Re: Asteroid Problem

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

Rick Tarara wrote:

> For quick calculations, it might be easier to look at slowing down or
> speeding up the asteroid.  While it depends on the angle of crossing
> relative to the earth's path, I get something like a two minute change in
> arrival time as enough to allow the earth to clear the asteroid path.  This
> is about a .0004% change (with 1 year lead time) which should require an
> average speed change of less than 1 m/s (if the asteroid is moving at less
> than a quarter-million meters/s).   Plug in the mass and at least you can
> easily get the force/energy needed.

1 m/s over 2.5*10^5 m/s is 4*10-6. The corresponding
relative change in kinetic energy is 16*10-12. Using
your speed and m=10^19 kg (Michael said that asteroids
are usually not as large as 10^20 kg) one gets the
kinetic energy of 3*10^29 joules. The required change
of kinetic energy must thus be equal to 16*10-12*3*10^29
or only 4.8*10^18 joules.

I would be more comfortable with the delay of two hours
or more (instead of two minutes). This would increase the
corresponding change in kinetic energy by many orders
of magnitude.

Compare the calculated above change in kinetic energy
(work = 4.8*10^18 J) with the energy released by one very
very large nuclear bomb (1000 megatons of TNT).

1 kg of TNT --> 5 MJ or 5*10^6 joules.
1000 megatons = 1,000,000,000 kg ---> 5*10^15 joules

4.8*10^18 / 5*10^15 = 960 bombs.

And ten times more for a ten times more massive asteroid.
Also note that one 1000 megaton bomb is like 100,000 of
the Hiroshima-type bombs (0.01 megaton of TNT each).
Keep in mind that this is for a delay of only two minutes.
Ludwik Kowalski