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*To*: dinosaur@usc.edu*Subject*: Re: new aspect of extinction-NO*From*: Russ Andersson <randersson@monmouth.com>*Date*: Tue, 04 Feb 1997 10:29:51 -0500*Reply-to*: randersson@monmouth.com*Sender*: owner-dinosaur@usc.edu

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To: dinosaur@usc.comSubject: Re: new aspect of extinction-NOFrom: Russ Andersson <randersson@monmouth.com>Date: Tue, 04 Feb 1997 10:08:58 -0500References: <9702040149.AA00229@mailhost.ElSegundoCA.NCR.COM>Stan Friesen wrote: > > On the other hand, an off-center *impact* might well have that as > one of its side effect. The force generated by such an event is > prders of magnitude larger than the gravitational effects. And if > the force vector were oriented correctly relative to the rotational > axis, it could have substantial effect on the rotation. OK, 'nuf talk. Maximum change in rotation rate is if asteroid impacts tangent to the earth's surface. Assume density of earth and asteroid are comparable. >From CRC Std Math Tables, Halliday&Resnick Mass of a sphere = 4/3 pi d r^3, d=its density, r=its radius Inertia I of a sphere= 2/5 m r^2, m=its mass, r=its radius Angular momentum L of a particle moving perpendicular to the center of rotation at L=r m v, r=radius to center, m=its mass, v=its velocity (ie mv= its linear momentum) Angular momentum= I w, I=rotational inertia, w=angular velocity Define Re = radius of earth, 3960 miles (sorry about English units) We = angular velocity of earth, 2 pi / 24 hrs (duh) Ra = radius of asteroid, say 25 miles Va = velocity of asteroid relative earth, say 10,000 mi/hr d = density of earth and asteroid, won't matter Following in the same notation Ie = 2/5 Me Re^2 = 2/5 (4/3 pi d Re^3) Re^2 = 8/15 pi d Re^5 Le = Ie We = 2/45 pi^2 d Re^5 La = Re Ma Va = 4/3 pi d Va Re Ra^3 Angular momentum is conserved in the collision, so the right equation is La + Le = (Ie + Ma Re) Wf where Wf is the final angular velocity of the earth. But we've run out of time for this, so we observe that Ma/Me = 2.5E-7 (= 2.5 * 10^-7 = 0.00000025) = not so much And that let's us neglect the change in the Earth's inertia and just think about La/Le = delta We/We = how much does the rotation rate change === La/Le = 8/(3*45*pi) Va Ra^3 / (Re^4) Now if you turn the crank and work this out, and convert to a percentage change in rotation rate so it looks like a bigger number, you still just get deltarate = 1.2E-6 PERCENT, ie 1.2E-8 = 1 millisecond So all in all, nobody's going to die :-) not even our favorite dinosaurs. If you're an astronomer, maybe you can tweak my radius and velocity numbers, but if not, let's not be thinking up the REALLY BIG ONE. Finis, Russ Andersson who once had some undergrad physics (admittedly this understates the case a tad, but addresses what's needed here)

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