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Path: utzoo!watmath!clyde!cbosgd!ihnp4!mhuxt!js2j
From: js2j@mhuxt.UUCP (sonntag)
Newsgroups: net.origins,net.physics
Subject: Lighter Gravity - the math
Message-ID: <1168@mhuxt.UUCP>
Date: Wed, 25-Sep-85 11:44:16 EDT
Article-I.D.: mhuxt.1168
Posted: Wed Sep 25 11:44:16 1985
Date-Received: Fri, 27-Sep-85 03:16:55 EDT
Organization: AT&T Bell Laboratories, Murray Hill
Lines: 51
Xref: watmath net.origins:2425 net.physics:3294


     Ted Holden has been arguing for weeks in net.origins about how
huge dinosaurs could never have existed in our current gravitational
field, due to square-cube types of problems.  His postulated cause for
the reduced felt-accelleration is that during the extreme past the Earth
orbited Saturn, producing extremely large tidal effects, (on a 'locked'
system so that one side of the earth faces Saturn all of the time.)
     I worked up the equations describing the situation, in the hopes that
they'll indicate that the earth would have to be grazing Saturn's atmosphere
(or inside it) for the free-fall accelleration at the ends of the earth
to be much lower.  I don't have handy a table full of the constants I'd
need to evaluate the results, however... perhaps someone out there would
be kind enough to evaluate my results.

     delta_a = G * M  * r1 (1/r0^3 - 1/r1^3)

     Where: G is the gravitational constant (as in F=(G*m1*m2)/r^2)
            r1 is the radius from the center of saturn to the inner side of
               the earth. (r1=r0 - radius of earth)
            r0 is the radius from the center of saturn to the center of earth.
            M  is the mass of Saturn
            delta_a is the changed in free-fall accelleration at the inner and
                    outer ends of the earth.  A delta_a of -5 meters would 
                    correspond to a 50% reduction in weight.             

    Derivation:
    
      	delta_a = a           - a
                   centripetal   Saturn's gravity
                 
 		= (2*pi*r1)^2 / (T^2 * r1) - G * M / r1^2 (where T=orbit time)

   Due to the fact that the earth is in orbit, we know that delta_a=0 when
evaluated with r0 instead of r1.  This allows us to solve for T^2:
  	T^2 = (4 * pi^2 * r0^3) / (G * M)
   Substituting this into the origional equation and rearranging, we get 
the result quoted above.  One last simplification we can make:
(1/X^3 - 1/(X - del)^3 = 3 * del * X^-4 (for large X/del)
   Using this in the original equation:
     
	delta_a = 3 * G * M * (radius of earth) * r0^-3

   This is simple enough that I hope that *someone* out there can find
the orbital radius required to cause say a 50% reduction in weight.  Please
compare this result to the radius of Saturn.  If it turns out to be
less than the radius of Saturn, do you think Ted will be quiet?
-- 
Jeff Sonntag
ihnp4!mhuxt!js2j
    Silly quote: "There are a few off-the-wall extremists, who are shunned
                  by us moderates." - Don Black


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