Subject: Re: [LPNY DISCUSS](OT) Mr. Science bad : End Game
Date: Thu, 2 Oct 2003 16:53:09 -0400
To:        <lpny_discuss@yahoogroups.com>
From:     Bob Armstrong <bob@cosy.com>
Just to finish the chapter :
>=A0Now that I think about it, though, the outer orbit has less=
 curvature
>=A0than the inner one, so just being there would already diminish=
 the
>=A0centifugal force, even at the same speed.

 Yes it's all about the reduced curvature . There is no such=
 thing
 as centrifugal force , just inertia . It's the centripetal=
 force
 of gravity which makes the objects fall towards each other .
 I read a little more Feynman last night , it is the reduction 
 in curvature which changes how far the orbiting object has move
 tangentially to match how much it falls per unit time that=
 makes
 the difference .

>=A0If I recall correctly, the curvature (k) of a line
>=A0varies inversely with the radius, IOW k =3D 1/r.

 Well , Feynman shows with an argument using similar triangles
 that the ratio of the distance an orbiting object must move
 tangentially , call it x , to match the distance it falls per
 unit time , call that d , is proportional to the ratio of
 diameter of the orbit to that tangent distance .  IE :

     ( x % d ) =3D  2 * r % x 

 or  ( x ^ 2 ) =3D 2 * r * d

 or  x =3D ( 2 * r * d ) ^ 1%2 

 But d , the distance fallen , is proportional to  1 % r ^2 .
 So :

     x  proportional to  ( r % r ^ 2 ) ^ 1%2 

 or  x =3D ( 1 % r ) ^ 1%2   , ie ,  r ^ -1%2  .


 Somehow I think you get can there directly from Blay's
 notion of curvature , but that would require a lot more
 remedial reading to take me back to homework I didn't do a
 quarter century ago .
 
--=A0
=A0Bob Armstrong -- http://CoSy.com -- 212-285-1864
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=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A02003/10/02 9:39:50 AM