From: ABlainey@aol.com
Date: Sat Mar 09 2002 - 17:13:43 MST
In a message dated 09/03/02 21:08:10 GMT Standard Time, mlorrey@datamann.com
writes:
> I'm not exactly sure this would occur, but I think that in a steep
> gravity gradient, the radius will NOT relate to the circumference by the
> same value of pi that we use. See below:
>
> circle from top:
>
>
> __----------__
> / \
> / \
> / \
> | |
> | C |
> | *------------+R
> | |
> | |
> \ /
> \ /
> \ /
> --____________--
>
> Would appear to be a normal circle, with a normal pi.
>
> Then assume that point C is a gravitational well, such that, looked at
> from the point of the gravity gradient:
>
>
> ---------------------------+R < circle, edge on
> \ /
> \ /
> \ /
> \ /
> \ /
> \ /
> \ /
> \ /
> \ /
> \ /
> C*
>
> Where radius CR does not relate to circumference R by the same ratio of
> pi that we use ....
>
>
But is this only true for the outside observer ? From within the curvature
Pi would still relate ?.
Alex
This archive was generated by hypermail 2.1.5 : Sat Nov 02 2002 - 09:12:53 MST