Re: Asteroid Problem

["John S. Denker" <jsd@MONMOUTH.COM>]

At 01:20 PM 8/31/01 -0400, Ludwik Kowalski wrote:
> (on my picture where the moon-earth line is horizontal).
> Thus lunar tidal forces (gradient of F) are essentially
> horizontal (on my picture).

Really?  Is this assertion based on a calculation, or just a hunch?

When I do the calculation, all indications are that tidal forces are quite
nicely balanced between "stretching" along one axis and "squashing" along
the two perpendicular axes.

When doing the calculation, remember that the force vector is a vector.  On
the earth-moon axis, the force differs in magnitude, and the difference is
fairly obvious.  Along the perpendicular axes, the force differs in
direction, which is perhaps less obvious, but no less important.

Note:  When I say "quite nicely balanced" I mean very very very nicely
balanced.  As in exactly balanced.  Proof:
        gravitational potential = 1/r
        Force = del (potential)
        Tide = del (Force) = del (del (potential))
        Average outward component of Tide
                = Integral_over_area normal dot del(del(potential)) dS
        and invoking the divergence theorem we see this is equal to
                = Integral_over_volume del dot del(del(potential)) dV
        and the integrand is zero anywhere outside the moon, since
        del dot del (1/r) = 0 in free space, and del commutes with del.