In the 1998 movie Armageddon a catastrophic asteroid collision is averted by exp
ID: 3163682 • Letter: I
Question
In the 1998 movie Armageddon a catastrophic asteroid collision is averted by exploding a nuclear warhead inside the Texas sized rock. In reality, the largest asteroid we know of is about 64% this large (in diameter) and could have a mass around 1021 kg. Barring the requisite suspension of disbelief associated with all Michael Bay movies, we're going to play with this. According to IMDB, the asteroid was moving at 22,000 mph and, when discovered, had 18 days before impact. Consider the explosion to split the asteroid into two pieces - one 3 times 10^20 kg, one 7 times 10^20 kg. Assume no acceleration of the asteroid as it falls toward the Earth (very poor assumption), and the asteroid is on a path toward the center of the Earth. (a) If the warhead is set off when the asteroid is 12 hours from impact, what velocity does the larger portion need to have to just miss the Earth? What velocity does the smaller piece have? (b) Suppose, instead of breaking it apart the plan is to deflect it. A rocket is launched and "docks" with the asteroid. Assuming the asteroid is at a distance equal to the average lunar orbit, what impulse must be delivered to it so that it just misses the Earth? (c) How do the answers to (a) and (b) change if their distances are flipped (break it up at the lunar orbit, deflect it at the 12 hour mark)?Explanation / Answer
a) the mass=10^21kg
the velocity=22000mph=9.83km/s
now kinetic energy of the asteroid part in outer space(before collision)=1/2*7*10^20*9.83^2Joules=338.2*10^20joules
lets consider the orbital plane is same for earth, sun and the asteroid large part
then earth is in circular orbit of R=1 AU (suppose) while asteroid and its belt are in eliptical orbit.
so we need apogee and perigee in this case. suppose apogee R1=3R
now at R due to earth's gravitation asteroid has high velocityV and for asteroid belt it is v
then using conversation of energy we get
1/2mV^2-GMm/R=1/2mv^2-GMm/3R
now taking earth mass and velocity me and Ve we get
collision speed V={v^2+16*3.14^2*R^2/3*p^2}^1/2 where p is the scale time=3600*24*365=3.15*10^7sec for circular orbit
we know that asteroid V=9.83km/s
R=6371km earth radius
using we get asteroid belt velocity v= 9.8km/s
then using Kepler's 3rd law p^2=(3R^3)
then at belt pb=sqrt(27)p=
using this we get impact speed= 8.2km/s
to miss earth the impact speed must be greater than 8.2km/s
now using conservation of energy
1/2*10^21*9.83= 1/2*3*10^20*Vs+1/2*7*10^20*8.2
then Vs=13.63km/s