# Completely Nuking Threatening Asteroids, and Star Wars

Completely Nuking Threatening Asteroids, and Star Wars

Recently, simulations showed that blowing up a very large asteroid would only result in the asteroid reconstituting itself by gravitational attraction.  The asteroid only took hours to reconstitute itself.

We can put a constraint on this without using simulations by just calculating the gravitational potential energy which binds the asteroid together in units of megatons of TNT.  That would show what size nuclear explosions would be needed to totally break it apart such that its components reach escape velocity.

Oddly, while contemplating what to write about the continuing Star Wars series and its conflicts, I realize that this could also be applied to the presumptuousness of a Death Star.

The gravitational potential energy of a solid sphere of uniform density and mass M and radius R is:

U = (3/5) G M^2/R.

The mass M = (4/3) Pi R^3 d (g/cm^3), where d is the density number in g/cm^3.  The density d has three values for different sets of asteroids, d=(1.38, 2.71, and 5.32) for Carbonaceous asteroids, Rocky silicone based, and Metallic asteroids, respectivly.  With M^2, we will be using

d^2 = (1.90, 7.34, and 28.3), respectively.

A megaton of TNT equals 4.18 petajoules, or 4.18 x 10^15 joules.

Converting to the asteroid diameter D = 2R, and conveniently scaling that by 7 km, we get:

U = (0.168, 0.645, 2.45) (D/7 km)^5 megatons, for the Carbon, Silicon, and Metallic cases.

The power of D^5 come from:  three powers from each M, and divide by R.  This greatly scales up the megatonnage needed with increasing D to totally disintegrate the asteroid.

In terms of launching a nuclear weapon to an asteroid, a 1.2 megaton weapon was given as weighing 1,100 kg, or 1.2 tons.

The largest diameter near earth object on the JPL risk list at Sentry:  Earth Impact Monitoring, has a diameter of 1.3 km, a cumulative impact probability off 1.2e-4, and is not expected until 2880.  The next largest is 0.49 km.

Of course, the destructive power of an asteroid collision with earth involves the asteroid’s kinetic energy in the earth’s frame upon collision.

The Chicxulub impactor in Mexico which ended the age of the dinosaurs and 2/3 of life 65 million years ago was 11-81 km or 7-50 miles in diameter, with a mass of 1.0×10^15 to 4.6×10^17 kg, and was a rare carbonaceous chondrite.  Even at a 14 km diameter, the 0.168x (14/7)^2 = 0.168 x 32 = 5.4 megatons, would have required multiple simultaneous weapon hits.

The nearly mile-wide Meteor Crater in Arizona was made by only a 30-50 meter diameter iron asteroid, around 50,000 years ago.

The Tunguska event in 1908 was supposed to be an air explosion by a meteoroid of 50-190 meters in size.

The Chelyabinsk meteor of 2013 was about 20 meters in size.

The asteroid deflection strategy is considered the safest method, and at the sizes we are considering, they should now be detected at a great distance.  Still, we need a defensive preparation for such possibilities.

For the Death Star, we now know that the Earth is one of the lower range of masses for rocky planets.  If we take its radius R = 6,400 km, and density of d = 5.51 gm/cm^3, we get for its binding energy, using an intermediate result:

U=1.68 x 10^(-4) d^2 (R/km)^5 megatons,

U = 1.68 x 10^(-4) (5.51)^2 (6.4×10^3)^5 megatons,

U = 51.0×10^(-4)x10,737×10^15

U = 5.48 x 10^16 megatons to disintegrate an earth sized planet.

Clearly, the Death Star is not able to generate and hold that much energy, in order to release it in a few seconds.  Whew!

Another problem for the Death Star, is that instead of the momentum going like the square root of the energy in non-relativistic physics, in the relativity of electromagnetic radiation, momentum is linear in energy:  P=E/c.  So the Death Star firing radiation at the planet, is going to cause a massive recoil, probably breaking it apart.