From: Michael M. Butler (butler@comp-lib.org)
Date: Sat Feb 12 2000 - 01:01:31 MST
Gee Whillikers, Spike, you use kitchen arithmetic better'n... well, I
can't say that, on account of the little'uns. But you're a pro, make no
mistake about it. "Me 'at's off to the Duke!" I always used 90 minutes,
so my numbers were always off. Now I see why. D'oh! :)
Spike Jones wrote:
> If you are in a boring meeting and need to derive the mass of the
> earth, one need only remember the orbit time of 85 minutes, and it
> can be backed out with only that information. The lowest satellites
> take a little longer, since they need to be a couple hundred km above
> the deck so they dont get dragged down immediately by the atmosphere.
>
> There is another way, if you need the earth's mass to only one
> digit of precision: you know the volume of a sphere is 4/3pi*r^3,
> but for single digit accuracy, pi~3, so V~4*r^3, and you know
> that the km was initially proposed as 1/10,000 the distance from
> the equator to the pole, so the radius of the earth can be calculated
> to be 10,000/pi*2~6370 km. Cube that and quadruple it, and the
> volume of the earth comes out to close enough to 1E12 km^3, and
> since the density of iron and nickel are in the 8000 kg/m^3 range
> and granite and other rocky stuff is in the 4000 range, then assume
> an average density of about 6000 kg/m^3 and you get an earth
> mass of 6E24 kg, which is quite close enough.
>
> >From that, knowing that a 1 kg mass weighs a force of 9.8 newtons,
> one can calculate the universal gravitation constant without having
> any reference material handy, assuming you remember that
> F=GMm/R^2, since you know F, M, m and R.
>
> But I digress. {8-]
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