Re: Asteroid Problem (2)

[Jack Uretsky <jlu@HEP.ANL.GOV>]

        Ahh, but we are told that tides are generated by the difference
between lunar and solar forces, so that 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.

                                Regards,
                                        Jack
On Wed, 29 Aug 2001, brian whatcott wrote:

> [Presentation in reverse time order...]
>
> Hmmmm,
> supposing the tidal force due to the Moon mass (Mm) at 60 earth r away
> is proportional to the rate of change of Gravity force at the Earth
> F = G.m1.m2.r^-2
> i.e
> Ftidal = -2.G.m1.m2.r^-3
>
> and wishing the planetaid's close encounter to no more than equal this
> tidal force, we have
> -2G m1 m2 r^-3 = -2 G m1/1000 m2 x^-3
> r^-3 = x^-3/1000
> so x = 6 Earth radii for r = 60 Re
>
> So Rick's estimate (4 Re) was evidently in the ball park.
> Brian W
>
>
> At 14:09 8/28/01 -0500, Jack Uretsky added helpfully:
> > Well, that's about 10^{-3} times the lunar mass, and the moon
> > is about 60 earth radii away.  Does that help?
> > --
> > Franz Kafka...
>
>
> > On Tue, 28 Aug 2001, Ludwik Kowalski responded:
>
> > > This begs for a question. How far should the m=10^19 kg
> > > asteroid pass to avoid undesirable tides and earthquakes?
> > > Ludwik Kowalski
>
> > > Earlier, Rick Tarara wrote:
> > >
> > > > With the earth moving at 30 km/s along its orbital path,
> > > > 15 minutes will move the earth 2 diameters [from the
> > > > trajectory of the asteroid].

--
Franz Kafka's novels and novella's are so Kafkaesque that one has to
wonder at the enormity of coincidence required to have produced a writer
named Kafka to write them.
                        Greg Nagan from "The Metamorphosis" in
                     <The 5-MINUTE ILIAD and Other Classics>