Re: new aspect of extinction-NO

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```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
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,