about 2 x 10^24 J and about 1 x 10^21 kg m/s. That is amazing. Sure it doesn't really factor in in any significant way (after all it is just 3 microseconds for crying out loud) 2 x 10^24 J still is a lot of energy to loose by mere mortal standards!Mood Data: nerdyMusical Data Set: New Math Tom Lehrer From: March 3rd, 2005 03:32 am (UTC)
I knew today felt off for some reason!! :) From: March 3rd, 2005 05:04 am (UTC)  