Re: Asteroid Problem (2)

[brian whatcott <inet@INTELLISYS.NET>]

Jack wrote a computation for a force from a nearby planetoid
equal to the gravity force due the Moon.
And in that respect, it is correct.

This gravity force is not in fact the tidal motivator, but his
calculation is often given as an initial response to the puzzle
 (as I know well!)

The shape of the Earth's equipotential surface is not intuitively
obvious.

Brian W

At 12:54 8/29/01 -0700, Stephen Murray wrote:
> ... the tides are due to the gradient of the
> gravitational field of the Moon and Sun...  The tidal force from
> an object varies, therefore, as M/r^3.
> ================================
> Stephen D. Murray

> >      .... the force equivalent to that of
> > the moon is generated by a satellite having 10^{-3} moon masses at a
> > distance (1/r_{moon})^{2} =10^{-3)/r^{2},  or
> >        r = 10^{-3/2}r_{moon} = .03x60 earth radii or about 1.8 earth
> > radii.

Jack Uretsky

brian whatcott <inet@intellisys.net>  Altus OK
                Eureka!