Mark Rand wrote:
On Tue, 23 Dec 2003 18:20:06 -0800, "Ken Davey"... wrote:
....
Now; For a clue as to why I think this is true I would like to pose another
question.
At what distance from the center of the earth, would the block be
'heaviest'?
There has to be a 'simple' answer for this.

Objects are heaviest at the surface of the earth. IIRC Gravity is proportional
to distance-from-centre/radius below the centre and proportional to
distance-from-centre/radius^2 above the centre.
....

If by "below the centre" you mean southern hemisphere, then your
brain has let you down. There is no difference in gravitational
formulae between north and south hemispheres of earth.

Anyhow, for items above the surface of the earth, gravitational
attraction is inversely proportional to the square of distance
from the center of the Earth, of course by Newton's F=G*m1*m2/r^2
formula. Below the surface, say at distance t from center, if
m2=mass of Earth, R=radius of Earth, then F=G*m1*m2*(t/R)^3/t^2
=G*m1*m2*t/R^3, which as you say is proportional to distance
to center.[*]
-jiw
[*] (t/R)^3 = the proportion of Earth's mass inside a sphere of
radius t. The mass outside that sphere is a hollow shell, and
its inverse square field cancels out at all points inside.