Re: Asteroid Problem (2)

[brian whatcott <inet@INTELLISYS.NET>]

[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].