Monday, December 14, 2009

Deep Impact? Relativistic Collision part 2

This is a continuation of yesterday's post which attempted to analyse the effects of a relativistic object impacting the earth. Yesterday we considered things from the point of view of energy, and we failed to answer the most interesting question which is how deep does an impact go?

Today we are going to do momentum.

Take a look at the diagram below (click on it to enlarge).

The impactor (as yesterday) of mass m hits the earth with velocity u m/s and immediately becomes a plasma. As it meets earth material (we assume molten rock) it shares its momentum with the new material. The result is a growing cone of tunneling material as shown in the diagram below. The semi-angle of the cone is α, its depth is d metres and its base radius is r = d * α.

Multiply the resulting volume by the density of molten rock ρ which is 5.5 tonnes per cubic metre and decide the final velocity v at which we believe further tunnelling will stop. Perhaps 10 metres per second?

Click to enlarge

Since γmu = ρα2d3v by conservation of momentum,

d = (γmu/ρα2v)1/3.

Now for some numbers. For the 1 kg impactor at 0.999c considered yesterday, with α = half a degree and with a final plasma velocity of 10 m/s, the crater-depth d = 1.2 km.

This 1.2 km deep crater is nothing like the journey to the centre of the earth we speculated about yesterday. If we increase the cone angle for a wider dissipation - say 2.5 degrees - then the depth decreases to only 400 metres.

Suppose we increase the mass of the impactor to a cubic metre of ice weighing one tonne, at the original half-degree spread angle? The penetration depth d is now around 12 km. Notice that the increase of mass of a thousandfold has only increased the penetration depth by a factor of ten. This is because of the cube law we see in the equation above.

What would get us to the centre of the earth? A million tons (109 kg) at 0.999999c would dig a crater 3,700 km deep. I reckon this would make a bit of a mess of the earth.

Here's the spreadsheet to play with (includes yesterday's).

A relativistic impactor is like a nuclear detonation

How does this compare with nuclear weapons? An approximate formula for the crater diameter d (km) of a nuclear explosion of M Megatons is simply:

d3 = M.

So a 27 Megaton detonation would produce a 3 km diameter crater. A rule of thumb states that the crater depth is around 1/5 of its diameter, so for the 27 Megaton device, the crater depth would be around 600 metres.

This is almost identical to the result given by the spreadsheet for a 1 kg impactor at 0.9c (kinetic energy = 27.84 Mt).

According to the Wikipedia article on the Orion spacecraft (powered by nuclear detonation) the initial plasma velocity of a 1 Megaton bomb is 10,000 km/sec. The velocity seems to scale linearly with the bomb yield suggesting that a 30 Megaton bomb might create a plasma shockfront expanding at a speed close to that of light. When this hits the ground, it might be indistinguishable from a relativistic impactor and therefore the cratering effects could be very similar.