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[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].