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Re: Asteroid Problem



[Presentation in reverse time order...]

Hmmmm,
estimating the force due to the Moon mass (Mm) at 60 earth r away
proportional to Mm/60^2
force due to the moon at 61 earth r away
proportional to Mm/61^2
Force driving the Earth's tide is roughly proportional to
(61^2 - 60^2)/(60^2*61^2)
= 121/13395600 = 9E-6

To limit the tidal effects to less than the current moon tides
we need
1E-3([x+1]^2 - x^2)/([x+1]^2*x^2) = 9E-6
so
x, the minimal desired planetoid separation = 5.8 Earth radii about

...at least it is if I churned the numbers right!

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].



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