From: mbs@adastra.cvl.va.us (Michael B. Smith)
Newsgroups: comp.sys.amiga.advocacy
Subject: Lets talk about gravity (was Re: Intel HW)
Distribution: world
Reply-To: adastra!mbs@Virginia.Edu
Date: 12 Apr 92 13:29:45 EDT
Organization: Private UUCP Node

In article <44893@mentor.cc.purdue.edu> kodak@mentor.cc.purdue.edu (KodaK) writes:
>>mean anything. How dare you claim that the speed of a 386 is any different
>>than an A3000 when dropped from a 2nd story window...they both accelerate
>>at 9.8 meters per second, which is around 32 feet per second!
>    ^^^^^^^^^^^^^^^^^^^^^	     ^^^^^^
>>(:
>
>Now, come on, if you wanna get technical, it's -9.81 meters per second
>squared!					^
>^^^^^^^

Actually, acceleration is a vector quantity, not a scalar (if you want to get
technical). So, as he presented it, it is fine. It simply depends on your
frame of reference. Remember that:

		2
	       d x	 dv
	a  =  -----  =	----
		 2	 dt
	       dt

where 'x' is the 4-space vector quantity x'(x,y,z,t); and 'a' is the
experienced acceleration.

Also, if you want to get technical about it, the mean acceleration of gravity
at the equator is 9.780573 m/s^2, and the mean acceleration of gravity at the
poles is 9.832251 m/s^2. The difference in the mean acceleration of gravity
between the north pole and the south pole is .00030 m/s^2. (Remember that the
Earth is not a perfect sphere -- it is "squashed" slightly at the poles.)

The mean acceleration of gravity for the entire Earth is 9.797830 m/s^2. And
the average is 9.806412 m/s^2. (This, BTW, is an excellent example of why you
might use the "mean" value, instead of the "average" value, depending on the
application of your figure.)

So, given that SUNY Stonybrook (where the original author was) is approximately
half-way in-between the north pole and the equator, his estimate is probably
just as close as yours.

This comes from the "Universal Law of Gravitation", which for sufficiently
large bodies, states:

			 m
		g = G --------
			r^2

where G = 6.67 x 10-11 N.m^2/kg^2 and is the so-called gravitational constant,
m is the mass of the Earth and r is the radius of the Earth at a certain point,
and the force 'g', is the acceleration of gravity.

(constants from _The_CRC_Handbook_of_Chemistry_and_Physics_, 59th ed., p. F-193)

>And "around"?  What kind'a sci-en-tific term is that?  You should say
>"jus' 'bout".  Sound's MUCH better.

Let us see....

	1 m = 3.281 feet, therefore 1 m/s^2 = 3.281 ft/s^2
	 .
	. . 9.8      m/s^2 = 32.1 ft/s^2	  (watch your sig fig!)
	 .
	. . 9.780573 m/s^2 = 32.090 ft/s^2	  (ditto!)

So, the original authors assertion of about 32 ft/s^2 has an error value of
(assume 32 ft/s^2 is "exact", and the dimensions cancel, so the answer is
dimensionless):

	(32.090 - 32.000)
	-----------------   =	0.280 %   (or just about around this number)
	       32.090

Which is certainly close enough for most applications.

(Conversion factors from _The_Principles_of_Physics_, Hurley and Garrod, 1978)

>:)

:)

>>	//   David Tiberio  SUNY Stony Brook 2-3662  AMIGA  DDD-MEN
>Jason Balicki

Last but not least, ALL of these numbers are approximations, since none of
them consider air resistance (can you say "Reynolds number"? I knew you
could....). Which, granted, would be minimal for most second story windows --
but not for a 10th. And to come to any precision at all, we need to know the
EXACT (multiple decimals places of accuracy) height from which the object was
dropped. And to figure out if it breaks when it hits, we need to know what it
lands on, and the elasticity (or viscosity) of that substance.

Think about it.

(That was fun! :))
--
  //   Michael B. Smith
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