Armageddon Physics
Saturday September 18th 2010, 7:29 pm
Filed under: Uncategorized

The movie Armageddon’s solution for saving the world was completely farfetched and impossible. There was no way, under any circumstances, NASA’s plan could have worked in the time slot they had and with the asteroid that was approaching. Essentially, if this were to really happen it would be “GAME OVER” for Earth and its inhabitants.

However with some major tweaking of just about every variable given by the movie, I feel a more realistic attack could be avoided.

Question:

Is there any circumstance that the Earth could survive an asteroid attack similar to the one in Armageddon? (What is the distance separating the two asteroid halves when they reach Earth?)

Relative Quantities:

I decided to down size the asteroid to the size of Rhode Island instead of Texas, change the density from being similar to Earth’s to similar to Mars’(3940 kg/m3), and changed the velocity at which the asteroid was approaching in order to give NASA more time (2mph). In order to calculate these quantities I will also need to know the volume (to find the mass), the nuclear bomb yield, the distance the asteroid travels once the astronauts arrive (the distance from the moon), how long they have until the asteroid reaches the Earth, and the diameter of the Earth.

Calculated Quantities:

• Width of Rhode Island: 482,700m
• Volume of Asteroid: 5.88×1016 m3
• Density of Asteroid: 3940 kg/m3 (http://www.squidoo.com/which-planet-has-the-greatest-density)
• Mass of Asteroid: 2.32×1020 kg
• Velocity of Asteroid: 2mph= .89408 m/s
• Distance from asteroid initially: Dmoon= 3.844×108m
• Time to hit Earth (After drilling): 429,910,355 s.
• Diameter of Earth: 1.2756×107 m
• Nuclear Bomb Yield: 4.184×1017 J
• Vy: .0601m/s
• Distance Between 2 asteroids at Earth: 51675224.67m

Armageddon Physics: Save the World